Answer
Converges
Work Step by Step
Given
$$\sum_{n=1}^{\infty}n^3a_n,\ \ \ \ |a_{n+1}/a_n| =\frac{1}{3}$$
By using the Ratio test, let $b_n= n^3a_n$
\begin{align*}
\rho&=\lim _{n \rightarrow \infty}\left|\frac{b_{n+1}}{b_{n}}\right|\\
&=\lim _{n \rightarrow \infty}\left|\frac{(n+1)^3a_{n+1}}{n^3a_n} \right|\\
&= \lim _{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^3 \frac{a_{n+1}}{a_n} \\
&= \lim _{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^3 \lim _{n \rightarrow \infty}\frac{a_{n+1}}{a_n} \\
&= \frac{1}{3}<1
\end{align*}
Thus the series converges.