## University Calculus: Early Transcendentals (3rd Edition)

Since, $\dfrac{1}{n^{3/2}}$ is a convergent series with $p=\dfrac{3}{2} \gt 1$ and $n^2-1 \gt n$ or, $n^2(n^2-1) \gt n^3$ or, $\dfrac{1}{n^{3/2}} \gt \dfrac{1}{n \sqrt{n^2-1}}$ Thus the series converges by the direct comparison test.