Approximation to 2n Choose n

Theorem

 * $\displaystyle \lim_{n \mathop \to \infty} \dbinom {2 n} n = \dfrac {4^n} {\sqrt {n \pi} }$

Proof
From Approximation to x+y Choose y:


 * $\displaystyle \lim_{x, y \mathop \to \infty} \dbinom {x + y} y = \sqrt {\dfrac 1 {2 \pi} \left({\frac 1 x + \frac 1 y}\right)} \left({1 + \dfrac y x}\right)^x \left({1 + \dfrac x y}\right)^y$

Thus: