Bounds on $\binom{2n}{n}$

Question

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.

Answer 1

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.