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: 23


Conditionally 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. $\Sigma |(-1)^{n-1}n^{-1/3}|=\Sigma _{n=1}^\infty n^{-1/3}=\Sigma _{n=1}^\infty \frac{1}{n^{1/3}}$ This is a p-series with $p=\frac{1}{3} \lt 1$ Hence, the series diverges and the original series given in the question is not absolutely converges. 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 $p\leq 1$. Now we want to see whether the series is conditionally convergent or divergent. Alternating Series Test Consider the series $\Sigma a_{n}$ where, $a_{n}=(-1)^{n}b^n$ or$(-1)^{n+1}b^n$ Then the series is convergent if 1. $\lim\limits_{n \to \infty} b_{n}=0$ 2. $b_{n}$ is positive. 3. $b_{n}$ is decreasing. Note that $\lim\limits_{n \to \infty}\frac{1}{n^{1/3}}=\frac{1}{\infty}=0$ and $\frac{1}{n^{1/3}}$ is always positive for $n\geq 1$ Also, $\frac{1}{n^{1/3}}$ is decreasing because $n^{1/3}$ is increasing for $n\geq 1$. Therefore, by AST , the given series converges conditionally. Hence, the given series is conditionally convergent.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.