# On the convergence of $\sum\left(\sqrt{n+1}-\sqrt{n}\right)^p$

I have to study the values of $p\in\mathbb{R}$ such that $$\sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^p$$ is a convergent series. So far I have found that the main term is equivalent to $n^{-p/2}$, that way $p$ should be $>2$ in order to ensure convergence. Still I cannot use the limit comparison test to prove it, as it does not lead to the constant after dividing.

• $$\left(\sqrt{n+1}-\sqrt n\right)^p=\frac1{\left(\sqrt{n+1}+\sqrt n\right)^p}\sim\frac1{2^p}\frac1{n^{p/2}}$$ – Did Apr 10 '18 at 13:41

Convergence for $p>2$ is trivial since $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n}+\sqrt{n+1}}$ is bounded between $\frac{1}{2\sqrt{n+1}}$ and $\frac{1}{2\sqrt{n}}$ for any $n\geq 1$, so the $p$-test together with squeezing completely solve the problem. The following identities (courtesy of Ramanujan) might be more challenging to prove: $$\sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^3 = \frac{3}{2\pi}\sum_{n\geq 1}\frac{1}{n\sqrt{n}}\tag{p=3}$$ $$\sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^5 = \frac{15}{2\pi^2}\sum_{n\geq 1}\frac{1}{n^2\sqrt{n}}.\tag{p=5}$$ You may have a look at this related question. $$\sum_{n\geq 0}\left(\sqrt{n+1}-\sqrt{n}\right)^4 = \frac{4\pi}{3}\int_{0}^{1}\frac{x^{3/2}(1-x)^{3/2}}{\sin^2(\pi x)}\,dx\tag{p=4}$$ is pretty nice, too.