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Let $$S= \sum_{n=1}^\infty \frac{n!}{n^n}$$

(Does anybody know of a closed form expression for $S$?)

It is easy to show that the series converges.

Prove that $S$ is irrational.

I tried the sort of technique that works to prove $e$ is irrational, but got bogged down.

GPhys
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Mark Fischler
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    As an aside, $$\sum_{n=0}^\infty \frac{n!}{n^n} \quad = \quad \int_0^\infty \frac{E(x)}{e^x} dx$$ where $\displaystyle\lim_{n \to 0} n^n = 1,$ and $$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad \text{and} \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ – Lucian Mar 11 '15 at 20:49
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    Also note that: $$\int_{0}^1 n(-x\log(x))^{n-1} dx=\frac{n!}{n^n}$$ $$\int_{0}^1\frac{1}{(x\ln(x)+1)^2}dx=\int_{0}^1 \sum_{n=1}^\infty n(-x\log(x))^{n-1} dx=\sum_{n=1}^\infty\frac{n!}{n^n}$$ – Ethan Splaver Mar 11 '15 at 21:15
  • A094082 – g.kov Mar 11 '15 at 21:17
  • @Lucian: Does the first integral converge? I believe $E(x)$ is the exponential integral. – science Mar 11 '15 at 21:18
  • For similar identities see: http://en.wikipedia.org/wiki/Sophomore%27s_dream – Ethan Splaver Mar 11 '15 at 21:28
  • @science: Yes, it converges, and no, $E(x)$ is not the exponential integral. – Lucian Mar 11 '15 at 21:42
  • See also OEIS A$094082$. – Lucian Mar 11 '15 at 21:45
  • @Lucian:Thank you for the clarification. – science Mar 11 '15 at 21:49
  • Alternately, $I=\displaystyle\int_0^\infty\frac{e^x}{\big(e^x-x\big)^2}~dx$ – Lucian Mar 11 '15 at 21:59
  • The last one I know! – science Mar 11 '15 at 22:00
  • Mathematica doesn't know of a closed form for the sum, by the way. – Patrick Stevens May 11 '16 at 10:07
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    Continued fraction: $\displaystyle\DeclareMathOperator*{\K}{K}S=b_0+\K_{n=1}^{\infty}\frac{a_n}{b_n}$ where $b_0=1$, $b_1=2$, $a_1=2$, and for $n\geq 2$, $a_n=-n^{2n-1}$ and $b_n=n^n+(n+1)^n$. – GPhys Jun 16 '17 at 16:58
  • (typo above: $a_1=1$) Here's also Mathematica (+WolframAlpha) code to reproduce the continued fraction approximations N[1+1/(2+ContinuedFractionK[-n^(2*n-1),n^n+(n+1)^n,{n,2,100}])]. There is enough out there about continued fractions that I thought I could make something work, but I'm not very experienced with manipulating them quickly. – GPhys Jun 16 '17 at 18:09
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    Is there any context to this problem, i.e. is it well known that the series is an irrational number? – Ivan Jun 17 '17 at 17:39
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    Computation seems to suggest a simple approach based on a Liouville approximation theorem is actually hopeless regardless of how clever your approximations are (at least when using the series as given): Table[N[Sum[n!/n^n,{n,m+1,Infinity}]-c/Denominator[Sum[n!/n^n,{n,1,m}]]],{m,1,20}] – GPhys Jun 18 '17 at 14:08
  • @ivan I don't think it is "well known" that the series sum is irrational; in fact, although I would be surprised to learn it is rational, it is not out of the question. – Mark Fischler Jun 19 '17 at 22:17
  • the ratio of consecutive terms in the sum is $n\over{n\big ({n\over n-1}\big)^{n-1}}$ –  Jul 09 '17 at 20:56
  • Now asked on MO: https://mathoverflow.net/questions/277969 – Watson Aug 05 '17 at 08:40

2 Answers2

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I couldn't mimic the proof of irrationality of $e$ but here goes my atempt ...

To prove the irrationality of $S=\sum_{n=0}^{\infty}\frac{n!}{n^n}$ it would be enough to find an infinite continued fraction expansion. For convenience, I am redefining $S$ adding a 0th term equal to one $\frac{0!}{0^0}:=1$

We first write $S$ in the form $S=\sum_{n=0}^{\infty} a_0 \cdot a_1 \cdot a_2 \cdots a_n $. If we calculate the ratio of two consecutive terms in the original definition of $S$ we find that, for $n\ge 2$ $$\frac{\frac{n!}{n^n}}{\frac{(n-1)!}{(n-1)^{n-1}}}=(1-\frac{1}{n})^{n-1}$$

So we can write $$\frac{n!}{n^n}=(1-\frac{1}{n})^{n-1} \cdot (1-\frac{1}{n-1})^{n-2} \cdots (1-\frac{1}{2})^1$$

Seting $a_0:=1, a_1:=1$ and $a_n=(1-\frac{1}{n})^{n-1}$ for $n\ge 2$ we find that $$ S=\sum_{n=0}^{\infty} a_0 \cdot a_1 \cdot a_2 \cdots a_n $$ Now we make use of Euler's continued fraction formula: $$a_0 + a_0a_1 + a_0a_1a_2 + \cdots + a_0a_1a_2\cdots a_n = \cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{1 + a_2 - \cfrac{\ddots}{\ddots \cfrac{a_{n-1}}{1 + a_{n-1} - \cfrac{a_n}{1 + a_n}}}}}}\,$$

But this is not yet a proper continued fraction $b_0 + \cfrac{1}{b_1 + \cfrac{1}{b_2 + \cfrac{1}{ \ddots + \cfrac{1}{b_n} }}}$ and we can't make use of the result of irrationality stated above.

We can apply an equivalence transformation: $$\cfrac{x_1}{y_1 + \cfrac{x_2}{y_2 + \cfrac{x_3}{y_3 + \cfrac{x_4}{y_4 + \ddots\,}}}} = \cfrac{z_1x_1}{z_1y_1 + \cfrac{z_1z_2x_2}{z_2z_2 + \cfrac{z_2z_3x_3}{z_3z_3 + \cfrac{z_3z_4x_4}{z_4z_4 + \ddots\,}}}}$$ (which holds if all the x's, y's and z's are nonzero)

by setting $x_1:=a_0,y_1=1$ and, for $n \ge 1$ $x_n:= -a_{n-1}, y_n:=1+a_{n-1}$ and choosing the sequence of z's satisfy the relation $z_{n+1}=\frac{1}{x_{n+1} z_n}$.

So, at the end we get an infinite proper continued fraction expansion.

Hope this helps.

Arundo Donax
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  • This looks promising but I'm trying to see if there are flaws. The first thing that comes to mind is: If $r\in \Bbb R$ can be expressed as an infinite continued fraction, why does that imply that $r$ is irrational? Perhaps it is possible for $r$ to be expressed as a finite continued fraction, or in some other rational form, as well as the infinite continued fraction. Of course, if $r$ can be expressed as a finite continued fraction then $r$ is rational but that is not the same as the statement you have used. – Mark Fischler Sep 09 '17 at 20:31
  • True. The assertion that a number is irrational if a contitued fraction expansion is infinite is not completely true. It only holds if the expansion is obtained by a certain method that involves finding the integer part of the reciprocal of the residue of the last aproximation. Thats why I tried to give it the standard form [b0;,b1,b2,...] with the b's being positive integers. – Arundo Donax Sep 10 '17 at 17:13
  • To be more precise, if $x$ is a positive real number and we define $x_1=x, b_1=[x_1]$ and, for $n \ge 1$, $b_{n+1}=[x_n], x_{n+1}=\frac{1}{x_n-b_n}$ then the sequence of b's becomes eventually equal to one if and only if $x$ is rational; and these b's are the terms of the contiuned fraction expansion of $x$. – Arundo Donax Sep 10 '17 at 17:24
  • If the $b_n$ are eventually equal to $1$ doesn't that give an infinite CF expansion (related to $\sqrt{2}$)? At any rate, I'd be interested to see how this statement was proven, although perhaps it requires advanced methods. – Mark Fischler Sep 11 '17 at 16:59
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This doesn't answer the question.

But.

Wolfy calculates it as

$1.8798538621752585334863061450709600388198734004892899048296176691222963$
$866612142113617650197389123532397...$

and the inverse symbolic calculator doesn't know, so I would be surprised if there is a reasonable formula.

Darío A. Gutiérrez
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marty cohen
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