# Exercise C.1.13

$\star$ Use Stirling's approximation to prove that

$$\binom{2n}{n} = \frac{2^{2n}}{\sqrt{\pi n}}(1 + \mathcal{O}(1/n))$$

So:

\begin{align} \binom{2n}{n} &= \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2} \\ &= \frac{\sqrt{2 \pi 2 n}\big(\frac{2n}{e}\big)^{2n}\Big(1 + \Theta(\frac{1}{n})\Big)} {2 \pi n \big(\frac{n}{e}\big)^{2n}\Big(1 + \Theta(\frac{1}{n})\Big)^2} \\ &= \frac{1}{\sqrt{\pi n}} \frac{2^{2n}n^{2n}}{n^{2n}}(1 + \mathcal{O}(1/n)) \\ &= \frac{2^{2n}}{\sqrt{\pi n}}(1 + \mathcal{O}(1/n)) \end{align}

There is a little hand-waving at the end, but it is good enough.