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: 28


$R=2$ ; interval of convergence is $(-2,2)$

Work Step by Step

Let $a_{n}=\frac{n!x^{n}}{1.3.5.....(2n-1)}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(n+1)!x^{n+1}}{1.3.5.....(2(n+1)-1)}}{\frac{n!x^{n}}{1.3.5.....(2n-1)}}|$ $=\lim\limits_{n \to \infty}|\frac{n+1}{2n+1}(x)|\lt 1$ $=|\frac{x}{2}|\lt 1$ Hence, $R=2$ ; interval of convergence is $(-2,2)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.