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: 16


The interval of convergence is $(-\infty,\infty)$.

Work Step by Step

We apply the ratio test $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ 8^{n+1}x^{n+1}/(n+1)! }{ 8^{n}x^{n}/n! }|=8|x|\lim _{n \rightarrow \infty} \frac{1}{(n+1)}=8|x|\times0 $$ Hence, the series $\Sigma_{n=0}^{\infty} 8^{n}x^{n}/n!$ converges if and only if $|x|\times0=0\lt1$ which is true for any value of $x$. That is, the interval of convergence is $(-\infty,\infty)$.
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