91Ó°ÊÓ

An alternating series is a type of series where the terms change sign with each successive element. This usually occurs due to a factor like (-1)\^n or (-1)\^(n+1) that alternates the sign. In the original exercise, the series given \( \sum_{n=1}^{\infty} (-1)^n \frac{\sin n}{n^2} \) has terms alternating in sign due to the (-1)\^n factor.

When dealing with alternating series, we often apply the Alternating Series Test to check for convergence. This test states that an alternating series \( \sum (-1)^n a_n \) converges if the terms \( a_n \) are:

• Decreasing in magnitude, and
• Converging to zero as \( n \to \infty \)

In this exercise, though the series converges absolutely, these conditions can confirm convergence in many alternating series scenarios.

Absolute convergence of a series involves looking at the sum of the absolute values of its terms. If \( \sum_{n=1}^{\infty} \left| a_n \right| \) converges, then the series \( \sum_{n=1}^{\infty} a_n \) is absolutely convergent.

In the given exercise, we examine the series \( \sum_{n=1}^{\infty} \left| \frac{\sin n}{n^2} \right| \). Since \( \left| \sin n \right| \) is always less than or equal to 1 due to the nature of the sine function, it follows that \( \left| \frac{\sin n}{n^2} \right| \leq \frac{1}{n^2} \).

We know that \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a convergent series. Therefore, by the Comparison Test, \( \sum_{n=1}^{\infty} \left| \frac{\sin n}{n^2} \right| \) also converges. This establishes the absolute convergence of the original series.

The Comparison Test for series is a useful tool to determine the convergence of a series by comparing it to another series whose convergence status is known. The test is straightforward:

• If \( 0 \leq a_n \leq b_n \) for all \( n \), and \( \sum_{n=1}^{\infty} b_n \) converges, then \( \sum_{n=1}^{\infty} a_n \) converges.
• If \( a_n \geq b_n \geq 0 \) and \( \sum_{n=1}^{\infty} b_n \) diverges, then \( \sum_{n=1}^{\infty} a_n \) diverges.

In the original problem, to determine absolute convergence, we compare \( \sum_{n=1}^{\infty} \left| \frac{\sin n}{n^2} \right| \) with the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). Because \( \left| \sin n \right| \leq 1 \), we have \( \left| \frac{\sin n}{n^2} \right| \leq \frac{1}{n^2} \). Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is convergent (a known p-series with \( p=2 \)), the Comparison Test shows that \( \sum_{n=1}^{\infty} \left| \frac{\sin n}{n^2} \right| \) also converges.