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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.

Trevor Gunn
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barmanthewise
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1 Answers1

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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.

Dr Richard Clare
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  • How can I use L'Hopital rule on the logarithm of a binamial coefficient, a function only defined for natural numbers? – barmanthewise Jul 30 '17 at 23:56
  • You can always see this as a function of x because they will converge to the same limit. This sequence of function converges to the same value as the real function as n goes to infinity. However if you prefer compute the limit directly and consider the telescopic sum of the logarithm. – Dr Richard Clare Jul 31 '17 at 00:04
  • Remember that you can generalize the factorial for the real numbers with the gamma function! – Dr Richard Clare Jul 31 '17 at 00:06
  • Verify the answer I make an edit and add some additional details. – Dr Richard Clare Jul 31 '17 at 00:54