Generating function for gamma function (or factorial)

Question

Does exist a generating function for the gamma function or for the factorial?

$$F(x)=\sum\limits_{k=0}^\infty k!x^k$$ or $$G(x)=\sum\limits_{k=0}^\infty \Gamma (k)x^k$$

I'm assuming it does not, as I did not find any by searching online. But maybe there is a way to get a suitable approximation.

Thank you, Janus

Answer 1

The divergent séries $\;f(x):=\sum\limits_{k=0}^\infty k!\,(-x)^k\;$ was considered by Euler and revisited by Hardy in his book Divergent Series (p. $26$). Defining $\,\phi(x):=x\,f(x)\,$ Euler obtained the differential equation : $$\tag{1}x^2\phi(x)'+\phi(x)=x$$ which may be solved using the integrating factor $\,x^{-2}e^{-1/x}\,$ with the solution (cf the link) : $$\tag{2}f(x)=\int_0^\infty\frac {e^{-w}}{1+xw}\,dw$$ that we may rewrite (from the definition of the exponential integral) as : $$\tag{3}f(x)=-\frac{e^{1/x}}x\operatorname{Ei}\left(-\frac 1x\right)$$ Replace $\,x\,$ by $-x\,$ to get your wished (regularized) function $F$.

Answer 2

A bit similar to Raymond's answer in spirit,

Start off with

$$F(x)=\sum_{k=0}^\infty k!x^k$$

$$f(x)=xF(x)=\sum_{k=0}^\infty k!x^{k+1}$$

$$f'(x)=\sum_{k=0}^\infty k!(k+1)x^k=\sum_{k=0}^\infty(k+1)!x^k=\frac1x\left(-1+\sum_{k=0}^\infty k!x^k\right)$$

$$xf'(x)=-1+f(x)$$

This is a rather easy DE to solve, and it gives,

$$f(x)=cx+1$$

Now, this is a bit awkward, since we would imagine $f(0)=0$, but this cannot be the case. I guess this would mean that we would have $f(0)=1$. Weird, huh? Anyways, this gives us

$$F(x)=c+\frac1x$$

Since both definitions of $F(x)$ and $f(x)$ are divergent series, I can't really tell you what $c$ is.