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: 26


$R=1$ ; interval of convergence is $[-1,1]$

Work Step by Step

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