Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{10 n^{1.1}+1} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{10n^{1.1}+1} \) is conditionally convergent.

Step 1: Analyze Absolute Convergence

To determine absolute convergence, we analyze the series \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{n}{10 n^{1.1} + 1} \right| = \sum_{n=1}^{\infty} \frac{n}{10 n^{1.1} + 1} \). Compare \( n \) and \( n^{1.1} \). As \( n \rightarrow \infty \), \( n^{1.1} \) grows faster than \( n \), hence the denominator grows faster than the numerator. Use a comparison test against the series \( \sum_{n=1}^{\infty} \frac{1}{n^{0.1}} \), which is known to diverge. Since \( \frac{n}{10n^{1.1}+1} \geq \frac{1}{10n^{0.1}} \) for large \( n \), the series diverges by comparison. Thus, the original series is not absolutely convergent.

Step 2: Analyze Conditional Convergence

Now apply the Alternating Series Test to analyze conditional convergence. Consider the series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{10n^{1.1}+1} \). For the alternating series test, check if the terms \( \frac{n}{10n^{1.1}+1} \) decrease and tend to zero as \( n \rightarrow \infty \). First, \( b_n = \frac{n}{10n^{1.1}+1} \) is positive and decreasing, since the derivative \( b'_n \) is negative: Calculate \( b'_n = \frac{(10n^{1.1}+1)(1) - n \cdot (11n^{0.1})}{(10n^{1.1}+1)^2} \), which is negative. Lastly, \( \lim_{n \to \infty} \frac{n}{10n^{1.1} + 1} = 0 \). Hence, the series converges conditionally by the alternating series test.

Key Concepts

Absolute Convergence

Absolute convergence of a series is when the series of absolute values converges. The series \( \sum_{n=1}^{\infty} a_n \) is said to be absolutely convergent if \( \sum_{n=1}^{\infty} |a_n| \) converges. If a series is absolutely convergent, then it is convergent in the traditional sense.

To determine if our series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{10n^{1.1}+1} \) is absolutely convergent, we first consider \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{n}{10n^{1.1}+1} \right| = \sum_{n=1}^{\infty} \frac{n}{10n^{1.1}+1} \).

- We compare this series to another one: \( \sum_{n=1}^{\infty} \frac{1}{n^{0.1}} \), which is known to diverge.- Since our series behaves similarly by growing just as slowly, it also diverges.This means the original series is not absolutely convergent.

Conditional Convergence

If a series converges but does not converge absolutely, it is conditionally convergent. This can occur for alternating series (where terms alternate in sign).

For the given series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{10 n^{1.1}+1} \), we use the concept of conditional convergence to classify it. Even though the absolute series diverges, the alternating nature allows a different kind of convergence.

- We apply the Alternating Series Test to this series.- For a series to meet the test, its terms must decrease and approach zero as \( n \rightarrow \infty \).

The term \( \frac{n}{10n^{1.1}+1} \) fulfills these conditions. As a result, our series is conditionally convergent.

Alternating Series Test

The alternating series test is a method to check convergence for series whose terms switch between positive and negative. It applies to series of the form \( \sum_{n=1}^{\infty} (-1)^n b_n \) or \( \sum_{n=1}^{\infty} (-1)^{n+1} b_n \), where \( b_n \) are positive terms.

To use this test effectively, two main conditions need to be satisfied:

• The sequence \( b_n \) should decrease monotonically.
• The limit \( \lim_{n \to \infty} b_n = 0 \).

In the context of our series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{10n^{1.1}+1} \), we find:

• The terms \( \frac{n}{10n^{1.1}+1} \) indeed decrease as \( n \) increases.
• The limit of these terms as \( n \) approaches infinity is zero.

These checks confirm that the alternating series test indicates convergence, explaining why the series is conditionally convergent despite the divergence of the absolute sum.