Finding the interval convergence power series? How would I find the interval of convergence of this power series?
$\sum\frac{1x^k}{k^22^k}$
I performed the ratio test and did.
$\frac{x^{k+1}}{(k+1)^2(2)^{k+1}}$*$\frac{k^2(2^k)}{x^k}$
Then I got
$k\rightarrow\infty$
$x\frac{k^2}{2(k+1)^2}$
$-1<\frac{1}{2}x<1$
$x=2$
$x=-2$
$\sum(-2)^k\frac{1}{k^2)(2)^k}$ converge becuase alternate series limit zero the converge.
$\sum\frac{2^k}{k^2(2)^k}$
But I am not sure if x=2 converge or diverge. I tried ratio and root they were inconclusive.

The series converges for $x=2$. You may ask why...
Why not use the integral test, i.e. 
$\lim_{n\rightarrow \infty} \int_{1}^{n} x^{-2}\; dx = \lim_{n \rightarrow \infty} 1 - \frac{1}{n} = 1$.
