Answer
Divergent
Work Step by Step
Given $$\sum_{n=2}^{\infty} \frac{n^{3}}{n^{4}-1}$$
Use the Limit $c$ Comparison Test with $a_n =\dfrac{n^{3}}{n^{4}-1}$ and $b_n=\dfrac{1}{n}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^{4}}{n^{4}-1}\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ is divergent , then $\displaystyle\sum_{n=2}^{\infty} \frac{n^{3}}{n^{4}-1}$ is also divergent