Exponent of Convergence is Less Than Order

Theorem
Let $f: \C \to \C$ be an entire function.

Let $\omega$ be its order.

Let $\tau$ be its exponent of convergence.

Then $\tau\leq\omega$.

Proof
We may assume $f(0)\neq0$.

Let $f$ have finitely many zeroes.

Then its exponent of convergence $\tau=0\leq\omega$.

Let $f$ have infinitely many zeroes.

Let $\left\langle{a_n}\right\rangle$ be the sequence of nonzero zeroes of $f$, repeated according to multiplicity and ordered by increasing modulus.

Let $r_n=|a_n|$ and $R_n=2|a_n|$.

By Jensen's Inequality:
 * $n \leq \dfrac{\log(\max_{|z|\leq R_n}|f|) - \log|f(0)|}{\log 2}$

for all $n\in\N$.

Let $\epsilon>0$.

Because $f$ has order $\omega$, $n\ll_\epsilon|a_n|^{\omega+\epsilon}$.

Thus $|a_n|^{-(\omega+\epsilon)}\ll_\epsilon n^{-1}$.

By P-Series Converges Absolutely, $\omega+\epsilon\geq\tau$.

Because $\epsilon$ is arbitrary, $\omega\geq\tau$.

Also see

 * Order is Maximum of Exponent of Convergence and Degree