Approximation to 2n Choose n

Theorem

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

Proof
From Approximation to $\dbinom {x + y} y$:


 * $\ds \lim_{x, y \mathop \to \infty} \dbinom {x + y} y = \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 x + \frac 1 y} } \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y$

Thus: