Convergence of $\sum_{n=1}^\infty \cos(\pi n^2)(\sqrt{n+1}-\sqrt{n})$ I want to test the convergence of $$\sum_{n=1}^\infty \cos(\pi n^2)(\sqrt{n+1}-\sqrt{n})$$
First of all, $\cos(\pi n^2)=-1$ if $n$ is odd, and $\cos(\pi n^2)=1$ if $n$ is even. That is, $\cos(\pi n^2)=(-1)^n$. So the summation reduces to $$\sum_{n=1}^\infty (-1)^n(\sqrt{n+1}-\sqrt{n})$$
and I don't know what to do from here. I tried the ratio test, root test and comparison test, got nothing. and I don't feel it would converge since $$\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})=\sum_{n=1}^\infty\int_n^{n+1}\frac23 x^\frac32 dx=\int_1^\infty\frac23 x^\frac32 dx=\infty$$ 
Maybe I can use this result but I don't know how, or is this approach wrong?

Our sum is $\sum_{k=1}^\infty\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)$, where we have paired the $n=2k-1$ and $n=2k$ terms. Now $\sqrt{2k\pm 1}=\sqrt{2k}\left(1\pm\frac{1}{4k}-\frac{1}{32k^2}+\cdots\right)$ so $\sqrt{2k+ 1}-2\sqrt{2k}+\sqrt{2k-1}\approx-\frac{\sqrt{2k}}{16k^2}\in\mathcal{O}(k^{-3/2})$, which establishes convergence.
