Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 786: 50



Work Step by Step

$xe^{2x}=x\Sigma_{n=0}^\infty(2)^{n}\frac{x^{n}}{n!}=\Sigma_{n=0}^\infty(2)^{n}\frac{x^{n+1}}{n!}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(2)^{n+1}\frac{x^{n+2}}{n+1!}}{(2)^{n}\frac{x^{n+1}}{n!}}|$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{2x}{n+1}|$ $=0$ Thus, the series has a radius of convergence of $\infty$ .
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