TITLE: Limit involving binomial coefficients: $\lim_{n\to\infty}\left(\binom{n}{0}\binom{n}{1}\dots\binom{n}{n}\right)^{\frac{1}{n(n+1)}}$
QUESTION [10 upvotes]: I am facing difficulty with the following limit.
$$ \lim_{n\to\infty}\left(\binom{n}{0}\binom{n}{1}\dots\binom{n}{n}\right)^{\frac{1}{n(n+1)}} $$
I tried to take log both sides but I could not simplify the resulting expression.
Please help in this regard. Thanks.

REPLY [6 votes]: Let limit be denoted as $L$,then $$\prod_{r=0}^{n}\binom{n}{r}= (n!)^{n+1}\left(\prod_{k=0}^n k!\right)^{-2} =(n!)^{n+1}\left(\prod_{k=0}^{n} k^{k-n-1}\right)^2= \dfrac{1}{(n!)^{n+1}}\left(\prod_{r=1}^n r^r\right)^2$$ Also we have the approximation  for hyperfactorial( the latter expression above )
$$\dfrac{1}{(n!)^{n+1}}\left(\prod_{r=1}^{n} r^r\right)^2 \\ \approx\dfrac{1}{(n!)^{n+1}}\left( A n^{\frac{6n^2+6n+1}{12}}e^{-\frac{n^2}{4}}\right)^2$$ using the Stirling approximation for $n!$ and simplifying we have $$L\approx \dfrac{A^2e^{\frac{n^2+2n}{2}}}{(2\pi)^{\frac{n+1}{2}}n^{\frac{3n+2}{6}}}$$ where $ A$ is  Glashier-Kinkelin Constant
and required we have and hence $$\lim_{n\to \infty} \sqrt[n(n+1)]{L} = \lim_{n\to\infty}e^{\frac{n(n+1)}{2n(n+1)}} =\sqrt{e}$$