# Tricky radius of convergence: $\sum\limits_{n=0}^\infty\cos\left(\alpha\sqrt{1+n^2}\right)z^n$

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?

What is the radius of convergence of the power series $$\sum_{n=0}^\infty\cos\left(\alpha\sqrt{1+n^2}\right)z^n,$$ where $\alpha$ is any real number? What if $\alpha$ is a complex number?

Hint: $\sqrt{1+n^2} = n + 1/(2n) + O(1/n^3)$.
• Out of curiosity, how did you derive this? I'm still not getting anywhere with $n$th roots or ratios (I suspect that the ratio version won't even apply, here). Could you expand at all on what I could do, in particular for the case $\alpha\in\Bbb R$? Jun 16 '12 at 20:41
• $\sqrt{1+n^2} = n \sqrt{1+1/n^2}$ and use the Taylor series for $\sqrt{1+t}$. If $\alpha$ is real, we end up with $\cos(\alpha \sqrt{1+n^2}) = \cos(\alpha n) + O(1/n)$, and $\lim \sup_{n \to \infty} |\cos(\alpha n)| = 1$ (but all you need is that it is $> 0$). Jun 17 '12 at 5:47
• Ah! I was trying to directly use the Taylor series of $\sqrt{1+x^2}$. Many thanks! Jun 17 '12 at 14:51
• @RobertIsrael, can you explain how $\cos(\alpha \sqrt{1+n^2})=\cos(\alpha n) + O(1/n)$? Mar 27 '15 at 16:50
• This is for the real case. Given that $\sqrt{1+n^2} = n \sqrt{1+1/n^2} = n + t$ where $t = O(1/n)$, $\cos(\alpha \sqrt{1+n^2}) = \cos(\alpha n) \cos(\alpha t) - \sin(\alpha n) \sin(\alpha t)$. $\cos(\alpha t) = 1 + O(t^2)$ and $\sin(\alpha t) = O(t)$. Mar 27 '15 at 19:06