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.
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.
We can rewrite the identity as $$ \lim_{n\to\infty} \Biggl({n\choose1}\dots{n\choose n}\Biggr)^{1\over1+\dots+n} = e. $$ This says that, for large $n$, we have $$ {n\choose 0}{n\choose1}\dots{n\choose n}\approx e^{0+1+\dots+n}. $$ (I added the trivial terms ${n\choose0}=e^0=1$ so that the left-hand side would be the product of a row of Pascal's triangle.) Write $s_n={n\choose0}\dots{n\choose n}$. Then, since ${n\choose k}={n!\over k!(n-k)!}$, we can calculate $$ \begin{align*} s_n &={n!\over0!n!}{n!\over1!(n-1)!}\dots{n!\over(n-1)!1!}{n!\over n!0!} \\ &={n!^{n+1}\over0!^21!^2\dots n!^2}. \end{align*} $$ And so $$ \begin{align*} {s_n\over s_{n-1}} &={n!^{n+1}\over0!^21!^2\dots n!^2} \Bigg/{(n-1)!^n\over0!^2 1!^2\dots(n-1)!^2} \\ &={n!^{n+1}\over 0!^2 1!^2\dots(n-1)!^2 n!^2} \cdot{0!^21!^2\dots{(n-1)!^2}\over(n-1)!^n} \\ &={n!\over n!^2}{n!^n\over(n-1)!^n}\\ &={n^n\over n!}. \end{align*} $$
Aha! This means that $$ \begin{align*} {s_{n+1}/s_n\over s_n/s_{n-1}} &={(n+1)^{n+1}/(n+1)!\over n^n/n!} \\ &={n!(n+1)^{n+1}\over n^n(n+1)!} \\ &={n+1\over n+1}\Bigl({n+1\over n}\Bigr)^n\\ &=\Bigl(1+{1\over n}\Bigr)^n\\ &\approx e, \end{align*} $$ so that $s_{n+1}\approx es_n^2/s_{n-1}$. We can now deduce the result inductively. Clearly $s_0={0\choose 0}=1=e^0$. Suppose inductively that $s_k\approx e^{1+\dots +k}$ for $k\le n$. Then $$ \begin{align*} s_{n+1} &\approx{es_n^2\over s_{n-1}} \\ &\approx{e(e^{1+\dots+n})^2\over e^{1+\dots+(n-1)}} \\ &={e(e^{1+\dots+n})(e^{1+\dots+n})\over e^{1+\dots+(n-1)}} \\ &={e e^ne^{1+\dots+n}} \\ &=e^{1+\dots+n+(n+1)}, \end{align*} $$ and we are done.
Brothers, Harlan J., “Math Bite: Finding $e$ in Pascal’s Triangle”, Mathematics Magazine 85(1) (2012), 51. https://doi.org/10.4169/math.mag.85.1.51
