TITLE: Showing probability no husband next to wife converges to $e^{-1}$
QUESTION [9 upvotes]: Inspired by these questions:

Probability of Couples sitting next to each other (Sitting in a Row)
Probability question about married couples
Four married couples, eight seats. Probability that husband sits next to his wife?
In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?
No husband can sit next to his wife in this probability question

the more general question of the probability that seating $n$ couples (i.e. $2n$ individuals) in a row at random means that no couples are sat together can be expressed using inclusion-exclusion as 
$$\displaystyle\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$$
which for small $n$ takes the values: 
n   Probability of no couple together           
1   0           0   
2   1/3         0.3333333
3   1/3         0.3333333
4   12/35       0.3481481
5   47/135      0.3428571   
6   3655/10395  0.3516114
7   1772/5005   0.3540460
8   20609/57915 0.3558491

This made me wonder whether it converges to $e^{-1} \approx  0.3678794$ as $n$ increases, like other cases such as the secretary/dating problem and $\left(1-\frac1n\right)^n$ do.  So I tried the following R code (using logarithms to avoid overflows)
couples  <- 1000000
together <- 0:couples
sum( (-1)^together * exp( log(2)*together + lchoose(couples,together) + 
     lfactorial(2*couples - together) - lfactorial(2*couples) ) ) 

which indeed gave a figure of $0.3678794$.
How might one try to prove this limit?

REPLY [3 votes]: Let $a_n=\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$, let $b_n=(1-\frac{1}{n})^n$, and let $c_n=\sum_{i=0}^n (-1)^i \frac{1}{i!}$.
We will show below that $b_n\le a_n\le c_n$ for $n\ge2$, so then we can conclude that
$\displaystyle\lim_{n\to\infty}a_n=\frac{1}{e}$
by the Squeeze Theorem since $\displaystyle\lim_{n\to\infty}b_n=\frac{1}{e}$ and $\displaystyle\lim_{n\to\infty}c_n=\frac{1}{e}$.
$\textbf{1)}$ To show that $a_n\le c_n$ for $n\ge 2$,  let $$c_n-a_n=\sum_{i=0}^n (-1)^i \frac{1}{i!}-\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$$
$$=\sum_{i=0}^n (-1)^i\frac{1}{i!}\bigg[1-\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}\bigg]=\sum_{i=0}^n(-1)^it_i,$$where $t_i=\frac{1}{i!}\bigg[1-\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}\bigg]>0$ for $i\ge 2$ since 
$\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}=\frac{(2n)(2n-2)(2n-4)\cdots(2n-2i+2)}{(2n)(2n-1)(2n-2)\cdots(2n-i+1)}<1$.
Therefore to show that $c_n-a_n\ge0$ for $n\ge2$, it suffices to show that
$t_{i}-t_{i+1}\ge0$ for any even integer $i\ge2$:
$$\displaystyle t_{i}-t_{i+1}=\frac{1}{i!}\bigg[1-\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}\bigg]-\frac{1}{(i+1)!}\bigg[1-\frac{2^{i+1}n!}{(n-i-1)!}\frac{(2n-i-1)!}{(2n)!}\bigg]$$
$$\displaystyle=\frac{1}{(i+1)!}\bigg[i+1-\frac{(i+1)2^{i}n!(2n-i)!}{(n-i)!(2n)!}-1+\frac{2^{i+1}n!(2n-i-1)!}{(n-i-1)!(2n)!}\bigg]$$
$$\displaystyle=\frac{1}{(i+1)!}\bigg[i-\frac{(i+1)2^{i}n!(2n-i)!}{(n-i)!(2n)!}+\frac{2^{i+1}(n-i)n!(2n-i-1)!}{(n-i)!(2n)!}\bigg],$$
so $t_{i}-t_{i+1}\ge0 \iff$  $$ i(n-i)!(2n)!\ge(i+1)2^{i}n!(2n-i)!-2^{i+1}(n-i)n!(2n-i-1)!$$
$$\\\ \;\;\;\;\; =\  2^{i}n!(2n-i-1)!\big[(i+1)(2n-i)-2(n-i)\big]$$
$$\\\ \;\;=2^{i}n!(2n-i-1)!(i)(2n-i+1)\iff$$
$$(n-i)!(2n)!\ge2^{i}n!(2n-i-1)!(2n-i+1)\iff$$
$$\big[(2n)(2n-1)(2n-2)\cdots(2n-i+2)\big](2n-i)\ge2^{i}n(n-1)(n-2)\cdots(n-i+1)\iff$$
$$\big[(2n)(2n-1)(2n-2)\cdots(2n-i+2)\big](2n-i)\ge(2n)(2n-2)(2n-4)\cdots(2n-2i+2),$$
which is clearly true for any even integer $i\ge2$.
$-------------------------------------------$
$\textbf{2)}$ To show that $a_n\ge b_n$ for $n\ge2$, we can use the Binomial Formula and then proceed as above:
Let $$a_n-b_n=\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}-\sum_{i=0}^n {n\choose i}\big(\frac{-1}{n}\big)^i$$
$$=\sum_{i=0}^n (-1)^{i} \frac{1}{i!}\bigg[\frac{2^{i}n!(2n-i)!}{(n-i)!(2n)!}-\frac{n!}{(n-i)!(n^{i})}\bigg]=\sum_{i=0}^n(-1)^i s_i,$$
where $s_i=\frac{1}{i!}\bigg[\frac{2^{i}n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i+1)}-\frac{n(n-1)\cdots(n-i+1)}{n^i}\bigg]>0$ for $i\ge2$ since $\;\;\;\;2^{i}n^{i}=(2n)^{i}\ge(2n)(2n-1)(2n-2)\cdots(2n-i+1).$
Therefore to show that $a_n-b_n\ge0$ for $n\ge2$, 
it suffices to show that $s_{i}-s_{i+1}\ge0$ for any even integer $i\ge2$: 
$$s_{i}-s_{i+1}=\frac{1}{i!}\bigg[\frac{2^{i}n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i+1)}-\frac{n(n-1)\cdots(n-i+1)}{n^i}\bigg]-\frac{1}{(i+1)!}\bigg[\frac{2^{i+1}n(n-1)\cdots(n-i)}{(2n)(2n-1)\cdots(2n-i)}-\frac{n(n-1)\cdots(n-i)}{n^{i+1}}\bigg]$$
$=\frac{1}{(i+1)!}\bigg[\frac{2^{i}(i+1)n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i+1)}-\frac{(i+1)n(n-1)\cdots(n-i+1)}{n^i}-\frac{2^{i+1}n(n-1)\cdots(n-i)}{(2n)(2n-1)\cdots(2n-i)}+\frac{n(n-1)\cdots(n-i)}{n^{i+1}}\bigg]$
$=\frac{1}{(i+1)!}\bigg[\frac{2^{i}(i+1)(2n-i)n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i)}-\frac{(n-1)(n-2)\cdots(n-i+1)[n(i+1)-(n-i)]}{n^i}\bigg]$,so
$s_i-s_{i+1}\ge0\iff$ 
$2^{i}n^{i}(i+1)(2n-i)n(n-1)(n-2)\cdots(n-i+1)\ge i(n+1)(n-1)(n-2)\cdots(n-i+1)(2n)(2n-1)\cdots(2n-i)\iff$
$$2^{i}n^{i}(i+1)(2n-i)n\ge i(n+1)(2n)(2n-1)(2n-2)\cdots(2n-i)\iff$$
$$(2n)^{i}(i+1)n\ge i(n+1)(2n)(2n-1)(2n-2)\cdots(2n-i+1),$$
and this inequality is valid since $(i+1)n\ge i(n+1)$ and $(2n)^{i}\ge (2n)(2n-1)(2n-2)\cdots(2n-i+1)$.