## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty}\frac{\ln n}{n^2}$$ Since $f(n)=\frac{\ln n}{n^2} ,\ \ f'(n) =\frac{1-2\ln \left(n\right)}{n^3}$, is positive, decreasing and continuous for $n\geq 1$, then by using the integral test \begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{\ln n}{n^2}d n \ \ \ \text{integrate by parts } \\ &= \lim _{t \rightarrow \infty} \left(\frac{-1}{n^3} +\int_{1}^{t} \frac{1}{n^4}d n\right)\\ &=- \lim _{t \rightarrow \infty}\frac{1}{t^3} (\ln t+1) -1 \\ &=1\end{aligned} Thus the series converges.