I want to show that $\sum_{n=1}^\infty cos(n^2\pi)\cdot(\sqrt{n+11}-\sqrt{n+2}) $ converges or diverges. I want to prove that $\sum_{n=1}^\infty cos(n^2\pi)\cdot(\sqrt{n+11}-\sqrt{n+2}) $ is convergent (or divergent).
I am lost on how to prove it. Could someone please show me? I assume I would need to use one of the criterias for determining convergence? 
I know that $cos(n^2\pi)$ is basicly the same as $(-1)^n$.
I tried to multiplicate it by $ \frac {\sqrt{n+11}+\sqrt{n+2}}{\sqrt{n+11}+\sqrt{n+2}} $ to see if I could get something more useful, which got me: $a_n=(-1)^n(\sqrt{n+11}-\sqrt{2+n})=\frac{9(-1)^n}{\sqrt{11+n}+\sqrt{2+n}}$. 
Here I guess I would want to see both cases - odd $n$ and even $n$. 
For the even case - the limit would be $\frac{9}{\sqrt{\frac{11}{n}+1}+\sqrt{\frac{2}{n}+1}}=\frac{9}{2}$ and odd case would result in $-\frac{9}{2}$. 
Is this enough? Is it even correct? So it is divergent, because there are 2 limits?

You can use the Leibnz's rule: showing that $$\lim_{n\to +\infty} \frac{9}{\sqrt{11+n}+\sqrt{2+n}}=0
$$ and $$\frac{9}{\sqrt{12+n}+\sqrt{3+n}}<
\frac{9}{\sqrt{11+n}+\sqrt{2+n}}
$$ and this is obviuos. Therefore the series converges
