Bourbaki

Example

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

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

Consider the hypergeometric series, with general term

un=α(α+1)…(α+n−1) β(β+1)…(β+n−1)1.2…n.γ(γ+1)…(γ+n−1) u_n = \frac{\alpha(\alpha+1)\ldots(\alpha+n-1)\,\beta(\beta+1)\ldots(\beta+n-1)}{1.2\ldots n.\gamma(\gamma+1)\ldots(\gamma+n-1)}

where α,β,γ\alpha, \beta, \gamma are arbitrary real numbers, not integers ≤0\leq 0: it is clear that unu_n is >0> 0 from a certain stage, or <0< 0 from a certain stage on. One has

un+1un=(α+n)(β+n)(n+1)(γ+n)=(1+α+βn+αβn2)(1+γ+1n+γn2)−1=1+α+β−γ−1n+αβ−(α+β)(γ+1)+γ2+γ+1n2+o(1n2). \begin{align*} \frac{u_{n+1}}{u_n} &= \frac{(\alpha+n)(\beta+n)}{(n+1)(\gamma+n)} \\ &= \left(1 + \frac{\alpha+\beta}{n} + \frac{\alpha\beta}{n^2}\right) \left(1 + \frac{\gamma+1}{n} + \frac{\gamma}{n^2}\right)^{-1} \\ &= 1 + \frac{\alpha+\beta-\gamma-1}{n} \\ &\quad + \frac{\alpha\beta-(\alpha+\beta)(\gamma+1)+\gamma^2+\gamma+1}{n^2} + o\left(\frac{1}{n^2}\right). \end{align*}

Raabe's test shows that the series converges for α+β<γ\alpha + \beta < \gamma, and has infinite sum for α+β>γ\alpha + \beta > \gamma; when α+β=γ\alpha + \beta = \gamma the series also has an infinite sum, as is shown by prop. 8.

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

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Transcribed from the printed volume. Not translated.