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, $limlimits_{n to infty} frac{n}{sqrt[n]{n!}} = e$. But I can't find such sequence $y_n$ that $x_n le y_n$ and $limlimits_{n to infty} y_n = e$.

There is proof for lower bound:

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

robjohn · Accepted Answer

Here is a completely different approach.
We will use
$$
lim_{ntoinfty}a_n^{1/n}=lim_{ntoinfty}frac{a_{n+1}}{a_n}tag{1}
$$
when the limit on the right exists.
Furthermore,
$$
begin{align}
prod_{k=0}^{n+1}binom{n+1}{k}
&=prod_{k=0}^nbinom{n}{k}frac{n+1}{n-k+1}\
&=prod_{k=0}^nbinom{n}{k}prod_{k=1}^{n+1}frac{n+1}{k}tag{2}
end{align}
$$
Thus,
$$
begin{align}
lim_{ntoinfty}prod_{k=0}^nbinom{n}{k}^{frac2{n(n+1)}}
&=lim_{ntoinfty}left(prod_{k=0}^nbinom{n}{k}^{frac2n}right)^{frac1{n+1}}tag{3}\
&=lim_{ntoinfty}prod_{k=1}^{n+1}left.binom{n+1}{k}^{frac2{n+1}}middle/prod_{k=1}^nbinom{n}{k}^{frac2n}right.tag{4}\
&=lim_{ntoinfty}left(prod_{k=1}^{n+1}left.binom{n+1}{k}middle/prod_{k=1}^nbinom{n}{k}right.right)^{frac2{n+1}}prod_{k=0}^nbinom{n}{k}^{frac{-2}{n(n+1)}}tag{5}\
&=lim_{ntoinfty}left(prod_{k=1}^{n+1}left.binom{n+1}{k}middle/prod_{k=1}^nbinom{n}{k}right.right)^{frac1{n+1}}tag{6}\
&=lim_{ntoinfty}left(prod_{k=1}^{n+1}frac{n+1}{k}right)^{frac1{n+1}}tag{7}\
&=lim_{ntoinfty}left.prod_{k=1}^{n+1}frac{n+1}{k}middle/prod_{k=1}^nfrac{n}{k}right.tag{8}\
&=lim_{ntoinfty}left(1+frac1nright)^ntag{9}\[12pt]
&=etag{10}
end{align}
$$
$(4)$: apply $(1)$
$(5)$: redistribute $prod_{k=1}^nbinom{n}{k}^{-frac2n}$
$(6)$: multiply both sides by the left side and take the square root
$(7)$: apply $(2)$
$(8)$: apply $(1)$
$(9)$: algebra

robjohn · Answer

Using Riemann Sums,
$$
begin{align}
&lim_{ntoinfty}logleft(prod_{k=0}^nbinom{n}{k}^{frac2{n(n+1)}}right)\
&=lim_{ntoinfty}frac2{n(n+1)}sum_{k=0}^nleft(sum_{j=1}^klog(n-j+1)-sum_{j=1}^klog(j)right)tag{1}\
&=lim_{ntoinfty}2sum_{k=0}^nleft(sum_{j=1}^klogleft(1-frac{j}{n+1}right)frac1{n+1}-sum_{j=1}^klogleft(frac{j}{n+1}right)frac1{n+1}right)frac1ntag{2}\
&=2int_0^1left(int_0^xlog(1-t),mathrm{d}t-int_0^xlog(t),mathrm{d}tright),mathrm{d}xtag{3}\
&=2int_0^1Big(-(1-x)log(1-x)-xlog(x)Big),mathrm{d}xtag{4}\
&=-4int_0^1xlog(x),mathrm{d}xtag{5}\[6pt]
&=1tag{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 $frac1{n+1}timesfrac1{n+1}$ grid, for this double integral
$(4)$: apply $intlog(x),mathrm{d}x=xlog(x)-x+C$ twice
$(5)$: separate integral; change variables $xmapsto1-x$; combine integrals

Therefore,
$$
lim_{ntoinfty}prod_{k=0}^nbinom{n}{k}^{frac2{n(n+1)}}=etag{7}
$$

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

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