Chapter 10.5, Problem 28E

To determine

To check: whether the Series ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sin n}{n^2}}$ converges absolutely, converges conditionally, or diverges.

Answer to Problem 28E

The given series ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sin n}{n^2}}$ converges absolutely.

Explanation of Solution

Given:

The series is ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sin n}{n^2}}$ .

Calculation:

Given, ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sin n}{n^2}}$

Taking nth-term test,

  ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ diverges if $\underset{n\to \infty }{\lim }{a}_n$ fails to exist or is different from zero.

Then,

  $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left[\frac{\sin n}{n^2}\right]$

As, $-1\le \sin n\le 1=0$

Therefore, ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{\sin n}{n^2}}$ converges absolutely.