Answer
Convergent
Work Step by Step
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
Use Limit Comparison Test with $a_{n}=\frac{1}{{n\sqrt {(n^{2}-1)}}}$ and $b_{n}=\frac{1}{n\sqrt {n^{2}}}$
$\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{\frac{1}{{n\sqrt {(n^{2}-1)}}}}{\frac{1}{n\sqrt {n^{2}}}}$
$=\lim\limits_{n \to \infty}\frac{\sqrt {n^{2}}}{\sqrt {n^{2}-1}}$
$=\lim\limits_{n \to \infty}\frac{n}{n\sqrt {1-\frac{1}{n^{2}}}}$
$=\lim\limits_{n \to \infty}\frac{1}{\sqrt {1-\frac{1}{n^{2}}}}$
$=\lim\limits_{n \to \infty}\frac{1}{\sqrt {1-0}}$
$=1$
Since,$\Sigma_{n=1}^{\infty}\frac{1}{n\sqrt {n^{2}}}$ is convergent because $p-$ series with $p=2 \gt 1$ is convergent, thus the series $\Sigma_{n=1}^{\infty}\frac{1}{{n\sqrt {(n^{2}-1)}}}$ also converges.
