## Calculus (3rd Edition)

The series $\Sigma_{n=2}^{\infty} \frac{1}{(\ln n)^2}$ diverges.
Using the property that $\ln n \lt n^{1/2}$, then $(\ln n)^2\lt n$ and hence we have $$\frac{1}{(\ln n)^2}\geq \frac{ 1}{n }$$ Since $\Sigma_{n=2}^{\infty}\frac{ 1}{n}$ is a divergent p-series ( $p=1$) then the series $\Sigma_{n=2}^{\infty} \frac{1}{(\ln n)^2}$ diverges by the comparison test.