Test for convergence $\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1})$ Test for convergence $$\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1})$$
So far I attempted to use the ratio test, but I'm stuck on what to do after.
$$\dfrac{\dfrac{\sqrt{n+2}-\sqrt{n}}{n+1}}{\dfrac{\sqrt{n+1}-\sqrt{n-1}}{n}} = \dfrac{n(\sqrt{n+2}-\sqrt{n})}{(n+1)(\sqrt{n+1}-\sqrt{n-1})}$$

First note that
$$
\frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt n} > \frac{\sqrt{n+1}}{\sqrt n} > 1.
$$
Therefore
$$
\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1}) < \sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1}) \frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt n}= \sum_{n=1}^\infty \frac{2}{n^{3/2}}
$$
which converges; hence the original sum converges by the comparison test.
The intuition, if it helps, is that
\begin{multline*}
\sqrt{n+1}-\sqrt{n-1} = \sqrt{n(1+O(\tfrac1n))} - \sqrt{n(1+O(\tfrac1n))} \\
= \sqrt n(1+O(\tfrac1n)) - \sqrt n(1+O(\tfrac1n)) = O(\tfrac1{\sqrt n}).
\end{multline*}
and so we want to compare the given sum with $\sum \frac1{n\sqrt n}$.
