TITLE: Probability that no couple sits together in a circle
QUESTION [6 upvotes]: Suppose $n$ couples are seated in a circle, and let $P_n$ be the probability that no couple is sitting together.
Using Inclusion-Exclusion, I believe it can be shown that 
$\hspace{.2 in}\displaystyle P_n=1-\sum_{i=1}^n (-1)^{i+1}\binom{n}{i}2^i\frac{(2n-1-i)!}{(2n-1)!},$
and I would like to find out how to prove that  $\displaystyle\lim_{n\to\infty}P_n=\frac{1}{e}\;\;$ (or show that this is not the case).

Here are some numerical values:
$P_3\approx.267,\;P_4\approx.295,\;P_5\approx.310\;, P_6\approx.320,\;P_7\approx.327,\;P_8\approx.332,\;P_9\approx.336,\;\;P_{10}\approx.340$
For a related question, see Showing probability no husband next to wife converges to $e^{-1}$

REPLY [2 votes]: Verification of the Sum
How many ways for $k$ of the $n$ couples to sit together:
$$
\overbrace{2^kk!\binom{n}{k}}^1\left[\vphantom{\binom{2n-k}{k}}\right.\overbrace{\binom{2n-k}{k}}^2+\overbrace{\binom{2n-k-1}{k-1}}^3\left.\vphantom{\binom{2n-k}{k}}\right]\overbrace{(2n-2k)!\vphantom{\binom{2n-k}{k}}}^4\tag{1}
$$
Explanation:
$1$: the number of ways to select the $k$ couples: $\binom{n}{k}$
$\phantom{1\text{: }}$times the number of ways to arrange them among themselves: $2^kk!$
$2$: the number of ways to arrange $k$ couples and $2n-2k$ singles
$3$: the number of ways to arrange $k-1$ couples and $2n-2k$ singles
$\phantom{1\text{: }}$this is the number of ways to have a couple across the end of the line
$4$: the number of ways to arrange the $2n-2k$ singles among themselves
Noting that $\binom{2n-k}{k}+\binom{2n-k-1}{k-1}=\frac{2n}{2n-k}\binom{2n-k}{k}$, Inclusion-Exclusion gives the probability that no couple sits together to be
$$
\begin{align}
&\frac1{(2n)!}\sum_{k=0}^n(-1)^k2^kk!\binom{n}{k}\binom{2n-k}{k}\frac{2n}{2n-k}(2n-2k)!\\
&=\sum_{k=0}^n(-2)^k\binom{n}{k}\frac{(2n-k-1)!}{(2n-1)!}\tag{2}
\end{align}
$$

Verification of the Limit
$$
\begin{align}
\sum_{k=0}^n(-2)^k\binom{n}{k}\frac{(2n-k-1)!}{(2n-1)!}
&=\sum_{k=0}^n\frac{(-1)^k}{k!}\frac{2^k\frac{n!}{(n-k)!}}{\frac{(2n-1)!}{(2n-k-1)!}}\\
&=\sum_{k=0}^n\frac{(-1)^k}{k!}\frac{2n(2n-2)\cdots(2n-2k+2)}{(2n-1)(2n-2)\cdots(2n-k)}\\
&\to\sum_{k=0}^\infty\frac{(-1)^k}{k!}\\[3pt]
&=\frac1e\tag{3}
\end{align}
$$