## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty}\frac{\ln n}{n^3}$$ Since for $n\geq 1$ \begin{align*} \frac{\ln n}{n^3} \leq \frac{ n^a}{n^3}\\ &\leq \frac{n n^{a-1}}{n^3}\\ &\leq \frac{1}{n^2} \end{align*} Compare with $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$, a convergent $p-$series $(p=2)$; then $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^3}$ also converges.