Stirling approximation of $\binom{2n}{n}$ How do I approximate ${2n \choose n}$ using Stirling's formula (which approximates ${n!}$ with $\pi$ and $e$? 
 A: We can almost avoid Stirling's approximation in providing tight bounds for the central binomial coefficient. The key ingredient is the following identity:
$$ \frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}. $$
If we consider the square of the general term of the last product, we get something that behaves like $\left(1-\frac{1}{k}\right)$ for large $k$s, but the product of such terms leads to a telescopic product. That is the motivation behind the following manipulation:
$$ \left(\frac{1}{4^n}\binom{2n}{n}\right)^2=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4n}\prod_{k=1}^{n-1}\left(1-\frac{1}{(2k+1)^2}\right)^{-1}.\tag{2}$$
By the Weierstrass product for the cosine function/ the Wallis product we have:
$$\prod_{k=1}^{+\infty}\left(1-\frac{1}{(2k+1)^2}\right)^{-1}=\frac{4}{\pi},\tag{3}$$
hence it follows that:
$$ \left(\frac{1}{4^n}\binom{2n}{n}\right)^2=\frac{1}{\pi n}\cdot\prod_{k\geq n}\left(1-\frac{1}{(2k+1)^2}\right)=\frac{1}{\pi n}\cdot\prod_{k\geq n}\left(1+\frac{1}{2k(2k+2)}\right)^{-1}\tag{4} $$
and the last product is clearly $1+O\left(\frac{1}{n}\right)$. By exploiting the inequality $\frac{1}{1+x}\geq e^{-x}$ over $[0,1]$ and a telescopic sum, from $(4)$ we get:

$$ \frac{1}{\sqrt{\pi n}}\geq \frac{1}{4^n}\binom{2n}{n}\geq \frac{1}{\sqrt{\pi n}}\,\exp\left(-\frac{1}{8n}\right).\tag{5}$$

A: Exactly as one would expect. Since $$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$$ you can use Stirling's approximation as follows:
$$
(2n)! \operatorname*{\sim}_{n\to\infty} 2\sqrt{\pi n}\frac{(2n)^{2n}}{e^{2n}}
$$
$$
(n!)^2 \operatorname*{\sim}_{n\to\infty} 2\pi n\frac{n^{2n}}{e^{2n}}
$$
so that
$$
\binom{2n}{n} \operatorname*{\sim}_{n\to\infty} 
\frac{2\sqrt{\pi n}}{2\pi n}\cdot  \frac{e^{2n}}{n^{2n}} \cdot \frac{(2n)^{2n}}{e^{2n}} = 
\frac{2^{2n}}{\sqrt{\pi n}}
$$
A: Obviously:
$$ \binom{2n}{n}=\frac{(2n)!}{n!n!}$$
I will apply the approximation obtained by H. Robbins in his A Remark on Stirling’s Formula (American Mathematical
Monthly 62, 26-29, 1955). Define two functions:
$$ f(x)=\sqrt{2\pi}x^{x+\frac{1}{2}}e^{-x}e^{\frac{1}{12x}}$$
$$ g(x)=\sqrt{2\pi}x^{x+\frac{1}{2}}e^{-x}e^{\frac{1}{12x+1}}$$
From the paper I've mentioned we see that $$g(n)<n!<f(n).$$ Thus:
$$\frac{g(2n)}{f(n)f(n)}<\binom{2n}{n}=\frac{(2n)!}{n!n!}<\frac{f(2n)}{g(n)g(n)}$$
After substituting with the formulas we see that:
$$\frac{\sqrt{2\pi}(2n)^{2n+\frac{1}{2}}e^{-2n}e^{\frac{1}{24n+1}}}{\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}e^{\frac{1}{12n}}}<\binom{2n}{n}$$
I think that the simplification and the upper bound you can obtain by yourself.
