Stirling And Stolz United!

We have $\prod_{r=1}^n \binom{n}{r}\;=\;\frac{(n!)^n}{\prod_{r=1}^n r! \prod_{r=1}^{n-1} r!}\;=\;\frac{(n!)^n}{n\$ (n-1)\$}\;=\;\frac{(n!)^n}{G(n+2)G(n+1)}$ using the Sloane/Plouffe superfactorial $n\$$ and the Barnes G-function. Using Stirling's approximation, and similar asymptotics for $G$ , we have $\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n} = e^{-1} \hspace{2cm} \lim_{n\to\infty}\frac{\sqrt[n^2]{G(n+1)}}{\sqrt{n}}= \lim_{n\to\infty}\frac{\sqrt[n^2]{G(n+2)}}{\sqrt{n}}= e^{-\frac34}$ so that the desired limit is $\sqrt{e} = \boxed{1.648721271}$ .

Let $\lim\limits_{n\to\infty} \sqrt[n^2]{{n \choose1}{n \choose 2}\ldots{n \choose n}} = L$ . Taking the natural logarithm of both sides, we obtain $\ln\left(L\right) = \lim\limits_{n\to\infty} \dfrac{1}{n^2}\sum\limits_{i=1}^{n}\ln\left(\binom{n}{i}\right) = \lim\limits_{n\to\infty} \dfrac{1}{n}\cdot\dfrac{1}{n}\sum\limits_{i=1}^{n}\left(\ln\left(n!\right)-\ln\left(i!\right)-\ln\left(\left(n-i\right)!\right)\right)$ By Stirling's approximation, this is equivalent to $\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{i=1}^{n}\left(\ln\left(n\right)+\dfrac{1}{2n}\ln\left(2\pi n\right)-1-\dfrac{i}{n}\ln\left(i\right)-\dfrac{1}{2n}\ln\left(2\pi i\right)+\dfrac{i}{n}-\left(1-\dfrac{i}{n}\right)\ln\left(n-i\right)-\dfrac{1}{2n}\ln\left(2\pi \left(n-i\right)\right)+1-\dfrac{i}{n}\right)$ By taking limits and cancelling terms, it is clear that this simplifies to $\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{i=1}^{n}\left(\ln\left(n\right)-\dfrac{i}{n}\ln\left(i\right)-\left(1-\dfrac{i}{n}\right)\ln\left(n-i\right)\right) = \lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{i=1}^{n}\left(-\ln\left(1-\dfrac{i}{n}\right)+\dfrac{i}{n}\ln\left(\dfrac{n}{i}-1\right)\right)$ Which by the definiton of a Riemann Sum is equivalent to $\int_0^1 \! \left(-\ln\left(1-x\right)+x\ln\left(\dfrac{1}{x}-1\right)\right) \, \mathrm{d}x$ Which by standard methods evaluates to $\dfrac{1}{2}$ , and our limit is equal to $\sqrt e \approx \boxed{1.64872127}$ .

Here is my solution.

It is rather simple to prove the following identity, but for those who cannot prove it there is a proof here $\prod_{r=1}^{n} \binom{n}{r}=\prod_{r=1}^{n}r^{2r-n-1}$ Note that $\ln \prod_{r=1}^{n} \binom{n}{r}= 2\sum_{r=1}^{n} r \ln r - (n+1) \sum_{r=1}^{n} \ln r$

Now, note that by integration by parts $\sum_{k=1}^n k \ln k\sim \int^n_0 x \log x dx= \frac{1}{2}n^2 \ln n - \frac{n^2}{4}$

Using this and Stirling's approximation that $\ln n! \sim n\ln n - n$ we find $2\sum_{r=1}^{n} r \ln r - (n+1) \sum_{r=1}^{n} \ln r \approx n^2 \ln n - \frac{1}{2}n^2 - (n^2+n) \ln n +n^2+n=\frac{n^2}{2}-n\ln n +n$

So $\lim_{x \to \infty}\frac{ \ln \prod_{r=1}^{n} \binom{n}{r}}{n^2}= \lim_{x \to \infty} \frac{n^2-2n \ln n+2n}{2n^2} =\frac{1}{2}$

So the desired limit is $e^{\frac{1}{2}}=1.648721271....$ .

Relevant wikis.- Stirling's formula and Cesaro-Stolz theorem

Stirling's formula is $\Gamma[x + 1] \sim \sqrt{2\pi x} \left(\frac{x}{e}\right)^{x}, \text{ as } x \to \infty, x \in \mathbb{R}$ and this means

if $n \in \mathbb{N}$ and $n$ is "big" then, $n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^{n}$ .

Ok, let's call $\displaystyle \Delta = \lim_{n\to\infty} \sqrt[n^2]{{n \choose 1} {n \choose 2} ... {n \choose n}} \Rightarrow$ $\displaystyle \ln \Delta = \lim_{n\to\infty} \frac{\ln {n \choose 1} {n \choose 2} ... {n \choose n}}{n^2} \Rightarrow$ applying Stolz theorem $\displaystyle \ln \Delta = \lim_{n\to\infty} \frac{\ln {n \choose 1} {n \choose 2} ... {n \choose n} - \ln {n-1 \choose 1} {n - 1 \choose 2} ... {n - 1 \choose n - 1}}{n^2 - (n - 1)^2} =$ $\displaystyle \ln \Delta = \lim_{n\to\infty} \frac{\ln \frac{{n \choose 1} {n \choose 2} ... {n \choose n}}{{n-1 \choose 1} {n - 1 \choose 2} ... {n - 1 \choose n - 1}}}{2n - 1} = \lim_{n\to\infty} \frac{\ln \left(\frac{n}{n -1}\times\frac{n}{n - 2}\ldots\frac{n}{2}\times\frac{n}{1}\right)}{2n -1} =$ $\displaystyle \lim_{n\to\infty} \frac{\ln \frac{n^{n -1}}{(n - 1)!}}{2n - 1} = \lim_{n\to\infty} \frac{\ln \frac{n^{n}}{n!}}{2n - 1} =$ Applying Stirling $\displaystyle = \lim_{n\to\infty} \frac{ \ln \frac{n^{n}}{n^{n} \cdot e^{-n} \cdot \sqrt{2\pi n}}}{2n - 1} = \lim_{n\to\infty} \frac{\ln e^n - \ln \sqrt{2\pi n}}{2n - 1} = \frac{1}{2}$ Therefore, $\ln \Delta = \frac{1}{2} \Rightarrow \Delta = e^{\frac{1}{2}}$