## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 577: 19

#### Answer

The radius of convergence is $R = 1$, and the series converges absolutely on the interval $−1 \lt x \leq 1$.

#### Work Step by Step

Since \begin{aligned} \rho &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n+1} x^{n+1}}{\sqrt{n^{2}+2 n+2}} \cdot \frac{\sqrt{n^{2}+1}}{(-1)^{n} x^{n}}\right| \\ &=\lim _{n \rightarrow \infty}\left|x \frac{\sqrt{n^{2}+1}}{\sqrt{n^{2}+2 n+2}}\right|\\ &=\lim _{n \rightarrow \infty}|x \sqrt{\frac{n^{2}+1}{n^{2}+2 n+2}}|\\ &=\lim _{n \rightarrow \infty}|x \sqrt{\frac{1+1 / n^{2}}{1+2 / n+2 / n^{2}}}| \\ &=|x| \end{aligned} Thus the radius of convergence is $R = 1$, and the series converges absolutely on the interval $−1 \lt x \lt 1$ For the endpoint $x=1,$ the series becomes $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n^{2}+1}},$ which converges by the Leibniz Test. For $x= -1$ the series becomes $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{2}+1}},$ which diverges by the Limit Comparison Test, comparing with the $\sum_{n=1}^{\infty} \frac{1}{n}$.

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