## Calculus 8th Edition

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 convergent.