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.

Solution 1:

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

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

Solution 2:

$\sqrt{1+1/n} < 1+1/(2n)$ (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.

Solution 3:

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

Solution 4:

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

$$\lim_{n \to \infty} \frac{\sqrt{2^{n+1}+1}-\sqrt{2^{n+1}}}{\sqrt{2^n+1}-\sqrt{2^n}}=\lim_{n \to \infty} \frac{1}{(\sqrt{2^{n+1}+1}+\sqrt{2^{n+1}})(\sqrt{2^n+1}-\sqrt{2^n})}<1$$