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 - Review - Exercises - Page 804: 44

Answer

$\frac{1}{4}$

Work Step by Step

Root test: $R=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=|\frac{\frac{(2n+2)!(x)^{n+1}}{((n+1)!)^{2}}}{\frac{2n!(x)^{n}}{(n!)^{2}}}|$ $=|\lim\limits_{n \to \infty}\frac{(2n+2)(2n+1}{(n+1)^{2}}.x|$ $=|\lim\limits_{n \to \infty}\frac{4n^{2}+6n+2}{n^{2}+2n+1}.x|$ $=|\frac{4.x}{1}|$ $=|4x|$ The series will converge when $|4x|\lt 1$ or $|x|\lt \frac{1}{4}$ Thus, the series has radius of convergence $\frac{1}{4}$ .
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