## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 737: 35

#### Answer

The series is divergent and $b_n$ is not decreasing so, we cannot apply the Alternating Series test.

#### Work Step by Step

We are given that $\Sigma_{n=1}^\infty (-1)^{n+1} b_n$ For odd values, we have $b_n=\dfrac{1}{n}$ For even values, we have $b_n=\dfrac{-1}{n^2}$ It has been seen that $\dfrac{-1}{n^2}\lt \dfrac{-1}{n+1}$ (For every $\dfrac{1}{n^2}$ term, which is less than the absolute value). Therefore, $b_n$ is not decreasing and thus AST can not be applied. Further, $\Sigma_{n=1}^\infty (-1)^{n+1} b_n=\Sigma_{n=1}^\infty \dfrac{1}{2n-1}+\Sigma_{n=1}^\infty\dfrac{-1}{(2n)^2}$ But $\Sigma_{n=1}^\infty \dfrac{1}{2n-1} \gt \Sigma_{n=1}^\infty\dfrac{-1}{(2n)^2}$ The series $\Sigma_{n=1}^\infty \dfrac{1}{2n-1}$ will diverge by the comparison test. Hence, it has been verified that the series is divergent and $b_n$ is not decreasing so, we cannot apply the Alternating Series test.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.