Does the series $\sum_{n=1}^\infty$ ${\sqrt{n+1}-\sqrt{n}}\over n$ converge or diverge?

Note that $$\dfrac{\sqrt{n+1} - \sqrt{n}}n = \dfrac1{n(\sqrt{n+1} + \sqrt{n})} < \dfrac1{2n\sqrt{n}}$$

Hence, $$\sum\limits_{n=1}^{\infty}\dfrac1{n(\sqrt{n+1} + \sqrt{n})} < \sum\limits_{n=1}^{\infty}\dfrac1{2n\sqrt{n}} = \dfrac12{\zeta(3/2)} < \infty$$

$\sqrt{1+1/n} < 1+1/(2n)$ (by squaring both sides), so $\displaystyle\frac{\sqrt{n+1} - \sqrt{n}}{n} = \frac{\sqrt{n}(\sqrt{1+\frac{1}{n}} - 1)}{n} < \frac{\sqrt{n}\frac{1}{2n}}{n} = \frac1{2n^{3/2}} $.

Since $\sum \frac1{n^{1+c}}$ converges for any $c > 0$ (by the integral test or many other ways), the sum converges.

Hint: $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^x}$ converges for $x>1$. Compare it with your simplified form.