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 if and only if $f$ can be given by an everywhere convergent power series:

$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n; \quad \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$

Proof

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Sufficient Condition

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.

$\Box$

Necessary Condition

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 analyticeverywhere is entire.

Note that by the Cauchy-Hadamard theorem:

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

if and only if this series has an infinite radius of convergence.

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

$\blacksquare$