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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 a problem from M.Stein's book, Complex Analysis. Yet I don't know how to get this. Will someone give me some hints on it? Thank you very much.

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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. $

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    This is good. But I think this only proves the order is no larger than $1/a$2013-03-29
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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|}}}. $

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    @Andrew Thanks I also found the proof in Levin's book *Zeros of Entire Functions*.2014-09-15