Is the closed form of $\sum_{n=0}^\infty n!x^n$ unique?

Question

So, I've been studying the power series $$f(x)=\sum_{n=0}^\infty n!x^n$$ and through algebraic manipulation I've found two ways to get a closed form for it.

The first notices that $$1+xf(x)+x^2f'(x)=f(x)$$ and then proceeds to solve this linear ODE assuming that $\displaystyle\lim_{x\to\pm 0}f(x)=1$.

The answer here is $f(x)=\frac{1}{x}e^{-\frac{1}{x}}\text{Ei}(x)$, where Ei is defined here.

The second method notices that $$n!=\int_0^\infty e^{-t}t^n dt$$ so $$f(x)=\int_0^\infty e^{-t}\sum_{n=0}^\infty (tx)^n dt=\int_0^\infty \frac{e^{-t}}{1-xt}dt$$ which again eventually resolves to $$f(x)=\frac{1}{x}e^{-\frac{1}{x}}\text{Ei}(x)$$ (assuming $\int_{-a}^a \frac{1}{t} dt=0)$.

It seems that the power series $\sum_{n=0}^\infty n!x^n$ is truly equal to $\frac{1}{x}e^{-\frac{1}{x}}\text{Ei}(x)$, in an analytic continuation kind of sense. However, the series has a radius of convergence equal to $0$, so it shouldn't be possible to fit any function to it. What is going on here? Is it possible to find another function that fits this power series (that doesn't just add a term with power series zero like $e^{-x^{-2}}$)?

What closed form are you talking about? You cannot consider $\sum_{n=0}^\infty n!x^n$ as a function on $\mathbb{R}$ or $\mathbb{C}$ since (as you know) it diverges for $x\neq 0$. (But we can interpret it as a function on [a subset of] $\mathbb{Q}_p$, say, or as a formal power series. This is where your definition, or at least a clarification, of "closed form" would be welcome.) — metamorphy, Nov 09 '20 at 03:15

score 1 · Answer 1 · answered Nov 09 '20 at 03:48

Too long for a comment.

The solution of $$1+xf(x)+x^2f'(x)=f(x)$$ is $$f(x)=c\frac{ e^{-1/x}}{x}+\frac{e^{-1/x} }{x}\text{Ei}\left(\color{red}{\frac{1}{x}}\right)$$ and, for any value of $c$, $\lim_{x\to 0} \, f(x)=1$ wich makes that we can use $c=0$ for a particular solution.

Effectively $$\frac{e^{-1/x} }{x}\text{Ei}\left(\frac{1}{x}\right)=\sum_{n=0}^\infty n!x^n$$

Is the closed form of $\sum_{n=0}^\infty n!x^n$ unique?

1 Answers1