TITLE: Radius of convergence of $\sum_{n = 0}^{\infty} (a_1^n + \dots + a_k^n)z^n$, where $|a_1| = |a_2| = \dots = |a_k| = 1$
QUESTION [6 upvotes]: Here's the problem: Find the radius of convergence of $f(z) = \sum_{n = 0}^{\infty} (a_1^n + \dots + a_k^n)z^n$, where $|a_1| = |a_2| = \dots = |a_k| = 1$, and $a_i \in \mathbb{C}$.
Since the series in question is the sum of as $j$ ranges from $1$ to $k$ of the series $f_j(z) = \sum_{n = 0}^{\infty} a_j^n z^n$ you see right away that the radius of convergence $R$ of $f(z)$ is at least $\displaystyle 1$.
I'm guessing that maybe the radius of convergence $R$ is exactly $\displaystyle 1$. 
I think you might be able to get the answer if you can show that the set $\{a_1^n + \dots + a_k^n : n \ge 0\}$ either has a non-zero limit point inside $\{ z : |z| \le k \}$, or else attains the same non-zero value infinitely many times, because then in either case you can find a subsequence $\{n_j\}$ with $|a_1^{n_j} + \dots + a_k^{n_j}|$ bounded away from $0$, and so $|a_1^{n_j} + \dots + a_k^{n_j}|^{1/n_j} \to 1$ as $n_j \to \infty$.
Any thoughts about how to proceed with this idea? Or, is there a far better way to look at this?
Edit

(*) For $a_j$ and any $N > 0$, you can find a $n > N$ such that $a_j^n$ is arbitrarily close to $1$. You can do this by writing $a_j^n = e^{i n \theta_j}$ and looking at $n \theta _j$ modulo $2 \pi$.
Let $b_i = a_i / a_1$.
It is not true that we have any of the limits $\lim \limits_{n \to \infty}a_1^n + a_2^n = 0, -1, -2, \dots$. This is because we can find a subsequence $\{n_k\}$ such that $\lim \limits_{n \to \infty}a_1^{n_k} = 1$. Assuming to the contrary for a contradiction, for this subsequence one would have to have $\lim \limits_{n \to \infty} a_1^{n_k}(1 + b_2^{n_k}) = 0, -1, -2, \dots$ so that $\lim \limits_{n \to \infty} (1 + b_2^{n_k}) = 0, -1, -2, \dots$ and therefore $\lim \limits_{n \to \infty} b_2^{n_k} = -1, -2, -3, \dots$ which is impossible by (*).
Similarly, it is not true that we have any of the limits $\lim \limits_{n \to \infty}a_1^n + a_2^n + a_3^n= 0, -1, -2, \dots$. Because otherwise, we can find a subsequence $\{n_k\}$ such that $\lim \limits_{n \to \infty}a_1^{n_k} = 1$, and then $\lim \limits_{n \to \infty} (1 + b_2^{n_k} + b_3^{n_k}) = 0, -1, -2, \dots$, and in this case $\lim \limits_{n \to \infty} b_2^{n_k} + b_3^{n_k}= -1, -2, -3, \dots$ which is impossible.
Continuing like this, we see that it is not true that we have any of the limits $\lim \limits_{n \to \infty}a_1^n + a_2^n + \dots + a_k^n= 0, -1, -2, \dots$. In particular we either have a non-zero limit point, or a non-zero value is attained infinitely many times.
Does this appear to be correct?

REPLY [3 votes]: The radius of convergence is $1$.  Note that
$$
\sum_{n = 0}^{\infty} (a_1^n + \dots + a_k^n)z^n \;=\; \frac{1}{1-a_1z} + \cdots + \frac{1}{1-a_kz}
$$
whenever $|z| < 1$.  Since this function has poles at $z = 1/a_i$ for each $i$, it cannot be extended holomorphically to a disk of radius larger than one, so the radius of convergence is precisely one.