## Calculus (3rd Edition)

The series $\Sigma_{n=3}^{\infty} \frac{1}{n(\ln n-1)}$ diverges.
We use the integral test as follows $$\int_3^{\infty}\frac{dn}{n(\ln n-1)}=\int_3^{\infty}\frac{d \ln n}{ (\ln n-1)}\\ =\ln(\ln n-1)|_3^{\infty}=\infty$$ Hence, the series $\Sigma_{n=3}^{\infty} \frac{1}{n(\ln n-1)}$ diverges.