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: 23

Answer

$R=2$ ; interval of convergence is $[\frac{1}{2},\frac{1}{2}]$

Work Step by Step

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