Determine the radius of convergence of $\sum_{n=0}^\infty \cos{(\alpha \sqrt{1+n^2})}z^n.$ Determine the radius of convergence of
$$\sum_{n=0}^\infty \cos{(\alpha \sqrt{1+n^2})}z^n,$$
if $\alpha$ is a real constant. What about if $\alpha \in \mathbb{C}$?
My attempt: at infinity, $\alpha \sqrt{n^2 + 1}$ should behave rather like $\alpha n$, and we know from advanced calculus that $\sum_{n=1}^N \cos{(\alpha n)}$ is bounded unless $\cos{\alpha} =  1$, but in general does not converge. Therefore, by summation by parts of the product $\cos{(\alpha \sqrt{1+n^2})}z^n$, since the coefficients of $z^n$ are bounded, the series should have radius of convergence $r = 1$. 
Would you say this is the right approach to take? I still have to figure out how to formalize the argument that we can treat $\alpha \sqrt{n^2 + 1}$ like $\alpha n$ at $\infty$, but before I do so, I want to know that I'm on the right track.
As for the second part, I would say that in general the ROC must be zero, because if $\alpha$ has nonzero imaginary part, $\cos{(n\alpha)}$ will go crazy as $n\to \infty$. 
Thanks stack for your help
 A: The first observation is $\cos(\alpha\sqrt{1+n^2})$ is bounded.
When $|z| < 1$, we can compare the series with a geometric series and
conclude the series converges.
The second observation is there are infinitely many integer pairs 
$(m,n) \in \mathbb{Z} \times \mathbb{Z}_{+}$ such that 
$\displaystyle \left|\frac{\alpha}{\pi} - \frac{m}{n}\right| < \frac{1}{n^2}$.
For such pairs, we have
$$
\left|\alpha\sqrt{1+n^2} - m\pi\right| 
\le \left|\alpha (\sqrt{1+n^2} - n )\right| + |\alpha n - m\pi|
<  \left|\frac{\alpha}{\sqrt{1+n^2}+n}\right| + \frac{\pi}{n}
<  \frac{|\alpha|+2\pi}{2n}
$$
This means there are infinitely many $\alpha\sqrt{1+n^2}$ which is close to a multiple of $\pi$ and hence the sequence $\cos(\alpha\sqrt{1+n^2})$ has a sub-sequence bounded away from $0$. This implies the series doesn't converge at $z = 1$.
Combine these two observations, we can conclude the radius of convergence is $1$.
The proof for the case of complex $\alpha$ is very similar. Writing $\alpha$ as $\mu + \nu i$ with $\nu \ne 0$, we can split the series into two pieces:
$$\begin{align}
  &\sum_{n=0}^{\infty}\cos((\mu + i\nu)\sqrt{1+n^2}) z^n\\
= &
\frac12\sum_{n=0}^{\infty}\left[ e^{-\frac{\nu}{\sqrt{1+n^2}+n} + i\mu\sqrt{1+n^2}}
(z e^{-\nu} )^n
+
e^{\frac{\nu}{\sqrt{1+n^2} + n} - i\mu\sqrt{1+n^2}}
(z e^{\nu} )^n
\right]
\end{align}$$
Notice 
$\displaystyle \lim_{n\to\infty} e^{\pm\frac{\nu}{\sqrt{1+n^2}+n}} = 1$,
similar arguments like the real case will show the radius of convergence for the
first piece is $e^{\nu}$ and the second peice is $e^{-\nu}$. Since these two radii of
convergences are different, the radius of convergence of the original series is the minimum
of these two, i.e equal to $\displaystyle e^{-|\nu|} = e^{-|\Im\alpha|}$.
