Bourbaki

Example

Tag 08G7, permanent. Cite it as [Bourbaki, Tag 08G7].

Functions of a Real Variable, chapter V, § 4. APPLICATION TO SERIES WITH POSITIVE TERMS

Stirling’s formula. Let us seek an asymptotic expansion for n!n!; this leads us to expand sn=∑p=1nlog⁡ps_n = \sum_{p=1}^{n} \log p, and then exp⁡(sn)\exp(s_n). The method of n 2 gives successively

sn=∑p=1nlog⁡p∼∫1nlog⁡t dt=nlog⁡n−n+1 s_n = \sum_{p=1}^{n} \log p \sim \int_{1}^{n} \log t \, dt = n \log n - n + 1

then

log⁡n−∫n−1nlog⁡t dt=log⁡n−(nlog⁡n−(n−1)log⁡(n−1)−1)∼12n \log n - \int_{n-1}^{n} \log t \, dt = \log n - (n \log n - (n-1) \log(n-1) - 1) \sim \frac{1}{2n}
whence
sn=nlog⁡n−n+12log⁡n+o(log⁡n). s_n = n \log n - n + \frac{1}{2} \log n + o(\log n).
Then
log⁡n−∫n−1nlog⁡t dt−12(log⁡n−log⁡(n−1))∼−112n2 \log n - \int_{n-1}^{n} \log t \, dt - \frac{1}{2} (\log n - \log(n-1)) \sim -\frac{1}{12n^2}
whence
sn=nlog⁡n−n+12log⁡n+k+112n+o1(1n)(k constant) s_n = n \log n - n + \frac{1}{2} \log n + k + \frac{1}{12n} + o_1 \left( \frac{1}{n} \right) \quad (k \text{ constant})
and one finally deduces (V, p. 226)
n!=ek nn+1/2 e−n(1+112n+o2(1n)).(5) n! = e^k \, n^{n+1/2} \, e^{-n} \left( 1 + \frac{1}{12n} + o_2 \left( \frac{1}{n} \right) \right). \tag{5}
We shall show in VII, p. 322, that ek=2πe^k = \sqrt{2\pi}. The formula (5) (with this value of kk) is called Stirling's formula. In the same way, for every real number aa not an integer >0> 0, one shows that
(a+1)(a+2)…(a+n)∼K(a) nn+a+12 e−n.(6) (a+1)(a+2)\ldots(a+n) \sim K(a) \, n^{n+a+\frac{1}{2}} \, e^{-n}. \tag{6}
We shall determine the function K(a)K(a) too (VII, p. 18). From formulae (5) and (6) one derives in particular that
(an)∼(−1)nφ(a) n−a−1(7) \binom{a}{n} \sim (-1)^n \varphi(a) \, n^{-a-1} \tag{7}
for every real number aa not an integer >0> 0, where φ(a)\varphi(a) is a function of aa which will be specified in VII, p. 322.

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

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Cites

Transcribed from the printed volume. Not translated.