Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.6 Absolute Convergence and the Radio and Root Tests - 11.6 Exercises - Page 782: 2

Answer

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ is conditionally convergent

Work Step by Step

We have, \[\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}\] Compare $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ with $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}a_n$ \[\Rightarrow a_n=\frac{1}{\sqrt{n}}\] $a_n$ is positive term and monotonically decreasing, and \[\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}=0\] By Alternating Series Test, $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ is convergent.$\;\;\;\;\;\;\;\;\;\;\ldots(1)$ Compare $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ with $\displaystyle\sum_{n=1}^{\infty} b_n$ \[\Rightarrow b_n=\frac{(-1)^{n-1}}{\sqrt{n}}\] \[|b_n|=\left|\frac{(-1)^{n-1}}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}(=c_n)\] Also $\displaystyle\sum_{n=1}^{\infty}c_n=\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ is divergent by P-Series Test. $\Rightarrow \displaystyle\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{\sqrt{n}}\right|$ is divergent.$\;\;\;\;\;\;\;\;\;\;\;\ldots (2)$ Using (1) and (2) $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ is conditionally convergent.
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