The integral test and Bertrand series

The test. If $f\ge0$ is decreasing on $[2,\infty)$, then $\sum f(n)$ and $\int_2^\infty f$ have the same nature.

Case $p>0$. With $u=\ln x$, $du=\frac{dx}x$:
$$\int_2^R\frac{dx}{x(\ln x)^p}=\int_{\ln2}^{\ln R}\frac{du}{u^p}.$$
As $R\to\infty$ this converges iff $p>1$ (the $p$-test for integrals).

Case $p=0$. $\sum\frac1n$ diverges.

Case $p<0$. $\frac1{n(\ln n)^p}=\frac{(\ln n)^{|p|}}n\ge\frac1n$ for $n\ge3$: diverges.

Answer. The series converges iff $p>1$.

Remark. $\sum\frac1{n\ln n}$ diverges, yet so slowly that the partial sum up to $10^{100}$ is only about $5.4$: $\ln\ln10^{100}\approx5.4$. By contrast $\sum\frac1{n(\ln n)^2}$ converges.