Why is $\binom{2n}{n} \asymp \Theta \big(\frac{2^{2n}}{\sqrt{n}}\big)$? I saw this statement : $$\binom{2n}{n} \asymp \Theta \bigg(\frac{2^{2n}}{\sqrt{n}}\bigg) \asymp \Theta\bigg(\frac{4^n}{\sqrt{n}}\bigg)$$
How did we go from the first statement to the second? I tried Stirling's aproximation, but that didn't get me anywhere.
 A: Stirling's approximation gives
$$\binom{2n}{n} = \frac{(2n)!}{n!^2} \sim \frac{\sqrt{2\pi \cdot 2n} \left( \frac{2n}{e} \right)^{2n}}{\left[\sqrt{2\pi n} \left( \frac{n}{e} \right)^n \right]^2} = \frac{2\sqrt{\pi n} \cdot 2^{2n} \cdot \left( \frac{n}{e} \right)^{2n}}{2 \pi n \cdot \left( \frac{n}{e} \right)^{2n}} = \frac{1}{\sqrt{\pi n}} 2^{2n}$$
A: If we set:
$$ I_n = \int_{0}^{\pi/2}\sin(x)^n\,dx \tag{1}$$
integration by parts gives:
$$ \forall n\geq 2,\qquad I_n = \left(1-\frac{1}{n}\right) I_{n-2} \tag{2}$$
hence, by induction:
$$ I_{2n}=\frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n},\qquad I_{2n-1}=\frac{1}{2n}\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}\tag{3}$$
and since $\{I_n\}_{n\geq 0}$ is trivially a decreasing sequence from $(1)$, $I_{2n}\leq I_{2n-1}$ gives:
$$ \binom{2n}{n}\frac{1}{4^n}\leq\frac{1}{\sqrt{\pi n}}\tag{4} $$
that together with the consequences of $I_{2n}\geq I_{2n+1}$ proves:
$$ \binom{2n}{n}\frac{1}{4^n}=\Theta\!\left(n^{-1/2}\right)\tag{5} $$
as wanted. We avoided Stirling's approximation. 
If we want to avoid integrals, too, we may consider that:
$$ \frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{6} $$
hence:
$$\begin{align*}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2&=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\\&=\frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{7}\end{align*}$$
but the product
$$ \prod_{k=2}^{+\infty}\left(1+\frac{1}{4k(k-1)}\right) $$
is converging since the series
$$ \sum_{k\geq 2}\frac{1}{4k(k-1)}=\frac{1}{4}\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k}\right)=\frac{1}{4}$$
is converging, proving $(5)$, again.
