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: 13


The series converges for all $x$ in the interval $[-1,1]$.

Work Step by Step

We apply the ratio test $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ x^{ n+1}/(n+1)^{5} }{x^{ n}/n^{5}}|=|x|\lim _{n \rightarrow \infty} \frac{n^5}{(n+1)^5}=|x| $$ Hence, the series $\Sigma_{n=4}^{\infty} x^{ n}/n^{5}$ converges if and only if $|x|\lt1$. That is, the interval of convergence is $(- 1,1)$. Now, we check the end points: At $x=-1$, then $\Sigma_{n=4}^{\infty} x^{ n}/n^{5}=\Sigma_{n=4}^{\infty} (-1)^{ n}/n^{5}$, converges (Alternating Series Test) At $x=1$, then $\Sigma_{n=4}^{\infty} x^{ n}/n^{5}=\Sigma_{n=4}^{\infty} 1/n^{5}$, converges (P- Series Test) Hence, the series converges for all $x$ in the interval $[-1,1]$.
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