Answer
Converges Absolutely
Work Step by Step
Consider $a_n=\dfrac{1}{n\sqrt {n^2-1}}$
or, $a_n=\dfrac{1}{n\sqrt {n^2-1}}=\dfrac{1}{n\sqrt {(n-1)(n+1)}}$
Since, we see that $\Sigma_{n=2}^\infty \dfrac{1}{n\sqrt {(n-1)(n+2)}} \lt \Sigma_{n=2}^\infty \dfrac{1}{\sqrt {(n-1)^2(n-1)(n-1)}}=\Sigma_{n=2}^\infty \dfrac{1}{{(n-1)^2}}$
Now, $\Sigma_{n=2}^\infty \dfrac{1}{{(n-1)^2}}=\Sigma_{n=1}^\infty \dfrac{1}{n^{2}}$
The series $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}}$ is a convergent p-series .
Hence, the given series Converges Absolutely by the direct comparison test.
