Corollary

Tag 08JG, permanent. Cite it as [Bourbaki, Tag 08JG].

Along the filter $\mathfrak{F}$ one has

\Gamma(z) \sim \sqrt{2\pi} \exp(z \log z - z - \frac{1}{2} \log z).

In particular, for $x$ real tending to $+\infty$ the formula (20) can be written

\Gamma(x) \sim \sqrt{2\pi} \, x^{x-1/2} e^{-x},

whence, as the integer $n$ tends to $+\infty$ ,

n! \sim \sqrt{2\pi} \, n^{n+1/2} \, e^{-n}

(cf. V, p. 244).

One can deduce numerous formulae from this. For example, for every complex number $\alpha$ and every integer $n$ one has, as $n$ tends to $+\infty$ ,

\frac{\Gamma(n+\alpha)}{\Gamma(n)} \sim n^{\alpha} \quad (= e^{\alpha \log n}).

Similarly, for every complex number $a$ not an integer $\leqslant 0$ one has

a(a+1)(a+2)\ldots(a+n) = \frac{\Gamma(n+a+1)}{\Gamma(a)} \sim \frac{\sqrt{2\pi}}{\Gamma(a)} \, n^{n+a+1/2} \, e^n

and for every complex number $a$ not an integer $\geqslant 0$

\binom{a}{n} = \frac{(-1)^n}{\Gamma(-a)} \frac{\Gamma(n-a)}{\Gamma(n+1)} \sim \frac{(-1)^n}{\Gamma(-a)} \, n^{-a-1}.

Finally, for every real constant $k > 1$ one has

\binom{kn}{n} = \frac{\Gamma(kn+1)}{\Gamma(n+1) \Gamma((k-1)n+1)} \sim \sqrt{\frac{k}{2\pi(k-1)n}} \left( \frac{k^k}{(k-1)^{k-1}} \right)^n.

The same argument leads to the following analogous proposition:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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