## Calculus (3rd Edition)

Given $$\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{3}}$$ Since $f(n)= \frac{1}{n(\ln n)^{3}},\ \ f'(n) =-\frac{\ln \left(n\right)+3}{n^2\ln ^4\left(n\right)}$ which is positive, decreasing, and continuous for $n\geq 2$. Then by using the integral test, we get: \begin{align*} \int_{2}^{\infty} f(n)dn&= \int_{2}^{\infty} \frac{1}{n(\ln n)^{3}}dn\\ &= \frac{-1}{2}(\ln n)^{-2}\bigg|_{2}^{\infty}\\ &=\frac{1}{2\ln ^2\left(2\right)}\lt\infty \end{align*} Thus $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{3}}$ converges.