Answer
Converges
Work Step by Step
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.
