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

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closed as off-topic by Cameron Buie, Tom-Tom, Claude Leibovici, calculus, Lord_Farin Sep 18 at 19:25

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$$\sum_{n>=1} n! x^n$$ is a generating function. –  Rudy the Reindeer Jul 5 '12 at 16:08
I think he means "I want a closed form expression for the following series:" –  Alex Nelson Jul 5 '12 at 16:13
yes. That's what I meant. Sorry for the confusion. –  ami Jul 5 '12 at 16:16
@MattN. : Please: write "$\ge$" if that's what you mean. –  Michael Hardy Jul 5 '12 at 17:57
Dear @MichaelHardy, I'm sorry, I do normally. In this case it is a mistake introduced by a careless copy-paste. –  Rudy the Reindeer Jul 5 '12 at 18:00

5 Answers 5

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 $R=0$.

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Generating functions don't need to converge, they're formal series whose coefficients are important data... –  Alex Nelson Jul 5 '12 at 16:01
but the point is that OP presumably wants some nice closed form, and there's no nice closed form that's analytic. –  user29743 Jul 5 '12 at 16:03
@Alex Please find generating function anywhere in my answer. –  Did Jul 5 '12 at 16:04
@did: you are correct, you didn't use the word once. But the OP did ask for the generating function for the series... –  Alex Nelson Jul 5 '12 at 16:05
@AlexNelson The OP asked a question that has no meaning - he gave a power series an asked for a power series. This answer is an answer to what many of us assumed the OP actually meant... –  Thomas Andrews Jul 5 '12 at 16:12

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 $\text{Ei}(z) = \int_{-\infty}^z e^u/u\ du$

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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 $$ \mathrm{Ei}(1/x) = -\int^{\infty}_{-1/x}\frac{e^{-t}}{t}dt.$$

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What is the numerical value of $f(1)$? –  Did Jul 5 '12 at 16:38
The power series is convergent only at $x=0$, of course; still, the series can be treated as an asymptotic series for the function that Alex has derived here... –  J. M. is back. Jul 5 '12 at 16:45
@Cocopuffs differentiating won't work. I tried it. <br> –  ami Jul 5 '12 at 16:45
For $x \ne 0$ you can't subtract $f(x)$ from both sides as you did above, since $\infty - \infty$ isn't defined. For $x = 0$ you can't divide by $x^2$. I find this derivation strange. –  Cocopuffs Jul 5 '12 at 16:48
FWIW: repeated integration by parts of a relevant integral will net you this cute little asymptotic series, which can then be formally (emphasized for a reason ;P ) manipulated to look like OP's series... so in a sense, Alex did manage to come up with a closed form for the ordinary generating function of the factorial. –  J. M. is back. Jul 5 '12 at 16:58

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

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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?

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