Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 24


Absolutely convergent.

Work Step by Step

First we will check whether the series is absolutely convergent. Because If the series is absolutely convergent then the series is convergent. But if the series is not absolutely convergent, then we will check whether the series is convergent or divergent. Remember that $\Sigma a_{n}$ is said to be absolutely convergent if $\Sigma |a_{n}|$ is convergent. Therefore, the given series will be absolutely convergent when the following series converges. $\Sigma |(-1)^{n-1}n^{-3}|=\Sigma _{n=1}^\infty n^{-3}=\Sigma _{n=1}^\infty \frac{1}{n^{3}}$ This is a p-series with $p=3 \gt 1$ Hence, the series converges and the original series given in the question is not absolutely convergent. The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Hence, the given series is absolutely convergent.
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