## Thomas' Calculus 13th Edition

Let $a_n=\dfrac{1}{n\sqrt {n^2-1}}=\dfrac{1}{n\sqrt {(n-1)(n+1)}}$ Apply comparison, we have $\Sigma_{n=2}^\infty \dfrac{1}{n\sqrt {(n-1)(n+2)}} \lt \Sigma_{n=2}^\infty \dfrac{1}{\sqrt {(n-1)^2(n-1)(n-1)}}=\Sigma_{n=2}^\infty \dfrac{1}{{(n-1)^2}}$ $\implies \Sigma_{n=2}^\infty \dfrac{1}{{(n-1)^2}}=\Sigma_{n=1}^\infty \dfrac{1}{n^{(2)}}$ Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}}$ shows a convergent p-series with $p=2 \gt 1$ Thus, the series Converges Absolutely by the comparison test.