So I was interested in formally assigning values to the completely divergent series $G(x) = \sum\_{n=0}^{\infty} n!x^n $. I guess the question COULD end here if you already have an idea of how to tackle this but feel free to continue reading for a strategy i think MIGHT work.


We start by consider a different totally divergent series $$F(s) = \sum\_{n=1}^{\infty} \log(n)^s $$


This doesn't converge for any choice of $s \in \mathbb{C}$. Now its worth observing that the series


$$ \zeta(-s) = 1 + 2^z + 3^z + 4^z ... $$


Has the property that for positive integers $k$ one "formally" has


$$ \frac{d^k}{ds^k} \left[ \zeta(-s) \right]\_{@(s = 0)} = \sum\_{n=1}^{\infty} \log(n)^k $$


So its natural then to define our diverging logarithmic series everywhere by writing


$$F(z) = \sum\_{n=1}^{\infty} \log(n)^z = \frac{d^z}{ds^z} \left[ \zeta(-s) \right]\_{@(s = 0)}$$


Where $z$ is taken to be an arbitrary complex number and we use the standard cauchy definition of the fractional derivative. Let's call this operator $Q$. To be explicit


$$ Q[f] = \frac{d^{\alpha}}{dx^{\alpha}} \left[ f(x) \right]\_{@(x=0)} $$


So your domain begins with $x$ and ends with $\alpha$ after applying our "Q-transform".


From here one can see easily that


$$ Q\left[ \sum\_{n=0}^{\infty} n! x^n \right] = \Gamma(\alpha+1)^2 $$


So it might be fruitful to consider then the expression


$$ Q^{-1} \left[ \Gamma(\alpha+1)^2 \right] $$


Unfortunately I don't know how to define the inverse Q-transform and perhaps that is as hard (or harder) than summing this series in the first place but I think its worth a shot.



Evaluating the series $\sum_{n=0}^{\infty} n! x^n$ and inverse variable-fractional-derivatives
$$
G(x) = \sum\_{n=0}^\infty n!x^n
\tag1$$
Another approach is to observe that the series $G(x)$ formally satisfies the differential equation
$$
x^2 G'(x) + (x-1) G(x) + 1 = 0 .
\tag2$$
The unique solution of $(2)$ with $\lim\_{x\to 0}G(x) = 1$ is
$$
\widetilde{G}(x) = -\frac{1}{x}\;e^{-1/x}\;\operatorname{Ei}\_1\left(-\frac{1}{x}\right) ,
\tag3$$
where $\operatorname{Ei}\_1$ is the [exponential integral function](https://en.wikipedia.org/wiki/Exponential_integral).  

For $x<0$, the series $(1)$ is [Borel summable](https://en.wikipedia.org/wiki/Borel_summation) to $(3)$.


