Complex Function is Entire iff it has Everywhere Convergent Power Series

Theorem
Let $f : \C \to \C$ be a complex function.

Then $f$ is an entire function $f$ can be given by an everywhere convergent power series:
 * $\displaystyle \map f z = \sum_{n \mathop = 0}^\infty a_n z^n; \quad \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$

Proof
From Holomorphic Function is Analytic, if $f$ is holomorphic on some open ball $D$ in $\C$, then $f$ is complex analytic on $D$.

However, as, by the definition of an entire function, $f$ is holomorphic everywhere, the radius of $D$ can be made arbitrarily large, meaning that $f$ is complex analytic everywhere.

This implies that an entire function is complex analytic everywhere.

It remains to prove that if $f$ is complex analytic everywhere, $f$ is entire.

By Power Series is Termwise Differentiable within Radius of Convergence, if $f$ is complex analytic everywhere, $f$ is holomorphic everywhere.

That is, a function complex analytic everywhere is entire.

Note that by the Cauchy-Hadamard theorem:


 * $\displaystyle \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$

this series has an infinite radius of convergence.

That is, the series converges for all $z \in \C$, satisfying our second demand.