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: 20

Answer

$R=\frac{5}{2}$ ; interval of convergence is $[-2,3)$

Work Step by Step

Let $a_{n}=\frac{(2x-1)^{n}}{5^{n}\sqrt n}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(2x-1)^{n+1}}{5^{n+1}\sqrt n}}{\frac{(2x-1)^{n}}{5^{n}\sqrt n}}|$ $=\frac{|2x-1|}{5}$ $=\frac{|2x-1|}{5}\lt 1$ $-2\lt x\lt 3$ When $x=3$, the series becomes a divergent p-series (p=1/2). Hence, $R=\frac{5}{2}$ ; interval of convergence is $[-2,3)$
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