Problem 19 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{n^{3}+1} $$

Step-by-Step Solution

Answer

The series converges conditionally, but not absolutely.

Step 1: Define Absolute Convergence

A series \( \sum a_n \) converges absolutely if \( \sum |a_n| \) converges. For the given series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^3+1} \), this means we need to check if \( \sum_{n=1}^{\infty} \frac{n}{n^3+1} \) converges.

Step 2: Apply Limit Comparison Test for Absolute Convergence

Consider the series \( \sum_{n=1}^{\infty} \frac{n}{n^3+1} \) for comparison. Use the term \( \frac{1}{n^2} \) as a comparison series because for large \( n \):\[\frac{n}{n^3+1} \approx \frac{n}{n^3} = \frac{1}{n^2}\]Perform the Limit Comparison Test:\[\lim_{n \to \infty} \frac{\frac{n}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^3}{n^3+1} = 1\]Since \( \frac{1}{n^2} \) is a convergent \( p \)-series (\( p=2 > 1 \)), \( \sum_{n=1}^{\infty} \frac{n}{n^3+1} \) also converges, implying the series does not converge absolutely.

Step 3: Test for Convergence using Alternating Series Test

Apply the Alternating Series Test to the original series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^3+1} \). Check if the conditions hold:1. The sequence \( b_n = \frac{n}{n^3+1} \) is decreasing.2. \( \lim_{n \to \infty} b_n = 0 \).For decreasing:\( b_{n+1} = \frac{n+1}{(n+1)^3+1} \) is smaller than \( b_n = \frac{n}{n^3+1} \) for sufficiently large \( n \).For the limit:\[\lim_{n \to \infty} \frac{n}{n^3+1} = \lim_{n \to \infty} \frac{1}{n^2}\]which equals 0.Both conditions are satisfied, so the original series converges.

Key Concepts

Absolute Convergence

When we talk about absolute convergence, we consider whether the series converges when we take the absolute values of its terms. For a series like \( \sum a_n \), it converges absolutely if the series of its absolute values, \( \sum |a_n| \), also converges.
In the case of the given series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^3+1} \), to test for absolute convergence, you need to look at \( \sum_{n=1}^{\infty} \frac{n}{n^3+1} \). This removes the alternating sign, focusing purely on the size of the terms.
The key reason absolute convergence is important is that if a series converges absolutely, it behaves much like a finite sum. It converges under any conditions, and many powerful mathematical tools can be applied to it. However, as determined in the solution, in this case, the series does not converge absolutely as the absolute series does not converge. This requires us to further analyze it for conditional convergence.

Limit Comparison Test

The Limit Comparison Test is an essential tool for determining the convergence of a series, especially when direct comparison is tough. Let's break it down:

• We start by comparing the series \( \sum_{n=1}^{\infty} \frac{n}{n^3+1} \) with a known convergent series, \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).
• We then compute the limit of the ratio of the terms from the original series and the comparison series as \( n \) approaches infinity: \[ \lim_{n \to \infty} \frac{\frac{n}{n^3+1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^3}{n^3+1} = 1 \]

The result of 1 tells us that both series behave similarly as \( n \to \infty \). Since the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a convergent \( p \)-series with \( p=2 \), the comparison implies that \( \sum_{n=1}^{\infty} \frac{n}{n^3+1} \) converges as well. However, this result tells us that the given series does not converge absolutely, leaving us to check for conditional convergence.

Alternating Series Test

When each term in a series alternates in sign, like the original series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^3+1} \), the Alternating Series Test can be a practical way to check for convergence.
This test uses two simple conditions:

• The terms \( b_n = \frac{n}{n^3+1} \) must decrease to zero.
• The limit of \( b_n \) as \( n \to \infty \) must be zero: \[ \lim_{n \to \infty} \frac{n}{n^3+1} = \lim_{n \to \infty} \frac{1}{n^2} = 0 \]

Both these conditions are satisfied for our series. As \( n \) increases, the terms get smaller and approach zero, confirming convergence by the Alternating Series Test.
Thus, although the series does not converge absolutely, it does converge conditionally. This means that while the series converges due to the alternating signs, it wouldn't converge if the terms did not alternate.