Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 773: 23

Answer

Conditionally convergent

Work Step by Step

Let $b_n=\frac{1}{\sqrt[3]{n^2}}$. We have: $\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{n^{2/3}}\right|=\sum_{n=1}^{\infty}\frac{1}{n^{2/3}}$ This is a $p$-series with $p=2/3<1$, so the series diverges, which means the given series in not absolutely convergent. We have: $b_n>0$ $\lim_{n\rightarrow\infty}b_n=0$ $b_{n+1}=1/(n+1)^{2/3}<1/n^{2/3}=b_n$ We apply the Alternating Series Test and we obtain that the given series is conditionally convergent.
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