Limit involving binomial coefficients: $lim_{n to infty} {left(binom{n}{0}.binom{n}{1}…binom{n}{n}right)}^{frac{1}{n(n+1)}}$

I am facing difficulty with the following limit.

$$\lim_{n \to \infty} {\left(\binom{n}{0}.\binom{n}{1}…\binom{n}{n}\right)}^{\frac{1}{n(n+1)}}$$

I tried to take log both sides but i couldnot simplify the resulting expression .

Please help in this regard.thanks.

Answer

We see that

$$\prod_{k=0}^n\binom{n}{k}=\frac{n!^{n+1}}{\prod_{k=0}^nk!^2}=\frac{n!^{n+1}}{\left(\prod_{k=0}^nk^{n+1-k}\right)^2}=\frac{H(n)^2}{n!^{n+1}}.$$

where $H(n)=\prod_{k=1}^nk^k$. Now we see that
$$\log(H(n))=\sum_{k=1}^nk\log(k)≥\int_{1}^nx\log(x)dx=\frac{n^2}{2}\log(n)-\frac{n^2}{4}$$
as well as
$$\log(H(n))=\sum_{k=1}^nk\log(k)≤\int_{1}^{n+1}x\log(x)dx=\frac{(n+1)^2}{2}\log(n+1)-\frac{(n+1)^2}{4}$$
This gives
$$-\frac{\log(n)}{2(n+1)}-\frac{n}{4(n+1)}≤\frac{1}{n(n+1)}\log(H(n))-\frac{1}{2}\log(n)=\frac{1}{n(n+1)}\log(H(n))-\frac{1}{2}\log(n+1)+\frac{1}{2}\log(1+1/n)≤\frac{\log(n+1)}{2n}-\frac{n+1}{4n}+\frac{1}{2}\log(1+1/n).$$
As both the lower and the upper bound tend to $-\frac{1}{4}$ as $n\to\infty$ we get by the squeeze theorem
$$\lim_{n\to\infty}\left[\frac{1}{n(n+1)}\log(H(n))-\frac{1}{2}\log(n)\right]=-\frac{1}{4}\iff\\ \lim_{n\to\infty}\frac{H(n)^{\frac{1}{n(n+1)}}}{\sqrt{n}}=e^{-\frac{1}{4}}$$
Using Stirlings approximation we notice
$$\lim_{n\to\infty}\frac{n!^{\frac{1}{n}}}{n}=e^{-1}$$
and thus
$$\lim_{n\to\infty}\left[\prod_{k=0}^n\binom{n}{k}\right]^{\frac{1}{n(n+1)}}=\lim_{n\to\infty}\frac{H(n)^{\frac{2}{n(n+1)}}}{n!^{\frac{1}{n}}}=\lim_{n\to\infty}\left(\frac{H(n)^{\frac{1}{n(n+1)}}}{\sqrt{n}}\right)^2\left(\frac{n}{n!^{\frac{1}{n}}}\right)=(e^{-1/4})^2\cdot\frac{1}{e^{-1}}=\sqrt{e}$$