W9-W12. Sequences, Infinite Series, Convergence Tests

Author

Mohammad Alkousa

Published

November 6, 2025

1. Summary

1.1 Sequences and Limits

A sequence is an ordered list of numbers \(\{a_n\}_{n \geq 1} = a_1, a_2, a_3, \dots\) where each \(a_n\) is determined by some rule or formula. Understanding the limit of a sequence is fundamental: we say \(\lim_{n \to \infty} a_n = L\) if the terms \(a_n\) get arbitrarily close to the value \(L\) as \(n\) increases without bound.

1.1.1 Computing Limits Involving \(e\)

Many limits involve the fundamental constant \(e = \lim_{n \to \infty} (1 + \frac{1}{n})^n \approx 2.71828\). More generally, we have the crucial identity:

\[ \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x \]

When computing limits, we often need to manipulate expressions to match this form. Key techniques include:

Factorial simplification: Remember that \(\frac{(n+1)!}{n!} = n+1\)
Power manipulation: Express ratios of powers as a single power, e.g., \(\frac{(n+1)^n}{(n+2)^n} = \left(\frac{n+1}{n+2}\right)^n\)
Rewriting to standard form: Transform expressions like \((1 - \frac{a}{n})^n\) into the form \((1 + \frac{1}{m})^m\) by appropriate substitutions
Exponent manipulation: Use the property \((1 + \frac{a}{n})^{bn} = ((1 + \frac{a}{n})^n)^b\) to isolate the standard limit

1.1.2 The Squeeze Theorem

The Squeeze Theorem (or Sandwich Theorem) is a powerful tool for evaluating limits of sequences that oscillate or are difficult to compute directly. If we have three sequences satisfying \(a_n \leq b_n \leq c_n\) for all sufficiently large \(n\), and if \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L\), then we can conclude that \(\lim_{n \to \infty} b_n = L\).

This is particularly useful for dealing with oscillating terms like \((-1)^n\), since we can bound them: \(-1 \leq (-1)^n \leq 1\).

1.1.3 Bounded and Monotonic Sequences

A sequence \(\{a_n\}\) is:

Bounded above if there exists \(M\) such that \(a_n \leq M\) for all \(n\)
Bounded below if there exists \(m\) such that \(a_n \geq m\) for all \(n\)
Bounded if it is both bounded above and below
Monotonic increasing if \(a_{n+1} \geq a_n\) for all \(n\)
Monotonic decreasing if \(a_{n+1} \leq a_n\) for all \(n\)

The Bolzano-Weierstrass Theorem (also called the Monotone Convergence Theorem) states that every bounded monotonic sequence converges. This is extremely useful: if we can prove a sequence is bounded and monotonic, we immediately know it converges, even if finding the exact limit requires additional work.

1.1.4 Proving Properties by Induction

To prove that a recursively defined sequence satisfies certain properties (like being bounded or monotonic), we often use mathematical induction:

Base case: Verify the property holds for \(n = 1\) (or the first few terms)
Inductive hypothesis: Assume the property holds for \(n = k\)
Inductive step: Prove that if the property holds for \(n = k\), it must also hold for \(n = k+1\)

Once boundedness and monotonicity are established, we can find the limit by treating the recursive formula as an equation where both \(a_n\) and \(a_{n+1}\) approach the same limit \(L\).

1.2 Infinite Series

1.2.1 Basic Definitions

An infinite series (or simply series) is the sum of infinitely many terms from a sequence. We write this as:

\[ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots \]

While we cannot literally add infinitely many numbers, we can define what we mean by such a sum using the concept of partial sums. The \(n\)-th partial sum is:

\[ s_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + \dots + a_n \]

The series converges if the sequence of partial sums \(\{s_n\}\) converges to a finite limit \(s\). In that case, we write \(\sum_{n=1}^{\infty} a_n = s\) and call \(s\) the sum of the series. If the sequence of partial sums does not converge to a finite limit, we say the series diverges.

1.2.2 Telescoping Series

A telescoping series is one where most terms cancel out when we expand the partial sum. Formally, if we can write the general term as \(a_n = b_n - b_{n+1}\), then:

\[ s_n = (b_1 - b_2) + (b_2 - b_3) + \dots + (b_n - b_{n+1}) = b_1 - b_{n+1} \]

The series converges if and only if \(\lim_{n \to \infty} b_n\) exists, and in that case:

\[ \sum_{n=1}^{\infty} a_n = b_1 - \lim_{n \to \infty} b_n \]

A common technique is to use partial fraction decomposition to rewrite terms in a form that telescopes. For example, \(\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\) leads to a telescoping series.

1.2.3 Geometric Series

A geometric series has the form:

\[ \sum_{n=0}^{\infty} aq^n = a + aq + aq^2 + aq^3 + \dots \]

where \(a \neq 0\) is the first term and \(q\) is the common ratio. The convergence behavior depends entirely on \(q\):

If \(|q| < 1\): the series converges to \(\frac{a}{1-q}\)
If \(|q| \geq 1\): the series diverges

This is one of the most important series in mathematics, as many other series can be compared to or expressed in terms of geometric series.

1.2.4 The Harmonic Series and \(p\)-Series

The harmonic series is:

\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots \]

Despite the terms approaching zero, this series diverges. This is a crucial counterexample showing that \(\lim_{n \to \infty} a_n = 0\) is necessary but not sufficient for convergence.

More generally, a \(p\)-series has the form:

\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]

This series:

Converges if \(p > 1\)
Diverges if \(p \leq 1\)

The \(p\)-series serves as a standard benchmark for comparison tests.

1.2.5 Necessary Condition for Convergence

The \(n\)-th Term Test (or Divergence Test) states: If \(\sum a_n\) converges, then \(\lim_{n \to \infty} a_n = 0\).

The contrapositive is more useful: If \(\lim_{n \to \infty} a_n \neq 0\) or the limit does not exist, then \(\sum a_n\) diverges.

Important: This test can only prove divergence. If \(\lim_{n \to \infty} a_n = 0\), we cannot conclude anything about convergence (the harmonic series is the classic example).

1.3 Convergence Tests for Series with Non-Negative Terms

When \(a_n \geq 0\) for all \(n\), the partial sums form a non-decreasing sequence. Such series either converge (if the partial sums are bounded) or diverge to infinity (if unbounded). Several tests help determine which case applies.

1.3.1 Direct Comparison Test

Theorem (Direct Comparison Test): Let \(\sum a_n\) and \(\sum b_n\) be two series with \(0 \leq a_n \leq b_n\) for all \(n \in \mathbb{N}\) (or at least for all sufficiently large \(n\)). Then:

If \(\sum b_n\) converges, then \(\sum a_n\) converges
If \(\sum a_n\) diverges, then \(\sum b_n\) diverges

Proof:

(1) Assume \(\sum b_n\) converges. Let \(s_n = \sum_{i=1}^{n} a_i\) and \(t_n = \sum_{i=1}^{n} b_i\) be the partial sums.

Since \(0 \leq a_i \leq b_i\) for all \(i\), we have: \[s_n = \sum_{i=1}^{n} a_i \leq \sum_{i=1}^{n} b_i = t_n\]

The sequence \(\{s_n\}\) is non-decreasing (since all terms are non-negative) and bounded above by \(\lim_{n \to \infty} t_n\) (which exists since \(\sum b_n\) converges). By the Monotone Convergence Theorem, \(\{s_n\}\) converges, so \(\sum a_n\) converges.

(2) This is the contrapositive of (1): if \(\sum a_n\) diverges to \(+\infty\) and \(a_n \leq b_n\), then the partial sums of \(\sum b_n\) must also diverge to \(+\infty\). \(\square\)

Strategy: To use this test, find a series you already know converges or diverges (often a geometric series or \(p\)-series) and compare term-by-term.

1.3.2 Limit Comparison Test

Theorem (Limit Comparison Test): Let \(\sum a_n\) and \(\sum b_n\) be two series with positive terms (\(a_n > 0\), \(b_n > 0\)). Define:

\[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \]

Then:

If \(0 < L < \infty\), both series have the same behavior (both converge or both diverge)
If \(L = 0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges
If \(L = \infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges

Proof:

(1) Suppose \(0 < L < \infty\). Then for any \(\epsilon > 0\), there exists \(N\) such that for all \(n > N\): \[L - \epsilon < \frac{a_n}{b_n} < L + \epsilon\]

Choose \(\epsilon = L/2\). Then for \(n > N\): \[\frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2}\]

This gives us: \[\frac{L}{2} b_n < a_n < \frac{3L}{2} b_n\]

If \(\sum b_n\) converges, then by the Direct Comparison Test (with the right inequality), \(\sum a_n\) converges.
If \(\sum b_n\) diverges, then by the Direct Comparison Test (with the left inequality), \(\sum a_n\) diverges.

(2) If \(L = 0\), then for any \(\epsilon > 0\), there exists \(N\) such that for \(n > N\): \[\frac{a_n}{b_n} < \epsilon\]

Thus \(a_n < \epsilon b_n\). If \(\sum b_n\) converges and we choose \(\epsilon = 1\), then \(\sum a_n\) converges by the Direct Comparison Test.

(3) If \(L = \infty\), then \(\lim_{n \to \infty} \frac{b_n}{a_n} = 0\). By part (2) applied to \(\sum b_n\) and \(\sum a_n\), if \(\sum a_n\) converges, then \(\sum b_n\) converges. Contrapositive: if \(\sum b_n\) diverges, then \(\sum a_n\) diverges. \(\square\)

Strategy: Choose \(b_n\) to be a simpler series (geometric or \(p\)-series) that resembles the dominant behavior of \(a_n\) for large \(n\). The limit comparison test is often easier to apply than the direct comparison test since you don’t need an explicit inequality.

1.3.3 Ratio Test (D’Alembert’s Test)

Theorem (Ratio Test): Let \(\sum a_n\) be a series with positive terms, and let:

\[ L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \]

Then:

If \(L < 1\), the series converges
If \(L > 1\), the series diverges
If \(L = 1\), the test is inconclusive

Proof:

(1) Suppose \(L < 1\). Choose \(r\) such that \(L < r < 1\). Then there exists \(N\) such that for all \(n > N\): \[\frac{a_{n+1}}{a_n} < r\]

This gives us: \[a_{N+1} < ra_N, \quad a_{N+2} < ra_{N+1} < r^2a_N, \quad a_{N+3} < r^3a_N, \ldots\]

In general, \(a_{N+k} < r^k a_N\) for \(k \geq 1\). Therefore: \[\sum_{n=N+1}^{\infty} a_n < a_N \sum_{k=1}^{\infty} r^k = a_N \cdot \frac{r}{1-r}\]

Since this is a convergent geometric series (with \(r < 1\)), the tail of \(\sum a_n\) converges, and thus \(\sum a_n\) converges.

(2) Suppose \(L > 1\). Then there exists \(N\) such that for all \(n > N\): \[\frac{a_{n+1}}{a_n} > 1\]

This means \(a_{n+1} > a_n\) for all \(n > N\), so the terms are eventually increasing. In particular, \(a_n\) does not approach 0, so by the \(n\)-th term test, \(\sum a_n\) diverges.

(3) When \(L = 1\), the test gives no information. Both convergent and divergent series can have \(L = 1\) (for example, \(\sum \frac{1}{n}\) diverges and \(\sum \frac{1}{n^2}\) converges, but both have \(L = 1\)). \(\square\)

Application: The ratio test is particularly effective for series involving factorials or exponentials, where the ratio simplifies nicely.

1.3.4 Root Test (Cauchy’s Test)

Theorem (Root Test): Let \(\sum a_n\) be a series with non-negative terms, and let:

\[ L = \lim_{n \to \infty} \sqrt[n]{a_n} \]

Then:

If \(L < 1\), the series converges
If \(L > 1\), the series diverges
If \(L = 1\), the test is inconclusive

Proof:

(1) Suppose \(L < 1\). Choose \(r\) such that \(L < r < 1\). Then there exists \(N\) such that for all \(n > N\): \[\sqrt[n]{a_n} < r\]

Thus \(a_n < r^n\) for all \(n > N\). Since \(\sum r^n\) is a convergent geometric series (with \(r < 1\)), by the Direct Comparison Test, \(\sum a_n\) converges.

(2) Suppose \(L > 1\). Then there exists \(N\) such that for all \(n > N\): \[\sqrt[n]{a_n} > 1\]

Thus \(a_n > 1\) for all \(n > N\), so \(\lim_{n \to \infty} a_n \neq 0\). By the \(n\)-th term test, \(\sum a_n\) diverges.

(3) When \(L = 1\), the test is inconclusive (same reasoning as for the ratio test). \(\square\)

Application: The root test is particularly useful when \(a_n\) involves \(n\)-th powers, such as \(a_n = (\text{expression})^n\).

1.3.5 Comparison Between Ratio and Root Tests

Remark: There is an important relationship between the ratio and root tests:

\[\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \to \infty} \sqrt[n]{a_n} \leq \limsup_{n \to \infty} \sqrt[n]{a_n} \leq \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}\]

In particular, if \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n}\) exists, then \(\lim_{n \to \infty} \sqrt[n]{a_n}\) also exists and they are equal:

\[\lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}\]

This means every series whose convergence can be determined by the ratio test can also be determined by the root test, but not vice versa (the root test is more powerful). However, the converse is not always true—some series can be analyzed by the root test but not by the ratio test.

In practice: Use the ratio test for series with factorials, and the root test for series with \(n\)-th powers.

1.4 Alternating Series

An alternating series has the form:

\[ \sum_{n=1}^{\infty} (-1)^n a_n \quad \text{or} \quad \sum_{n=1}^{\infty} (-1)^{n+1} a_n \]

where \(a_n > 0\). The signs alternate between positive and negative.

1.4.1 Leibniz Test

Theorem (Leibniz Test): Let \(\sum (-1)^{n+1} a_n\) be an alternating series where \(a_n > 0\). If:

The sequence \(\{a_n\}\) is non-increasing (i.e., \(a_{n+1} \leq a_n\) for all \(n\)), and
\(\lim_{n \to \infty} a_n = 0\)

Then the alternating series converges.

Proof:

Let \(s_n = \sum_{k=1}^{n} (-1)^{k+1} a_k\) be the \(n\)-th partial sum. We consider the even and odd partial sums separately.

Even partial sums: Consider \(s_{2n}\): \[s_{2n} = (a_1 - a_2) + (a_3 - a_4) + \cdots + (a_{2n-1} - a_{2n})\]

Since \(a_k \geq a_{k+1}\), each parenthesis is non-negative, so: \[s_{2n+2} = s_{2n} + (a_{2n+1} - a_{2n+2}) \geq s_{2n}\]

Thus \(\{s_{2n}\}\) is non-decreasing.

We can also write: \[s_{2n} = a_1 - (a_2 - a_3) - (a_4 - a_5) - \cdots - (a_{2n-2} - a_{2n-1}) - a_{2n}\]

Since each parenthesis is non-negative and \(a_{2n} \geq 0\): \[s_{2n} \leq a_1\]

So \(\{s_{2n}\}\) is bounded above. By the Monotone Convergence Theorem, \(\{s_{2n}\}\) converges to some limit \(s\).

Odd partial sums: We have: \[s_{2n+1} = s_{2n} + a_{2n+1}\]

Taking limits: \[\lim_{n \to \infty} s_{2n+1} = \lim_{n \to \infty} s_{2n} + \lim_{n \to \infty} a_{2n+1} = s + 0 = s\]

Since both subsequences converge to the same limit \(s\), the full sequence \(\{s_n\}\) converges to \(s\). \(\square\)

Application: This test is particularly powerful because the conditions are easy to check, and it applies to many series that don’t meet the criteria for other tests.

1.5 Absolute and Conditional Convergence

1.5.1 Absolute Convergence

A series \(\sum a_n\) converges absolutely if the series of absolute values \(\sum |a_n|\) converges.

Theorem (The Absolute Convergence Test): If \(\sum |a_n|\) converges, then \(\sum a_n\) converges. In other words, absolute convergence implies convergence.

Proof:

We use the Cauchy criterion for convergence. Since \(\sum |a_n|\) converges, for any \(\epsilon > 0\), there exists \(N\) such that for all \(m > n > N\): \[\left| \sum_{k=n+1}^{m} |a_k| \right| = \sum_{k=n+1}^{m} |a_k| < \epsilon\]

By the triangle inequality: \[\left| \sum_{k=n+1}^{m} a_k \right| \leq \sum_{k=n+1}^{m} |a_k| < \epsilon\]

This shows that the partial sums of \(\sum a_n\) form a Cauchy sequence, and therefore \(\sum a_n\) converges. \(\square\)

Application: This is powerful because it allows us to apply tests designed for positive series (like the ratio test, root test, and comparison tests) to series with mixed signs by considering \(\sum |a_n|\) instead.

1.5.2 Conditional Convergence

A series is conditionally convergent if it converges, but does not converge absolutely. In other words, \(\sum a_n\) converges but \(\sum |a_n|\) diverges.

Example: The alternating harmonic series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) converges (by Leibniz test) to \(\ln(2)\), but \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges. Hence it is conditionally convergent.

Important distinction: Absolutely convergent series have nice properties (they can be rearranged without changing the sum), while conditionally convergent series are more delicate (rearranging terms can change the sum or even make the series diverge).

1.5.3 Applying Tests for Absolute Convergence

To test for absolute convergence:

Ratio test for absolute convergence: Compute \(L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\)
Root test for absolute convergence: Compute \(L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\)

If \(L < 1\), the series converges absolutely. If \(L > 1\), the series diverges.

1.6 Power Series

1.6.1 Definition and Convergence

A power series centered at \(c\) is a series of the form:

\[ \sum_{n=0}^{\infty} a_n(x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \dots \]

Unlike the series we’ve studied so far, a power series is a function of \(x\). The key question is: For which values of \(x\) does the series converge?

1.6.2 The Convergence Theorem for Power Series

Theorem (The Convergence Theorem for Power Series): If a power series \(\sum_{n=0}^{\infty} a_n x^n\) converges at a point \(x = c \neq 0\), then it converges absolutely for all \(x\) with \(|x| < |c|\). If the series diverges at \(x = d\), then it diverges for all \(x\) with \(|x| > |d|\).

Proof:

Suppose \(\sum a_n c^n\) converges. Then \(\lim_{n \to \infty} a_n c^n = 0\), so the sequence \(\{a_n c^n\}\) is bounded: there exists \(M > 0\) such that \(|a_n c^n| \leq M\) for all \(n\).

Now let \(x\) be any value with \(|x| < |c|\). Then: \[|a_n x^n| = |a_n c^n| \cdot \left|\frac{x}{c}\right|^n \leq M \left|\frac{x}{c}\right|^n\]

Since \(\left|\frac{x}{c}\right| < 1\), the series \(\sum M \left|\frac{x}{c}\right|^n\) is a convergent geometric series. By the Direct Comparison Test, \(\sum |a_n x^n|\) converges, so \(\sum a_n x^n\) converges absolutely.

The second statement follows by contrapositive: if \(\sum a_n x^n\) converged for some \(|x| > |d|\), then by the first part, it would have to converge at \(d\), contradicting the assumption that it diverges there. \(\square\)

1.6.3 The Domain of Convergence of a Power Series

Theorem: For any power series \(\sum_{n=0}^{\infty} a_n(x-c)^n\), exactly one of the following holds:

The series converges only at \(x = c\) (radius of convergence \(R = 0\))
The series converges absolutely for all \(x \in \mathbb{R}\) (radius of convergence \(R = \infty\))
There exists a finite \(R > 0\) such that:
- The series converges absolutely for \(|x-c| < R\)
- The series diverges for \(|x-c| > R\)
- At the endpoints \(x = c \pm R\), the series may either converge or diverge (must be checked separately)

The value \(R\) is called the radius of convergence.

Proof:

Define: \[S = \{r \geq 0 : \sum a_n r^n \text{ converges}\}\]

This set is non-empty (it contains 0) and is an interval \([0, R)\) or \([0, R]\) or \([0, \infty)\) by the Convergence Theorem.

If \(S = \{0\}\), then \(R = 0\).
If \(S = [0, \infty)\), then \(R = \infty\).
Otherwise, \(R = \sup S\) is finite and positive. \(\square\)

The set of all \(x\) for which the series converges is called the interval of convergence. Depending on convergence at the endpoints, this can be any of: \((c-R, c+R)\), \([c-R, c+R)\), \((c-R, c+R]\), or \([c-R, c+R]\).

1.6.4 The Radius of Convergence of a Power Series

Theorem (Radius of Convergence by Ratio Test): For the power series \(\sum a_n(x-c)^n\), if the limit:

\[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\]

exists (or equals \(\pm\infty\)), then the radius of convergence \(R\) is given by:

\(R = 0\) if \(L = +\infty\)
\(R = +\infty\) if \(L = 0\)
\(R = \frac{1}{L}\) if \(0 < L < +\infty\)

Proof:

Apply the ratio test to the series \(\sum a_n(x-c)^n\): \[\lim_{n \to \infty} \left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = |x-c| \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L|x-c|\]

The series converges absolutely when \(L|x-c| < 1\), i.e., \(|x-c| < \frac{1}{L} = R\). \(\square\)

Theorem (Hadamard Formula: Radius of Convergence by Root Test): For the power series \(\sum a_n(x-c)^n\), if the limit:

\[L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\]

exists (or equals \(\pm\infty\)), then the radius of convergence is given by the same formulas as in the ratio test.

Proof:

Apply the root test to \(\sum a_n(x-c)^n\): \[\lim_{n \to \infty} \sqrt[n]{|a_n(x-c)^n|} = |x-c| \lim_{n \to \infty} \sqrt[n]{|a_n|} = L|x-c|\]

The series converges absolutely when \(L|x-c| < 1\). \(\square\)

Important: These formulas give the radius. To find the interval of convergence, you must separately check convergence at the endpoints \(x = c - R\) and \(x = c + R\).

1.6.5 Operations on Power Series

Theorem (Operations on Power Series):

Addition and Subtraction: If \(f(x) = \sum a_n x^n\) has radius \(R_1\) and \(g(x) = \sum b_n x^n\) has radius \(R_2\), then \(f(x) \pm g(x) = \sum (a_n \pm b_n) x^n\) has radius at least \(\min\{R_1, R_2\}\).
Scalar Multiplication: If \(f(x) = \sum a_n x^n\) has radius \(R\), then \(cf(x) = \sum (ca_n) x^n\) has the same radius \(R\).
Cauchy Product of Power Series: If \(f(x) = \sum a_n x^n\) and \(g(x) = \sum b_n x^n\) both have positive radius of convergence, then their product is: \[f(x) \cdot g(x) = \sum_{n=0}^{\infty} c_n x^n \quad \text{where} \quad c_n = \sum_{k=0}^{n} a_k b_{n-k}\]

The product series has radius at least \(\min\{R_1, R_2\}\) where \(R_1\) and \(R_2\) are the radii of \(f\) and \(g\).

1.7 Properties of Absolutely Convergent Series

1.7.1 Rearrangement Theorem

Definition: A rearrangement of a series \(\sum a_n\) is a series \(\sum b_n\) where \(b_n = a_{\sigma(n)}\) for some bijection \(\sigma: \mathbb{N} \to \mathbb{N}\) (i.e., the terms are the same but in a different order).

Theorem (The Rearrangement Theorem for Absolutely Convergent Series): If \(\sum a_n\) converges absolutely to a sum \(A\), and \(\sum b_n\) is any rearrangement of \(\sum a_n\), then \(\sum b_n\) also converges absolutely to \(A\).

Proof:

Since \(\sum a_n\) converges absolutely, \(\sum |a_n|\) converges. Let \(A = \sum_{n=1}^{\infty} a_n\) and \(\alpha = \sum_{n=1}^{\infty} |a_n|\).

Let \(\epsilon > 0\). Since \(\sum |a_n|\) converges, there exists \(N\) such that: \[\sum_{n=N+1}^{\infty} |a_n| < \frac{\epsilon}{2}\]

Let \(M\) be large enough so that \(\{a_1, a_2, \ldots, a_N\} \subseteq \{b_1, b_2, \ldots, b_M\}\) (i.e., the first \(N\) terms of the original series all appear among the first \(M\) terms of the rearranged series).

For \(m > M\), the partial sum \(s_m^{(b)} = \sum_{k=1}^{m} b_k\) differs from \(A\) by at most terms from \(\{a_{N+1}, a_{N+2}, \ldots\}\), so: \[|s_m^{(b)} - A| \leq \sum_{n=N+1}^{\infty} |a_n| < \frac{\epsilon}{2} < \epsilon\]

Therefore, \(\lim_{m \to \infty} s_m^{(b)} = A\), proving that the rearranged series converges to the same sum. \(\square\)

Remark: In contrast, conditionally convergent series do not have this property. The Riemann Rearrangement Theorem states that if \(\sum a_n\) converges conditionally, then for any extended real number \(L \in [-\infty, +\infty]\), there exists a rearrangement of \(\sum a_n\) that converges to \(L\) (or diverges if \(L = \pm\infty\))!

1.7.2 Cauchy Product

Theorem: If \(\sum a_n\) and \(\sum b_n\) are two absolutely convergent series with sums \(A\) and \(B\) respectively, then their Cauchy product:

\[ \sum_{n=0}^{\infty} c_n \quad \text{where} \quad c_n = \sum_{k=0}^{n} a_k b_{n-k} \]

converges absolutely to \(AB\).

Theorem (Merten’s Theorem): Suppose \(\sum_{n=0}^{\infty} a_n = A\) and \(\sum_{n=0}^{\infty} b_n = B\) are two convergent series, and at least one of them converges absolutely. Then their Cauchy product \(\sum_{n=0}^{\infty} c_n\) converges to \(AB\), where:

\[ c_n = \sum_{k=0}^{n} a_k b_{n-k} = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0 \]

Proof (outline):

Without loss of generality, assume \(\sum a_n\) converges absolutely. Let \(A_n = \sum_{k=0}^{n} a_k\), \(B_n = \sum_{k=0}^{n} b_k\), and \(C_n = \sum_{k=0}^{n} c_k\) be the partial sums.

We can write: \[C_n = a_0 B_n + a_1 B_{n-1} + \cdots + a_n B_0\]

We want to show that \(\lim_{n \to \infty} C_n = AB\).

Consider: \[C_n - A_n B = \sum_{k=0}^{n} a_k (B_{n-k} - B)\]

Since \(B_n \to B\), we have \(B_{n-k} - B \to 0\) as \(n \to \infty\). Using the absolute convergence of \(\sum |a_k|\) and careful estimates, we can show: \[\lim_{n \to \infty} (C_n - A_n B) = 0\]

Since \(A_n \to A\) and \(B\) is constant: \[\lim_{n \to \infty} C_n = \lim_{n \to \infty} A_n B = AB \quad \square\]

Remark: If both series are only conditionally convergent (neither converges absolutely), their Cauchy product may diverge. For example, the Cauchy product of \(\sum \frac{(-1)^n}{\sqrt{n+1}}\) with itself (which converges conditionally by Leibniz) results in a divergent series.

1.8 Asymptotic Notation: Big O, little o, and Asymptotic Equivalence

Asymptotic notation provides a convenient way to compare the behavior of sequences or functions as \(n \to \infty\) (or as \(x \to a\)).

1.8.1 Big O Notation

Definition: We write \(a_n = O(b_n)\) (read “\(a_n\) is big O of \(b_n\)”) if there exists a constant \(C > 0\) and \(N \in \mathbb{N}\) such that: \[|a_n| \leq C|b_n| \quad \text{for all } n > N\]

This means \(a_n\) grows no faster than \(b_n\) (up to a constant multiple).

Example: \(3n^2 + 5n + 1 = O(n^2)\) because for large \(n\), the \(n^2\) term dominates.

1.8.2 Little o Notation

Definition: We write \(a_n = o(b_n)\) (read “\(a_n\) is little o of \(b_n\)”) if: \[\lim_{n \to \infty} \frac{a_n}{b_n} = 0\]

This means \(a_n\) grows much slower than \(b_n\).

Example: \(5n = o(n^2)\) because \(\lim_{n \to \infty} \frac{5n}{n^2} = 0\).

1.8.3 Asymptotic Equivalence

Definition: We write \(a_n \sim b_n\) (read “\(a_n\) is asymptotically equivalent to \(b_n\)” or “\(a_n\) is asymptotic to \(b_n\)”) if: \[\lim_{n \to \infty} \frac{a_n}{b_n} = 1\]

This means \(a_n\) and \(b_n\) have the same asymptotic behavior.

Example: \(3n^2 + 5n + 1 \sim 3n^2\) because \(\lim_{n \to \infty} \frac{3n^2 + 5n + 1}{3n^2} = 1\).

1.8.4 Properties and Applications

Key Properties:

If \(a_n = o(b_n)\), then \(a_n = O(b_n)\)
If \(a_n \sim b_n\), then \(a_n = O(b_n)\)
If \(a_n \sim b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) converges
If \(a_n \sim b_n\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges

Application to Series: When analyzing series, we can often replace complicated terms with simpler asymptotically equivalent expressions.

Example: Consider \(\sum_{n=1}^{\infty} \frac{e^n + n^4}{3^n + \ln^2(n+1)}\).

For large \(n\):

Numerator: \(e^n + n^4 \sim e^n\) (exponentials dominate polynomials)
Denominator: \(3^n + \ln^2(n+1) \sim 3^n\) (exponentials dominate logarithms)

Therefore: \[\frac{e^n + n^4}{3^n + \ln^2(n+1)} \sim \frac{e^n}{3^n} = \left(\frac{e}{3}\right)^n\]

Since \(\sum \left(\frac{e}{3}\right)^n\) is a convergent geometric series (with ratio \(\frac{e}{3} < 1\)), the original series also converges.

2. Definitions

Sequence: An ordered list of numbers \(\{a_n\}_{n \geq 1}\) where each term \(a_n\) is determined by a rule or formula.
Limit of a Sequence: \(\lim_{n \to \infty} a_n = L\) means that the terms \(a_n\) get arbitrarily close to \(L\) as \(n\) increases.
Convergent Sequence: A sequence that has a finite limit.
Divergent Sequence: A sequence that does not converge to a finite limit.
Bounded Sequence: A sequence \(\{a_n\}\) is bounded if there exist numbers \(m\) and \(M\) such that \(m \leq a_n \leq M\) for all \(n\).
Monotonic Increasing Sequence: A sequence where \(a_{n+1} \geq a_n\) for all \(n\).
Monotonic Decreasing Sequence: A sequence where \(a_{n+1} \leq a_n\) for all \(n\).
Bolzano-Weierstrass Theorem: Every bounded monotonic sequence converges.
Squeeze Theorem: If \(a_n \leq b_n \leq c_n\) for large \(n\) and \(\lim a_n = \lim c_n = L\), then \(\lim b_n = L\).
Series: The sum of infinitely many terms from a sequence, denoted \(\sum_{n=1}^{\infty} a_n\).
Partial Sum: The \(n\)-th partial sum is \(s_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + \dots + a_n\).
Convergent Series: A series whose sequence of partial sums converges to a finite limit \(s\), called the sum of the series.
Divergent Series: A series whose sequence of partial sums does not converge to a finite limit.
Telescoping Series: A series of the form \(\sum (b_n - b_{n+1})\) where consecutive terms cancel.
Geometric Series: A series of the form \(\sum_{n=0}^{\infty} aq^n\) where \(a\) is the first term and \(q\) is the common ratio.
Harmonic Series: The series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which diverges.
\(p\)-Series: A series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), which converges if \(p > 1\) and diverges if \(p \leq 1\).
Alternating Series: A series with terms that alternate in sign, of the form \(\sum (-1)^n a_n\) where \(a_n > 0\).
Absolutely Convergent Series: A series \(\sum a_n\) for which \(\sum |a_n|\) converges.
Conditionally Convergent Series: A series that converges but does not converge absolutely.
Power Series: A series of the form \(\sum_{n=0}^{\infty} a_n(x-c)^n\) where \(x\) is a variable.
Radius of Convergence: The value \(R\) such that a power series converges absolutely for \(|x-c| < R\) and diverges for \(|x-c| > R\).
Interval of Convergence: The set of all values of \(x\) for which a power series converges.
Cauchy Product: For two series \(\sum a_n\) and \(\sum b_n\), the Cauchy product is \(\sum c_n\) where \(c_n = \sum_{k=0}^{n} a_k b_{n-k}\).
Big O Notation (\(O\)): We write \(a_n = O(b_n)\) if there exists \(C > 0\) and \(N\) such that \(|a_n| \leq C|b_n|\) for all \(n > N\).
Little o Notation (\(o\)): We write \(a_n = o(b_n)\) if \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\).
Asymptotic Equivalence (\(\sim\)): We write \(a_n \sim b_n\) if \(\lim_{n \to \infty} \frac{a_n}{b_n} = 1\).

3. Formulas

Limit Definition of \(e\): \[e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\]
Generalized \(e\) Limit: \[\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x\]
Squeeze Theorem: If \(a_n \leq b_n \leq c_n\) and \(\lim a_n = \lim c_n = L\), then \(\lim b_n = L\).
Recursive Limit Evaluation: If \(\lim_{n \to \infty} a_n = L\) and \(a_{n+1} = f(a_n)\) where \(f\) is continuous, then \(L = f(L)\).
\(n\)-th Partial Sum: \[s_n = \sum_{i=1}^{n} a_i\]
Series Convergence: \(\sum_{n=1}^{\infty} a_n = s\) if \(\lim_{n \to \infty} s_n = s\).
Telescoping Series Sum: If \(a_n = b_n - b_{n+1}\), then \[\sum_{n=1}^{\infty} a_n = b_1 - \lim_{n \to \infty} b_n\]
Geometric Series Sum: For \(|q| < 1\), \[\sum_{n=0}^{\infty} aq^n = \frac{a}{1-q}\]
Mengoli Series: \[\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1\]
Generalized Telescoping Sum: \[\sum_{n=1}^{\infty} \frac{1}{n(n+1)\cdots(n+r)} = \frac{1}{r \cdot r!}\]
Harmonic Series Divergence: \[\sum_{n=1}^{\infty} \frac{1}{n} = \infty\]
\(p\)-Series Convergence: \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) converges if and only if \(p > 1\).
Linear Property of Series: \[\sum_{n=1}^{\infty} (\alpha a_n + \beta b_n) = \alpha \sum_{n=1}^{\infty} a_n + \beta \sum_{n=1}^{\infty} b_n\]
\(n\)-th Term Test: If \(\sum a_n\) converges, then \(\lim_{n \to \infty} a_n = 0\).
Direct Comparison Test: If \(0 \leq a_n \leq b_n\), then: (1) \(\sum b_n\) converges \(\Rightarrow\) \(\sum a_n\) converges; (2) \(\sum a_n\) diverges \(\Rightarrow\) \(\sum b_n\) diverges.
Limit Comparison Test: If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\) where \(0 < L < \infty\), then \(\sum a_n\) and \(\sum b_n\) have the same behavior.
Ratio Test: Let \(L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\). If \(L < 1\), the series converges absolutely; if \(L > 1\), the series diverges; if \(L = 1\), the test is inconclusive.
Root Test: Let \(L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\). If \(L < 1\), the series converges absolutely; if \(L > 1\), the series diverges; if \(L = 1\), the test is inconclusive.
Leibniz Test: If \(\{a_n\}\) is non-increasing and \(\lim a_n = 0\), then \(\sum (-1)^{n+1} a_n\) converges.
Absolute Convergence Test: If \(\sum |a_n|\) converges, then \(\sum a_n\) converges.
Alternating Harmonic Series Sum: \[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln(2)\]
Radius of Convergence (Ratio Test): \[R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|\]
Radius of Convergence (Root Test): \[R = \frac{1}{\lim_{n \to \infty} \sqrt[n]{|a_n|}}\]
Cauchy Product: \((\sum a_n)(\sum b_n) = \sum c_n\) where \[c_n = \sum_{k=0}^{n} a_k b_{n-k}\]

4. Examples

4.1. Find the Limit with Factorials and Powers (Lab 9, Task 1)

Find the limit:

\[\lim_{n \to \infty} \left[\frac{(n+1)!}{n!}\right]^n \cdot \frac{1}{(n+2)^n}\]

Click to see the solution

Key Concept: Simplify factorials and manipulate the expression to match the standard form \((1 + \frac{1}{m})^m \to e\).

Simplify the factorial ratio: \[\frac{(n+1)!}{n!} = \frac{1 \cdot 2 \cdot 3 \cdots n \cdot (n+1)}{1 \cdot 2 \cdot 3 \cdots n} = n+1\]
Combine the powers: \[(n+1)^n \cdot \frac{1}{(n+2)^n} = \frac{(n+1)^n}{(n+2)^n} = \left(\frac{n+1}{n+2}\right)^n\]
Rewrite the numerator: \[\frac{n+1}{n+2} = \frac{(n+2)-1}{n+2} = 1 - \frac{1}{n+2}\]
Express in standard form: \[\left(1 - \frac{1}{n+2}\right)^n = \left(1 + \frac{1}{-(n+2)}\right)^n\]
Manipulate the exponent: Let \(m = -(n+2)\). We want to rewrite this in the form \((1 + \frac{1}{m})^m\): \[\left(1 + \frac{1}{-(n+2)}\right)^n = \left(1 + \frac{1}{-(n+2)}\right)^{-(n+2) \cdot \frac{-n}{n+2}}\] \[= \frac{1}{\left[\left(1 + \frac{1}{-(n+2)}\right)^{-(n+2)}\right]^{\frac{n}{n+2}}}\]
Evaluate the limit: As \(n \to \infty\), we have \(-(n+2) \to -\infty\), so: \[\lim_{n \to \infty} \left(1 + \frac{1}{-(n+2)}\right)^{-(n+2)} = e\]

Also: \[\lim_{n \to \infty} \frac{n}{n+2} = 1\]

Therefore: \[\lim_{n \to \infty} \left(1 + \frac{1}{-(n+2)}\right)^n = \frac{1}{e^1} = \frac{1}{e}\]

Answer: \(\frac{1}{e}\)

4.2. Convergence of Geometric Series (a) (Lab 9, Task 1a)

Determine the convergence or divergence of the following geometric series:

\[1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}\]

Click to see the solution

Key Concept: A geometric series \(\sum aq^n\) converges if \(|q| < 1\) and its sum is \(\frac{a}{1-q}\).

Identify the first term and common ratio:
- First term: \(a = 1\) (when \(n = 1\), we get \((1/2)^0 = 1\))
- Common ratio: \(q = \frac{1}{2}\)
Check the convergence condition: \[|q| = \left|\frac{1}{2}\right| = \frac{1}{2} < 1\]

Therefore, the series converges.
Calculate the sum: \[\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1} = \frac{a}{1-q} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2\]

Answer: The series converges to \(2\).

4.3. Convergence of Geometric Series (b) (Lab 9, Task 1b)

Determine the convergence or divergence of the following geometric series:

\[\pi - e + \frac{e^2}{\pi} - \frac{e^3}{\pi^2} + \cdots = \sum_{n=1}^{\infty} \pi\left(-\frac{e}{\pi}\right)^{n-1}\]

Click to see the solution

Identify the first term and common ratio:
- First term: \(a = \pi\)
- Common ratio: \(q = -\frac{e}{\pi}\)
Check the convergence condition: \[|q| = \left|-\frac{e}{\pi}\right| = \frac{e}{\pi} \approx \frac{2.718}{3.142} \approx 0.865 < 1\]

Therefore, the series converges.
Calculate the sum: \[\sum_{n=1}^{\infty} \pi\left(-\frac{e}{\pi}\right)^{n-1} = \frac{a}{1-q} = \frac{\pi}{1 - \left(-\frac{e}{\pi}\right)} = \frac{\pi}{1 + \frac{e}{\pi}} = \frac{\pi}{\frac{\pi + e}{\pi}} = \frac{\pi^2}{\pi + e}\]

Answer: The series converges to \(\frac{\pi^2}{\pi + e}\).

4.4. Convergence of Geometric Series (c) (Lab 9, Task 1c)

Determine the convergence or divergence of the following geometric series:

\[1 + \sqrt{2} + 2 + 2^{3/2} + \cdots = \sum_{n=1}^{\infty} (\sqrt{2})^{n-1}\]

Click to see the solution

Identify the first term and common ratio:
- First term: \(a = 1\)
- Common ratio: \(q = \sqrt{2}\)
Check the convergence condition: \[|q| = |\sqrt{2}| = \sqrt{2} \approx 1.414 > 1\]

Therefore, the series diverges.

Answer: The series diverges.

4.5. Limit with Exponential and Rational Terms (Lab 9, Task 2)

Find the limit:

\[\lim_{n \to \infty} \left[2 \cdot \left(\frac{1}{3}\right)^n + \frac{\sqrt{5}}{(1 - \frac{3}{n})^{2n}}\right]\]

Click to see the solution

Key Concept: Evaluate each term separately, recognizing that exponentials with base less than 1 approach 0, and transform the second term to match the \(e\) limit form.

Evaluate the first term: \[\lim_{n \to \infty} 2 \cdot \left(\frac{1}{3}\right)^n = \lim_{n \to \infty} \frac{2}{3^n} = 0\]

(since \(\frac{1}{3} < 1\), the powers approach 0)
Rewrite the denominator of the second term: \[\left(1 - \frac{3}{n}\right)^{2n} = \left[\left(1 - \frac{3}{n}\right)^n\right]^2\]
Transform to standard form: \[\left(1 - \frac{3}{n}\right)^n = \left(1 + \frac{1}{-n/3}\right)^n = \left[\left(1 + \frac{1}{-n/3}\right)^{-n/3}\right]^{-3}\]
Evaluate the limit: As \(n \to \infty\), we have \(-n/3 \to -\infty\), so: \[\lim_{n \to \infty} \left(1 + \frac{1}{-n/3}\right)^{-n/3} = e\]

Therefore: \[\lim_{n \to \infty} \left(1 - \frac{3}{n}\right)^n = e^{-3}\]

And: \[\lim_{n \to \infty} \left(1 - \frac{3}{n}\right)^{2n} = (e^{-3})^2 = e^{-6}\]
Evaluate the second term: \[\lim_{n \to \infty} \frac{\sqrt{5}}{(1 - \frac{3}{n})^{2n}} = \frac{\sqrt{5}}{e^{-6}} = \sqrt{5} \cdot e^6\]
Combine both terms: \[\lim_{n \to \infty} \left[2 \cdot \left(\frac{1}{3}\right)^n + \frac{\sqrt{5}}{(1 - \frac{3}{n})^{2n}}\right] = 0 + \sqrt{5} \cdot e^6 = \sqrt{5} \cdot e^6\]

Answer: \(\sqrt{5} \cdot e^6\)

4.6. Parametric Convergence of Geometric Series (a) (Lab 9, Task 2a)

For which values of \(x \in \mathbb{R}\) does the following series converge, and what is its sum?

\[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 + x^2)^n}\]

Click to see the solution

Key Concept: Rewrite as a geometric series and find when \(|q| < 1\).

Rewrite as a geometric series: \[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 + x^2)^n} = \sum_{n=0}^{\infty} \left(\frac{-1}{2 + x^2}\right)^n\]

This is a geometric series with first term \(a = 1\) and common ratio \(q = \frac{-1}{2 + x^2}\).
Find the convergence condition: The series converges when \(|q| < 1\): \[\left|\frac{-1}{2 + x^2}\right| < 1\] \[\frac{1}{2 + x^2} < 1\] \[1 < 2 + x^2\] \[x^2 > -1\]

This is true for all \(x \in \mathbb{R}\) since \(x^2 \geq 0 > -1\) and \(2 + x^2 \geq 2 > 0\).
Calculate the sum: \[\sum_{n=0}^{\infty} \left(\frac{-1}{2 + x^2}\right)^n = \frac{1}{1 - \frac{-1}{2 + x^2}} = \frac{1}{\frac{2 + x^2 + 1}{2 + x^2}} = \frac{2 + x^2}{3 + x^2}\]

Answer: The series converges for all \(x \in \mathbb{R}\), and its sum is \(\frac{2 + x^2}{3 + x^2}\).

4.7. Parametric Convergence of Geometric Series (b) (Lab 9, Task 2b)

For which values of \(x \in \mathbb{R}\) does the following series converge, and what is its sum?

\[\sum_{n=0}^{\infty} \frac{(2x)^n}{(1 + x^2)^n}\]

Click to see the solution

Rewrite as a geometric series: \[\sum_{n=0}^{\infty} \frac{(2x)^n}{(1 + x^2)^n} = \sum_{n=0}^{\infty} \left(\frac{2x}{1 + x^2}\right)^n\]

This is a geometric series with \(a = 1\) and \(q = \frac{2x}{1 + x^2}\).
Find the convergence condition: The series converges when \(|q| < 1\): \[\left|\frac{2x}{1 + x^2}\right| < 1\] \[|2x| < 1 + x^2\] \[-1 - x^2 < 2x < 1 + x^2\]
Analyze both inequalities:
- Right inequality: \(2x < 1 + x^2\) \[x^2 - 2x + 1 > 0\] \[(x - 1)^2 > 0\] This is true for all \(x \neq 1\).
- Left inequality: \(2x > -1 - x^2\) \[x^2 + 2x + 1 > 0\] \[(x + 1)^2 > 0\] This is true for all \(x \neq -1\).
Combine conditions: The series converges for all \(x \in \mathbb{R} \setminus \{-1, 1\}\).
Calculate the sum: \[\sum_{n=0}^{\infty} \left(\frac{2x}{1 + x^2}\right)^n = \frac{1}{1 - \frac{2x}{1 + x^2}} = \frac{1}{\frac{1 + x^2 - 2x}{1 + x^2}} = \frac{1 + x^2}{1 + x^2 - 2x} = \frac{1 + x^2}{(x-1)^2}\]

Answer: The series converges for all \(x \in \mathbb{R} \setminus \{-1, 1\}\), and its sum is \(\frac{1 + x^2}{(x-1)^2}\).

4.8. Limit with Factorial, Exponential, and Oscillating Terms (Lab 9, Task 3)

Find the limit:

\[\lim_{n \to \infty} \left[\frac{3^n}{n!} + \frac{(-1)^n}{n} \cdot \left(1 + \frac{1}{n}\right)^{-n}\right]\]

Click to see the solution

Key Concept: Use the squeeze theorem for the oscillating term and recognize that factorials grow faster than exponentials.

Evaluate the first term: Factorials grow much faster than exponentials, so: \[\lim_{n \to \infty} \frac{3^n}{n!} = 0\]

(This can be proven using the ratio test: \(\frac{a_{n+1}}{a_n} = \frac{3}{n+1} \to 0\))
Evaluate the second term’s power: \[\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{-n} = \frac{1}{e}\]
Handle the oscillating factor using the squeeze theorem: We have \(-1 \leq (-1)^n \leq 1\), so dividing by \(n > 0\): \[-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}\]
Apply limits to the bounds: \[\lim_{n \to \infty} -\frac{1}{n} = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{n} = 0\]
Conclude by the squeeze theorem: \[\lim_{n \to \infty} \frac{(-1)^n}{n} = 0\]
Evaluate the second term: \[\lim_{n \to \infty} \frac{(-1)^n}{n} \cdot \left(1 + \frac{1}{n}\right)^{-n} = 0 \cdot \frac{1}{e} = 0\]
Combine both terms: \[\lim_{n \to \infty} \left[\frac{3^n}{n!} + \frac{(-1)^n}{n} \cdot \left(1 + \frac{1}{n}\right)^{-n}\right] = 0 + 0 = 0\]

Answer: \(0\)

4.9. Ratio Test (a) (Lab 9, Task 3a)

Determine the convergence of the following series using the ratio test:

\[\sum_{n=1}^{\infty} \frac{99^n}{n!}\]

Click to see the solution

Key Concept: For series with factorials, the ratio test is usually effective.

Set up the ratio: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{99^{n+1}}{(n+1)!}}{\frac{99^n}{n!}}\]
Simplify: \[= \lim_{n \to \infty} \frac{99^{n+1}}{(n+1)!} \cdot \frac{n!}{99^n}\] \[= \lim_{n \to \infty} \frac{99 \cdot 99^n}{99^n} \cdot \frac{n!}{(n+1)!}\] \[= \lim_{n \to \infty} 99 \cdot \frac{1}{n+1}\] \[= \lim_{n \to \infty} \frac{99}{n+1} = 0\]
Apply the ratio test: Since \(L = 0 < 1\), the series converges (absolutely).

Answer: The series converges.

4.10. Ratio Test (b) (Lab 9, Task 3b)

Determine the convergence of the following series using the ratio test:

\[\sum_{n=1}^{\infty} \frac{n^5}{2^n}\]

Click to see the solution

Set up the ratio: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{(n+1)^5}{2^{n+1}}}{\frac{n^5}{2^n}}\]
Simplify: \[= \lim_{n \to \infty} \frac{(n+1)^5}{2^{n+1}} \cdot \frac{2^n}{n^5}\] \[= \lim_{n \to \infty} \frac{(n+1)^5}{n^5} \cdot \frac{2^n}{2 \cdot 2^n}\] \[= \lim_{n \to \infty} \frac{1}{2} \cdot \left(\frac{n+1}{n}\right)^5\] \[= \frac{1}{2} \cdot \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^5\] \[= \frac{1}{2} \cdot 1^5 = \frac{1}{2}\]
Apply the ratio test: Since \(L = \frac{1}{2} < 1\), the series converges.

Answer: The series converges.

4.11. Convergence of a Recursive Sequence (Lab 9, Task 4)

Let \(\{a_n\}_{n \geq 1}\) be a sequence such that \(a_1 = 2\) and \(a_{n+1} = 2 - \frac{2}{2 + a_n}\) for \(n \geq 1\).

(a) Prove by induction that the sequence is bounded below by 1, i.e., \(a_n \geq 1\) for all \(n \geq 1\).

(b) Prove that the sequence is decreasing, i.e., \(a_{n+1} - a_n \leq 0\) for all \(n \geq 1\).

(c) Conclude that the sequence converges and find its limit.

Click to see the solution

Key Concept: Use induction to establish boundedness, analyze the difference \(a_{n+1} - a_n\) for monotonicity, then apply the Bolzano-Weierstrass theorem.

(a) Prove \(a_n \geq 1\) by induction:

Base case (\(n = 1\)): \[a_2 = 2 - \frac{2}{2 + 2} = 2 - \frac{2}{4} = 2 - 0.5 = 1.5 \geq 1 \quad \checkmark\]
Inductive hypothesis: Assume \(a_k \geq 1\) for some \(k \geq 1\).
Inductive step: We need to show \(a_{k+1} \geq 1\). From the recursive formula: \[a_{k+1} = 2 - \frac{2}{2 + a_k}\]

We want to show: \[2 - \frac{2}{2 + a_k} \geq 1\] \[1 \geq \frac{2}{2 + a_k}\] \[2 + a_k \geq 2\] \[a_k \geq 0\]

This is true since \(a_k \geq 1 > 0\) by the inductive hypothesis.

Therefore, \(a_n \geq 1\) for all \(n \geq 1\).

(b) Prove the sequence is decreasing:

Compute the difference: \[a_{n+1} - a_n = 2 - \frac{2}{2 + a_n} - a_n\]
Find a common denominator: \[= \frac{2(2 + a_n) - 2 - a_n(2 + a_n)}{2 + a_n}\] \[= \frac{4 + 2a_n - 2 - 2a_n - a_n^2}{2 + a_n}\] \[= \frac{2 - a_n^2}{2 + a_n}\]
Analyze the sign: The denominator \(2 + a_n > 0\) since \(a_n \geq 1\). For the numerator:
- If \(a_n > \sqrt{2}\), then \(a_n^2 > 2\), so \(2 - a_n^2 < 0\)
- If \(a_n = \sqrt{2}\), then \(2 - a_n^2 = 0\)
- If \(1 \leq a_n < \sqrt{2}\), then \(a_n^2 < 2\), so \(2 - a_n^2 > 0\)
Determine the behavior: Since \(a_1 = 2 > \sqrt{2} \approx 1.414\) and \(a_2 = 1.5 > \sqrt{2}\), we can verify that once \(a_n > \sqrt{2}\), we have \(a_{n+1} - a_n < 0\). The sequence decreases from \(a_1 = 2\) toward \(\sqrt{2}\).

Therefore, the sequence is decreasing (at least after the first term and while \(a_n > \sqrt{2}\)).

(c) Find the limit:

Apply Bolzano-Weierstrass theorem: The sequence is bounded below (by 1) and decreasing, so it converges.
Let \(L = \lim_{n \to \infty} a_n\): Taking limits on both sides of the recursive formula: \[\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \left(2 - \frac{2}{2 + a_n}\right)\] \[L = 2 - \frac{2}{2 + L}\]
Solve for \(L\): \[L = \frac{2(2 + L) - 2}{2 + L} = \frac{4 + 2L - 2}{2 + L} = \frac{2 + 2L}{2 + L}\] \[L(2 + L) = 2 + 2L\] \[2L + L^2 = 2 + 2L\] \[L^2 = 2\] \[L = \pm\sqrt{2}\]
Choose the correct value: Since \(a_n \geq 1 > 0\) for all \(n\), we must have \(L \geq 0\). Therefore, \(L = \sqrt{2}\).

Answer:

(a) The sequence is bounded below by 1 (proven by induction).

(b) The sequence is decreasing (proven by analyzing the sign of \(a_{n+1} - a_n\)).

(c) The sequence converges to \(\sqrt{2}\).

4.12. Root Test (a) (Lab 9, Task 4a)

Determine the convergence of the following series using the root test:

\[\sum_{n=1}^{\infty} \frac{2^{n+1}}{n^n}\]

Click to see the solution

Key Concept: The root test is ideal for series where the general term involves \(n\)-th powers.

Compute the \(n\)-th root: \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{2^{n+1}}{n^n}}\]
Simplify: \[= \lim_{n \to \infty} \frac{\sqrt[n]{2^{n+1}}}{\sqrt[n]{n^n}}\] \[= \lim_{n \to \infty} \frac{2 \cdot \sqrt[n]{2^n}}{n}\] \[= \lim_{n \to \infty} \frac{2 \cdot 2}{n}\] \[= \lim_{n \to \infty} \frac{4}{n} = 0\]
Apply the root test: Since \(L = 0 < 1\), the series converges.

Answer: The series converges.

4.13. Root Test (b) (Lab 9, Task 4b)

Determine the convergence of the following series using the root test:

\[\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^{n^2}\]

Click to see the solution

Compute the \(n\)-th root: \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}\]
Simplify using exponent rules: \[= \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n^2/n}\] \[= \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n\]
Rewrite the base: \[\frac{n}{n+1} = \frac{1}{1 + \frac{1}{n}}\]

Therefore: \[\left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + \frac{1}{n}}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}\]
Evaluate the limit: \[L = \lim_{n \to \infty} \frac{1}{\left(1 + \frac{1}{n}\right)^n} = \frac{1}{e}\]
Apply the root test: Since \(L = \frac{1}{e} \approx 0.368 < 1\), the series converges.

Answer: The series converges.

4.14. Telescoping Series with Partial Fractions (a) (Lab 10, Part I, Example 1)

Find the sum of the series:

\[\sum_{n=1}^{\infty} \frac{1}{n(n+3)}\]

Click to see the solution

Key Concept: Use partial fraction decomposition to create a telescoping series.

Decompose using partial fractions: \[\frac{1}{n(n+3)} = \frac{A}{n} + \frac{B}{n+3}\]
Solve for \(A\) and \(B\): \[1 = A(n+3) + Bn\]

Setting \(n = 0\): \(1 = 3A \Rightarrow A = \frac{1}{3}\)

Setting \(n = -3\): \(1 = -3B \Rightarrow B = -\frac{1}{3}\)
Rewrite the series: \[\sum_{n=1}^{\infty} \frac{1}{n(n+3)} = \sum_{n=1}^{\infty} \frac{1}{3}\left(\frac{1}{n} - \frac{1}{n+3}\right) = \frac{1}{3}\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+3}\right)\]
Compute the partial sum: \[s_N = \frac{1}{3}\left[\left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{1}{6}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+3}\right)\right]\]

Rearranging: \[s_N = \frac{1}{3}\left[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} - \frac{1}{N+1} - \frac{1}{N+2} - \frac{1}{N+3}\right]\]
Take the limit: \[\lim_{N \to \infty} s_N = \frac{1}{3}\left(1 + \frac{1}{2} + \frac{1}{3} - 0 - 0 - 0\right) = \frac{1}{3} \cdot \frac{11}{6} = \frac{11}{18}\]

Answer: \(\frac{11}{18}\)

4.15. Telescoping Series with Square Roots (a) (Lab 10, Part I, Homework H1)

Determine whether the series converges and find its sum if possible:

\[\sum_{n=2}^{\infty} (\sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n})\]

Click to see the solution

Key Concept: Recognize that this is a second-order telescoping series.

Rewrite the general term: Let \(a_n = \sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n}\). Notice this can be written as: \[a_n = (\sqrt{n+2} - \sqrt{n+1}) - (\sqrt{n+1} - \sqrt{n})\]
Define differences: Let \(b_n = \sqrt{n+1} - \sqrt{n}\). Then: \[a_n = b_{n+1} - b_n\]
This is a telescoping series of differences: \[s_N = \sum_{n=2}^{N} (b_{n+1} - b_n) = b_{N+1} - b_2 = (\sqrt{N+2} - \sqrt{N+1}) - (\sqrt{3} - \sqrt{2})\]
Evaluate the limit: \[\lim_{N \to \infty} (\sqrt{N+2} - \sqrt{N+1}) = \lim_{N \to \infty} \frac{(N+2) - (N+1)}{\sqrt{N+2} + \sqrt{N+1}} = \lim_{N \to \infty} \frac{1}{\sqrt{N+2} + \sqrt{N+1}} = 0\]

Therefore: \[\lim_{N \to \infty} s_N = 0 - (\sqrt{3} - \sqrt{2}) = \sqrt{2} - \sqrt{3}\]

Answer: The series converges to \(\sqrt{2} - \sqrt{3}\)

4.16. \(n\)-th Term Test Example (a) (Lab 10, Part II, Example 1)

Study the convergence of:

\[\sum_{n=1}^{\infty} \left(\frac{n+2}{n+3}\right)^n\]

Click to see the solution

Key Concept: Transform the expression to recognize the limit involving \(e\).

Rewrite the general term: \[a_n = \left(\frac{n+2}{n+3}\right)^n = \left(\frac{(n+3)-1}{n+3}\right)^n = \left(1 - \frac{1}{n+3}\right)^n\]
Find the limit: \[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+3}\right)^n\]

Let \(m = -(n+3)\). As \(n \to \infty\), we have \(m \to -\infty\), and: \[\left(1 + \frac{1}{m}\right)^{-m-3} = \frac{\left(1 + \frac{1}{m}\right)^{-m}}{\left(1 + \frac{1}{m}\right)^3}\]

As \(m \to -\infty\): \[\lim_{n \to \infty} \left(1 - \frac{1}{n+3}\right)^n = \frac{1}{e^1 \cdot 1} = \frac{1}{e} \neq 0\]
Apply the \(n\)-th term test: Since \(\lim_{n \to \infty} a_n = \frac{1}{e} \neq 0\), the series diverges.

Answer: The series diverges.

4.17. Asymptotic Equivalence Example (a) (Lab 10, Part II, Homework H1)

Study the convergence of:

\[\sum_{n=1}^{\infty} n \sin\left(\frac{1}{n}\right)\]

Click to see the solution

Key Concept: Use the asymptotic behavior \(\sin(x) \sim x\) as \(x \to 0\).

Apply asymptotic equivalence: For large \(n\), \(\frac{1}{n}\) is small, so: \[\sin\left(\frac{1}{n}\right) \sim \frac{1}{n}\]

Therefore: \[n \sin\left(\frac{1}{n}\right) \sim n \cdot \frac{1}{n} = 1\]
Find the limit: \[\lim_{n \to \infty} n \sin\left(\frac{1}{n}\right) = \lim_{n \to \infty} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} = 1 \neq 0\]

(using L’Hôpital’s rule or the fact that \(\lim_{x \to 0} \frac{\sin x}{x} = 1\))
Apply the \(n\)-th term test: Since \(\lim_{n \to \infty} a_n = 1 \neq 0\), the series diverges.

Answer: The series diverges.

4.18. Direct Comparison Test with Logarithm (Lab 10, Part III, Example 1)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{1}{\ln(e^n - 1)}\]

Click to see the solution

Key Concept: Simplify the logarithm and compare with a \(p\)-series.

Simplify the logarithm: For large \(n\), \(e^n - 1 \approx e^n\), so: \[\ln(e^n - 1) \approx \ln(e^n) = n\]
Find the asymptotic behavior: More precisely: \[\ln(e^n - 1) = \ln(e^n(1 - e^{-n})) = n + \ln(1 - e^{-n})\]

Since \(\ln(1 - e^{-n}) \to 0\) as \(n \to \infty\): \[\ln(e^n - 1) \sim n\]
Compare with the harmonic series: \[\frac{1}{\ln(e^n - 1)} \sim \frac{1}{n}\]

Since \(\sum \frac{1}{n}\) diverges, by the limit comparison test, \(\sum \frac{1}{\ln(e^n - 1)}\) also diverges.

Answer: The series diverges.

4.19. \(p\)-Series Example (Lab 10, Part III, Homework H1)

Study the convergence of:

\[\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}\]

Click to see the solution

Key Concept: Recognize this as a \(p\)-series.

Identify the \(p\)-series: This is a \(p\)-series with \(p = \frac{1}{2}\): \[\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=2}^{\infty} \frac{1}{n^{1/2}}\]
Apply the \(p\)-series test: A \(p\)-series \(\sum \frac{1}{n^p}\) converges if and only if \(p > 1\).

Here, \(p = \frac{1}{2} < 1\), so the series diverges.

Answer: The series diverges.

4.20. Ratio Test with Factorial (a) (Lab 10, Part IV, Example 1)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{n!}{2n+1}\]

Click to see the solution

Key Concept: For series involving factorials, the ratio test is usually effective.

Apply the ratio test: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{(n+1)!}{2(n+1)+1}}{\frac{n!}{2n+1}}\]
Simplify: \[= \lim_{n \to \infty} \frac{(n+1)! \cdot (2n+1)}{n! \cdot (2n+3)}\] \[= \lim_{n \to \infty} \frac{(n+1) \cdot (2n+1)}{2n+3}\] \[= \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{2n+3}\] \[= \lim_{n \to \infty} \frac{n^2(2 + \frac{3}{n} + \frac{1}{n^2})}{n(2 + \frac{3}{n})}\] \[= \lim_{n \to \infty} \frac{n(2 + \frac{3}{n} + \frac{1}{n^2})}{2 + \frac{3}{n}} = \infty\]
Apply the ratio test: Since \(L = \infty > 1\), the series diverges.

Answer: The series diverges.

4.21. Ratio Test with Powers and Factorials (a) (Lab 10, Part IV, Homework H1)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{n^n}{(n!)^2}\]

Click to see the solution

Apply the ratio test: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{(n+1)^{n+1}}{((n+1)!)^2}}{\frac{n^n}{(n!)^2}}\]
Simplify: \[= \lim_{n \to \infty} \frac{(n+1)^{n+1} \cdot (n!)^2}{n^n \cdot ((n+1)!)^2}\] \[= \lim_{n \to \infty} \frac{(n+1)^{n+1} \cdot (n!)^2}{n^n \cdot (n+1)^2 \cdot (n!)^2}\] \[= \lim_{n \to \infty} \frac{(n+1)^{n+1}}{n^n \cdot (n+1)^2}\] \[= \lim_{n \to \infty} \frac{(n+1)^{n-1}}{n^n}\] \[= \lim_{n \to \infty} \frac{(n+1)^{n-1}}{n^n}\] \[= \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^n \cdot \frac{1}{(n+1)}\] \[= e \cdot 0 = 0\]
Apply the ratio test: Since \(L = 0 < 1\), the series converges.

Answer: The series converges.

4.22. Root Test with Double Exponential (Lab 10, Part V, Example 1)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{n^n}{3^{n^2}}\]

Click to see the solution

Key Concept: The root test is ideal when the general term has \(n\)-th powers.

Apply the root test: \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{n^n}{3^{n^2}}}\]
Simplify: \[= \lim_{n \to \infty} \frac{\sqrt[n]{n^n}}{\sqrt[n]{3^{n^2}}}\] \[= \lim_{n \to \infty} \frac{n}{3^n}\] \[= \lim_{n \to \infty} \frac{n}{3^n} = 0\]

(since exponentials grow much faster than polynomials)
Apply the root test: Since \(L = 0 < 1\), the series converges.

Answer: The series converges.

4.23. Root Test with Exponential Powers (a) (Lab 10, Part V, Homework H1)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{2^{n^2-1}}{3^{n^2} \cdot \sqrt{n}}\]

Click to see the solution

Apply the root test: \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{2^{n^2-1}}{3^{n^2} \cdot \sqrt{n}}}\]
Simplify: \[= \lim_{n \to \infty} \frac{\sqrt[n]{2^{n^2-1}}}{\sqrt[n]{3^{n^2} \cdot \sqrt{n}}}\] \[= \lim_{n \to \infty} \frac{2^{n-1/n}}{3^n \cdot n^{1/(2n)}}\] \[= \lim_{n \to \infty} \frac{2^n \cdot 2^{-1/n}}{3^n \cdot n^{1/(2n)}}\] \[= \lim_{n \to \infty} \left(\frac{2}{3}\right)^n \cdot \frac{2^{-1/n}}{n^{1/(2n)}}\]
Evaluate limits:
- \(\lim_{n \to \infty} 2^{-1/n} = 1\)
- \(\lim_{n \to \infty} n^{1/(2n)} = 1\) (taking log: \(\frac{\ln n}{2n} \to 0\))
- \(\lim_{n \to \infty} \left(\frac{2}{3}\right)^n = 0\)
Combine: \[L = 0 \cdot \frac{1}{1} = 0\]
Apply the root test: Since \(L = 0 < 1\), the series converges.

Answer: The series converges.

4.24. Series with Factorial and Odd Numbers (Lab 10, Part I, Example 2)

Find the sum of the series:

\[\sum_{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}\]

Click to see the solution

Key Concept: Express the general term as a difference to create a telescoping series.

Denote the product of odd numbers: Let \(P_n = 1 \cdot 3 \cdot 5 \cdots (2n+1)\).
Find a telescoping form: Notice that: \[\frac{n}{P_n} = \frac{2n+1-1}{2P_n} = \frac{1}{2}\left(\frac{2n+1}{P_n} - \frac{1}{P_n}\right)\]

Since \(P_n = 1 \cdot 3 \cdot 5 \cdots (2n+1)\), we have \(P_{n-1} = 1 \cdot 3 \cdot 5 \cdots (2n-1)\), so: \[\frac{2n+1}{P_n} = \frac{1}{P_{n-1}}\]

Therefore: \[\frac{n}{P_n} = \frac{1}{2}\left(\frac{1}{P_{n-1}} - \frac{1}{P_n}\right)\]
Write the partial sum: \[s_N = \frac{1}{2}\sum_{n=1}^{N}\left(\frac{1}{P_{n-1}} - \frac{1}{P_n}\right)\]

This telescopes: \[s_N = \frac{1}{2}\left(\frac{1}{P_0} - \frac{1}{P_N}\right) = \frac{1}{2}\left(1 - \frac{1}{P_N}\right)\]

(where \(P_0 = 1\))
Take the limit: \[\lim_{N \to \infty} s_N = \frac{1}{2}(1 - 0) = \frac{1}{2}\]

Answer: \(\frac{1}{2}\)

4.25. Series with Factorial in Denominator (Lab 10, Part I, Homework H2)

Determine whether the series converges and find its sum if possible:

\[\sum_{n=0}^{\infty} \frac{n^2+3}{n!}\]

Click to see the solution

Key Concept: Break the series into simpler parts and use known series involving factorials.

Split the series: \[\sum_{n=0}^{\infty} \frac{n^2+3}{n!} = \sum_{n=0}^{\infty} \frac{n^2}{n!} + 3\sum_{n=0}^{\infty} \frac{1}{n!}\]
Evaluate the second sum: The Taylor series for \(e^x\) gives us: \[\sum_{n=0}^{\infty} \frac{1}{n!} = e\]
Evaluate the first sum: For \(\sum_{n=0}^{\infty} \frac{n^2}{n!}\), write \(n^2 = n(n-1) + n\): \[\sum_{n=0}^{\infty} \frac{n^2}{n!} = \sum_{n=0}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=0}^{\infty} \frac{n}{n!}\]

For the first part (note terms with \(n=0,1\) contribute 0): \[\sum_{n=2}^{\infty} \frac{n(n-1)}{n!} = \sum_{n=2}^{\infty} \frac{1}{(n-2)!} = \sum_{k=0}^{\infty} \frac{1}{k!} = e\]

For the second part (note term with \(n=0\) contributes 0): \[\sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = \sum_{k=0}^{\infty} \frac{1}{k!} = e\]

Therefore: \[\sum_{n=0}^{\infty} \frac{n^2}{n!} = e + e = 2e\]
Combine the results: \[\sum_{n=0}^{\infty} \frac{n^2+3}{n!} = 2e + 3e = 5e\]

Answer: The series converges to \(5e\)

4.26. \(n\)-th Term Test Example (b) (Lab 10, Part II, Example 2)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{n^2}{2n^2+1}\]

Click to see the solution

Find the limit of the general term: \[\lim_{n \to \infty} \frac{n^2}{2n^2+1} = \lim_{n \to \infty} \frac{n^2}{n^2(2+\frac{1}{n^2})} = \lim_{n \to \infty} \frac{1}{2+\frac{1}{n^2}} = \frac{1}{2} \neq 0\]
Apply the \(n\)-th term test: Since \(\lim_{n \to \infty} a_n = \frac{1}{2} \neq 0\), the series diverges by the \(n\)-th term test.

Answer: The series diverges.

4.27. Comparison Test Example (b) (Lab 10, Part II, Homework H2)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}-\sqrt{n+1}}\]

Click to see the solution

Key Concept: Rationalize the denominator to find an equivalent form.

Rationalize the denominator: \[\frac{1}{\sqrt{n}-\sqrt{n+1}} = \frac{1}{\sqrt{n}-\sqrt{n+1}} \cdot \frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n}+\sqrt{n+1}}\] \[= \frac{\sqrt{n}+\sqrt{n+1}}{n-(n+1)} = \frac{\sqrt{n}+\sqrt{n+1}}{-1} = -(\sqrt{n}+\sqrt{n+1})\]
Analyze the term: \[|a_n| = \sqrt{n}+\sqrt{n+1} \sim 2\sqrt{n}\]
Compare with a known series: Since \(a_n \sim 2\sqrt{n}\) and \(\sum \sqrt{n}\) diverges (it’s a \(p\)-series with \(p = -\frac{1}{2} < 1\), or equivalently, the terms grow without bound), the series diverges.

Answer: The series diverges.

4.28. Direct Comparison Test with Geometric Series (Lab 10, Part III, Example 2)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{2n-1}{(\sqrt{2})^n}\]

Click to see the solution

Key Concept: For series with exponentials in the denominator, compare with a geometric series.

Bound the numerator: For all \(n \geq 1\), we have \(2n - 1 \leq 2n\), so: \[\frac{2n-1}{(\sqrt{2})^n} \leq \frac{2n}{(\sqrt{2})^n}\]
Apply the ratio test: Let \(a_n = \frac{n}{(\sqrt{2})^n}\). Then: \[L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{n+1}{(\sqrt{2})^{n+1}}}{\frac{n}{(\sqrt{2})^n}} = \lim_{n \to \infty} \frac{n+1}{n} \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} < 1\]

Therefore, \(\sum \frac{n}{(\sqrt{2})^n}\) converges, and so does \(\sum \frac{2n}{(\sqrt{2})^n}\).
Apply the comparison test: By the direct comparison test, \(\sum \frac{2n-1}{(\sqrt{2})^n}\) also converges.

Answer: The series converges.

4.29. Integral Test or Comparison (Lab 10, Part III, Homework H2)

Study the convergence of:

\[\sum_{n=3}^{\infty} \frac{\ln(n)}{n}\]

Click to see the solution

Key Concept: Use the integral test or comparison with known series.

Apply the integral test: Consider the function \(f(x) = \frac{\ln(x)}{x}\) for \(x \geq 3\). This function is positive and decreasing for \(x \geq e\).

Evaluate the improper integral: \[\int_3^{\infty} \frac{\ln(x)}{x} dx\]
Use substitution: Let \(u = \ln(x)\), so \(du = \frac{1}{x}dx\): \[\int_3^{\infty} \frac{\ln(x)}{x} dx = \int_{\ln 3}^{\infty} u \, du = \left[\frac{u^2}{2}\right]_{\ln 3}^{\infty} = \infty\]
Conclude: Since the integral diverges, by the integral test, the series diverges.

Answer: The series diverges.

4.30. Ratio Test with Exponential (b) (Lab 10, Part IV, Example 2)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{n!}{e^{n^2}}\]

Click to see the solution

Apply the ratio test: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{(n+1)!}{e^{(n+1)^2}}}{\frac{n!}{e^{n^2}}}\]
Simplify: \[= \lim_{n \to \infty} \frac{(n+1)! \cdot e^{n^2}}{n! \cdot e^{(n+1)^2}}\] \[= \lim_{n \to \infty} (n+1) \cdot e^{n^2 - (n+1)^2}\] \[= \lim_{n \to \infty} (n+1) \cdot e^{n^2 - n^2 - 2n - 1}\] \[= \lim_{n \to \infty} (n+1) \cdot e^{-2n-1}\] \[= \lim_{n \to \infty} \frac{n+1}{e^{2n+1}} = 0\]
Apply the ratio test: Since \(L = 0 < 1\), the series converges.

Answer: The series converges.

4.31. Ratio Test with Stirling’s Approximation (Lab 10, Part IV, Homework H2)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{2^n \cdot n!}{n^n}\]

Click to see the solution

Apply the ratio test: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{2^{n+1} \cdot (n+1)!}{(n+1)^{n+1}}}{\frac{2^n \cdot n!}{n^n}}\]
Simplify: \[= \lim_{n \to \infty} \frac{2 \cdot 2^n \cdot (n+1) \cdot n! \cdot n^n}{2^n \cdot n! \cdot (n+1)^{n+1}}\] \[= \lim_{n \to \infty} \frac{2(n+1) \cdot n^n}{(n+1)^{n+1}}\] \[= \lim_{n \to \infty} \frac{2 \cdot n^n}{(n+1)^n}\] \[= 2 \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n\] \[= 2 \lim_{n \to \infty} \left(\frac{1}{1+\frac{1}{n}}\right)^n\] \[= \frac{2}{e}\]
Apply the ratio test: Since \(L = \frac{2}{e} \approx 0.736 < 1\), the series converges.

Answer: The series converges.

4.32. Root Test with Rational Powers (Lab 10, Part V, Example 2)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{(2n+1)^n}{n^{2n}}\]

Click to see the solution

Apply the root test: \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{(2n+1)^n}{n^{2n}}}\]
Simplify: \[= \lim_{n \to \infty} \frac{\sqrt[n]{(2n+1)^n}}{\sqrt[n]{n^{2n}}}\] \[= \lim_{n \to \infty} \frac{2n+1}{n^2}\] \[= \lim_{n \to \infty} \frac{2n}{n^2}\] \[= \lim_{n \to \infty} \frac{2}{n} = 0\]
Apply the root test: Since \(L = 0 < 1\), the series converges.

Answer: The series converges.

4.33. Root Test with Pattern (Lab 10, Part V, Homework H2)

Study the convergence of:

\[\left(\frac{3}{4}\right)^{1/2} + \left(\frac{5}{7}\right)^{3/2} + \left(\frac{7}{10}\right)^{5/2} + \dots\]

Click to see the solution

Key Concept: Identify the pattern and express the general term.

Identify the pattern: Looking at the terms:
- \(n=1\): \(\left(\frac{3}{4}\right)^{1/2} = \left(\frac{2(1)+1}{3(1)+1}\right)^{1/2}\)
- \(n=2\): \(\left(\frac{5}{7}\right)^{3/2} = \left(\frac{2(2)+1}{3(2)+1}\right)^{3/2}\)
- \(n=3\): \(\left(\frac{7}{10}\right)^{5/2} = \left(\frac{2(3)+1}{3(3)+1}\right)^{5/2}\)
The general term is: \[a_n = \left(\frac{2n+1}{3n+1}\right)^{(2n-1)/2}\]
Apply the root test: This is challenging directly. Instead, note that: \[a_n = \left(\frac{2n+1}{3n+1}\right)^{(2n-1)/2}\]
Find the limit: \[\lim_{n \to \infty} a_n^{1/n} = \lim_{n \to \infty} \left(\frac{2n+1}{3n+1}\right)^{(2n-1)/(2n)}\]

As \(n \to \infty\):
- \(\frac{2n+1}{3n+1} \to \frac{2}{3}\)
- \(\frac{2n-1}{2n} \to 1\)
Therefore: \[L = \left(\frac{2}{3}\right)^1 = \frac{2}{3} < 1\]
Apply the root test: Since \(L = \frac{2}{3} < 1\), the series converges.

Answer: The series converges.

4.34. Series with Trigonometric Functions (Lab 10, Part I, Example 3)

Find the sum of the series:

\[\sum_{n=1}^{\infty} \frac{\sin\left(\frac{1}{n(n+1)}\right)}{\cos\left(\frac{1}{n}\right)\cos\left(\frac{1}{n+1}\right)}\]

Click to see the solution

Key Concept: Use the trigonometric identity \(\sin(A-B) = \sin A \cos B - \cos A \sin B\) to create a telescoping series.

Apply the sine difference formula: Note that: \[\sin\left(\frac{1}{n} - \frac{1}{n+1}\right) = \sin\left(\frac{1}{n}\right)\cos\left(\frac{1}{n+1}\right) - \cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n+1}\right)\]
Recognize the pattern: Since \(\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\), and for small \(x\), \(\sin x \approx x\), we can use: \[\frac{\sin\left(\frac{1}{n(n+1)}\right)}{\cos\left(\frac{1}{n}\right)\cos\left(\frac{1}{n+1}\right)} = \tan\left(\frac{1}{n}\right) - \tan\left(\frac{1}{n+1}\right)\]

This follows from the tangent difference formula.
Write the partial sum: \[s_N = \sum_{n=1}^{N}\left[\tan\left(\frac{1}{n}\right) - \tan\left(\frac{1}{n+1}\right)\right]\]

This telescopes: \[s_N = \tan(1) - \tan\left(\frac{1}{N+1}\right)\]
Take the limit: \[\lim_{N \to \infty} s_N = \tan(1) - \tan(0) = \tan(1)\]

Answer: \(\tan(1)\)

4.35. Series with Oscillating Sine Function (Lab 10, Part II, Example 3)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{1}{\sin^4 n}\]

Click to see the solution

Key Concept: The terms \(\sin^4 n\) oscillate and can be very small, making \(\frac{1}{\sin^4 n}\) very large.

Analyze the general term: The values \(\sin n\) for \(n \in \mathbb{N}\) are dense in \([-1, 1]\) and can get arbitrarily close to 0. When \(\sin n\) is close to 0, \(\frac{1}{\sin^4 n}\) becomes arbitrarily large.
Consider the limit: The sequence \(a_n = \frac{1}{\sin^4 n}\) does not have a limit as \(n \to \infty\) (it oscillates), but more importantly, there are infinitely many terms that are very large.
Comparison argument: By Dirichlet’s approximation theorem, there exist infinitely many integers \(n\) such that \(|n - k\pi| < \frac{1}{n}\) for some integer \(k\). For such \(n\), \(|\sin n| < C/n\) for some constant \(C\), giving \(\frac{1}{\sin^4 n} > \frac{n^4}{C^4}\), which would make the series diverge.

Answer: The series diverges (the terms do not approach zero consistently, and infinitely many terms are unbounded).

4.36. Limit Comparison Test with \(p\)-Series (Lab 10, Part III, Example 3)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{2n+1}{(n+1)^2(n+2)^2}\]

Click to see the solution

Key Concept: Identify the dominant behavior for large \(n\) and use limit comparison.

Determine asymptotic behavior: For large \(n\):
- Numerator: \(2n + 1 \sim 2n\)
- Denominator: \((n+1)^2(n+2)^2 \sim n^2 \cdot n^2 = n^4\)
Therefore: \[\frac{2n+1}{(n+1)^2(n+2)^2} \sim \frac{2n}{n^4} = \frac{2}{n^3}\]
Apply limit comparison test: Let \(a_n = \frac{2n+1}{(n+1)^2(n+2)^2}\) and \(b_n = \frac{1}{n^3}\).

\[L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(2n+1)n^3}{(n+1)^2(n+2)^2}\]

Dividing numerator and denominator by \(n^4\): \[L = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{(1+\frac{1}{n})^2(1+\frac{2}{n})^2} = \frac{2}{1 \cdot 1} = 2\]
Conclude: Since \(0 < L = 2 < \infty\) and \(\sum \frac{1}{n^3}\) converges (\(p\)-series with \(p = 3 > 1\)), the original series also converges.

Answer: The series converges.

4.37. Logarithmic Series (Lab 10, Part III, Homework H3)

Study the convergence of:

\[\sum_{n=2021}^{\infty} \frac{1}{(\ln(n))^{\ln(n)}}\]

Click to see the solution

Key Concept: Use the root test or comparison with exponential decay.

Apply the root test: \[L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{(\ln(n))^{\ln(n)}}}\] \[= \lim_{n \to \infty} \frac{1}{(\ln(n))^{\ln(n)/n}}\]
Evaluate the exponent: \[\lim_{n \to \infty} \frac{\ln(n)}{n} = 0\]

Therefore: \[\lim_{n \to \infty} (\ln(n))^{\ln(n)/n} = \lim_{n \to \infty} e^{\frac{\ln(n) \cdot \ln(\ln(n))}{n}} = e^0 = 1\]
Root test is inconclusive: Since \(L = 1\), the root test is inconclusive.
Alternative approach - comparison: For large \(n\), \(\ln(n) > 2\), so \((\ln(n))^{\ln(n)} > 2^{\ln(n)} = n^{\ln 2}\).

Therefore: \[\frac{1}{(\ln(n))^{\ln(n)}} < \frac{1}{n^{\ln 2}}\]

Since \(\ln 2 \approx 0.693 > 0\), and we need \(p > 1\) for convergence, this comparison doesn’t immediately help. However, the term \((\ln(n))^{\ln(n)}\) grows much faster than any polynomial, so the series converges rapidly.
More precise analysis: Since \((\ln(n))^{\ln(n)} = e^{\ln(n) \ln(\ln(n))}\) grows faster than \(n^k\) for any fixed \(k\), we have: \[\frac{1}{(\ln(n))^{\ln(n)}} < \frac{1}{n^2}\]

for sufficiently large \(n\). By comparison with the convergent \(p\)-series \(\sum \frac{1}{n^2}\), the series converges.

Answer: The series converges.

4.38. Comparison Test Analysis (Lab 10, Part III, Example 4)

Study the convergence of:

\[\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)\sqrt{n}}\]

Click to see the solution

Simplify the expression: \[\frac{\sqrt{n}}{(n+1)\sqrt{n}} = \frac{1}{n+1}\]
Compare with harmonic series: The series becomes: \[\sum_{n=1}^{\infty} \frac{1}{n+1} = \sum_{k=2}^{\infty} \frac{1}{k}\]

This is the harmonic series starting from \(k=2\), which diverges.

Answer: The series diverges.

4.39. Epsilon-Delta Definition: Limit of Square Root (Lab 11, Task 1)

Prove by epsilon-delta definition that \(\lim_{x \to 5} \sqrt{x - 1} = 2\)

Click to see the solution

Key Concept: Use the definition of a limit to find a suitable \(\delta\) for any given \(\varepsilon > 0\) such that \(|\sqrt{x-1} - 2| < \varepsilon\) whenever \(0 < |x - 5| < \delta\).

Set up the inequality: For any given \(\varepsilon > 0\), we need to find \(\delta > 0\) such that: \[\left| \sqrt{x - 1} - 2 \right| < \varepsilon \quad \text{whenever} \quad 0 < |x - 5| < \delta\]
Manipulate the inequality: For all \(x \neq 5\): \[\left| \sqrt{x - 1} - 2 \right| < \varepsilon \implies -\varepsilon < \sqrt{x - 1} - 2 < \varepsilon\] \[\implies 2 - \varepsilon < \sqrt{x - 1} < 2 + \varepsilon\] \[\implies (2 - \varepsilon)^2 < x - 1 < (2 + \varepsilon)^2\] \[\implies 1 + (2 - \varepsilon)^2 < x < 1 + (2 + \varepsilon)^2\]
Find \(\delta\): We need the interval \((5 - \delta, 5 + \delta)\) centered at \(x = 5\) to be inside \((1 + (2 - \varepsilon)^2, 1 + (2 + \varepsilon)^2)\).

The distance from \(x = 5\) to the endpoints is:
- Left endpoint: \(5 - (1 + (2 - \varepsilon)^2) = 4 - (2 - \varepsilon)^2 = 4\varepsilon - \varepsilon^2\)
- Right endpoint: \((1 + (2 + \varepsilon)^2) - 5 = -4 + (2 + \varepsilon)^2 = 4\varepsilon + \varepsilon^2\)
For \(\varepsilon < 4\), take: \[\delta = \min \{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\} = 4\varepsilon - \varepsilon^2\]
For \(\varepsilon \geq 4\): Take \(\delta = \min \{5, -4 + (2 + \varepsilon)^2\}\)

For example:
- If \(\varepsilon = 1\): \(\delta = \min\{3, 5\} = 3\)
- If \(\varepsilon = 5\): \(\delta = \min\{5, 45\} = 5\)
Conclusion: With this choice of \(\delta\), the inequality \(0 < |x - 5| < \delta\) automatically places \(x\) in the desired interval, making \(\left| \sqrt{x - 1} - 2 \right| < \varepsilon\).

Answer: For any \(\varepsilon > 0\), choose \(\delta = \min \{4\varepsilon - \varepsilon^2, 4\varepsilon + \varepsilon^2\}\) when \(\varepsilon < 4\), or \(\delta = \min \{5, -4 + (2 + \varepsilon)^2\}\) when \(\varepsilon \geq 4\). This proves \(\lim_{x \to 5} \sqrt{x - 1} = 2\).

4.40. Epsilon-Delta Definition: Piecewise Function (Lab 11, Task 2)

Let \[f(x) = \begin{cases} x^2, & x \neq 2; \\ 1, & x = 2. \end{cases}\] Prove by epsilon-delta definition that \(\lim_{x \to 2} f(x) = 4\).

Click to see the solution

Key Concept: The limit depends on the behavior near \(x = 2\), not at \(x = 2\). Since \(f(x) = x^2\) for all \(x \neq 2\), we work with \(|x^2 - 4| < \varepsilon\).

Set up the inequality: For any given \(\varepsilon > 0\), we need to find \(\delta > 0\) such that: \[|f(x) - 4| < \varepsilon \quad \text{whenever} \quad 0 < |x - 2| < \delta\]
Work with the inequality (assuming \(\varepsilon < 4\)): For \(x \neq 2\), we have \(f(x) = x^2\): \[|x^2 - 4| < \varepsilon \implies -\varepsilon < x^2 - 4 < \varepsilon\] \[\implies 4 - \varepsilon < x^2 < 4 + \varepsilon\] \[\implies \sqrt{4 - \varepsilon} < x < \sqrt{4 + \varepsilon}\]
Find \(\delta\) for \(\varepsilon < 4\): We want the interval \((2 - \delta, 2 + \delta)\) centered at \(x = 2\) to lie inside \((\sqrt{4 - \varepsilon}, \sqrt{4 + \varepsilon})\).

Take \(\delta\) to be the distance from \(x = 2\) to the nearer endpoint: \[\delta = \min \{2 - \sqrt{4 - \varepsilon}, \sqrt{4 + \varepsilon} - 2\}\]
For \(\varepsilon \geq 4\): Take: \[\delta = \min \{2, \sqrt{4 + \varepsilon} - 2\}\]
Conclusion: With this choice of \(\delta\), the inequality \(0 < |x - 2| < \delta\) ensures \(x\) is in the interval where \(|f(x) - 4| = |x^2 - 4| < \varepsilon\).

Answer: For any \(\varepsilon > 0\), choose \(\delta = \min \{2 - \sqrt{4 - \varepsilon}, \sqrt{4 + \varepsilon} - 2\}\) when \(\varepsilon < 4\), or \(\delta = \min \{2, \sqrt{4 + \varepsilon} - 2\}\) when \(\varepsilon \geq 4\). This proves \(\lim_{x \to 2} f(x) = 4\).

4.41. Leibniz Test Application (a) (Lab 11, Task 3a)

Use the Leibniz test to show that the following series converges: \[\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\]

Click to see the solution

Key Concept: Apply the Leibniz test by checking that the terms \(a_k = \frac{1}{k}\) are positive, decreasing, and approach zero.

Verify positivity: \[a_k = \frac{1}{k} > 0 \quad \forall k \geq 1 \quad \checkmark\]
Verify decreasing: \[a_k = \frac{1}{k} > \frac{1}{k + 1} = a_{k+1} \quad \forall k \geq 1 \quad \checkmark\]
Verify limit is zero: \[\lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{1}{k} = 0 \quad \checkmark\]
Apply Leibniz test: All three conditions are satisfied, so the series converges.

Answer: The series converges by the Leibniz test. (It converges to \(\ln(2)\).)

4.42. Leibniz Test Application (b) (Lab 11, Task 3b)

Use the Leibniz test to show that the following series converges: \[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k + 1}\]

Click to see the solution

Verify positivity: \[a_k = \frac{1}{2k + 1} > 0 \quad \forall k \geq 1 \quad \checkmark\]
Verify decreasing: \[a_k = \frac{1}{2k + 1} > \frac{1}{2k + 3} = a_{k+1} \quad \forall k \geq 1 \quad \checkmark\]
Verify limit is zero: \[\lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{1}{2k + 1} = 0 \quad \checkmark\]
Apply Leibniz test: All conditions of the alternating series test are satisfied.

Answer: The series converges by the Leibniz test.

4.43. Leibniz Test Application (c) (Lab 11, Task 3c)

Use the Leibniz test to show that the following series converges: \[\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k + 3}{k(k + 1)}\]

Click to see the solution

Verify positivity: \[a_k = \frac{k + 3}{k(k + 1)} > 0 \quad \forall k \geq 1 \quad \checkmark\]
Verify decreasing: Check if \(a_{k+1} < a_k\) by comparing the ratio: \[\frac{a_{k+1}}{a_k} = \frac{k + 4}{(k + 1)(k + 2)} \cdot \frac{k(k + 1)}{k + 3} = \frac{(k + 4) \cdot k(k + 1)}{(k + 1)(k + 2)(k + 3)}\] \[= \frac{k(k + 4)}{(k + 2)(k + 3)} = \frac{k^2 + 4k}{k^2 + 5k + 6}\]

For \(k \geq 1\): \(k^2 + 4k < k^2 + 5k + 6\), so \(\frac{a_{k+1}}{a_k} < 1\), thus \(a_{k+1} < a_k\). ✓
Verify limit is zero: \[\lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{k + 3}{k(k + 1)} = \lim_{k \to \infty} \frac{k + 3}{k^2 + k} = 0 \quad \checkmark\]
Apply Leibniz test: All conditions are satisfied.

Answer: The series converges by the Leibniz test.

4.44. Leibniz Test Application (d) (Lab 11, Task 3d)

Use the Leibniz test to show that the following series converges: \[\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k}{3^k}\]

Click to see the solution

Verify positivity: \[a_k = \frac{k}{3^k} > 0 \quad \forall k \geq 1 \quad \checkmark\]
Verify decreasing: Check the ratio: \[\frac{a_{k+1}}{a_k} = \frac{k + 1}{3^{k+1}} \cdot \frac{3^k}{k} = \frac{k + 1}{3k}\]

For \(k \geq 1\): \(\frac{k + 1}{3k} = \frac{k + 1}{3k} \leq \frac{2}{3} < 1\) (equality when \(k = 1\))

Thus \(a_{k+1} < a_k\) for all \(k \geq 1\). ✓
Verify limit is zero: \[\lim_{k \to \infty} a_k = \lim_{k \to \infty} \frac{k}{3^k} = 0\]

(Exponentials grow faster than polynomials) ✓
Apply Leibniz test: All conditions are satisfied.

Answer: The series converges by the Leibniz test.

4.45. Absolute Convergence Classification (a) (Lab 11, Task 4a)

Classify the series as absolutely convergent, conditionally convergent, or divergent: \[\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k!}\]

Click to see the solution

Key Concept: Test for absolute convergence first using the ratio test.

Test for absolute convergence: Consider: \[\sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1}}{k!} \right| = \sum_{k=1}^{\infty} \frac{1}{k!}\]
Apply the ratio test: \[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \frac{\frac{1}{(k+1)!}}{\frac{1}{k!}} = \lim_{k \to \infty} \frac{k!}{(k + 1)!} = \lim_{k \to \infty} \frac{1}{k + 1} = 0\]
Conclude: Since \(L = 0 < 1\), the series of absolute values converges.

Answer: The series converges absolutely.

4.46. Absolute Convergence Classification (b) (Lab 11, Task 4b)

Classify the series as absolutely convergent, conditionally convergent, or divergent: \[\sum_{k=1}^{\infty} \frac{\sin k}{k^3}\]

Click to see the solution

Test for absolute convergence: Consider: \[\sum_{k=1}^{\infty} \left| \frac{\sin k}{k^3} \right|\]
Use comparison test: Since \(|\sin k| \leq 1\) for all \(k\): \[\left| \frac{\sin k}{k^3} \right| \leq \frac{1}{k^3}\]
Compare with \(p\)-series: The series \(\sum_{k=1}^{\infty} \frac{1}{k^3}\) is a convergent \(p\)-series with \(p = 3 > 1\).
Apply comparison test: By the comparison test, \(\sum \left| \frac{\sin k}{k^3} \right|\) converges.

Answer: The series converges absolutely.

4.47. Absolute Convergence Classification (c) (Lab 11, Task 4c)

Classify the series as absolutely convergent, conditionally convergent, or divergent: \[\sum_{k=3}^{\infty} \frac{(-1)^k \ln k}{k}\]

Click to see the solution

Key Concept: Check absolute convergence first, then conditional convergence if absolute convergence fails.

Test for absolute convergence: Consider: \[\sum_{k=3}^{\infty} \left| \frac{(-1)^k \ln k}{k} \right| = \sum_{k=3}^{\infty} \frac{\ln k}{k}\]
Use comparison with harmonic series: For \(k \geq 3\): \(\ln k > 1\), so: \[\frac{\ln k}{k} > \frac{1}{k}\]

Since \(\sum_{k=1}^{\infty} \frac{1}{k}\) diverges (harmonic series), by comparison test, \(\sum_{k=3}^{\infty} \frac{\ln k}{k}\) also diverges.
The series does NOT converge absolutely.
Test for conditional convergence using Leibniz test: Let \(b_k = \frac{\ln k}{k}\).
- Positivity: \(b_k > 0\) for all \(k \geq 3\) ✓
- Limit is zero: Using L’Hôpital’s rule: \[\lim_{k \to \infty} \frac{\ln k}{k} = \lim_{k \to \infty} \frac{1/k}{1} = 0 \quad \checkmark\]
- Decreasing: We need to verify \(b_{k+1} < b_k\). Consider: \[k(k + 1)(b_{k+1} - b_k) = \ln(k+1) - (k+1)\ln\frac{k+1}{k} - \ln k + k\ln\frac{k+1}{k}\] \[= \ln\frac{k+1}{k} \cdot k - \ln k = k\ln\left(1 + \frac{1}{k}\right) - \ln k\]
  
  As \(k \to \infty\), this approaches \(-\infty\), so eventually \(b_{k+1} < b_k\). ✓
Apply Leibniz test: The series converges by the Leibniz test.

Answer: The series converges conditionally.

4.48. Absolute Convergence Classification (d) (Lab 11, Task 4d)

Classify the series as absolutely convergent, conditionally convergent, or divergent: \[\sum_{k=1}^{\infty} \frac{k \cos \pi k}{k^2 + 1}\]

Click to see the solution

Key Concept: Note that \(\cos(\pi k) = (-1)^k\), so this is an alternating series.

Rewrite the series: \[\sum_{k=1}^{\infty} \frac{k \cos \pi k}{k^2 + 1} = \sum_{k=1}^{\infty} \frac{(-1)^k k}{k^2 + 1}\]
Test for absolute convergence: Consider: \[\sum_{k=1}^{\infty} \left| \frac{(-1)^k k}{k^2 + 1} \right| = \sum_{k=1}^{\infty} \frac{k}{k^2 + 1}\]
Use limit comparison test with \(\sum \frac{1}{k}\): \[\lim_{k \to \infty} \frac{\frac{k}{k^2 + 1}}{\frac{1}{k}} = \lim_{k \to \infty} \frac{k^2}{k^2 + 1} = 1\]

Since \(\sum \frac{1}{k}\) diverges and the limit is positive and finite, \(\sum \frac{k}{k^2 + 1}\) also diverges.
The series does NOT converge absolutely.
Test for conditional convergence using Leibniz test: Let \(b_k = \frac{k}{k^2+1}\).
- Positivity: \(b_k > 0\) for all \(k \geq 1\) ✓
- Limit is zero: \[\lim_{k \to \infty} \frac{k}{k^2 + 1} = \lim_{k \to \infty} \frac{1}{k + \frac{1}{k}} = 0 \quad \checkmark\]
- Decreasing: Check the ratio: \[\frac{b_{k+1}}{b_k} = \frac{(k + 1)(k^2 + 1)}{k((k + 1)^2 + 1)} = \frac{(k + 1)(k^2 + 1)}{k(k^2 + 2k + 2)}\]
  
  We need to show this is less than 1: \[(k + 1)(k^2 + 1) < k(k^2 + 2k + 2)\] \[k^3 + k^2 + k + 1 < k^3 + 2k^2 + 2k\] \[1 < k^2 + k \quad \text{(true for all } k \geq 1\text{)}\]
  
  So \(b_{k+1} < b_k\). ✓
Apply Leibniz test: The series converges by the Leibniz test.

Answer: The series converges conditionally.

4.49. Power Series: Radius and Interval (a) (Lab 11, Task 5a)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints. \[\sum_{k=1}^{\infty} \frac{k!x^k}{k^k}\]

Click to see the solution

Key Concept: Use the ratio test to find the radius of convergence, then check endpoints separately.

Apply the ratio test: Let \(a_k = \frac{k!}{k^k}\). Then: \[\rho = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| |x| = \lim_{k \to \infty} \left| \frac{\frac{(k+1)!}{(k+1)^{k+1}}}{\frac{k!}{k^k}} \right| |x|\]
Simplify: \[= \lim_{k \to \infty} \frac{(k + 1)!}{(k + 1)^{k+1}} \cdot \frac{k^k}{k!} |x|\] \[= \lim_{k \to \infty} \frac{(k + 1) \cdot k!}{k!} \cdot \frac{k^k}{(k + 1)^{k+1}} |x|\] \[= \lim_{k \to \infty} \frac{k^k}{(k + 1)^k} |x|\] \[= \lim_{k \to \infty} \left(\frac{k}{k + 1}\right)^k |x|\] \[= \lim_{k \to \infty} \left(\frac{1}{1 + \frac{1}{k}}\right)^k |x|\] \[= \frac{1}{e} |x|\]
Find the radius: The series converges absolutely when \(\frac{1}{e}|x| < 1\), i.e., \(|x| < e\).

Radius of convergence: \(R = e\)
Check the endpoint \(x = e\): \[\sum_{k=1}^{\infty} \frac{k!e^k}{k^k}\]

By Stirling’s approximation or induction, we can show that \(\frac{k!e^k}{k^k} \geq e\) for all \(k \geq 1\).

Since the terms don’t approach zero, the series diverges.
Check the endpoint \(x = -e\): \[\sum_{k=1}^{\infty} \frac{k!(-e)^k}{k^k} = \sum_{k=1}^{\infty} (-1)^k \frac{k!e^k}{k^k}\]

Since \(\frac{k!e^k}{k^k} \geq e\), the terms don’t approach zero, so the series diverges.

Answer: Radius of convergence \(R = e\). Interval of convergence: \((-e, e)\).

4.50. Power Series: Radius and Interval (b) (Lab 11, Task 5b)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints. \[\sum_{k=0}^{\infty} kx^k\]

Click to see the solution

Apply the ratio test (for \(x \neq 0\)): \[\rho = \lim_{k \to \infty} \left| \frac{u_{k+1}}{u_k} \right| = \lim_{k \to \infty} \left| \frac{(k + 1)x^{k+1}}{kx^k} \right| = |x| \lim_{k \to \infty} \frac{k + 1}{k} = |x|\]
Find the radius: The series converges when \(\rho < 1\), i.e., \(|x| < 1\).

Radius of convergence: \(R = 1\)
Check the endpoints:
- At \(x = 1\): \[\sum_{k=0}^{\infty} k = 0 + 1 + 2 + 3 + \cdots \to \infty\] Diverges.
- At \(x = -1\): \[\sum_{k=0}^{\infty} k(-1)^k = 0 - 1 + 2 - 3 + \cdots\] Since \(|k(-1)^k| = k \to \infty\), the series diverges.

Answer: Radius of convergence \(R = 1\). Interval of convergence: \((-1, 1)\). Absolute convergence occurs in \((-1, 1)\).

4.51. Power Series: Radius and Interval (c) (Lab 11, Task 5c)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints. \[\sum_{k=2}^{\infty} \frac{x^k}{\ln k}\]

Click to see the solution

Apply the ratio test: \[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{x^{k+1}}{\ln(k + 1)} \cdot \frac{\ln k}{x^k} \right| = |x| \lim_{k \to \infty} \frac{\ln k}{\ln(k + 1)}\]
Evaluate the limit: \[\lim_{k \to \infty} \frac{\ln k}{\ln(k + 1)} = 1\]

So \(L = |x|\).
Find the radius: The series converges absolutely when \(L < 1\), i.e., \(|x| < 1\).

Radius of convergence: \(R = 1\)
Check the endpoint \(x = 1\): \[\sum_{k=2}^{\infty} \frac{1}{\ln k}\]

Since \(\ln k < k\) for \(k \geq 2\), we have \(\frac{1}{\ln k} > \frac{1}{k}\).

The harmonic series \(\sum \frac{1}{k}\) diverges, so by comparison, \(\sum \frac{1}{\ln k}\) diverges.
Check the endpoint \(x = -1\): \[\sum_{k=2}^{\infty} \frac{(-1)^k}{\ln k}\]

Check the Alternating Series Test conditions:
- \(\frac{1}{\ln k} > 0\) for \(k \geq 2\) ✓
- \(\frac{1}{\ln k}\) is decreasing ✓
- \(\lim_{k \to \infty} \frac{1}{\ln k} = 0\) ✓
By the Leibniz test, the series converges.

Answer: Radius of convergence \(R = 1\). Interval of convergence: \([-1, 1)\). The series converges conditionally at \(x = -1\).

4.52. Power Series: Radius and Interval (d) (Lab 11, Task 5d)

Find the radius and interval of convergence of the following series. Investigate what happens at the endpoints. \[\sum_{k=1}^{\infty} \frac{(-1)^{k-1} x^k}{\sqrt{k}}\]

Click to see the solution

Apply the ratio test: \[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{x^{k+1}}{\sqrt{k + 1}} \cdot \frac{\sqrt{k}}{x^k} \right| = |x| \lim_{k \to \infty} \sqrt{\frac{k}{k + 1}} = |x|\]
Find the radius: The series converges absolutely when \(L < 1\), i.e., \(|x| < 1\).

Radius of convergence: \(R = 1\)
Check the endpoint \(x = 1\): \[\sum_{k=1}^{\infty} (-1)^{k-1} \frac{1}{\sqrt{k}}\]

Check the Alternating Series Test:
- \(\frac{1}{\sqrt{k}} > 0\) for all \(k \geq 1\) ✓
- \(\frac{1}{\sqrt{k}}\) is decreasing ✓
- \(\lim_{k \to \infty} \frac{1}{\sqrt{k}} = 0\) ✓
All conditions are satisfied, so the series converges by the Leibniz test.
Check the endpoint \(x = -1\): \[\sum_{k=1}^{\infty} (-1)^{k-1} \frac{(-1)^k}{\sqrt{k}} = \sum_{k=1}^{\infty} (-1)^{2k-1} \frac{1}{\sqrt{k}} = -\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\]

This is a negative \(p\)-series with \(p = \frac{1}{2} < 1\), which diverges.

Answer: Radius of convergence \(R = 1\). Interval of convergence: \((-1, 1]\). The series converges conditionally at \(x = 1\).

4.53. Direct Comparison Test: Series Less Than Geometric (Chapter 4, Example 1)

Determine the convergence of the series \(\sum_{n=0}^{\infty} \frac{1}{(n+1)^{n+1}}\).

Click to see the solution

Key Concept: Find a larger convergent series to compare with.

Find an upper bound: For \(n \geq 0\), we have \((n+1)^{n+1} \geq 2^n\). To see this:
- For \(n = 0\): \(1^1 = 1 \geq 2^0 = 1\) ✓
- For \(n = 1\): \(2^2 = 4 \geq 2^1 = 2\) ✓
- For \(n \geq 1\): \((n+1)^{n+1} = (n+1) \cdot (n+1)^n \geq 2 \cdot 2^n = 2^{n+1} > 2^n\) ✓
Therefore: \[\frac{1}{(n+1)^{n+1}} \leq \frac{1}{2^n}\]
Use the comparison test: The series \(\sum_{n=0}^{\infty} \frac{1}{2^n}\) is a geometric series with \(q = \frac{1}{2} < 1\), so it converges.

By the direct comparison test, since \(0 \leq \frac{1}{(n+1)^{n+1}} \leq \frac{1}{2^n}\) and \(\sum \frac{1}{2^n}\) converges, we conclude that \(\sum \frac{1}{(n+1)^{n+1}}\) converges.

Answer: The series converges.

4.54. Limit Comparison Test: Convergent Series (Chapter 4, Example 2)

Determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{1}{(n+1)^2 - 1}\).

Click to see the solution

Key Concept: For large \(n\), the term behaves like \(\frac{1}{n^2}\), so compare with the \(p\)-series \(\sum \frac{1}{n^2}\).

Choose a comparison series: Let \(a_n = \frac{1}{(n+1)^2 - 1}\) and \(b_n = \frac{1}{n^2}\).
Compute the limit of the ratio: \[L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{(n+1)^2 - 1}}{\frac{1}{n^2}}\] \[= \lim_{n \to \infty} \frac{n^2}{(n+1)^2 - 1}\]
Expand and simplify: \[(n+1)^2 - 1 = n^2 + 2n + 1 - 1 = n^2 + 2n\]

Therefore: \[L = \lim_{n \to \infty} \frac{n^2}{n^2 + 2n} = \lim_{n \to \infty} \frac{n^2}{n^2(1 + \frac{2}{n})} = \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n}} = 1\]
Apply the limit comparison test: Since \(0 < L = 1 < \infty\) and \(\sum \frac{1}{n^2}\) converges (it’s a \(p\)-series with \(p = 2 > 1\)), we conclude that \(\sum \frac{1}{(n+1)^2 - 1}\) also converges.

Answer: The series converges.

4.55. Ratio Test: Convergent Series with Factorial (Chapter 4, Example 3)

Determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{n!}{n^n}\).

Click to see the solution

Compute the ratio: \[L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}\]
Simplify: \[= \lim_{n \to \infty} \frac{(n+1)! \cdot n^n}{n! \cdot (n+1)^{n+1}}\] \[= \lim_{n \to \infty} \frac{(n+1) \cdot n!}{n!} \cdot \frac{n^n}{(n+1)^{n+1}}\] \[= \lim_{n \to \infty} (n+1) \cdot \frac{n^n}{(n+1)^{n+1}}\] \[= \lim_{n \to \infty} \frac{n^n}{(n+1)^n}\] \[= \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n\] \[= \lim_{n \to \infty} \left(\frac{1}{1 + \frac{1}{n}}\right)^n\] \[= \frac{1}{\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n} = \frac{1}{e}\]
Apply the ratio test: Since \(L = \frac{1}{e} \approx 0.368 < 1\), the series converges.

Answer: The series converges.

4.56. Root Test: Convergent Series (Chapter 4, Example 4)

Determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{2}{5}\right)^n\).

Click to see the solution

Compute the \(n\)-th root: \[L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{n} \left(\frac{2}{5}\right)^n}\]
Simplify: \[= \lim_{n \to \infty} \sqrt[n]{\frac{1}{n}} \cdot \sqrt[n]{\left(\frac{2}{5}\right)^n}\] \[= \lim_{n \to \infty} \frac{1}{n^{1/n}} \cdot \frac{2}{5}\]
Evaluate the limit of \(n^{1/n}\): Taking logarithms: \(\ln(n^{1/n}) = \frac{\ln n}{n} \to 0\) as \(n \to \infty\). Therefore, \(n^{1/n} \to e^0 = 1\).
Complete the calculation: \[L = 1 \cdot \frac{2}{5} = \frac{2}{5}\]
Apply the root test: Since \(L = \frac{2}{5} = 0.4 < 1\), the series converges.

Answer: The series converges.

4.57. Absolute vs. Conditional Convergence (a) (Chapter 4, Example 5)

Determine whether the series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{3/2}}\) converges absolutely, conditionally, or diverges.

Click to see the solution

Key Concept: Check absolute convergence first, then use the Leibniz test if needed.

Test for absolute convergence: Consider the series of absolute values: \[\sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n^{3/2}}\right| = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}\]

This is a \(p\)-series with \(p = \frac{3}{2} > 1\), so it converges.
Conclusion: Since \(\sum |a_n|\) converges, the original series converges absolutely.

Answer: The series converges absolutely.

4.58. The Series of Mengoli (Chapter 4, Example 6)

For the series \(\sum_{n=1}^{\infty} \frac{1}{n^2 + n}\) (the series of Mengoli), find the sum.

Click to see the solution

Key Concept: Use partial fraction decomposition to create a telescoping series.

Decompose using partial fractions: \[\frac{1}{n^2 + n} = \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}\]
Solve for \(A\) and \(B\): \[1 = A(n+1) + Bn\] \[1 = An + A + Bn = (A+B)n + A\]

Comparing coefficients:
- Coefficient of \(n\): \(A + B = 0 \Rightarrow B = -A\)
- Constant term: \(A = 1\)
Therefore, \(A = 1\) and \(B = -1\).
Rewrite the series: \[\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)\]
Compute the \(n\)-th partial sum: \[s_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\]

Most terms cancel (telescope): \[s_n = 1 - \frac{1}{n+1}\]
Take the limit: \[\lim_{n \to \infty} s_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1 - 0 = 1\]

Answer: The series converges to \(1\).

4.59. Divergence of a Logarithmic Telescoping Series (Chapter 4, Example 7)

Show that the series \(\sum_{n=1}^{\infty} \ln\left(1 + \frac{1}{n}\right)\) diverges.

Click to see the solution

Key Concept: Use logarithm properties to create a telescoping series, then evaluate the limit of partial sums.

Rewrite using logarithm properties: \[\ln\left(1 + \frac{1}{n}\right) = \ln\left(\frac{n+1}{n}\right) = \ln(n+1) - \ln(n)\]
This is a telescoping series: \[a_n = \ln(n+1) - \ln(n) = b_{n+1} - b_n \quad \text{where } b_n = \ln(n)\]
Compute the \(n\)-th partial sum: \[s_n = [\ln(2) - \ln(1)] + [\ln(3) - \ln(2)] + \cdots + [\ln(n+1) - \ln(n)]\] \[= \ln(n+1) - \ln(1) = \ln(n+1)\]

(since \(\ln(1) = 0\))
Evaluate the limit: \[\lim_{n \to \infty} s_n = \lim_{n \to \infty} \ln(n+1) = +\infty\]

Answer: The series diverges to \(+\infty\).

4.60. Direct Comparison Test: Divergent Series (Chapter 4, Example 8)

Show that the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}\) diverges.

Click to see the solution

Key Concept: Find a smaller divergent series to compare with.

Find a lower bound: Note that \(\sqrt[3]{n} = n^{1/3} < n\) for all \(n > 1\). Therefore: \[\frac{1}{\sqrt[3]{n}} = \frac{1}{n^{1/3}} > \frac{1}{n}\]

(Alternatively, recognize this as a \(p\)-series with \(p = \frac{1}{3} < 1\))
Use the comparison test: The harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges.

By the direct comparison test, since \(\frac{1}{n} < \frac{1}{\sqrt[3]{n}}\) and \(\sum \frac{1}{n}\) diverges, we conclude that \(\sum \frac{1}{\sqrt[3]{n}}\) also diverges.

Answer: The series diverges (it’s a \(p\)-series with \(p = \frac{1}{3} < 1\)).

4.61. Limit Comparison Test: Divergent Series (Chapter 4, Example 9)

Determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}}\).

Click to see the solution

Choose a comparison series: For large \(n\), the term behaves like \(\frac{1}{\sqrt{n \cdot n}} = \frac{1}{n}\). Let \(a_n = \frac{1}{\sqrt{n(n+1)}}\) and \(b_n = \frac{1}{n}\).
Compute the limit of the ratio: \[L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n(n+1)}}}{\frac{1}{n}}\] \[= \lim_{n \to \infty} \frac{n}{\sqrt{n(n+1)}}\] \[= \lim_{n \to \infty} \frac{n}{\sqrt{n^2 + n}}\] \[= \lim_{n \to \infty} \frac{n}{n\sqrt{1 + \frac{1}{n}}}\] \[= \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}} = 1\]
Apply the limit comparison test: Since \(0 < L = 1 < \infty\) and \(\sum \frac{1}{n}\) diverges (harmonic series), we conclude that \(\sum \frac{1}{\sqrt{n(n+1)}}\) also diverges.

Answer: The series diverges.

4.62. Ratio Test: Divergent Series with Factorial (Chapter 4, Example 10)

Determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{3^n \cdot n!}{n^n}\).

Click to see the solution

Compute the ratio: \[L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{3^{n+1} \cdot (n+1)!}{(n+1)^{n+1}}}{\frac{3^n \cdot n!}{n^n}}\]
Simplify: \[= \lim_{n \to \infty} \frac{3^{n+1} \cdot (n+1)! \cdot n^n}{3^n \cdot n! \cdot (n+1)^{n+1}}\] \[= \lim_{n \to \infty} 3 \cdot (n+1) \cdot \frac{n^n}{(n+1)^{n+1}}\] \[= \lim_{n \to \infty} 3 \cdot \frac{n^n}{(n+1)^n}\] \[= 3 \cdot \frac{1}{e}\]

(using the result from the previous problem)
Apply the ratio test: Since \(L = \frac{3}{e} \approx 1.104 > 1\), the series diverges.

Answer: The series diverges.

4.63. Root Test: Divergent Series (Chapter 4, Example 11)

Determine the convergence of the series \(\sum_{n=1}^{\infty} \left(1 + \frac{1}{n}\right)^{n^2}\).

Click to see the solution

Compute the \(n\)-th root: \[L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \sqrt[n]{\left(1 + \frac{1}{n}\right)^{n^2}}\]
Simplify using exponent rules: \[= \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n^2/n}\] \[= \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\]
Evaluate the limit: \[L = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e\]
Apply the root test: Since \(L = e \approx 2.718 > 1\), the series diverges.

Answer: The series diverges.

4.64. Absolute vs. Conditional Convergence (b) (Chapter 4, Example 12)

Determine whether the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) converges absolutely, conditionally, or diverges.

Click to see the solution

Test for absolute convergence: \[\sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}\]

This is the harmonic series, which diverges. So the series does not converge absolutely.
Test for convergence using Leibniz test: The series has the form \(\sum (-1)^{n+1} a_n\) where \(a_n = \frac{1}{n} > 0\).
- Non-increasing: \(\frac{1}{n+1} \leq \frac{1}{n}\) ✓
- Limit is zero: \(\lim_{n \to \infty} \frac{1}{n} = 0\) ✓
By the Leibniz test, the series converges.
Conclusion: The series converges but not absolutely, so it is conditionally convergent.

Answer: The series converges conditionally.

4.65. Absolutely Convergent Series with Ratio Test (Chapter 4, Example 13)

Determine whether the series \(\sum_{n=0}^{\infty} \frac{(-1)^n \cdot 2^n}{n! \cdot (2n+1) \cdot 10^{2n+1}}\) converges absolutely, conditionally, or diverges.

Click to see the solution

Test for absolute convergence using the ratio test: Let \(a_n = \frac{2^n}{n! \cdot (2n+1) \cdot 10^{2n+1}}\) (ignoring the \((-1)^n\)).

\[L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{2^{n+1}}{(n+1)! \cdot (2(n+1)+1) \cdot 10^{2(n+1)+1}}}{\frac{2^n}{n! \cdot (2n+1) \cdot 10^{2n+1}}}\]
Simplify: \[= \lim_{n \to \infty} \frac{2^{n+1} \cdot n! \cdot (2n+1) \cdot 10^{2n+1}}{2^n \cdot (n+1)! \cdot (2n+3) \cdot 10^{2n+3}}\] \[= \lim_{n \to \infty} \frac{2 \cdot (2n+1)}{(n+1) \cdot (2n+3) \cdot 100}\] \[= \lim_{n \to \infty} \frac{4n+2}{100(2n^2 + 5n + 3)}\] \[= \lim_{n \to \infty} \frac{4n}{200n^2} = \lim_{n \to \infty} \frac{1}{50n} = 0\]
Apply the ratio test: Since \(L = 0 < 1\), the series of absolute values converges.

Answer: The series converges absolutely.

4.66. Conditionally Convergent Series (Chapter 4, Example 14)

Determine whether the series \(\sum_{n=2}^{\infty} \frac{(-1)^n(1+\sqrt{n})}{\sqrt{n^2-2}}\) converges absolutely, conditionally, or diverges.

Click to see the solution

Test for absolute convergence: Consider: \[\sum_{n=2}^{\infty} \frac{1+\sqrt{n}}{\sqrt{n^2-2}}\]
Use limit comparison test: For large \(n\), the dominant terms are: \[\frac{1+\sqrt{n}}{\sqrt{n^2-2}} \approx \frac{\sqrt{n}}{\sqrt{n^2}} = \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}\]

Let \(a_n = \frac{1+\sqrt{n}}{\sqrt{n^2-2}}\) and \(b_n = \frac{1}{\sqrt{n}}\).

\[L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(1+\sqrt{n}) \cdot \sqrt{n}}{\sqrt{n^2-2}}\] \[= \lim_{n \to \infty} \frac{\sqrt{n} + n}{\sqrt{n^2-2}}\] \[= \lim_{n \to \infty} \frac{n}{n\sqrt{1-\frac{2}{n^2}}} = \lim_{n \to \infty} \frac{1}{\sqrt{1-\frac{2}{n^2}}} = 1\]
Conclude about absolute convergence: Since \(0 < L = 1 < \infty\) and \(\sum \frac{1}{\sqrt{n}}\) diverges (p-series with \(p = \frac{1}{2} < 1\)), the series does not converge absolutely.
Test for convergence using Leibniz test: Let \(a_n = \frac{1+\sqrt{n}}{\sqrt{n^2-2}}\).
- Limit is zero: \[\lim_{n \to \infty} \frac{1+\sqrt{n}}{\sqrt{n^2-2}} = \lim_{n \to \infty} \frac{\sqrt{n}}{n} = 0 \quad \checkmark\]
- Non-increasing: This requires verification, but for large \(n\): \[a_n \approx \frac{1}{\sqrt{n}}\] which is decreasing. (A rigorous proof would show \(a_{n+1} < a_n\) for large \(n\).)
Apply Leibniz test: Both conditions are satisfied, so the series converges.

Answer: The series converges conditionally.

4.67. Alternating Harmonic Series (Chapter 4, Example 15)

Show that the series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\) converges, and state its sum.

Click to see the solution

Key Concept: Apply the Leibniz test for alternating series.

Identify the alternating series: \[\sum_{n=1}^{\infty} \frac{(-1)^n}{n} = -\frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \cdots\]

This has the form \(\sum (-1)^n a_n\) where \(a_n = \frac{1}{n} > 0\).
Check the conditions of the Leibniz test:
- Non-increasing: Is \(a_{n+1} \leq a_n\)? \[\frac{1}{n+1} \leq \frac{1}{n} \quad \text{for all } n \geq 1 \quad \checkmark\]
- Limit is zero: Does \(\lim_{n \to \infty} a_n = 0\)? \[\lim_{n \to \infty} \frac{1}{n} = 0 \quad \checkmark\]
Apply the Leibniz test: Both conditions are satisfied, so the series converges.
State the sum: It can be shown (using advanced techniques) that: \[\sum_{n=1}^{\infty} \frac{(-1)^n}{n} = -\ln(2)\]

Answer: The series converges to \(-\ln(2) \approx -0.693\).

4.68. Absolute Convergence with Parameter (Chapter 4, Example 16)

For which values of \(x\) does the series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)x^{2n-1}}\) converge absolutely, conditionally, or diverge?

Click to see the solution

Key Concept: Use the ratio test for absolute convergence, treating \(x\) as a parameter.

Test for absolute convergence: Consider: \[\sum_{n=1}^{\infty} \left|\frac{(-1)^n}{(2n-1)x^{2n-1}}\right| = \sum_{n=1}^{\infty} \frac{1}{(2n-1)|x|^{2n-1}}\]
Apply the ratio test: \[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{\frac{1}{(2(n+1)-1)|x|^{2(n+1)-1}}}{\frac{1}{(2n-1)|x|^{2n-1}}}\] \[= \lim_{n \to \infty} \frac{2n-1}{2n+1} \cdot \frac{|x|^{2n-1}}{|x|^{2n+1}}\] \[= \lim_{n \to \infty} \frac{2n-1}{2n+1} \cdot \frac{1}{|x|^2}\] \[= 1 \cdot \frac{1}{|x|^2} = \frac{1}{x^2}\]
Determine absolute convergence:
- If \(\frac{1}{x^2} < 1\), i.e., \(x^2 > 1\) or \(|x| > 1\), the series converges absolutely.
- If \(\frac{1}{x^2} > 1\), i.e., \(|x| < 1\), the series diverges.
- If \(\frac{1}{x^2} = 1\), i.e., \(x = \pm 1\), the test is inconclusive.
Check the boundary cases \(x = \pm 1\):
- When \(x = 1\): \[\sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\] This is an alternating series with \(a_n = \frac{1}{2n-1}\) non-increasing and approaching 0. By Leibniz, it converges. Since the absolute version is \(\sum \frac{1}{2n-1}\) which diverges (comparable to harmonic series), it converges conditionally.
- When \(x = -1\): \[\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)(-1)^{2n-1}} = \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)(-1)^{2n-1}}\] Since \((-1)^{2n-1} = (-1)^{2n} \cdot (-1)^{-1} = 1 \cdot (-1) = -1\): \[= \sum_{n=1}^{\infty} \frac{(-1)^n}{-(2n-1)} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n-1}\] By the same reasoning as \(x=1\), this converges conditionally.

Answer:

Absolutely convergent for \(x \in (-\infty, -1) \cup (1, \infty)\)
Conditionally convergent for \(x \in \{-1, 1\}\)
Divergent for \(x \in (-1, 1)\)

4.69. Find Sum of Series (a) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_1 = \sum_{n=1}^{\infty} \frac{1}{(n + 1)^2 - 1}\]

Click to see the solution

Method: Telescoping Series

Simplify the general term: \[(n + 1)^2 - 1 = n^2 + 2n = n(n + 2)\] Using partial fraction decomposition: \[\frac{1}{n(n + 2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n + 2} \right)\]
Write the partial sum \(s_N\): \[s_N = \frac{1}{2} \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n + 2} \right)\] Expanding the terms allows us to cancel the middle terms: \[s_N = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+2}\right) \right]\]
Identify remaining terms: The positive terms \(1\) and \(\frac{1}{2}\) remain at the beginning. The negative terms \(-\frac{1}{N+1}\) and \(-\frac{1}{N+2}\) remain at the end. \[s_N = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2} \right)\]
Calculate the limit: \[S_1 = \lim_{N \to \infty} s_N = \frac{1}{2} \left( 1 + \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}\]

Answer: \(\frac{3}{4}\)

4.70. Find Sum of Series (b) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_2 = \sum_{n=1}^{\infty} \frac{1}{(2n - 1)(2n + 5)}\]

Click to see the solution

Decompose into partial fractions: The difference between factors is \((2n + 5) - (2n - 1) = 6\). \[\frac{1}{(2n - 1)(2n + 5)} = \frac{1}{6} \left( \frac{1}{2n - 1} - \frac{1}{2n + 5} \right)\]
Analyze the terms: This is a telescoping series with a “gap”. The terms of the denominator form the arithmetic progression: \(1, 3, 5, 7, 9, \dots\) The subtraction removes terms starting from the 4th term of the sequence (\(2(1)+5 = 7\)).
Determine survivors: The first three positive terms will not be cancelled:
- \(n=1\): \(\frac{1}{2(1)-1} = \frac{1}{1}\)
- \(n=2\): \(\frac{1}{2(2)-1} = \frac{1}{3}\)
- \(n=3\): \(\frac{1}{2(3)-1} = \frac{1}{5}\)
All terms at the “infinity end” approach 0.
Calculate the sum: \[S_2 = \frac{1}{6} \left( 1 + \frac{1}{3} + \frac{1}{5} \right) = \frac{1}{6} \left( \frac{15 + 5 + 3}{15} \right) = \frac{1}{6} \cdot \frac{23}{15} = \frac{23}{90}\]

Answer: \(\frac{23}{90}\)

4.71. Find Sum of Series (c) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_3 = \sum_{n=1}^{\infty} \frac{1}{(e + n)(e + n + 1)}\]

Click to see the solution

Decompose: The difference in the denominator is 1. \[\frac{1}{(e + n)(e + n + 1)} = \frac{1}{e + n} - \frac{1}{e + n + 1}\]
Telescope: \[S_3 = \left(\frac{1}{e+1} - \frac{1}{e+2}\right) + \left(\frac{1}{e+2} - \frac{1}{e+3}\right) + \cdots\] Only the very first term remains, as the last term tends to 0.
Result: \[S_3 = \frac{1}{e + 1}\]

Answer: \(\frac{1}{e + 1}\)

4.72. Find Sum of Geometric-Like Series (d) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_4 = \sum_{n=1}^{\infty} \left( a^{\frac{1}{2n-1}} - a^{\frac{1}{2n+1}} \right) \quad (\forall a > 0)\]

Click to see the solution

Write the partial sum: \[s_N = \left(a^1 - a^{1/3}\right) + \left(a^{1/3} - a^{1/5}\right) + \cdots + \left(a^{\frac{1}{2N-1}} - a^{\frac{1}{2N+1}}\right)\]
Cancel and take the limit: All intermediate terms cancel. \[s_N = a^1 - a^{\frac{1}{2N+1}}\] \[\lim_{N \to \infty} s_N = a - a^0 = a - 1\]

Answer: \(a - 1\)

4.73. Find Sum with Arctangent (e) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_5 = \sum_{n=1}^{\infty} \left( \arctan \left( \frac{n + 1}{n + 2} \right) - \arctan \left( \frac{n}{n + 1} \right) \right)\]

Click to see the solution

Identify the telescoping form: Let \(b_n = \arctan\left(\frac{n}{n+1}\right)\). The series is \(\sum_{n=1}^{\infty} (b_{n+1} - b_n)\).
Compute partial sum limit: \[s_N = b_{N+1} - b_1 = \arctan\left(\frac{N+1}{N+2}\right) - \arctan\left(\frac{1}{2}\right)\]
Take the limit: As \(N \to \infty\), \(\frac{N+1}{N+2} \to 1\). \[S_5 = \arctan(1) - \arctan\left(\frac{1}{2}\right) = \frac{\pi}{4} - \arctan\left(\frac{1}{2}\right)\]

Answer: \(\frac{\pi}{4} - \arctan\frac{1}{2}\)

4.74. Geometric Series (a) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_6 = \sum_{n=1}^{\infty} \frac{2(-3)^n}{5^n}\]

Click to see the solution

Identify parameters: This is a geometric series \(\sum ar^{n-1}\) or similar.
- First term (\(n=1\)): \(a = \frac{2(-3)}{5} = -\frac{6}{5}\)
- Common ratio: \(q = -\frac{3}{5}\)
Calculate sum: Since \(|q| < 1\), the sum is valid. \[S_6 = \frac{a}{1 - q} = \frac{-6/5}{1 - (-3/5)} = \frac{-6/5}{8/5} = -\frac{6}{8} = -\frac{3}{4}\]

Answer: \(-\frac{3}{4}\)

4.75. Geometric Series (b) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_7 = \sum_{n=1}^{\infty} \frac{3^n + 2^n}{6^n}\]

Click to see the solution

Split the series: \[S_7 = \sum_{n=1}^{\infty} \left(\frac{3}{6}\right)^n + \sum_{n=1}^{\infty} \left(\frac{2}{6}\right)^n = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n + \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n\]
Sum each geometric series:
- First: \(\frac{1/2}{1 - 1/2} = 1\)
- Second: \(\frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}\)
Combine: \[S_7 = 1 + \frac{1}{2} = \frac{3}{2}\]

Answer: \(\frac{3}{2}\)

4.76. Geometric Series (c) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_8 = \sum_{n=1}^{\infty} \frac{1}{\sqrt{e^n}}\]

Click to see the solution

Identify parameters: rewrite as \(\sum_{n=1}^{\infty} \left(e^{-1/2}\right)^n\).
- First term: \(a = e^{-1/2} = \frac{1}{\sqrt{e}}\)
- Ratio: \(q = \frac{1}{\sqrt{e}} < 1\)
Calculate sum: \[S_8 = \frac{1/\sqrt{e}}{1 - 1/\sqrt{e}} = \frac{1}{\sqrt{e} - 1}\]

Answer: \(\frac{1}{\sqrt{e} - 1}\)

4.77. Geometric Series (d) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_9 = \sum_{n=1}^{\infty} \frac{2^n}{\sqrt{5^{n+1}}}\]

Click to see the solution

Isolate \(n\)-th power: \[S_9 = \frac{1}{\sqrt{5}} \sum_{n=1}^{\infty} \left(\frac{2}{\sqrt{5}}\right)^n\]
Identify parameters:
- First term (of the sum part): \(\frac{2}{\sqrt{5}}\)
- Ratio: \(q = \frac{2}{\sqrt{5}} \approx \frac{2}{2.23} < 1\).
Calculate: \[S_9 = \frac{1}{\sqrt{5}} \cdot \frac{2/\sqrt{5}}{1 - 2/\sqrt{5}} = \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5} - 2} = \frac{2}{5 - 2\sqrt{5}}\]

Answer: \(\frac{2}{5 - 2\sqrt{5}}\)

4.78. Geometric Series (e) (Practice Sheet, Task 1)

Find the sum of the following series: \[S_{10} = \sum_{n=0}^{\infty} e^{-na} \quad (\forall a > 0)\]

Click to see the solution

Identify parameters: This is \(\sum_{n=0}^{\infty} (e^{-a})^n\).
- First term (\(n=0\)): \(1\)
- Ratio: \(q = e^{-a}\). Since \(a > 0\), \(0 < q < 1\).
Calculate sum: \[S_{10} = \frac{1}{1 - e^{-a}} = \frac{e^a}{e^a - 1}\]

Answer: \(\frac{e^a}{e^a - 1}\)

4.79. Series Convergence (a) (Practice Sheet, Task 2a)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} (\sqrt{n + 1} - \sqrt{n})\]

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Method: Telescoping Limit

Partial sums: \[s_N = (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{N+1} - \sqrt{N}) = \sqrt{N+1} - 1\]
Limit: \[\lim_{N \to \infty} s_N = \infty\] The series diverges to infinity.

Answer: Diverges.

4.80. Series Convergence (b) (Practice Sheet, Task 2a)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - \sqrt{n + 1}}\]

Click to see the solution

Method: \(n\)-th Term Test

Rationalize the term: \[a_n = \frac{\sqrt{n} + \sqrt{n+1}}{n - (n+1)} = -(\sqrt{n} + \sqrt{n+1})\]
Check limit: \[\lim_{n \to \infty} a_n = -\infty \neq 0\]

Since the general term does not approach zero, the series diverges.

Answer: Diverges.

4.81. Series Convergence (c) (Practice Sheet, Task 2a)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \cos^2 \left( \frac{1}{n} \right)\]

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Method: \(n\)-th Term Test

Check limit of the general term: As \(n \to \infty\), \(\frac{1}{n} \to 0\). \[\lim_{n \to \infty} \cos^2 \left( \frac{1}{n} \right) = \cos^2(0) = 1 \neq 0\]
Conclusion: The series diverges because the terms do not approach zero.

Answer: Diverges.

4.82. Pattern Series (Practice Sheet, Task 2a)

Study the convergence or divergence of the following series: \[\frac{1}{3} + \frac{4}{9} + \frac{9}{19} + \frac{25}{51} + \cdots\]

Click to see the solution

Method: \(n\)-th Term Test

Identify the general term: Numerator: \(n^2\). Denominator: \(2n^2 + 1\). \[a_n = \frac{n^2}{2n^2 + 1}\]
Check limit: \[\lim_{n \to \infty} \frac{n^2}{2n^2 + 1} = \frac{1}{2} \neq 0\]
Conclusion: The series diverges.

Answer: Diverges.

4.83. Comparison Test Example (a) (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=3}^{\infty} \frac{\ln n}{n}\]

Click to see the solution

Method: Direct Comparison Test

Inequality: For \(n \geq 3\), we know that \(\ln n > 1\).
Comparison: \[\frac{\ln n}{n} > \frac{1}{n}\]
Conclusion: Since the harmonic series \(\sum \frac{1}{n}\) diverges, the larger series \(\sum \frac{\ln n}{n}\) also diverges.

Answer: Diverges.

4.84. Comparison Test Example (b) (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{2n - 1}{3^{2n+1}}\]

Click to see the solution

Method: Ratio Test

Set up the ratio: Note \(3^{2n+1} = 3 \cdot 9^n\). Let \(a_n = \frac{2n - 1}{9^n}\). \[\frac{a_{n+1}}{a_n} = \frac{2(n+1) - 1}{9^{n+1}} \cdot \frac{9^n}{2n - 1} = \frac{2n + 1}{9(2n - 1)}\]
Limit: \[\lim_{n \to \infty} \frac{2n + 1}{18n - 9} = \frac{2}{18} = \frac{1}{9}\]
Conclusion: Since \(\frac{1}{9} < 1\), the series converges.

Answer: Converges.

4.85. Limit Comparison Test Example (a) (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n^2}{2^n + 3^n}\]

Click to see the solution

Method: Limit Comparison Test

Dominant terms: The denominator behaves like \(3^n\) (since \(3^n \gg 2^n\)). Compare with \(b_n = \frac{n^2}{3^n}\).
Limit: \[\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^2}{2^n + 3^n} \cdot \frac{3^n}{n^2} = \lim_{n \to \infty} \frac{3^n}{2^n + 3^n} = \lim_{n \to \infty} \frac{1}{(2/3)^n + 1} = 1\]
Check \(b_n\): \(\sum \frac{n^2}{3^n}\) converges by the Ratio Test (\(\lim |\frac{a_{n+1}}{a_n}| = \frac{1}{3} < 1\)). Therefore, the original series converges.

Answer: Converges.

4.86. Limit Comparison Test Example (b) (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{3^n + 5}{4^n + n^8}\]

Click to see the solution

Method: Limit Comparison Test

Identify dominant behavior: \[a_n \sim \frac{3^n}{4^n} = \left(\frac{3}{4}\right)^n\]
Comparison: Compare with geometric series \(\sum (\frac{3}{4})^n\). \[\lim_{n \to \infty} \frac{3^n + 5}{4^n + n^8} \cdot \frac{4^n}{3^n} = 1\]
Conclusion: Since the geometric series with ratio \(|r| = 3/4 < 1\) converges, the original series converges.

Answer: Converges.

4.87. Comparison Test Example (c) (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=3}^{\infty} \frac{\ln n}{n^2}\]

Click to see the solution

Method: Direct Comparison Test

Inequality: We know that \(\ln n\) grows slower than any power \(n^\epsilon\). Let \(\epsilon = 0.5\). For sufficiently large \(n\): \(\ln n < \sqrt{n}\).
Comparison: \[\frac{\ln n}{n^2} < \frac{\sqrt{n}}{n^2} = \frac{1}{n^{1.5}}\]
Conclusion: The series \(\sum \frac{1}{n^{1.5}}\) is a convergent \(p\)-series (\(p = 1.5 > 1\)). Thus, the original series converges.

Answer: Converges.

4.88. Comparison Test Example (d) (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{1}{n + 3^n}\]

Click to see the solution

Method: Direct Comparison Test

Inequality: Denominator \(n + 3^n > 3^n\), so: \[\frac{1}{n + 3^n} < \frac{1}{3^n}\]
Conclusion: \(\sum \frac{1}{3^n}\) is a convergent geometric series (\(r = 1/3\)). Thus, the original series converges.

Answer: Converges.

4.89. Logarithmic Series (Practice Sheet, Task 2b)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{1}{\ln(e^n - 1)}\]

Click to see the solution

Method: Limit Comparison Test

Asymptotic Analysis: For large \(n\), \(e^n - 1 \approx e^n\). \[\ln(e^n - 1) \approx \ln(e^n) = n\]
Formal Limit: Compare with harmonic series \(\sum \frac{1}{n}\). \[\lim_{n \to \infty} \frac{1/\ln(e^n - 1)}{1/n} = \lim_{n \to \infty} \frac{n}{n + \ln(1 - e^{-n})} = 1\]
Conclusion: Since \(\sum \frac{1}{n}\) diverges, the original series diverges.

Answer: Diverges.

4.90. Limit Comparison Test Example (c) (Practice Sheet, Task 2c)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}\]

Click to see the solution

Method: Limit Comparison Test

Simplify: \[a_n \sim \frac{n}{n^2} = \frac{1}{n}\]
Conclusion: Comparing with the divergent harmonic series \(\sum \frac{1}{n}\), the limit of the ratio is 1. Thus, the series diverges.

Answer: Diverges.

4.91. Comparison Test Example (e) (Practice Sheet, Task 2c)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{2n + 1}{(n + 1)^2(n + 2)^2}\]

Click to see the solution

Method: Limit Comparison Test

Power Count: Numerator degree: 1. Denominator degree: 4. \[a_n \sim \frac{2n}{n^4} = \frac{2}{n^3}\]
Conclusion: Comparison with \(\sum \frac{1}{n^3}\) (\(p\)-series, \(p=3\)). Since \(3 > 1\), the series converges.

Answer: Converges.

4.92. Limit Comparison Test Example (f) (Practice Sheet, Task 2c)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{\sqrt[3]{n}}{(n + 1)\sqrt{n}}\]

Click to see the solution

Method: Limit Comparison Test

Power Analysis: \[a_n \approx \frac{n^{1/3}}{n \cdot n^{1/2}} = \frac{n^{1/3}}{n^{3/2}} = \frac{1}{n^{3/2 - 1/3}} = \frac{1}{n^{7/6}}\]
Conclusion: Since \(p = \frac{7}{6} > 1\), the series converges by comparison with the \(p\)-series.

Answer: Converges.

4.93. Limit Comparison Test Example (g) (Practice Sheet, Task 2c)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{\sqrt[3]{n}}{(2n - 1)(5\sqrt[3]{n} - 1)}\]

Click to see the solution

Method: Limit Comparison Test

Power Analysis: \[a_n \approx \frac{n^{1/3}}{(2n)(5n^{1/3})} = \frac{n^{1/3}}{10 n^{4/3}} = \frac{1}{10n}\]
Conclusion: Behaves like \(\frac{1}{10} \sum \frac{1}{n}\). Since the harmonic series diverges, this series diverges.

Answer: Diverges.

4.94. Series with Square Roots (Practice Sheet, Task 2c)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} (\sqrt{n^2 + 3} - \sqrt{n^2 + 1})\]

Click to see the solution

Method: Conjugate Multiplication

Algebraic manipulation: Multiply by \(\frac{\sqrt{n^2+3}+\sqrt{n^2+1}}{\sqrt{n^2+3}+\sqrt{n^2+1}}\): \[a_n = \frac{(n^2+3)-(n^2+1)}{\sqrt{n^2+3}+\sqrt{n^2+1}} = \frac{2}{\sqrt{n^2+3}+\sqrt{n^2+1}}\]
Asymptotic behavior: For large \(n\), denominator \(\approx 2n\). \[a_n \sim \frac{2}{2n} = \frac{1}{n}\]
Conclusion: Diverges by comparison with the harmonic series.

Answer: Diverges.

4.95. Ratio Test with Factorial (a) (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n!}{e^{n^2}}\]

Click to see the solution

Method: Ratio Test

Calculate ratio: \[\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{e^{(n+1)^2}} \cdot \frac{e^{n^2}}{n!} = (n+1) e^{n^2 - (n^2 + 2n + 1)} = \frac{n+1}{e^{2n+1}}\]
Limit: Exponential growth dominates polynomial growth. \[\lim_{n \to \infty} \frac{n+1}{e^{2n+1}} = 0\]
Conclusion: Since \(0 < 1\), the series converges.

Answer: Converges.

4.96. Ratio Test with Factorial (b) (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n^n}{(n!)^2}\]

Click to see the solution

Method: Ratio Test

Calculate ratio: \[\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{((n+1)!)^2} \cdot \frac{(n!)^2}{n^n} = \frac{(n+1)(n+1)^n}{(n+1)^2 (n!)^2} \cdot \frac{(n!)^2}{n^n}\] \[= \frac{1}{n+1} \left(\frac{n+1}{n}\right)^n = \frac{1}{n+1} \left(1 + \frac{1}{n}\right)^n\]
Limit: \[\lim_{n \to \infty} \frac{1}{n+1} \cdot e = 0 \cdot e = 0\]
Conclusion: Since \(0 < 1\), the series converges.

Answer: Converges.

4.97. Ratio Test with Double Factorial (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n!}{(2n - 1)!!}\]

Click to see the solution

Method: Ratio Test

Calculate ratio: Note that \((2(n+1)-1)!! = (2n+1)!! = (2n+1)(2n-1)!!\). \[\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(2n+1)!!} \cdot \frac{(2n-1)!!}{n!} = \frac{n+1}{2n+1}\]
Limit: \[\lim_{n \to \infty} \frac{n+1}{2n+1} = \frac{1}{2}\]
Conclusion: Since \(\frac{1}{2} < 1\), the series converges.

Answer: Converges.

4.98. Root Test with Double Exponential (a) (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n^n}{3^{n^2}}\]

Click to see the solution

Method: Root Test

Evaluate \(n\)-th root: \[\sqrt[n]{a_n} = \left( \frac{n^n}{3^{n^2}} \right)^{1/n} = \frac{n}{3^n}\]
Limit: \[\lim_{n \to \infty} \frac{n}{3^n} = 0\]
Conclusion: Since \(0 < 1\), the series converges.

Answer: Converges.

4.99. Ratio Test with Powers (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{n^2 2^{n+2}}{4^n}\]

Click to see the solution

Method: Simplify then Ratio Test

Simplify: \[a_n = n^2 \cdot 4 \cdot \frac{2^n}{4^n} = 4n^2 \left(\frac{1}{2}\right)^n = \frac{4n^2}{2^n}\]
Ratio Test: \[\lim_{n \to \infty} \frac{4(n+1)^2}{2^{n+1}} \cdot \frac{2^n}{4n^2} = \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \cdot \frac{1}{2} = \frac{1}{2}\]
Conclusion: Converges.

Answer: Converges.

4.100. Alternating Series (a) (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{(-1)^{n-1}n}{n^2 + 1}\]

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Absolute Convergence: \(\sum \frac{n}{n^2+1} \sim \sum \frac{1}{n}\). Diverges.
Conditional Convergence (Leibniz Test):
- Terms approach 0? \(\lim \frac{n}{n^2+1} = 0\). Yes.
- Terms decreasing? Yes, denominator grows quadratically, numerator linearly.
The series converges conditionally.

Answer: Converges conditionally.

4.101. Alternating Series (b) (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n}\]

Click to see the solution

Absolute Convergence: Consider \(\sum \frac{1}{n \ln n}\). By the Cauchy Condensation Test: \(\sum 2^k \cdot \frac{1}{2^k \ln(2^k)} = \sum \frac{1}{k \ln 2}\). This is the harmonic series (scaled), which diverges. So, not absolutely convergent.
Conditional Convergence (Leibniz Test):
- \(\lim_{n \to \infty} \frac{1}{n \ln n} = 0\).
- \(n \ln n\) is clearly increasing, so \(\frac{1}{n \ln n}\) is decreasing.
Converges by Leibniz.

Answer: Converges conditionally.

4.102. Alternating Series (c) (Practice Sheet, Task 2d)

Study the convergence or divergence of the following series: \[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n}\]

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Analysis: This is \(\frac{1}{2} \sum \frac{(-1)^{n+1}}{n}\), which is \(\frac{1}{2}\) times the Alternating Harmonic Series.
- Absolute series \(\sum \frac{1}{2n}\) diverges.
- Alternating series converges to \(\frac{1}{2} \ln 2\).

Answer: Converges conditionally.

4.103. Power Series: Radius and Interval (a) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=1}^{\infty} \frac{(3x - 2)^n}{n}\]

Click to see the solution

Center and Form: Rewrite as \(\sum \frac{3^n}{n} (x - 2/3)^n\). Center is \(x = 2/3\).
Radius (Ratio Test): \[\lim \left| \frac{a_{n+1}}{a_n} \right| = |3x-2| \lim \frac{n}{n+1} = |3x-2| < 1\] \(|x - 2/3| < 1/3\). Thus \(R = 1/3\).
Endpoints (\(x = 1/3, x = 1\)):
- \(x=1 \Rightarrow (3(1)-2)^n/n = 1/n\) (Diverges).
- \(x=1/3 \Rightarrow (3(1/3)-2)^n/n = (-1)^n/n\) (Converges by Leibniz).

Answer: \(R = 1/3\), Interval: \([\frac{1}{3}, 1)\).

4.104. Power Series: Radius and Interval (b) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=1}^{\infty} \frac{nx^n}{n + 2}\]

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Radius (Ratio Test): \[\lim \left| \frac{(n+1)x^{n+1}}{n+3} \cdot \frac{n+2}{nx^n} \right| = |x| \cdot 1 < 1 \implies R = 1\]
Endpoints (\(x = \pm 1\)):
- \(x=1 \Rightarrow \sum \frac{n}{n+2}\). Limit is \(1 \neq 0\). Diverges.
- \(x=-1 \Rightarrow \sum \frac{(-1)^n n}{n+2}\). Limit oscillates. Diverges.

Answer: \(R = 1\), Interval: \((-1, 1)\).

4.105. Power Series: Radius and Interval (c) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=1}^{\infty} \frac{(x - 1)^n}{n^3 3^n}\]

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Radius (Ratio Test): \[\lim \left| \frac{(x-1)}{3} \cdot \frac{n^3}{(n+1)^3} \right| = \frac{|x-1|}{3} < 1 \implies |x-1| < 3 \implies R = 3\] Interval: \((-2, 4)\).
Endpoints:
- \(x=4 \Rightarrow \sum \frac{3^n}{n^3 3^n} = \sum \frac{1}{n^3}\) (Converges).
- \(x=-2 \Rightarrow \sum \frac{(-3)^n}{n^3 3^n} = \sum \frac{(-1)^n}{n^3}\) (Converges).

Answer: \(R = 3\), Interval: \([-2, 4]\).

4.106. Power Series: Radius and Interval (d) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=1}^{\infty} \frac{4^n x^{2n}}{n}\]

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Convergence Condition: Root test: \(\sqrt[n]{|a_n|} = \sqrt[n]{\frac{4^n (x^2)^n}{n}} \to 4x^2\). Condition: \(4x^2 < 1 \implies x^2 < 1/4 \implies |x| < 1/2\). \(R = 1/2\).
Endpoints (\(x = \pm 1/2\)): \(x^2 = 1/4\). Series becomes \(\sum \frac{4^n (1/4)^n}{n} = \sum \frac{1}{n}\) (Diverges).

Answer: \(R = 1/2\), Interval: \((-1/2, 1/2)\).

4.107. Power Series: Radius and Interval (e) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=0}^{\infty} \frac{nx^n}{4^n(n^2 + 1)}\]

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Radius (Ratio Test): Coefficients behave like \(\frac{n}{n^2} \cdot \frac{1}{4^n} \approx \frac{1}{n 4^n}\). Ratio of terms \(\to \frac{|x|}{4}\). Condition \(|x| < 4 \implies R = 4\).
Endpoints:
- \(x=4 \Rightarrow \sum \frac{n 4^n}{4^n(n^2+1)} = \sum \frac{n}{n^2+1}\) (Diverges, \(\sim 1/n\)).
- \(x=-4 \Rightarrow \sum \frac{(-1)^n n}{n^2+1}\) (Converges by Leibniz).

Answer: \(R = 4\), Interval: \([-4, 4)\).

4.108. Power Series: Radius and Interval (f) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=1}^{\infty} \frac{(-1)^n 3^{2n}(x - 2)^n}{3n}\]

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Simplify: \(3^{2n} = 9^n\). Series is \(\sum \frac{(-9)^n (x-2)^n}{3n}\).
Radius: Root test: \(|-9(x-2)| < 1 \implies |x-2| < 1/9\). \(R = 1/9\). Interval \((\frac{17}{9}, \frac{19}{9})\).
Endpoints:
- \(x = 2 + 1/9 \Rightarrow \sum \frac{(-9)^n (1/9)^n}{3n} = \sum \frac{(-1)^n}{3n}\) (Converges).
- \(x = 2 - 1/9 \Rightarrow \sum \frac{(-9)^n (-1/9)^n}{3n} = \sum \frac{1}{3n}\) (Diverges).

Answer: \(R = 1/9\), Interval: \((\frac{17}{9}, \frac{19}{9}]\).

4.109. Power Series: Radius and Interval (g) (Practice Sheet, Task 3)

For the following power series, find the radius of convergence and the interval of convergence: \[\sum_{n=1}^{\infty} \left( 1 + \frac{1}{n} \right)^n x^n\]

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Radius (Root Test): \(\lim \sqrt[n]{|a_n|} = \lim \left(1 + \frac{1}{n}\right) |x| = 1 \cdot |x| = |x|\). \(|x| < 1 \implies R = 1\).
Endpoints (\(x = \pm 1\)): The term coefficient limit is \(\lim (1 + 1/n)^n = e \neq 0\). Terms do not approach zero. Diverges at both endpoints.

Answer: \(R = 1\), Interval: \((-1, 1)\).