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 - 11.10 Exercises - Page 789: 17

Answer

$\Sigma_{n=0}^{\infty}\frac{(x-3)^{n}2^{n}e^{6}}{n!}$ $R=\infty$

Work Step by Step

$e^{2x}=\Sigma_{n=0}^{\infty}\frac{(x-3)^{n}2^{n}e^{6}}{n!}$ $\lim\limits_{n \to \infty}|\frac{\frac{(x-3)^{n+1}2^{n+1}}{(n+1)!}}{\frac{(x-3)^{n}2^{n}}{n!}}|$ $=|x-3|\lim\limits_{n \to\infty}|\frac{2}{n+1}|$ $=0\lt 1$ The the radius of convergence is $R=\infty$.
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