Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.9 Exercises: 28

Answer

The radius of convergence is $1$ $\int \frac{tan^{-1}x}{x}dx=c+\Sigma_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n+1}}{(2n+1)^{2}}$

Work Step by Step

The sum of a geometric series with initial term $a$ and common ratio $r$ is $S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$ Here, $\frac{tan^{-1}x}{x}=\frac{a}{1-r}$ Therefore, $f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n}}{2n+1}$ This is the power series representation of $f(x)$. We know that the power series converges when $r=|x^{2}|\lt 1$ The radius of convergence is $1$ $\int \frac{tan^{-1}x}{x}dx=c+\Sigma_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n+1}}{(2n+1)^{2}}$
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