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: 24


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

Work Step by Step

Let $a_{n}=\frac{n^{2}x^{n}}{2.4.6.....(2n)}=\frac{n^{2}x^{n}}{2^{n}n!}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(n+1)^{2}x^{n+1}}{2^{n+1}(n+1)!}}{\frac{n^{2}x^{n}}{2^{n}n!}}|$ $=0\lt 1$ Hence, $R=\infty$ ; interval of convergence is $(-\infty,\infty)$
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