Bound of power series coefficients of a growth-order-one entire function An entire function $f(z)$ satisfies
$$|f(z)| \leq A_\varepsilon e^{2\pi(M+\varepsilon)|z|}$$
for every positive $\varepsilon$. I want to show that
$$\limsup_{n \to \infty}\ [f^{(n)}(0)]^{1/n} \leq 2\pi M.$$
Alternatively, we can state the result as
$$\limsup_{n \to \infty}\ (n! |a_n|)^{1/n} \leq 2\pi M,$$
where $f(z) = \sum_{n=0}^\infty a_n z^n$ is the power series expansion.
I can prove this result with Paley-Wiener theorem when $f$ is of moderate decrease; I've no idea otherwise. Note that this is a pretty good bound, since equality holds when $f(z)=e^{2\pi Mz}$, just to give one example.

P.S. Sorry for the bad question title, since I can't really come up with an illuminating summarization of this problem. Please feel free to edit if you think of something better.

We have
$$\begin{align}
\left\lvert f^{(n)}(0)\right\rvert &= \frac{n!}{2\pi} \left\lvert\int_{\lvert z\rvert = R} \frac{f(z)}{z^{n+1}}\,dz\right\rvert\\
&\leqslant \frac{n!}{2\pi} \int_0^{2\pi} \frac{\lvert f(Re^{i\varphi})\rvert}{R^n}\,d\varphi\\
&\leqslant \frac{n! A_\varepsilon e^{2\pi(M+\varepsilon)R}}{R^n}
\end{align}$$
for all $R > 0$. The right hand side is minimised for $R = \dfrac{n}{2\pi(M+\varepsilon)}$, which gives the estimate
$$\left\lvert f^{(n)}(0)\right\rvert \leqslant A_\varepsilon \frac{n!e^n\bigl(2\pi(M+\varepsilon)\bigr)^n}{n^n},$$
and taking $n$-th roots,
$$\left\lvert f^{(n)}(0)\right\rvert^{1/n} \leqslant \sqrt[n]{A_\varepsilon}\sqrt[n]{n!}\frac{e}{n}2\pi(M+\varepsilon),$$
from which we deduce
$$\limsup_{n\to\infty} \left\lvert f^{(n)}(0)\right\rvert^{1/n} \leqslant 2\pi(M+\varepsilon).$$
Since that holds for all $\varepsilon > 0$, we have indeed
$$\limsup_{n\to\infty} \left\lvert f^{(n)}(0)\right\rvert^{1/n} \leqslant 2\pi M.$$
