proof that area of convergence is bounded by a circle I'm currently looking into the topic of holomorphic functions and their radii of convergence. While I do understand according to the Cauchy's integral formula why a Taylor series converges in a radius r, which is the distance to the nearest singularity from the centre, then I do not understand why the shape of the area has to be precisely a disk, bounded by a circle, contrary to e.g. an ellipse without encompassing the singularities. How to prove that the power series diverges at any point at a distance greater than r?

This happens because the region $C$ of convergence of a power series about $a$ is always is always such that $D(a,r)\subset C\subset\overline{D(a,r)}$, for some $r>0$, with two exceptions: when $C=\{a\}$ and when $C=\mathbb C$. This is so because if a power series $\sum_{n=0}^\infty a_n(z-a)^n$ converges at some $z_0\neq a$, then it converges (absolutely) at any $z$ such that $\lvert z-a\rvert<\lvert z_0-a\rvert$. And this is so because$$\bigl\lvert a_n(z-a)^n\bigr\rvert=\bigl\lvert a_n(z_0-a)^n\bigr\rvert.\left\lvert\frac{z-a}{z_0-a}\right\rvert^n,$$and $\left\lvert\frac{z-a}{z_0-a}\right\rvert<1$.
