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: 10


The series converges for all $x$ in the interval $[-1/ 2,1/2)$.

Work Step by Step

We apply the ratio test $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} \frac{2^{(n+1)}x^{n+1}/(n+1) }{2^nx^n/n}=2|x|\lim _{n \rightarrow \infty} \frac{n}{n+1}=2|x| $$ Hence, the series $\Sigma_{n=0}^{\infty} 2^nx^n/n$ converges if and only if $|x|\lt1/2$. That is, the interval of convergence is $(-1/2,1/2)$. Now, we check the end points: At $x=-1/ 2$, then $\Sigma_{n=0}^{\infty} 2^nx^n/n=\Sigma_{n=0}^{\infty} (-1)^n/n$ , converges (Alternating Series Test) At $x=1/ 2$, then $\Sigma_{n=0}^{\infty} 2^nx^n/n=\Sigma_{n=0}^{\infty}1/n$, diverges (P-Series Test) Hence, the series converges for all $x$ in the interval $[-1/ 2,1/2)$.
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