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Let $0<s<1$ and consider the power series $$\sum_{n=0}^{\infty}\frac{r^n}{(n!)^{1/s}}.$$ I need to show that for any given $\epsilon>0$, there exists $R>0$ such that for all $r>R$, $$\sum_{n=0}^{\infty}\frac{r^n}{(n!)^{1/s}}<\exp({r^{s+\epsilon}}).$$

This comes from a complex analysis problem, where I need to prove that the entire function $\sum_{n=0}^{\infty}\frac{z^n}{(n!)^{1/s}}$ has order at most $s$, which, in this context, translates to the above problem.

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You can show that the order $\rho$ of an entire function $f$ can be computed as $$ \rho = \limsup_{n\to\infty} \frac{n\log n}{\log(1/|a_n|)} $$ where $a_n$ are the Maclaurin coefficients of $f$.

In your case, you have $a_n = \dfrac{1}{(n!)^{1/s}}$, so $$ \rho = \limsup_{n\to\infty} \frac{s\, n\log n}{\log(n!)} = s $$ (using some standard estimates of $n!$).

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  • $\begingroup$ Thanks a lot! In fact I have been trying to prove the more general fact that you just stated: the order of $f$ equals $\alpha:=\limsup_n (n\log n)/\log(1/|a_n|)$. I have an estimate that for any $\epsilon>0$, $|a_n|\le n^{-n/(\alpha+\epsilon)}$ if $n$ is sufficiently large. But I could not estimate $\sum_{n=0}^{\infty}|a_n|r^n$ when $\alpha<1$. Do you have a reference to this result? $\endgroup$ – Chuwei Zhang Apr 29 '15 at 16:07
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    $\begingroup$ That result should be almost any textbook that talks about the order of entire functions. One of the standard references would be "Entire functions" by Boas. $\endgroup$ – mrf Apr 29 '15 at 19:20

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