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What is the nature of $\prod_{k=1}^{k=n}\binom{n}{k}$ as a function of n?

How fast does it grow in comparison with $2^n$, etc?

My intuition is that:

$$\prod_{k=1}^{k=n}\binom{n}{k}$$

Grows much faster than

$$2^n$$

Simply because the summation is $2^n$. How fast does it grow in comparison, though?

Martin Sleziak
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jcao219
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  • Related: http://math.stackexchange.com/questions/1571739/what-is-the-closed-form-approximation-of-the-asymptotic-growth-rate-of-the-super/1572440#1572440 – tired Nov 21 '16 at 11:21
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    Essentially because your product can be written as $\text{hyperfactorial}/\text{superfactorial}$ the asymptotic relation given by @Brian can be proved with help of the above link – tired Nov 21 '16 at 11:27
  • Thanks @tired, I have never heard of hyperfactorial before. – jcao219 Nov 21 '16 at 12:52

2 Answers2

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This sequence is OEIS A001142. The entry gives the asymptotic approximation

$$\frac{A^2e^{\frac{n^2}2+n-\frac{1}{12}}}{n^{\frac{n}2+\frac{1}2}(2\pi)^{\frac{n+1}2}}\;,$$

where $A\approx 1.282427129$ is the Glaisher-Kinkelin constant.

Brian M. Scott
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Hint

If you only look at the "middle term" $\binom{2n}{n}$ for the even case and you use Stirling's approximation, you'll find that just this term is growing in the range of $$\frac{2^{2n}}{\sqrt{ \pi n}}$$

mathcounterexamples.net
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  • Thank you for helping me understand and evaluate the middle term! I always wondered how to do it. – jcao219 Nov 21 '16 at 10:00