Prove that the series $\sum_{1}^{\infty}a^n \frac{logn}{n^2}$ is convergent only for $[-1,1]$ and divergent elsewhere. Prove that the series $\sum_{1}^{\infty}a^n \frac{logn}{n^2}$ is convergent only for $a \epsilon [-1,1]$and divergent elsewhere.
I have given it an honest attempt, I can see why this must be true graphically and I have established convergence on $[-1,1]$, But I am having troubles establishing the divergence part. 
 A: Applying the $\;n\,-$ th root test (Cauchy-Hadamard):
$$\sqrt[n]{\left|\frac{\log n}{n^2}\right|}\xrightarrow[n\to\infty]{}1\implies\;\text{the convergence radius is}\;\;R=1$$
and thus the series converges for $\;|a|<1\;$, yet at the extreme points $\;a=\pm1\;$ we get
$$\sum_{n=1}^\infty\frac{\log n}{n^2}\;,\;\;\sum_{n=1}^\infty(-1)^n\frac{\log n}{n^2}$$
and both series converge because the second one converges absolutely as it is a Leibniz series and, for example, $\;\log n\le n^{1/2}\implies \dfrac{\log n}{n^2}\le\dfrac1{n^{3/2}}\;$ (the series is positive for $\;n\ge3\;$) .
You already knew this.
Now, for $\;a>1\;$ we can write $\;a=1+h\;,\;\;h>0\;$ (for $\;a<-1\;$ it is similar), so
$$a^n=1+\sum_{k=1}^n\binom nkh^k\ge1+nh\implies a^n\frac{\log n}{n^2}\ge\frac{\log n}{n^2}+\frac{h\log n}n$$
and since
$$\sum_{n=1}^\infty\frac{\log n}{n^2}\;\;\;\text{converges but}\;\;\;\sum_{n=1}^\infty\frac{h\log n}n\;\;\;\text{diverges}$$
the right side as a whole diverges and thus the comparison test gives us our series diverges, too.
