## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 577: 21

#### Answer

The radius of convergence is $R = 1,$ and the series converges absolutely for $-1 \lt x \lt 1.$

#### Work Step by Step

Apply the ratio test: \begin{align*} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{x^{2 n+3}}{3 n+4} \cdot \frac{3 n+1}{x^{2 n+1}}\right|\\ &=\lim _{n \rightarrow \infty}\left|x^{2} \frac{3 n+1}{3 n+4}\right|\\ &=\left|x^{2}\right| \end{align*} The radius of convergence is $R = 1,$ and the series converges absolutely for $-1 \lt x \lt 1.$ For the endpoint $x = 1$, the series becomes $\sum_{n=15}^{\infty} \frac{1}{3 n+1},$ which diverges by the Limit Comparison Test. For $x=-1$, the series $\sum_{n=15}^{\infty} \frac{-1} {3 n+1},$ which also diverges by the Limit Comparison Test (comparing with the divergent harmonic series).

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