Answer
Converges
Work Step by Step
Let $u_n=\dfrac{1}{n^{3/2}}$
This shows a convergent $p$-series with $p=\dfrac{3}{2} \gt 1$
Now, $n^2-1 \gt n \implies n^2(n^2-1) \gt n^3$
Thus, we have $\dfrac{1}{n^{(3/2)}} \gt \dfrac{1}{n (\sqrt{n^2-1})}$
Hence, the given series converges by the direct comparison test.