# Evaluating the series $\sum_{n=0}^{\infty} n! x^n$ and inverse variable-fractional-derivatives

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.

• I asked a similar question involving complex order derivatives and Riemann Zeta Function mathoverflow.net/questions/362439/… (I am sorry I post this as an answer instead of comment due to lack of reputation) Thanks Nov 5, 2022 at 6:18
• Euler studied this series in 1746 ar5iv.labs.arxiv.org/html/2206.15434
– Nemo
Mar 16 at 20:30

$$G(x) = \sum_{n=0}^\infty n!x^n \tag1$$ 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 $$(3)$$.
• I had some questions about this function which might be a little elementary. Is it the case that that $\forall \epsilon > 0 \exists \delta \in \mathbb{C}$ such that $\widetilde{G} (\delta)$ is a pole or essential singularity? My understanding is a power series of radius of convergence 0 can ONLY occur if there is a natural boundary passing through the point. Jul 14 at 19:13
$$S_p=\sum_{n=0}^{p} n! \,x^n=-\frac 1 x \,e^{-\frac 1 x }\Bigg[\Gamma \left(0,-\frac{1}{x}\right)+(-1)^p \,\Gamma (p+2)\, \Gamma \left(-(p+1),-\frac{1}{x}\right) \Bigg]$$
Mathematica tells us that the Borel-regularized sum of $$G(-1/y)$$ is given by $$G(-1/y) \stackrel{\mathfrak B}{=} y \, \mathrm{e}^y\, \Gamma(0,y),$$ with the incomplete Gamma function $$\Gamma(a,z)$$. This is the first term in the answer by @Claude.