Does the series $\sum_{n=1}^\infty$ ${\sqrt{n+1}-\sqrt{n}}\over n$ converge or diverge? Does this series converge or diverge?
$$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$$
I tried the comparsion test with $\sum_{n=1}^\infty$ $1 \over n$ but this does not help.
I also tried to simplify the series to $\sum_{n=1}^\infty \frac{1}{n (\sqrt{n+1}+\sqrt{n}{}{}{})}$ but this become harder.
 A: $\sqrt{1+1/n} < 1+1/(2n)$ (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.
A: Hint: 
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^x}$ converges for $x>1$. Compare it with your simplified form.
A: If we use the Cauchy condensation test which says 
$\displaystyle \sum f(n)$ converges if and only if $\displaystyle \sum 2^n f(2^n)$ converges
then we can study the following summation
$$\sum 2^n \frac{\sqrt{2^n+1}-\sqrt{2^n}}{2^n}= \sum \sqrt{2^n+1}-\sqrt{2^n}$$
Which clearly converges by rationalizing the comparison test
$$\lim_{n \to \infty} \frac{\sqrt{2^{n+1}+1}-\sqrt{2^{n+1}}}{\sqrt{2^n+1}-\sqrt{2^n}}=\lim_{n \to \infty} \frac{1}{(\sqrt{2^{n+1}+1}+\sqrt{2^{n+1}})(\sqrt{2^n+1}-\sqrt{2^n})}<1$$
A: Note that
$$\dfrac{\sqrt{n+1} - \sqrt{n}}n = \dfrac1{n(\sqrt{n+1} + \sqrt{n})} < \dfrac1{2n\sqrt{n}}$$

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

