Showing probability no husband next to wife converges to $e^{-1}$ 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  &lt;- 1000000
together &lt;- 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?
I observe that each term with $i$ fixed approaches a nice limit. We have

$$ 2^i \frac{n(n-1)(n-2)\cdots(n-i+1)}{i!} \frac1{(2n-i+1)(2n-i+2)\cdots(2n)} $$

or

$$ \frac1{i!} \frac{2n}{2n} \frac{2(n-1)}{2n-1} \cdots \frac{2(n-i+1)}{(2n-i+1)} \sim \frac 1{i!} $$

This gives you the series, assuming the limits (defining terms with $i&gt;n$ to be zero) may be safely exchanged,

$$\lim_{n\to\infty} \sum_{i=0}^\infty [\cdots] =  \sum_{i=0}^\infty \lim_{n\to\infty} [\cdots] = \sum_{i=0}^\infty (-1)^i \frac1 {i!} \equiv e^{-1}$$

Justifying the limit interchange I haven't thought about but I suspect this can be shown to be fine without too much effort... Edit: You can probably use the Weierstrass M-test.
Without loss of generality, suppose everyone is sitting in a circle instead of a row and that there is one unoccupied seat.  For each wife, the probably that her husband is not sitting next to her on either side is $1 - \frac{2}{2 \cdot n}$ and there are $n$ wives, so the probability that no wife is sitting next to her husband is $(1- \frac{1}{n})^n$, whose limit is the definition of $\frac{1}{e}$.