Answer
Converges
Work Step by Step
Given $$ \sum_{n=2}^{\infty} \frac{n^2}{n^4-1}$$
Compare with the convergent series $\displaystyle \sum_{n=2}^{\infty}\frac{1}{n^2}$ and by using the Limit Comparison Test, we get:
\begin{align*}
\lim_{n\to \infty} \frac{a_n}{b_n}&=\lim_{n\to \infty} \frac{n^2}{n^4-1}n^2\\
&=1
\end{align*}
Then $\displaystyle \sum_{n=2}^{\infty}\frac{n^2}{n^4-1}$ also converges
