On the growth order of an entire function $\sum \frac{z^n}{(n!)^a}$ Here $a$ is a real positive number. The result is that $f(z)=\sum_{n=1}^{+\infty} \frac{z^n}{(n!)^a}$ has a growth order $1/a$ (i.e. $\exists A,B\in \mathbb{R}$ such that $|f(z)|\leq A\exp{(B|z|^{1/a})},\forall z\in \mathbb{C}$). It is Problem 3* from Chapter 5 of E.M. Stein's book, Complex Analysis, page 157. Yet I don't know how to get this. Will someone give me some hints on it? Thank you very much.
 A: This answer is self-contained, with all nuts and bolts thrown in. First of all, since 
$$
|f(z)| \leq \sum_n \frac{|z|^n}{(n!)^{a}},
$$
so it suffices to show that
$$
f(r) := \sum_n \frac{r^n}{(n!)^{a}} \leq A \exp(B r^{1/a}),
$$
for real $r \geq 0$. 
Define the threshold $N := (2r)^{1/a} e$, and verify that for $n \geq N$, we have $\frac{r^n}{(n!)^a} \leq 2^{-n}$. (Hopefully I got the calculation right; in any case, I'll leave it as a straightforward exercise :-).) Then we have:
$$\begin{eqnarray*}
f(r) 
&=& \sum_{n} \frac{r^n}{(n!)^a}  
\\ &=& \sum_{n < N} \left(\frac{(r^{1/a})^n}{n!} \right)^a + \sum_{n \geq N} \frac{r^n}{(n!)^a} 
\\ &\leq& N \left( \sum_{n < N} \frac{(r^{1/a})^n}{n!} \right)^a + \sum_{n \geq N} 2^{-n}
\\ &\leq& N \left( \sum_{n} \frac{(r^{1/a})^n}{n!} \right)^a + 2
\\ &=& N \exp(r^{1/a})^a + 2 = N \exp(a r^{1/a}) + 2.
\end{eqnarray*}$$
Here, we are using the loose inequality: $t_1^a + \ldots + t_N^a \leq N(t_1 + \ldots + t_N)^a$. To complete the estimate, plug in $N = e2^{1/a} r^{1/a} \leq e 2^{1/a} \exp(r^{1/a})$ to get:
$$
f(r) \leq e 2^{1/a} \exp((1+a) r^{1/a})+2.
$$
A: There is a formula expressing the growth order of entire function $f(z)=\sum_{n=0}^\infty c_nz^n\ $ in terms of its Taylor coeffitients:
$$
\rho=\limsup_{n\to\infty}\frac{\log n}{\log\frac{1}{\sqrt[n]{|c_n|}}}.
$$  
A: @Srivatsan's answer only proves the order of $f$ is at most $1 / \alpha$, so here's a way of proving that the order of $f$ is at least $1 / \alpha$. Consider the sequence $z_k := (ek)^\alpha$, and observe that
$$
|f(z_k)| = |f((ek)^\alpha)| = \sum_{n = 0}^\infty \frac{((ek)^\alpha)^n}{(n!)^\alpha}
\geq \frac{((ek)^\alpha)^k}{(k!)^\alpha} = \left(\frac{(ek)^k}{k!}\right)^\alpha.
$$
Using the inequality that $k! \leq k^k$, we get
$$
|f(z_k)|
\geq \left(\frac{(ek)^k}{k^k}\right)^\alpha
= e^{\alpha k}
= \exp\left(\frac{\alpha}{e} (z_k)^{1 / \alpha}\right).
$$
It then follows that $f$ cannot have order less than $1 / \alpha$.
