How to find closed form expression for a power series.

Question

How can I find a closed form expression for the following series:
$$ \sum_{n\geq 1} n! x^n $$

Answer 1

If the question is: How to find a function $f$ defined on an interval $(-a,a)$ with series expansion $f(x)=\sum\limits_{n=1}^{+\infty}n!\,x^n$ on $(-a,a)$?, the answer is that there can exist no such function since the radius of convergence of the series is $R=0$.

Answer 2

Perhaps you want to consider using an exponential generating function instead?

Answer 3

Okay, so let $$ f(x) = \sum^{\infty}_{n=1}n!x^{n}$$ Observe that $$ xf'(x) = x\sum^{\infty}_{n=1} n! n x^{n-1} = \sum^{\infty}_{n=1}n!nx^{n}.$$ Thus we can find $$ f(x)+xf'(x) = \sum^{\infty}_{n=1}n!(n+1)x^{n} = \sum^{\infty}_{n=2}n!x^{n-1}.$$ Thus we have $$ f(x)+xf'(x) = x^{-1}f(x)-1.$$ Rewriting this gives us a first order ODE $$ f'(x) =\frac{(1-x)f(x)-x}{x^{2}}$$ This can be solved (I'm ashamed to admit I used wolfram alpha): $$ f(x) = \frac{c_{0}e^{-1/x}}{x} + \frac{e^{-1/x}}{x}\mathrm{Ei}(1/x) - 1 = \frac{\bigl(c_{0}+\mathrm{Ei}(1/x)\bigr)e^{-1/x}}{x}-1$$ where $c_{0}$ is a constant of integration, and $\mathrm{Ei}$ is the exponential integral function given by $$ \mathrm{Ei}(1/x) = -\int^{\infty}_{-1/x}\frac{e^{-t}}{t}dt.$$

Answer 4

Well, $$\lim_{n\to\infty}\sqrt[n]{n!}=\infty,$$ so the radius of convergence of the series is $0$. Consequently, it is defined only for $x=0$, and takes a value there of $0$. Thus, your closed form is the only function $f:\{0\}\to\{0\}$. Kind of underwhelming, no?

Answer 5

A function with this asymptotic expansion as $x \to 0$ is $\dfrac{\text{Ei}(1/x) e^{-1/x}}{x}-1$.

One way to obtain this is as follows: the Borel transform of your series is $B(x) = \sum_{n=1}^\infty x^n = x/(1-x)$. This is analytic on the negative half-line, so for $x < 0$ (or using the Cauchy principal value for $x > 0$) we take $$\eqalign{f(x) &= \int_{0}^\infty e^{-t} B(tx)\ dt = \int_0^\infty e^{-t} \dfrac{tx}{1-tx}\ dt\cr &= -1 + \int_0^\infty e^{-t} \dfrac{dt}{1-tx}\cr &= -1 + \int_{-\infty}^{1/x} e^{-1/x + u} \dfrac{du}{xu}} $$ and note that $\text{Ei}(z) = \int_{-\infty}^z e^u/u\ du$