Answer
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 $[-\sqrt 2,\sqrt 2]$.
