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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $46$