# 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 $|c| < R$?

• 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 result is (with $x=exp(c \beta)$) $(2 \pi)^{(1-\beta)/2} \beta^{-1/2} x^{(1-\beta)/(2 \beta)} exp(\beta x^{1/\beta})$. – Johannes Trost Nov 29 '14 at 16:15
• The "result" in my above comment means the asymptotic expansion of the sum for large $x$. – Johannes Trost Dec 1 '14 at 12:04
• ... and of course I would have to have given away my copy of Bender and Orszag :). If you put that as an answer, I'll accept it. – AatG Dec 1 '14 at 18:54
• Buy one from Amazon ? Google ? However, I put my comment as an answer. Maybe you can accept it anyways. – Johannes Trost Dec 3 '14 at 8:55

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.

If $\beta$ is an integer $\ge 3$, it can be written as a hypergeometric $$\mbox{_0F_{\beta-1}}(\ ;\, 1,\ldots,1;,\exp(c\beta))$$ where there are $\beta-1$ ones.

• I think what you are saying is that this is how the hypergeometric is defined, but this does not really shed a huge amount of light on the matter :) – Igor Rivin Oct 31 '14 at 15:10
• I agree, it does not by itself shed a lot of light on the matter, but it does provide a name that can be looked up. For example, you can ask Maple's FunctionAdvisor about hypergeom([],[1,1,1],t). In this case, you'll find a differential equation ${\frac {{\rm d}^{4}}{{\rm d}{t}^{4}}}f \left( t \right) =-6\,{\frac {{ \frac {{\rm d}^{3}}{{\rm d}{t}^{3}}}f \left( t \right) }{t}}-7\,{ \frac {{\frac {{\rm d}^{2}}{{\rm d}{t}^{2}}}f \left( t \right) }{{t}^{ 2}}}-{\frac {{\frac {\rm d}{{\rm d}t}}f \left( t \right) }{{t}^{3}}}+{ \frac {f \left( t \right) }{{t}^{3}}}$ – Robert Israel Oct 31 '14 at 16:31