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

The series converges for all $x$ in the interval $[-\sqrt 2,\sqrt 2]$.

#### Work Step by Step

We apply the ratio test $$\rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{(-1)^{(n+1)}x^{2n+3}/(n+1)2^{n+1} }{(-1)^{n}x^{2n+1}/n2^{n} }|=\frac{|x^2|}{2}\lim _{n \rightarrow \infty} \frac{n}{n+1}=\frac{|x^2|}{2}.$$ Hence, the series $\Sigma_{n=1}^{\infty} (-1)^{n}x^{2n+1}/n2^{n}$ converges if and only if $\frac{|x^2|}{2}\lt1$. That is, the interval of convergence is $(-\sqrt 2,\sqrt 2)$. Now, we check the end points: At $x=-\sqrt 2$, then $\Sigma_{n=1}^{\infty} (-1)^{n}x^{2n+1}/n2^{n}=\Sigma_{n=1}^{\infty} (-1)^{n}2/n$, converges (Alternating Series Test) At $x=\sqrt 2$, then $\Sigma_{n=1}^{\infty} (-1)^{n}x^{2n+1}/n2^{n}=\Sigma_{n=1}^{\infty} (-1)^{n}2/n$, converges (Alternating Series Test) Hence, the series converges for all $x$ in the interval $[-\sqrt 2,\sqrt 2]$.

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