Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 577: 7


The radius of convergence is $\sqrt 3$.

Work Step by Step

We apply the ratio test: $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ x^{2n+2} /3^{n+1} }{ x^{2n} /3^{n} }|= x^2/3 $$ Hence, the series $\Sigma_{n=0}^{\infty} x^{2n} /3^{n}$ converges if and only if $\rho= x^2/3| \lt1$. That is, the interval of convergence is $(-\sqrt 3,\sqrt 3)$ and the radius of convergence is $\sqrt 3$.
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.