↑ stuart clark:
Basically, the problem can be solved easily...
Skrytý text:First, we solve an auxiliary corresponding limit
The value
L can be calculated by straightforward utilization of the Stolz-Cesàro theorem:
![$
L
&=\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1}\ln{n+1\choose k}-\sum_{k=1}^{n}\ln{n\choose k}}{2(n+1)}\\[1em]
&=\lim_{n\to\infty}\frac{\sum_{k=1}^{n}\ln\frac{{n+1\choose k}}{{n\choose k}}}{2(n+1)}\\[1em]
&=\lim_{n\to\infty}\frac{\ln\prod_{k=1}^{n}\frac{n+1}{n+1-k}}{2(n+1)}\\[1em]
&=\lim_{n\to\infty}\frac{n\,\ln(n+1)-\ln (n!)}{2(n+1)}.
$](/mathtex/22/222fea5d028f811c0d46b11751fbb49f.gif "kopírovat do textarea")
Now, using the Stirling approximation of the factorial
one gets for the value
L![$
L
&=\lim_{n\to\infty}\frac{n\,\ln(n+1)-\bigl [n\,\ln(n)-n+\mathcal{O}(\ln(n))\bigr ]}{2(n+1)}\\[1em]
&=\lim_{n\to\infty}\frac{\ln(n+1)-\ln(n)+1+\mathcal{O}\left (\frac{\ln (n)}{n}\right )}{2}\\[1em]
&=\frac{1}{2}.
$](/mathtex/41/41fefeb4d95620bf4f47e3882c76e397.gif "kopírovat do textarea")
Finally, the continuity of the exponential function implies that the limit in the OP is equal to exp (
L), i.e.,
![$
\boldsymbol{\lim_{n\to\infty}\sqrt[n^2+n]{\prod_{k=0}^{n}{n\choose k}}=\sqrt{\textnormal{\bfseries e}}},
$](/mathtex/d6/d64bdf02e04b54967fef8b3c88e78582.gif "kopírovat do textarea")
where
e denotes the Euler number.
![$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n^2+n]{\binom{n}{0}\binom{n}{1}\binom{n}{2}\cdots \cdots \binom{n}{n}}$](/mathtex/21/217f1a8db5eefae57bb1007dff4a2ec5.gif "kopírovat do textarea")
