Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n(1+\sqrt{n})} $$

Step-by-Step Solution

Answer

The series is absolutely convergent.

Step 1: Identify the Series Type

The series given is \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(1+\sqrt{n})} \). Recognize that it is an alternating series due to the \((-1)^{n+1}\) term. This indicates we should consider the alternating series test initially.

Step 2: Apply the Alternating Series Test

For the alternating series test, we consider the non-alternating part of the series, \( b_n = \frac{1}{n(1+\sqrt{n})} \). Check if \( \lim_{n \to \infty} b_n = 0 \) and if \( b_n \) is decreasing. Calculating the limit, we find \( \lim_{n \to \infty} \frac{1}{n(1+\sqrt{n})} = 0 \). To show it's decreasing, note that as \( n \) increases, \( n(1+\sqrt{n}) \) becomes larger, making \( b_n \) smaller. Thus, \( b_n \) is decreasing and the series meets the conditions of the alternating series test, proving it is convergent.

Step 3: Test for Absolute Convergence

Now, consider the absolute convergence by evaluating \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{n(1+\sqrt{n})} \right| = \sum_{n=1}^{\infty} \frac{1}{n(1+\sqrt{n})} \). Use the comparison test by comparing this series to \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \), which is a p-series with \( p = \frac{3}{2} > 1 \) and is therefore convergent (given that the terms of \( \frac{1}{n(1+\sqrt{n})} \) are smaller for all \( n \geq 1 \) than \( \frac{1}{n^{3/2}} \)). Hence, \( \sum_{n=1}^{\infty} \frac{1}{n(1+\sqrt{n})} \) is also convergent.

Step 4: Conclusion About Convergence

Since the series \( \sum_{n=1}^{\infty} \frac{1}{n(1+\sqrt{n})} \) converges, the original series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(1+\sqrt{n})} \) is absolutely convergent. For a series to be absolutely convergent, its absolute series must sum to a limit.

Key Concepts

Alternating Series Test

The alternating series test helps us determine the convergence of a series that alternates in sign. It applies to series of the form \( \sum (-1)^{n} a_n \), where the terms \( a_n \) are positive. This test requires checking two conditions:

• The limit of \( a_n \) as \( n \to \infty \) is zero, specifically \( \lim_{n \to \infty} a_n = 0 \).
• The sequence \( a_n \) is decreasing, meaning \( a_n \geq a_{n+1} \) for all \( n \).

When both conditions are met, the series converges. In our example, the alternating series' form is given by the term \( \frac{1}{n(1+\sqrt{n})} \). As \( n \) increases, its denominator grows, making the sequence decrease. The limit also approaches zero, satisfying the alternating series test.

Absolute Convergence

Absolute convergence refers to whether the series of absolute values converges. For absolute convergence, we consider \( \sum |a_n| \) instead of \( \sum a_n \).

If the series of absolute values converges, then the original series is absolutely convergent. Absolute convergence implies convergence of the series itself, but more strongly; if a series is absolutely convergent, it's guaranteed to converge, regardless of any alternation in signs.
In our example, we checked the series \( \sum \frac{1}{n(1+\sqrt{n})} \) and found it convergent. Therefore, the original alternating series \( \sum (-1)^{n+1} \frac{1}{n(1+\sqrt{n})} \) is absolutely convergent.

Comparison Test

The comparison test is a handy tool to determine convergence, by comparing a known series with another similar one. It requires another series, usually simpler, for comparison. To apply this test, you:

• Find a comparison series \( \sum b_n \), where \( b_n \geq 0 \).
• If \( a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges too.
• Conversely, if \( a_n \geq b_n \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges.

In the given exercise, we compare with the p-series \( \sum \frac{1}{n^{3/2}} \), which converges. Since \( \frac{1}{n(1+\sqrt{n})} \leq \frac{1}{n^{3/2}} \), the series converges by the comparison test.

P-Series

A p-series is one of the cornerstones in analyzing series convergence. It takes the form \( \sum \frac{1}{n^p} \), where \( p \) is a positive constant. Understanding p-series helps in utilizing the comparison test effectively:

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

In the original exercise, we use \( \sum \frac{1}{n^{3/2}} \), a p-series with \( p = \frac{3}{2} \). Because \( p > 1 \), it converges. Recognizing that it's a p-series helps simplify reasoning about the series' convergence.