# Radius of convergence and analicity

Is there power series of some function f with radius of convergence $0$ that is analytic on all of $\Bbb{C}$? I don't quite get the idea of function being analytic and its radius of convergence in power series expansion. Is there any link between them? Can somebody explain that?

• Did I answer your question sufficiently? Commented Nov 9, 2014 at 20:04

No. Any entire function $f$ is given uniquely by its Taylor series: $$f(\zeta) =\sum_{n=0}^{\infty}{a_n(\zeta-P)^n}$$ where, $$a_n = \frac{f^{(n)}(P)}{n!}\quad \forall n\in \Bbb Z^{\geq0}$$
which converges uniformly on compact sets. A power series defines a unique holomorphic function about a point $P$ s.t. the coefficients correspond with the entire function's derivatives of all order at the point $P$. It will also thus converge everywhere to the function $f$ and have $R=\infty :\Leftrightarrow \limsup \limits_{n\to\infty}{|a_n|^{\frac 1 n} = 0}$.
Additionally, any holomorphic function defined on some open simply-connected domain $D$, can be uniquely extended to a holomorphic function inside an open disc with a possibly larger radius of convergence containing the domain $D$. This can be made to be a disc of largest possible radius in the sense that on the boundary of the disc there is a singularity in the holomorphic function. One can in some way "avoid" this point though, by traversing around it, a process known as analytic continuation: http://mathworld.wolfram.com/AnalyticContinuation.html