Math Help - Non-removable singularity and radius of Taylor expansion

1. Non-removable singularity and radius of Taylor expansion

Hello,
I'm trying to prove the follwing:
For a function $f(z)$ analytic in $C$, the radius of convergence of its Taylor expantion around $\beta \in C$ equals r, where r is the minimal distance between $\beta$ and a non-removable singularity of $f$.

I have attempted to do so by first showing that the radius cannot be greater than r, for in that case the Taylor expantion will still be valid at the singuality.
However, I cannot find a way to show the radius cannot be smaller than r.

2. Originally Posted by dudyu
Hello,
I'm trying to prove the follwing:
For a function $f(z)$ analytic in $C$, the radius of convergence of its Taylor expantion around $\beta \in C$ equals r, where r is the minimal distance between $\beta$ and a non-removable singularity of $f$.

I have attempted to do so by first showing that the radius cannot be greater than r, for in that case the Taylor expantion will still be valid at the singuality.
However, I cannot find a way to show the radius cannot be smaller than r.
The basic idea is to use the Cauchy integral formula on any circle of radius less than r, centred at $\beta$.