Condition for the power series to have a single singularity on its circle of convergence

Question

I'm looking to the prove of the following statement:

If for the complex power series $\sum^\infty_{i=1} a_i z^i$ it holds that $$ \limsup_{n \to \infty} \sqrt[n]{\left| \frac{a_n}{a_{n+1}} -z_0 \right|}<1, $$ then $\sum^\infty_{i=1}a_nz^n$ converges to the holomorphic function $f$ inside the disc $D =D(0,|z_0|)$ and the only singularity is $z_0$.

In the context of this question by singularity I mean there is no such holomorphic function $\varphi$ defined in some neighborhood $U$ of $z_0$ such that $f_{|U \cap D} = \varphi_{|U \cap D}$.

It is easy to see that in case for any $c \in \mathbb{R}_{++}$ $$ \left|\left\{ n \in \mathbb{N} : \left| \frac{a_n}{a_{n+1}} - z_0 \right| \ge c\right\}\right| = \infty, $$ the limit superior will be not less then $1$: $$ \lim_{n\to\infty}\sup \left| \frac{a_n}{a_{n+1}} -z_0 \right|^{1/n} \ge \lim_{n \to \infty} \sqrt[n]{c} = 1. $$ This is a contradiction, hence there must be a convergence $$ \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = z_0, $$ implying that the radius of convergence for $\sum^\infty_{n=0}a_nz^n$ is indeed $|z_0|$.

However, I don't know how to prove that $z_0$ is a singularity and any other point of $\partial D$ is not.

Help me to prove that $z_0$ is a singular point and the other points in $\partial D$ are not.

p. s.

This question seems to be a converse to About the limit of the coefficient ratio for a power series over complex numbers

Answer 1

I think I got it:

The series define holomorphic function $f(z) = \sum^\infty_{n=0}a_n z^n$ inside the disc $D =D(0,|z_0|)$. Consider the function $g(z) = (z - z_0)f(z)$. It can be expanded at the $0$ as $g(z) = -z_0a_0 + \sum^\infty_{n=1}(a_{n-1} - z_0a_n)z^n$. Then it is possible to estimate a limit: $$ \limsup_{n \to \infty}\sqrt[n]{|a_{n-1} - z_0a_n |} = \limsup_{n\to\infty} \sqrt[n]{|a_n|}\sqrt[n]{\left| \frac{a_{n-1}}{a_n} - z_0 \right|} < |z_0|^{-1} $$ as it is known from before that $\lim_{n\to\infty}\sup \sqrt[n]{|a_n|} = |z_0|^{-1}$. Hence, $g(z)$ is holomorphic inside the disc with the radius extending over the one of $D$. This is enough to claim that the series $\sum^\infty_{n=1}a_nz_n$ has only a 1st order pole in $z_0$ and regular point everywhere else on $\partial D$, because function $\varphi(z) = \frac{g(z)}{z - z_0}$ will work as a sufficient analytic extension.