Problem 34 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}+2 n+1} $$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2 + 2n + 1} \). This is an alternating series due to the term \((-1)^{n-1}\), which causes the signs of the terms to alternate.

Step 2: Simplify the Denominator

Simplify the expression in the denominator. Notice that \(n^2 + 2n + 1 = (n+1)^2\). So, the terms of the series become \( \frac{(-1)^{n-1}}{(n+1)^2} \).

Step 3: Apply the Alternating Series Test

The alternating series test states that an alternating series \(\sum (-1)^{n} a_n\) converges if \(a_n\) is positive, decreasing, and \(\lim_{n \to \infty} a_n = 0\). So, consider \(a_n = \frac{1}{(n+1)^2}\).

Step 4: Check Decreasing Condition

The sequence \(a_n = \frac{1}{(n+1)^2}\) is decreasing because as \(n\) increases, \((n+1)^2\) increases, making \(\frac{1}{(n+1)^2}\) decrease.

Step 5: Check Limit Condition

Evaluate \(\lim_{n \to \infty} \frac{1}{(n+1)^2} = 0\). Since this limit is 0, the condition for the alternating series test is satisfied.

Step 6: Check Absolute Convergence with Comparison Test

To check for absolute convergence, consider the series \(\sum_{n=1}^{\infty} \frac{1}{(n+1)^2}\). Compare it with the \(p\)-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) where \(p=2\). The \(p\)-series converges when \(p>1\).

Step 7: Conclude on Absolute Convergence

Since \(\sum_{n=1}^{\infty} \frac{1}{(n+1)^2}\) is similar to \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) and both converge, the original series converges absolutely.

Key Concepts

Alternating Series Test

Many series have terms that alternate between positive and negative. The Alternating Series Test helps us determine whether an alternating series converges. This test is applicable to series of the form \( \sum (-1)^{n} a_n \), where \(a_n\) represents the non-alternating part of the series.

For the test to indicate convergence, two conditions must be satisfied:

• The terms \(a_n\) must be positive. This ensures that ignoring the sign, the terms themselves are always greater than zero.
• The terms \(a_n\) must form a decreasing sequence. If each term is smaller than the previous one, it suggests that they are moving toward zero.
• The limit as \(n\) approaches infinity of \(a_n\) must be zero, i.e., \( \lim_{n \to \infty} a_n = 0 \). This means the terms eventually become negligible.

If all these criteria are fulfilled, the alternating series converges. In our exercise, we found that the series meets all these criteria, hence it converges.

Absolute Convergence

Absolute convergence is a stronger form of convergence in series, providing certainty under specific conditions. A series \( \sum a_n \) is said to converge absolutely if the series formed by the absolute values of its terms \( \sum |a_n| \) also converges.

In simpler terms:

• If a series converges absolutely, it still converges when the plus and minus signs of its terms are ignored.
• Absolute convergence ensures the series converges regardless of the order of its terms, making it a particularly robust form of convergence.

To test for absolute convergence in our exercise, we considered the series with the absolute terms, \( \sum \frac{1}{(n+1)^2} \). By comparing it with the well-known \(p\)-series \( \sum \frac{1}{n^2} \), which converges for \(p > 1\), we concluded that our series also converges absolutely.

Comparison Test

The Comparison Test is a handy tool when dealing with series, allowing us to compare a tricky series with a familiar one to deduce convergence, divergence, or absolute convergence.

This test involves comparing a given series \( \sum a_n \) to another series \( \sum b_n \):

• If \( 0 \leq a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• Conversely, if \( a_n \geq b_n \geq 0 \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges as well.

In our exercise, we used the Comparison Test to compare \( \sum \frac{1}{(n+1)^2} \) with the \(p\)-series \( \sum \frac{1}{n^2} \). Since the \(p\)-series converges for \(p>1\), our series \( \sum \frac{1}{(n+1)^2} \) converges. It helped us confirm that the absolute convergence of the given series holds true.