TITLE: Prove that $S_n\sim2^{n+1}$ where $S_n=\sum_{p=0}^n\sum_{q=0}^n\binom{n}{p}\binom{n}{q} e^{-pq}$.
QUESTION [5 upvotes]: For $n\in\Bbb N$, define
  $$S_n=\sum_{p=0}^n\sum_{q=0}^n\binom{n}{p}\binom{n}{q} e^{-pq}$$ I
  want to prove that $$\lim_{n\to\infty}\frac{S_n}{2^{n+1}}=1$$

My attempts:

For $n\in\Bbb N^*$,
$$S_n=2^{n+1}-1+\sum_{p=1}^n\sum_{q=1}^n\binom{n}{p}\binom{n}{q} e^{-pq}$$
The role of $e$ is not important. In fact, we can replace $e$ with any constant $x\ge1$.
I tried to use some similar methods (such as CLT) of the limit below but failed.
$$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}=\frac12$$
Summation of $q$ gives
$$S_n=\sum_{p=0}^n\binom{n}{p}(1+e^{-p})^{n}$$
Some calculation results:
$$\frac{S_{10}}{2^{11}}\simeq1.23\qquad\frac{S_{20}}{2^{21}}\simeq1.01\qquad\frac{S_{30}}{2^{31}}\simeq1.00035$$

The answer of the link in my comment is satisfied. However, it will also be appreciated if there are any other ideas.

REPLY [4 votes]: If both $p$ and $q$ are $\geq 2$ then $pq\geq p+q$ and this allows an elementary estimation. We have
$$ S_n = 2^{n+1}+2n\left(1+\frac{1}{e}\right)^n-\left(1+2n+\frac{n^2}{e}\right) +\sum_{p,q\in[2,n]}\binom{n}{p}\binom{n}{q}e^{-pq}$$
and
$$\sum_{p,q\in[2,n]}\binom{n}{p}\binom{n}{q}e^{-pq}\leq \sum_{p,q\in[2,n]}\binom{n}{p}\binom{n}{q}e^{-(p+q)}=\left(\sum_{p=2}^{n}\binom{n}{p}e^{-p}\right)^2\leq\left(1+\frac{1}{e}\right)^{2n} $$
so the claim follows from the fact that $\left(1+\frac{1}{e}\right)^2<2$.

This is not really related to the current problem but I believe it is interesting, anyway.
The last inequality can be proved through Cauchy-Schwarz: for instance,
$$ \int_{0}^{1}\left(1+\frac{x}{2}\right)^2\,dx\int_{0}^{1}e^x\,dx \geq \left(\int_{0}^{1}\left(1+\frac{x}{2}\right) e^{x/2}\,dx \right)^2 $$
leads to $\frac{19}{12}(e-1)\geq e$, hence to $\color{red}{e\geq\frac{19}{7}}$ and
$$ \left(1+\frac{1}{e}\right)^2 \leq \left(\frac{26}{19}\right)^2<2. $$