The way I see it you can compare
$\sum_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) } < 1/n$
$1/n$ is a $p$-series in which $p = 1 \leq 1 $
So $1/n$ diverges.
Thus $\sum\limits_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) }$ diverges.
The way I see it you can compare
$\sum_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) } < 1/n$
$1/n$ is a $p$-series in which $p = 1 \leq 1 $
So $1/n$ diverges.
Thus $\sum\limits_{n=2}^{\infty} \frac{ 1}{ n^2 \log (n) }$ diverges.
As $\ln(n) > 1$ for $n > e$, $\frac{1}{n^2 \ln(n)} < \frac{1}{n^2}$. The latter is known to converge.
Use the Integral test for convergence: Since $\frac d{dt}\text{Ei}(-\log t)=\frac d{dt}\text{li}(\frac1t)=\frac1{t^2\log(t)}$, we get $ \begin{eqnarray} \int\limits_2^\infty \frac1{n^2\log(n)}dn&=&\text{Ei}(-\log n)\Biggr|_{2}^\infty&<&\infty\\&=&\underbrace{\text{Ei}(-\log \infty)}_{=0}-\text{Ei}(-\log 2)\\ &=&-\text{li}(\frac12)\\ &=&-\int_{0}^{1/2}\frac{dn}{\ln n}&&\hskip0.7in (*) \\ &=&0.378\dots&<&\infty \end{eqnarray} $ your sum converges: $(*)$ is finite, since $\underbrace{\text{li}(1)}_{-\infty}<\int_{0}^{1/2}\frac{dn}{\ln n}< \underbrace{\text{li}(0)}_{=0}$ and $0<\frac12<1.$