Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.8 - Power Series - 11.8 Exercises - Page 751: 19


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

Work Step by Step

Let $a_{n}=\frac{(x-2)^{n}}{n^{n}}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(x-2)^{n+1}}{n^{n+1}}}{\frac{(x-2)^{n}}{n^{n}}}|$ $=|x+2|\lim\limits_{n \to \infty}\frac{1}{n}$ $=0$ Hence, $R=\infty$ ; interval of convergence is $(-\infty,\infty)$
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