Limit of $\prod_{k=0}^n {n \choose{k}} ^{\frac{1}{n(n+1)}}$

Question

What is the limit of $$\prod_{k=0}^n {n \choose{k}} ^{\frac{1}{n(n+1)}}?$$ It should be $\sqrt e$. I've tried several approaches, including Stirling's formula (which I'd rather avoid), though I came close, I can't prove it.

Answer 1

Use the fact that $ \prod_{k=0}^n {n \choose{k}} ^{\frac{1}{n(n+1)}} = e^{\frac{1}{n(n+1)}\ln(\prod_{k=0}^n{n \choose{k}})} $

By the symmetry of the pascal coefficient, it is enough to consider the midpoint, we will assume that $n$ is even.

$ {n \choose 0}{n \choose 1}{n\choose 2}\times ... \times{n\choose n-1}{n\choose n} = \biggr(\frac{[n][n(n-1)][n(n-1)(n-2)]\times...\times[n(n-1)(n-2)\times...\times(n-(\frac{n}{2} -1))]}{[0!][1!][2!][3!]\times...\times[\frac{n}{2}]!}\biggr)^2 $

The last equation is equal to

$n^{\frac{n}{2}}(n-1)^{\frac{n}{2} - 1}\times...\times(n - (\frac{n}{2} - 1))^{\frac{n}{2} - (\frac{n}{2} - 1)}$

If you will consider the limit, the highest exponent is the only term that matters

That term using the Gauss formula for the sum is

$\displaystyle\sum_{i = 0}^{n}\frac{n}{2} - i = \frac{1}{4}[ (n)(n+1) - 4n]$

Those terms are inside the logarithm, then you can see that

$\displaystyle\lim_{n \to \infty}\frac{(\ln(\prod_{k=0}^n{n \choose{k}})}{n(n+1)} = \lim_{n \to \infty}2\frac{\frac{1}{4}[ (n)(n+1) - 4n]}{n(n+1)} = \frac{1}{2}$

Maybe there is a shorter approach, but this was the first that came to my mind.