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.
Best Answer
$$ G(x) = \sum_{n=0}^\infty n!x^n \tag1$$ 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.
For $x<0$, the series $(1)$ is Borel summable to $(3)$.