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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.

Aforest
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1 Answers1

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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 $$ \left(1+\frac{1}{e}\right)^2 \leq \left(\frac{26}{19}\right)^2<2. $$

Jack D'Aurizio
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