Approximation to x+y Choose y

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Theorem

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


Proof

It can be assumed that both $x$ and $y$ are integers.


\(\ds \dbinom {x + y} y\) \(=\) \(\ds \dfrac {\paren {x + y}!} {x! \, y!}\)
\(\ds \leadsto \ \ \) \(\ds \ds \lim_{x, y \mathop \to \infty} \dbinom {x + y} y\) \(=\) \(\ds \dfrac {\sqrt {2 \pi \paren {x + y} } \paren {\dfrac {x + y} e}^{\paren {x + y} } } {\sqrt {2 \pi x} \paren {\dfrac x e}^x \, \sqrt {2 \pi y} \paren {\dfrac y e}^y}\) Stirling's Formula
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {2 \pi} } \dfrac {\paren {\frac 1 e}^{x + y} } {\paren {\frac 1 e}^x \paren {\frac 1 e}^y} \sqrt {\dfrac {x + y} {x y} } \dfrac {\paren {x + y}^x \paren {x + y}^y} {x^x \, y^y}\) rearrangement
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {2 \pi} } \sqrt {\frac 1 x + \frac 1 y} \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y\) simplifying
\(\ds \) \(=\) \(\ds \sqrt {\dfrac 1 {2 \pi} \paren {\frac 1 x + \frac 1 y} } \paren {1 + \dfrac y x}^x \paren {1 + \dfrac x y}^y\)

$\blacksquare$


Sources

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