$\sum{z^n\over (n!)^\alpha}$ is an entire function of order $1/\alpha$

Question

The problem is from Stein Complex analysis Chapter 5 Problem 3.

Show that $\sum {z^n\over (n!)^\alpha}$ is an entire function of order $1/\alpha$.

The problem was already posted here, but it seems the definition of growth of order is different from mine.

Definition. Let $f$ be an entire function. If there exist a positive number $\rho$ and constants $A,B>0$ such that $$|f(z)|\leq Ae^{B|z|^\rho}\quad \text{for all}\ z\in\Bbb C,$$ then we say that $f$ has an order of growth $\leq\rho$. We define the order of growth of $f$ as $$\rho_f = \inf\rho,$$ where the infimum is over all $\rho>0$ such that $f$ has an order growth $\leq\rho$.

The fact that given series is entire can be shown using the ratio test. But how can I show the order of growth is $1/\alpha$? Any hint will be appreciated.

Answer 1

You can prove these claims: (for $a>0$)

$f(z)=\sum_n z^n / (n!)^a$ has the same order as $g(z)=\sum_n z^n / n^{an}$

and $g(z^a)=\sum_n z^{an} / n^{an}$ has the same order as $\sum_n z^n / n^n$ which has the same order as $\exp(z)$.

So $g(z^a)$ has order $1$ and $g$ has order $1/a$ and the same for $f$.