## Calculus (3rd Edition)

The series $\sum_{n=2}^{\infty}\frac{ n}{n^2-n}$ diverges.
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.