I have an interesting problem that I cannot make progress on over several days. It is as follows:
Suppose $f(z) = \sum a_nz^n$ is analytic around $0$ such that $a_n \geq 0$ for all $n$, in particular the $a_n$ are real. If the radius of convergence of $f$ is equal to $1$, then there is no analytic continuation of $f$ to the point $1$.
I have been trying it by assuming that there is an analytic continuation, and then show that the radius of convergence is actually larger than $1$, but I'm not sure if it's a good idea. If there is an analytic continuation then it's easy to show that $a_k < 1$ for all $k \geq N$ for some sufficiently large $N$, and that $$\sum_N^{\infty} a_k < \infty$$ So even if we don't want to use the contradiction we can safely assume this, because otherwise it will clearly not be possible. We have only the basic facts about analytic functions at our disposal.
Does anybody see a possible way to approach this?