Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 6


$R=1$ ; interval of convergence is $[-1,1]$

Work Step by Step

Let $a_{n}=(-1)^{n}\frac {x^{n}}{n^{2}}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{(-1)^{n+1}\frac {x^{n+1}}{(n+1)^{2}}}{(-1)^{n}\frac {x^{n}}{n^{2}}}|$ $=|x|\lt 1$ At $x=1$ , alternating series, thus converges and $x=-1$ the given series is a p-series , thus convergent. Hence, $R=1$ ; interval of convergence is $[-1,1]$
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