Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.8 - Power Series - 11.8 Exercises - Page 786: 13

Answer

$R=\infty$ ; interval of convergence is $(-\infty, \infty)$

Work Step by Step

Let $a_{n}=\frac {x^{n}}{n!}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac {x^{n+1}}{(n)!}}{\frac {x^{n}}{n!}}|$ $=0\lt 1$ Hence, $R=\infty$ ; interval of convergence is $(-\infty, \infty)$
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