If $x_n = (\prod_{k=0}^n \binom{n}{k})^\frac{2}{n(n+1)}$ then $\lim_{n \to \infty} x_n = e$

Question

I try to prove the following: $$x_n = \left(\prod_{k=0}^n \binom{n}{k}\right)^\frac{2}{n(n+1)}$$ $$\lim_{n \to \infty} x_n = e$$ I want to use double sided theorem, so I've proven that $$x_n \ge \frac{n}{\sqrt[n]{n!}}$$ As it known, $\lim\limits_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = e$. But I can't find such sequence $y_n$ that $x_n \le y_n$ and $\lim\limits_{n \to \infty} y_n = e$.

There is proof for lower bound:

$$x_n = \frac{(n!)^\frac{2}{n}}{(0!\cdot 1! \dots n!)^\frac{4}{n(n+1)}} \ge \frac{(n!)^\frac{2}{n}\cdot n}{(0!\cdot 1! \dots n!)^\frac{4}{n(n+1)}\cdot(n^n(n-1)^{n-1}(n-2)^{n-2}\dots2^2)^\frac{4}{n(n+1)}} = $$ $$= \frac{(n!)^\frac 2n \cdot n}{(n!)^\frac 4n} = \frac{n}{(n!)^\frac 2n} \ge \frac{n}{\sqrt[n]{n!}}$$

Answer 1

Here is a completely different approach.

We will use $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\tag{1} $$ when the limit on the right exists.

Furthermore, $$ \begin{align} \prod_{k=0}^{n+1}\binom{n+1}{k} &=\prod_{k=0}^n\binom{n}{k}\frac{n+1}{n-k+1}\\ &=\prod_{k=0}^n\binom{n}{k}\prod_{k=1}^{n+1}\frac{n+1}{k}\tag{2} \end{align} $$ Thus, $$ \begin{align} \lim_{n\to\infty}\prod_{k=0}^n\binom{n}{k}^{\frac2{n(n+1)}} &=\lim_{n\to\infty}\left(\prod_{k=0}^n\binom{n}{k}^{\frac2n}\right)^{\frac1{n+1}}\tag{3}\\ &=\lim_{n\to\infty}\prod_{k=1}^{n+1}\left.\binom{n+1}{k}^{\frac2{n+1}}\middle/\prod_{k=1}^n\binom{n}{k}^{\frac2n}\right.\tag{4}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^{n+1}\left.\binom{n+1}{k}\middle/\prod_{k=1}^n\binom{n}{k}\right.\right)^{\frac2{n+1}}\prod_{k=0}^n\binom{n}{k}^{\frac{-2}{n(n+1)}}\tag{5}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^{n+1}\left.\binom{n+1}{k}\middle/\prod_{k=1}^n\binom{n}{k}\right.\right)^{\frac1{n+1}}\tag{6}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^{n+1}\frac{n+1}{k}\right)^{\frac1{n+1}}\tag{7}\\ &=\lim_{n\to\infty}\left.\prod_{k=1}^{n+1}\frac{n+1}{k}\middle/\prod_{k=1}^n\frac{n}{k}\right.\tag{8}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{9}\\[12pt] &=e\tag{10} \end{align} $$ $(4)$: apply $(1)$
$(5)$: redistribute $\prod_{k=1}^n\binom{n}{k}^{-\frac2n}$
$(6)$: multiply both sides by the left side and take the square root
$(7)$: apply $(2)$
$(8)$: apply $(1)$
$(9)$: algebra

Answer 2

Using Riemann Sums, $$ \begin{align} &\lim_{n\to\infty}\log\left(\prod_{k=0}^n\binom{n}{k}^{\frac2{n(n+1)}}\right)\\ &=\lim_{n\to\infty}\frac2{n(n+1)}\sum_{k=0}^n\left(\sum_{j=1}^k\log(n-j+1)-\sum_{j=1}^k\log(j)\right)\tag{1}\\ &=\lim_{n\to\infty}2\sum_{k=0}^n\left(\sum_{j=1}^k\log\left(1-\frac{j}{n+1}\right)\frac1{n+1}-\sum_{j=1}^k\log\left(\frac{j}{n+1}\right)\frac1{n+1}\right)\frac1n\tag{2}\\ &=2\int_0^1\left(\int_0^x\log(1-t)\,\mathrm{d}t-\int_0^x\log(t)\,\mathrm{d}t\right)\,\mathrm{d}x\tag{3}\\ &=2\int_0^1\Big(-(1-x)\log(1-x)-x\log(x)\Big)\,\mathrm{d}x\tag{4}\\ &=-4\int_0^1x\log(x)\,\mathrm{d}x\tag{5}\\[6pt] &=1\tag{6} \end{align} $$ $(1)$: $\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots1}$
$(2)$: distribute the $\frac1n$ and $\frac1{n+1}$; subtract $\frac{k}{n+1}\log(n+1)$ from each of the inner sums
$(3)$: $(2)$ is $\frac{n+1}{n}$ times the Riemann sum, on a $\frac1{n+1}\times\frac1{n+1}$ grid, for this double integral
$(4)$: apply $\int\log(x)\,\mathrm{d}x=x\log(x)-x+C$ twice
$(5)$: separate integral; change variables $x\mapsto1-x$; combine integrals

Therefore, $$ \lim_{n\to\infty}\prod_{k=0}^n\binom{n}{k}^{\frac2{n(n+1)}}=e\tag{7} $$

Answer 3

Too late, more than likely too advanced but done for the fun of deriving asymptotics. $$\prod_{k=0}^n \binom{n}{k}=\frac{\big[\Gamma (n+1)\big]^n}{G(n+1)\, G(n+2)}$$ where $G(.)$ is the Barnes G-function.

Using Stirling-like approximations $$\log(G(p))=\frac{1}{4} p^2 (2 \log (p)-3)+\frac{1}{2} p (-2 \log (p)+2+\log (\pi )+\log (2))+$$ $$\frac{1}{12} \left(-6 \log \left(2 \pi A^2\right)+5 \log (p)+1\right)-\frac{1}{12 p}-\frac{1}{240 p^2}+O\left(\frac{1}{p^3}\right)$$ $$\log\Bigg[\prod_{k=0}^n \binom{n}{k} \Bigg]=\frac{n^2}{2}+\frac{1}{2} n (-\log (n)+2-\log (\pi )-\log (2))+\frac{1}{12} (24 \log (A)-4 \log (n)-1-6 \log (2 \pi ))-\frac{1}{12 n}+\frac{1}{180 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(x_n)=1-\frac 1 n \log \left(\frac{2 \pi n}{e}\right)+\frac{24 \log (A)+2 \log (n)-7}{6 n^2}+O\left(\frac{1}{n^3}\right)$$

$$\color{blue}{x_n=e\left(1-\frac 1 n \log \left(\frac{2 \pi n}{e}\right) \right)+O\left(\frac{1}{n^2}\right)}$$

The relative error is less than $0.1$% as soon as $n>135$.