Prove that $\sum_{n=1}^\infty \frac{n!}{n^n}$is irrational.

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.

Answer

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.

Attribution
Source : Link , Question Author : Mark Fischler , Answer Author : Arundo Donax

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