The basic idea is to use the Cauchy integral formula on any circle of radius less than r, centred at .
You will find a more detailed proof on this Wikipedia page.
Hello,
I'm trying to prove the follwing:
For a function analytic in , the radius of convergence of its Taylor expantion around equals r, where r is the minimal distance between and a non-removable singularity of .
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 .
You will find a more detailed proof on this Wikipedia page.