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: 24

Answer

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

Work Step by Step

We apply the ratio test and by using L'Hopital rule, we have $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ x^{3n+5}/\ln(n+1) }{ x^{3n+2}/\ln n}|\\ =|x^3|\lim _{n \rightarrow \infty} \frac{\ln n}{\ln(n+1)}=|x^3|\lim _{n \rightarrow \infty} \frac{n+1}{n}=|x^3|. $$ Hence, the series $\Sigma_{n=2}^{\infty} x^{3n+2}/\ln n$ converges if and only if $|x^3|\lt1$. That is, the interval of convergence is $(-1,1)$. Now, we check the end points: At $x=-1$, then $\Sigma_{n=2}^{\infty} x^{3n+2}/\ln n=\Sigma_{n=2}^{\infty}(-1)^{3n+2}/\ln n$, converges (Alternating Series Test). At $x=1$, then $\Sigma_{n=2}^{\infty} x^{3n+2}/\ln n=\Sigma_{n=2}^{\infty}1/\ln n$, diverges (by the comparison test with $\frac{1}{\ln n}\geq \frac{1}{n}$). Hence, the series converges for all $x$ in the interval $[-1,1)$.
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