A Combinatorial Geometric Mean

We have: $\displaystyle\binom{n}{0}\binom{n}{1}\binom{n}{2}\ldots\binom{n}{n}=\prod\limits_{k=0}^{n}{\frac{n!}{k!\left( n-k \right)!}}=\prod\limits_{k=1}^{n}{{{\left( n-k+1 \right)}^{n+1-2k}}}$

On the other hand, $\sum\limits_{k=1}^{n}{\left( n+1-2k \right)}=0$ or ${{\left( n+1 \right)}^{\sum\limits_{k=1}^{n}{\left( n+1-2k \right)}}}=1$

Thus, $\displaystyle\binom{n}{0}\binom{n}{1}\binom{n}{2}\ldots\binom{n}{n}=\prod\limits_{k=1}^{n}{{{\left( \frac{n-k+1}{n+1} \right)}^{n+1-2k}}}=\prod\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+1} \right)}^{n+1-2k}}}$

This implies that $\displaystyle G_n=\sqrt[n+1]{\prod\limits_{k=1}^{n}{{{\left( \frac{n-k+1}{n+1} \right)}^{n+1-2k}}}}=\prod\limits_{k=1}^{n}{{{\left( 1-\frac{k}{n+1} \right)}^{1-\frac{2k}{n+1}}}}$

$\begin{aligned} & \Rightarrow \frac{1}{n}\ln {{G}_{n}}=\frac{1}{n}\sum\limits_{k=1}^{n}{\left( 1-\frac{2k}{n+1} \right)\ln \left( 1-\frac{k}{n+1} \right)} \\ & \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\ln {{G}_{n}}=\int\limits_{0}^{1}{\left( 1-2x \right)\ln \left( 1-x \right)dx} \\ \end{aligned}$

Applying integral by part, we get: $\displaystyle\int\limits_{0}^{1}{\left( 1-2x \right)\ln \left( 1-x \right)dx}=\underset{a\to {{1}^{-}}}{\mathop{\lim }}\,\int\limits_{0}^{a}{\left( 1-2x \right)\ln \left( 1-x \right)dx}=\frac{1}{2}$

Hence, $\displaystyle\lim_{n\to \infty }\frac{1}{n}\ln {{G}_{n}}=\frac{1}{2}\Rightarrow \lim_{n\to \infty }\sqrt[n]{{{G}_{n}}}=\sqrt{e}\approx\boxed{1.649}$

Using Stirling's approximation of Binomial Coefficients , for $0\leq k \leq n$ $\binom{n}{k} \sim 2^{nH(\frac{k}{n})}$ Where $H(\cdot)$ is the Binary Entropy function . Hence, $\sqrt[n]{G_n} \sim 2^{\frac{1}{n}\sum_{k=0}^{n}H(\frac{k}{n})}\stackrel{(a)}{\to} 2^{\int_{0}^{1}H(x)dx}=\sqrt{e},$ where in (a) we have used the fundamental theorem of integral calculus.