expression for infinite series with powers of factorial in denominator The series
$$\sum_{k=0}^\infty \frac{\exp(c k \beta)}{(k!)^\beta} $$
has come up when I'm trying to apply the methodology in this paper (http://www.ism.ac.jp/~eguchi/pdf/Robustify_MLE.pdf) to Poisson regression.
When $\beta = 1,$ it is $e^{e^c}$ and when $\beta = 2,$ it is $I_0(2e^c),$ the modified Bessel function of the first kind, but have no idea what happens at other values, or how to approach the question.
Does this series have either a closed form expression in terms of special functions, or an asymptotic expansion that is numerically useful for $\beta \in [1, 5]$ and $|c| < R$?
 A: If $\beta$ is an integer $\ge 3$, it can be written as a hypergeometric
$$ \mbox{$_0$F$_{\beta-1}$}(\ ;\, 1,\ldots,1;,\exp(c\beta))$$
where there are $\beta-1$ ones. 
A: I copied this from my comments above:
This sum was discussed in the book "Advanced Mathematical Methods for Scientists and Engineers" by Carl M. Bender and Steven A Orszag in section 6.7 Example 4. The asymptotic result is (with $x=\exp(c \beta)$)  $(2 \pi)^{(1-\beta)/2} \beta^{-1/2} x^{(1-\beta)/(2\beta)}\exp(\beta x^{1/\beta})$.  
Edit: Sketch of a proof following Bender's & Orszag's lines: Let us look at a somewhat simpler sum
$$
\sum_{k=0}^{\infty} \frac{x^{k}}{k!^{\beta}}.
$$
By examining the ratio of two consecutive terms of this sum we find that the terms are increasing until $k=k_{max}:=\lfloor x^{1/\beta}\rfloor$ and then decreasing. Using Stirling's formula and relaxing the requirement of $k$ being integer, we can expand $k!$ around $k_{0}$ by inserting $k=x^{1/\beta}+t$
$$
k! \sim x^{k/\beta} \exp(-x^{1/\beta}+t^{2}x^{-1/\beta}/2) \sqrt{2 \pi} x^{1/(2 \beta)}
$$
for large $x$ and small $t$, i.e. $t^{3}\ll x^{2/\beta}$, which let us neglect higher terms in $t$. We find that the terms are sharply peaked for $k$ around $k_{0}$. Approximate the sum over $k$ by an integral over $t$, with the integration range extended to the whole real axis. (We can do this since the additional contributions to the integral are negligible.) Evaluating the integral gives the result.
Edit: Getting higher orders for any real $\beta$ is straight forward but tedious with the above (Laplace-) method. However, going after chapter 5.11 in Luke's "The Special Functions and their Approximation", Vol. 1, one can give a rather compact formula for any order, when restricting $\beta$ to positive integers. Actually, it is the asymptotic expansion of the hypergeometric function $_{0}F_{\beta-1}(;1,...,1;\exp(c \beta))$, which is the same as the OP's sum. Write $x=\exp(c \beta)$. Then
$$
_{0}F_{\beta-1}(;1,...,1;x)= (2 \pi)^{(1-\beta)/2} \beta^{-1/2} \exp(\beta x^{1/\beta}) x^{\frac{1}{2 \beta}-\frac{1}{2}} \sum_{k=0}^{\infty} c_{k} \beta^{-k} x^{-k/\beta}
$$
where $c_{k}$ is recursively defined as
$$
c_{k}=\frac{1}{\beta k} \sum_{s=1}^{\beta-1} c_{k-s} \sum_{r=0}^{s}\frac{(-1)^{s-r}}{r!(s-r)!} (r+k+\frac{1}{2}-\frac{\beta}{2})^{\beta}
$$ 
and $c_{0}=1$, $c_{k}=0$ for $k<0$. 
I checked the formula with Mathematica for several values of $\beta$ and orders and it works fine.
