Answer
The series converges for all $x$ in the interval $(-2,2)$.
Work Step by Step
We apply the ratio test
$$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{(-1)^{(n+1)}(n+1)x^{2n+1}/4^{n+1} }{(-1)^{n}(n)x^{2n}/4^{n}}|=\frac{|x^2|}{4}\lim _{n \rightarrow \infty} \frac{n+1}{n}=\frac{x^2}{4}. $$
Hence, the series $\Sigma_{n=0}^{\infty} (-1)^{n}(n)x^{2n}/4^{n}$ converges if and only if $\frac{x^2}{4}\lt1$. That is the interval of convergence is $(- 2, 2)$.
Now, we check the end points:
At $x=- 2$, then
$\Sigma_{n=0}^{\infty} (-1)^{n}(n)x^{2n}/4^{n}=\Sigma_{n=0}^{\infty} (-1)^{n}n$, diverges (limit Test)
At $x= 2$, then
$\Sigma_{n=0}^{\infty} (-1)^{n}(n)x^{2n}/4^{n}=\Sigma_{n=0}^{\infty} (-1)^{n}n$, diverges (Limit Test)
Hence, the series converges for all $x$ in the interval $(-2,2)$.
