Bourbaki

Example 1

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

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

Let aa be a finite real number >0> 0, and consider the sequence (ann!)\left( \frac{a^n}{n!} \right); let n0n_0 be the smallest integer such that a<n0a < n_0. Then for each n≥n0n \geq n_0 we have

ann!≤an0n0!⋅(an0)n−n0; \frac{a^n}{n!} \leq \frac{a^{n_0}}{n_0!} \cdot \left( \frac{a}{n_0} \right)^{n-n_0};
since q=an0<1q = \frac{a}{n_0} < 1, the sequence (qn−n0)(q^{n-n_0}) is summable, and therefore so is the sequence (ann!)\left( \frac{a^n}{n!} \right).

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.