Largest neighborhood on which $f(z)=\sum\limits_{p\text{ prime}} z^p$ defines an analytic function

Question

Find the largest $r$ such that $$f(z) = \sum\limits_{p\text{ prime}}z^p$$ defines an analytic function on $B_r(0).$

The series diverges for $|z|\geq 1$, since the terms don't go to zero, and converges absolutely for $|z|<1$ by comparison with the geometric series, so I think the answer is $r=1$. Am I missing something?

Answer 1

Your answer is right; if $r$ is the radius of convergence of the series, then $$\frac{1}{r}=\lim\sup |a_n|^{1/n}.$$ Here $a_n$ is either $0$ or $1$, so in the supremum, we consider only those $a_n$'s which are equal to $1$, so $\frac{1}{r}=1$.