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 826: 48


series has radius of convergence is $1$

Work Step by Step

$tan^{-1}x^{2}=\Sigma_{n=0}^\infty(-1)^{n}\frac{(x^{2})^{2n+1}}{2n+1}$ $=\Sigma_{n=0}^\infty(-1)^{n}\frac{x^{4n+2}}{2n+1}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{x^{4n+6}}{2n+3}}{\frac{x^{4n+2}}{2n+1}}|$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{x^{4}(2+1/n)}{(2+3/n)}|$ $=|\frac{x^{4}(2+0)}{(2+0)}|$ $=|x^{4}|$ This converges when $|x^{4}|\lt 1$ or $=|x| \lt 1$ Thus, the series has radius of convergence is $1$ .
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