Approximate formula for the series: $\sum_{k=1}^{+\infty}\dfrac{k^x}{(k!)^x}$ I found that this series:
$$S(x)=\sum_{k=1}^{+\infty}\dfrac{k^x}{(k!)^x}$$ can be very well approximated in this way:
$$S(x)=\dfrac{1}{\left(a+b\exp(cx)\right)^d}$$
with: $a=0.1876$, $b=-0.1895$, $c=-0.6567$, $d=0.4141$. I know the function $S(x)$ can be obtained in an uneasy way using hypergeometric functions.
Is there some deep reason for this result? Thanks.
 A: I played with your formula and I generated data for integer values of $x$ between $1$ and $12$ but I took into account the fact that $S(x)$ goes to $\infty$ when $x$ goes to $0$. On the other side, $S(x)$ goes to $2$ if $x$ goes to $\infty$. All of that made that, by analogy with your work, I used $$S(x)=\dfrac{2}{\left(1-\exp(cx)\right)^d}$$ for which I obtained $c=-0.661508$ and $d=0.422774$ which are very similar to your. However, the residuals I obtained seem to be better conditioned than your and they become almost unsignificant for large values of $x$.
Added later to this answer
If we force the equation to produce $S(1)=e$, we can have a quite good model using  $$d=\frac{\log \left(\frac{e}{2}\right)}{\log \left(1-e^c\right)}$$ and only one parameter $c$ is left for the regression. Moreover $c=-\frac{2}{3}$ seems to be a pretty good approximation which makes the model totally parameter free.
It would be very nice if some one could justify this.
A: For large $x$ the first two terms will dominate so you get the asymptotic behavior
$$S(x) \sim 2 + \frac{1}{2^x}$$
Your approximation on the other hand has
$$f(x) \sim \frac{1}{a^d}\left(1 - \frac{db}{a}e^{cx}\right) \approx 1.99 + \frac{0.836}{2^{0.94 x}}$$
If you consider $S(x)-2$ then the approximation you have found is really not that good for large $x$ (in terms of the relative error of $S(x)-2$).
If we consider $x\to 0$ will also see that the approximation is not that good. For intermediate values of $x$ it does fit very well, but you have four parameters to play with so that should not come as a big surprise.
