Approximation to 2n Choose n

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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:


\(\ds \lim_{n \mathop \to \infty} \dbinom {2 n} n\) \(=\) \(\ds \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 n + \frac 1 n} } \paren {1 + \dfrac n n}^n \paren {1 + \dfrac n n}^n\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac 1 {2 \pi} \paren {\frac 2 n} } \paren {1 + 1}^n \paren {1 + 1}^n\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac 1 {n \pi} } 2^{2 n}\)
\(\ds \) \(=\) \(\ds \dfrac {4^n} {\sqrt {n \pi} }\)

$\blacksquare$


Sources