Q:

Radius of Convergence for a complex sin function

I encountered the following power series, and while I know a couple of ways to determine radius of convergence, I wasn't able to figure out how to evaluate the appropriate limit to get said radius. Can anyone help?
$\sum_1^\infty\sin(\sqrt{n+1}+\sqrt{n})(x-2)^n$

A:

We can show that the fractional part of $a_n=\frac{1}{\pi}(\sqrt{n+1}+\sqrt{n-1})$ is dense in $(0,1)$. 
This is because $$a_{n+1}-a_{n}=\frac{1}{\pi}(\sqrt{n+2}-\sqrt{n})=\frac{1}{\pi}\frac{2}{\sqrt{n+2}+\sqrt{n}}$$
So $$\frac{1}{\pi\sqrt{n+2}}<a_{n+1}-a_n<\frac{1}{\pi\sqrt{n}}$$
In particular, then, there are infinitely many $n$ so that the fractional part of $a_n$ is less than $\frac{1}{2}$ and the fractional part of $a_{n+1}$ is greater than $\frac{1}{2}$ and thus the fractional part of $a_n$ can be made to be within   $\frac{1}{\pi\sqrt{n}}$ of $1/2$ inifinitely often. But that means that $\pi a_n$ can he made so that it is arbitrarily close to $k\pi + \frac{\pi}{2}$ for some integer $k$, and thus that $\sin(\sqrt{n+1}+\sqrt{n})$ can be made arbitrarily close to $\pm  1$. 
This means the power series must have radius of convergence $1$.

