Bourbaki

Example

Tag 0BTT, permanent. Cite it as [Bourbaki, Tag 0BTT].

General Topology, chapter IV, § 7. Infinite sums and products of real numbers

If we take u2n−1=−1/nu_{2n-1} = -1/n and u2n=1/nu_{2n} = 1/n for n≥2n \geq 2, the product (1+un)(1 + u_n) is not absolutely convergent, since the series (un)(u_n) is not absolutely convergent; but since

∏p=3n(1+up)=∏p=2n(1−1p2), \prod_{p=3}^{n} (1 + u_p) = \prod_{p=2}^{n} \left(1 - \frac{1}{p^2}\right),
and
∏p=32n+1(1+up)=(1−1n+1)∏p=2n(1−1p2), \prod_{p=3}^{2n+1} (1 + u_p) = \left(1 - \frac{1}{n+1}\right) \prod_{p=2}^{n} \left(1 - \frac{1}{p^2}\right),
it follows from Theorem 4 that the product is convergent and that its value is
∏n=2∞(1−1n2). \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right).

Moreover, it should be observed that the convergence of the series whose general term is unu_n is neither necessary nor sufficient for the convergence of the product whose general factor is 1+un1 + u_n (see Exercises 21 and 22).

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.