## Calculus (3rd Edition)

Given $$\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$$ Since $f(n)=\frac{1}{n(\ln n)^2},\ \ f'(n) =-\frac{\ln \left(n\right)+2}{n^2\ln ^3\left(n\right)}$, is positive, decreasing and continuous for $n\geq 2$, then by using the integral test \begin{aligned} \int_{2}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{2}^{t} \frac{1}{n(\ln n)^2}d n \\ &= \left.\lim _{t \rightarrow \infty}\frac{-1}{\ln n}\right|_{2} ^{t} \\ &= \lim _{t \rightarrow \infty}\frac{-1}{\ln n}+\frac{1}{\ln 2}\\ &=\frac{1}{\ln 2} \end{aligned} Thus the series converges.