Answer
The series $\sum_{n=2}^{\infty}\frac{ n}{n^2-n}$ diverges.
Work Step by Step
Using the limit comparison test with the divergent p-series (also known as the harmonic series) $\sum_{n=1}^{\infty} \frac{1}{n}$
Since we have
$$L=\lim_{n\to \infty } \frac{n/(n^2-n)}{1/n}=\lim_{n\to \infty } \frac{ n^2}{n^2-n}\\
=\lim_{n\to \infty } \frac{ 1}{1-1/n}=1\gt0$$
then the series $\sum_{n=2}^{\infty}\frac{ n}{n^2-n}$ diverges.
