Answer
Converges Absolutely
Work Step by Step
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.
