Radius of Convergence for a complex sin function

Question

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?

$\sum_1^\infty\sin(\sqrt{n+1}+\sqrt{n})(x-2)^n$

This isn't a power series. A power series has the form $\sum a_n x^n$. — user296113, Apr 11 '16 at 16:45
Why do you think the terms of that series go to zero as $n \rightarrow \infty$? — Eric Towers, Apr 11 '16 at 16:54
You can just replace $(x-2)^n$ with $x^n$ and still have the same radius of convergence, of course. I suspect you can make $\sin(\sqrt{n+1}+\sqrt{n})$ close enough to $1$ enough times to ensure the radius of convergence is $1$. — Thomas Andrews, Apr 11 '16 at 16:58
@MarkoSvraka : Maybe you should edit you question so that is closer to your actual question? — Eric Towers, Apr 11 '16 at 17:06

score 1 · Accepted Answer · answered Apr 11 '16 at 17:33

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 $1$.

Radius of Convergence for a complex sin function

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