Examples

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

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

1) For each number $q$ such that $0 \leqslant q < 1$ , the sequence $(q^n)$ ("geometric progression of ratio $q$ ") is summable in $\mathbf{R}$ , since

s_n = \frac{1 - q^{n+1}}{1 - q} \leqslant \frac{1}{1 - q}

the sum of this sequence is

\lim_{n \to \infty} s_n = \frac{1}{1 - q}

2) Let $a$ and $b$ be two numbers such that $0 \leqslant a < 1$ and $0 \leqslant b < 1$ ; then the family $(a^m b^n)_{(m, n) \in \mathbf{N} \times \mathbf{N}}$ is summable in $\mathbf{R}$ . For each finite subset of $\mathbf{N} \times \mathbf{N}$ is contained in a subset of the form $[0, p] \times [0, p]$ , and we have

\sum_{m=0}^p \sum_{n=0}^p a^m b^n = \left( \sum_{m=0}^p a^m \right) \left( \sum_{n=0}^p b^n \right) = \frac{1 - a^{p+1}}{1 - a} \cdot \frac{1 - b^{p+1}}{1 - b} \leqslant \frac{1}{(1 - a)(1 - b)}.

3) For each integer $p > 1$ , the sequence $(n^{-p})$ ( $n > 0$ ) is summable, since

s_{2^n + 1} - s_{2^n} = \sum_{k=1}^{2^n} (2^n + k)^{-p} < 2^n \cdot (2^n)^{-p}

and therefore, adding these inequalities together,

s_{2^n} < \frac{1}{1 - 2^{1-p}}.

4) The sequence $(1/n)$ ( $n > 0$ ) is not summable in $\mathbf{R}$ , since

s_{2^{n+1}} - s_{2^n} = \sum_{k=1}^{2^n} \frac{1}{2^n + k} > \frac{2^n}{2^{n+1}} = \frac{1}{2}

and therefore, adding these inequalities together,

s_{2^n} > n/2

so that the criterion of Theorem 1 is not satisfied.

5) Let $(I_n)$ be a sequence of mutually disjoint non-empty open intervals, all contained in an interval of finite length $l$ . The sum of the lengths of a finite number of intervals of this family is $\leq l$ ( $§ 1$ , no. 5) and therefore the family of lengths of the $I_n$ is summable in $\mathbf{R}$ , and its sum is $\leq l$ .

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

Cited by

Nothing.

Cites

Theorem 1