Answer
$\Sigma_{n=2}^{\infty}\frac{1}{n^{p} ln(n)}$ converges if $p\gt 1$
Work Step by Step
Clearly, if $p\lt 0$, then the series converges, since $\lim\limits_{n \to \infty}\frac{1}{n^{p}ln n}=\infty$ if $0\leq p\leq 1$, $n^{p}ln (n)\leq n ln (n)$
and $\Sigma_{n=2}^{\infty}\frac{1}{n ln(n)}$ diverges.
So, $\Sigma_{n=2}^{\infty}\frac{1}{n^{p} ln(n)}$ converges if $p\gt 1$
