91Ó°ÊÓ

An alternating series is a sequence of numbers where the signs of the terms alternate between positive and negative. This type of series often takes the form of \( \sum_{n=1}^{\infty} (-1)^n a_n \),where each term's sign alternates due to the \((-1)^n\) factor.

To determine if an alternating series converges, we typically apply the Alternating Series Test. Two conditions must be met:

• The absolute value of the terms \(a_n\) decreases steadily as \(n\) becomes larger.
• The limit of \(a_n\) as \(n\) approaches infinity is zero.

For a series that meets these criteria, it's said to be convergent. However, the focus here is on how the series behaves absolutely, so let's dive into absolute convergence.

Absolute convergence refers to the condition under which a series converges when you consider the absolute values of its terms.

In the context of our original exercise, the series \(\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{(n-4)^{2}+5} \)is considered for absolute convergence by examining its absolute version: \(\sum_{n=1}^{\infty} \left| (-1)^n \frac{1}{(n-4)^{2}+5} \right| = \sum_{n=1}^{\infty} \frac{1}{(n-4)^{2}+5} \).

A series is absolutely convergent if the series made up of the absolute values of its terms converges. Absolute convergence implies the original series is also convergent. Therefore, if a series is absolutely convergent, it is automatically classified as convergent, eliminating the need to check for conditional convergence.

For this series, we need to analyze if \( \sum_{n=1}^{\infty} \frac{1}{(n-4)^{2}+5} \) converges.

The Limit Comparison Test is a powerful method for determining the convergence or divergence of a series by comparing it to another series whose behavior (convergence or divergence) is already known.

We apply it here to compare the series \(\sum_{n=1}^{\infty} \frac{1}{(n-4)^{2}+5}\)with a well-known convergent series, the \(p\)-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\).

For the Limit Comparison Test, compute the limit:\[\lim_{n \to \infty} \frac{\frac{1}{(n-4)^{2}+5}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{(n-4)^{2}+5}.\]
As \(n\) approaches infinity, we find:\[\lim_{n \to \infty} \frac{n^2}{n^2} = 1.\]The limit is a finite, non-zero number. Thus, the two series \(\sum \frac{1}{(n-4)^{2}+5}\) and \(\sum \frac{1}{n^2}\)have the identical convergence behavior. Since \(\sum \frac{1}{n^2}\) is convergent, so is \(\sum \frac{1}{(n-4)^{2}+5}\).

A \(p\)-series is typically written in the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is a positive constant. A crucial property of \(p\)-series is that they converge when \(p > 1\) and diverge when \(p \leq 1\).

For instance, \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is a convergent \(p\)-series as \(p = 2 > 1\).

The series from our problem, \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{1}{(n-4)^{2}+5} \right|\),is similar to an adjusted version of a \(p\)-series, making it useful to compare them using the Limit Comparison Test. By relating a complicated series to a \(p\)-series, we can leverage the known behavior of \(p\)-series to determine convergence, which was precisely done in this exercise.