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.

  • $\begingroup$ Since $1\le \binom{n}{k}\le 2^n,$ and $\lim_{n\to\infty}2^{\frac{n^2}{n(n+1)}}=2,$ this limit exist and should be a number between $1$ and $2.$ $\endgroup$ – Bumblebee Nov 20 '16 at 20:38
  • 1
    $\begingroup$ Seems to be approaching $\sqrt{e} \approx 1.648721271...$ $\endgroup$ – Sil Nov 20 '16 at 21:24
  • $\begingroup$ @Bumblebee How to prove this identity,i.e,$1\le \binom{n}{k}\le 2^n$ $\endgroup$ – Anonymous Jul 14 '20 at 4:30
  • $\begingroup$ @VenkatAmith: Binomial expansion gives us $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\cdots+\binom{n}{k}x^k+\cdots \binom{n}{n}x^n.$ At $x=1,$ we can see that $2^n$ is the sum of all binomial coefficients, hence must be bigger (or equal when $n=0$) than all of them. $\endgroup$ – Bumblebee Jul 14 '20 at 5:31
  • $\begingroup$ oh, yes, I forgot about that, Thank you @Bumblebee $\endgroup$ – Anonymous Jul 14 '20 at 5:32

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 $$ \lim_{n\to\infty}\left[\prod_{k=0}^n\binom{n}{k}\right]^{\frac{1}{n(n+1)}}=\lim_{n\to\infty}\frac{H(n)^{\frac{2}{n(n+1)}}}{n!^{\frac{1}{n}}}=\lim_{n\to\infty}\left(\frac{H(n)^{\frac{1}{n(n+1)}}}{\sqrt{n}}\right)^2\left(\frac{n}{n!^{\frac{1}{n}}}\right)=(e^{-1/4})^2\cdot\frac{1}{e^{-1}}=\sqrt{e} $$


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 $$\lim_{n\to \infty} \sqrt[n(n+1)]{L} = \lim_{n\to\infty}e^{\frac{n(n+1)}{2n(n+1)}} =\sqrt{e}$$


This should help simplify this problem, but I don't know how to get an exact answer and this is too long for a comment.

$$\prod_{k=0}^n {n\choose k}=\prod_{k=0}^n\frac{n!}{k!(n-k)!}$$ Using $\prod_{k=0}^nk!(n-k)!=(\prod_{k=0}^nk!)*(\prod_{k=0}^n(n-k)!)$ and $\prod_{k=0}^n(n-k)!=\prod_{k=0}^nk!$ and $\prod_{k=0}^nk!=\prod_{k=1}^nk^{n+1-k}$, we can derive: $$\prod_{k=0}^n\frac{n!}{(k!)^2}=\frac{1^n*2^n*3^n*...}{1^{2n}*2^{2n-2}*3^{2n-4}*...}=\prod_{k=1}^nk^{2k-n-2}$$ which gets you a nasty result according to Wolfram Alpha.


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