Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 25


$R=\frac{1}{5}$ ; interval of convergence is $[\frac{3}{5},1)$

Work Step by Step

Let $a_{n}=\frac{(5x-4)^{n}}{n^{3}}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(5x-4)^{n+1}}{(n+1)^{3}}}{\frac{(5x-4)^{n}}{n^{3}}}|$ $=|5x-4|$ $=|5x-4|\lt 1$ Hence, $R=\frac{1}{5}$ ; interval of convergence is $[\frac{3}{5},1)$
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