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?
 A: This has nothing to do with the Cauchy Integral Formula or any other complex analysis, it's just simple  inequalities.
Let $R$ be the supremum of $|z|$ such that $\sum a_nz^n$ converges. The definition of $R$ shows that 


If $|z|>R$ then $\sum a_nz^n$ diverges.


In fact anything can happen for $|z|=R$, but 


If $|z|<R$ then $\sum a_nz^n$ converges.


Proof: The definition of $R$ shows that there exists $w$ with $|w|>|z|$ such that $\sum a_na^n$ converges. Hence $a_nw^n\to0$, so there exists $c$ with $|a_nw^n|\le c$ for all $n$. Now $$\sum|a_nz^n|\le c\sum|z/w|^n<\infty,$$since $|z/w|<1$.
A: 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 $\left\lvert\frac{z-a}{z_0-a}\right\rvert<1$.
