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 $p\leq 1$.
$a_{n}=\frac{\sqrt {n+1}-\sqrt {n-1}}{n}$
$=\frac{(\sqrt {n+1}-\sqrt {n-1})(\sqrt {n+1}+\sqrt {n-1})}{n(\sqrt {n+1}+\sqrt {n-1})}$
$=\frac{2}{n(\sqrt {n+1}+\sqrt {n-1})} \lt \frac{2}{n\sqrt {n+1}} \lt \frac{2}{n\sqrt {n}}=\frac{2}{n^{3/2}}$
Thus, the series is convergent.