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The following bounds can be found in the German wikipedia page of Central Binomial Coefficient $$\tag{$\star$}\label{star} \frac{1}{2} \frac{4^n}{\sqrt{\pi n}}<\binom{2n}{n}< \frac{4^n}{\sqrt{\pi n}}, \quad n\ge 1. $$ However, no proof nor reference is provided. So my question:

Is there a simple way to derive the bounds in \eqref{star}?

References are also very welcome.

Ludwig
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    The bounds remind me very much of Stirling's formula. – user159517 Oct 31 '18 at 22:55
  • I think you can use Wallis's product to prove this. Maybe this helpful: https://en.wikipedia.org/wiki/Wallis_product. – Batominovski Oct 31 '18 at 22:57
  • Are you interested in this exact bound or in any estimate on $\binom{2n}{n}$? Because there are very simple bounds that are off by a factor of $n$ from each other, and a similar inequality to yours but with slightly different constants has a proof using Chebyshev's inequality. But $\frac{1}{\sqrt \pi}$ is the true constant in the limit, and you need something as powerful as Stirling's formula to get there. – Misha Lavrov Oct 31 '18 at 23:38
  • @MishaLavrov: I was interested in this special bound, because it looked very simple and neat to me. Anyways, thanks for your the additional info! – Ludwig Oct 31 '18 at 23:41

1 Answers1

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By (one of the formulations of) Stirling approximation formula, you have $n!=(2\pi n)^{1/2}(n/e)^{n}e^{\theta(n)}$, with $\frac{1}{12n+1}\le \theta(n) \le \frac{1}{12n}$. Therefore, you can compute

$$(2n)!=2^{2n+1}(\pi n)^{1/2}(n/e)^{2n}e^{\theta(2n)}=2^{2n}\frac{n!^2}{(\pi n)^{1/2}} e^{\theta(2n)-2\theta(n)},$$

that is

$$ \binom{2n}{n}= \frac{2^{2n}}{(\pi n)^{1/2}} e^{\theta(2n)-2\theta(n)}.$$

Now, noting that

$\theta(2n)-2\theta(n)\le\frac{1}{24n}-\frac{2}{12n+1}<0$

and

$\theta(2n)-2\theta(n)\ge \frac{1}{24n+1}-\frac{2}{12n}\ge -\frac{1}{6}>-\log 2$,

you can conclude.

Maurizio Moreschi
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