# About the radius of convergence of power series representation of analytic functions

I had read that if $g$ is an entire function then the radius of convergence of any power series expansion of $g$ is infinite, because we can enlarge the radius of the disc where the Taylor series represent the function as we want.

Also I know that if a function is real analytic then it can be uniquely extended to an holomorphic function. Thus if the analytic extension to the complex plane of a real-analytic function is entire we can conclude that the radius of convergence of any power series of $f$ is infinite. This is right?

Using the same reasoning we also can think that for any real-analytic function the radius of convergence of any power series expansion around a point only depends on the minimum distance to any singularity of the extension of $f$ to the complex plane. It is this reasoning correct?

• Sounds right to me. – herb steinberg Apr 15 '18 at 21:55
• If $R$ is the radius of convergence of $\sum_n A_nz^n,$ then the sequence $(A_nz^n)_n$ does not even converge to $0$ when $|z|>R.$ This follows from the Cauchy-Hadanard Radius Formula $R=\frac {1}{\lim\sup_{n\to \infty}|A_n|^{1/n}},$ which is proved by elementary means. – DanielWainfleet Apr 15 '18 at 23:08
• The power series, centered at $0$, for the real function $f(x)=(x+1)\ln (1+x)$ (for $x>-1$ ), has radius of convergence $R=1,$ and the power series converges uniformly on $\{z\in \Bbb C: |z|\leq 1\}$ to a continuous function. But there is a non-removable singularity of the derivative $f'(z)$ at $z=-1.$ – DanielWainfleet Apr 15 '18 at 23:25
• Typo: It's the Cauchy-Hadamard formula. – DanielWainfleet Apr 15 '18 at 23:26