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.


3 Answers 3


$$ G(x) = \sum_{n=0}^\infty n!x^n \tag1$$ 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 $(3)$.

  • $\begingroup$ Per a conversation with Caleb Briggs here: math.stackexchange.com/questions/1901612/… is there a sign error in this answer? $\endgroup$ Jul 8 at 21:59
  • $\begingroup$ 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. $\endgroup$ Jul 14 at 19:13

May be, the partial sum could give some ideas

$$S_p=\sum_{n=0}^{p} n! \,x^n=-\frac 1 x \,e^{-\frac 1 x }\Bigg[\Gamma \left(0,-\frac{1}{x}\right)+(-1)^p \,\Gamma (p+2)\, \Gamma \left(-(p+1),-\frac{1}{x}\right) \Bigg]$$


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.


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