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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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  • $\begingroup$ A very nice approach,indeed!Thank you very much. $\endgroup$ – user14242 Aug 7 '11 at 9:37
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    $\begingroup$ This is good. But I think this only proves the order is no larger than $1/a$ $\endgroup$ – Roun Mar 29 '13 at 13:58
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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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  • $\begingroup$ :you are right!Yet here logically I can't employ this fomula since this fomula you give is a result of the growth order of the above entire function in the book of Stein's.Thank you very much. $\endgroup$ – user14242 Aug 7 '11 at 9:29
  • $\begingroup$ @Andrew Can you please provide a reference to this formula? $\endgroup$ – Wilson of Gordon Sep 12 '14 at 12:08
  • $\begingroup$ @WilsonofGordon in equivalent form here: en.wikipedia.org/wiki/Entire_function#Order_and_type $\endgroup$ – Andrew Sep 13 '14 at 10:41
  • $\begingroup$ @Andrew Thanks I also found the proof in Levin's book Zeros of Entire Functions. $\endgroup$ – Wilson of Gordon Sep 15 '14 at 9:35

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