If (ai)1≤i≤p is a finite sequence of numbers >0, then
i=1∏p(1+ai)≥1+i=1∑pai. (ii)
If also ai<1 for each i,
then i=1∏p(1−ai)≥1−i=1∑pai.
These inequalities are clear if p=1, and are proved by induction on p. If
i=1∏p−1(1+ai)≥1+i=1∑p−1ai, we have
i=1∏p(1+ai)≥(1+ap)(1+i=1∑p−1ai)=1+i=1∑pai+ap⋅i=1∑p−1ai≥1+i=1∑pai. Again, if
i=1∏p−1(1−ai)≥1−i=1∑p−1ai, we have
i=1∏p(1−ai)≥(1−ap)(1−i=1∑p−1ai)=1−i=1∑pai+ap⋅i=1∑p−1ai≥1−i=1∑pai.
Having proved the lemma, we note that if the family (1+ui) is multipliable then so are the families (1+ui+) and (1−ui−), since R∗ is a complete group (Chapter III, § 5, no. 3, Proposition 2); and conversely, if the families (1+ui+) and (1−ui−) are multipliable, then so is (1+ui) (Chapter III, § 5, no. 3, Proposition 3). We need therefore consider only the cases in which all the ui are ≥0 and in which they are all ≥0.
Suppose first that ui≥0 for all i. If the family (1+ui) is multipliable, then for each ε>0 there is a finite subset J of the index set in I such that, for each finite subset H of I disjoint from J, we have 1≤∏i∈H(1+ui)≤1+ε; by (2) it follows that ∑i∈Hui≤ε, which shows that (ui) is summable in R by virtue of Cauchy’s criterion (Chapter III, § 5, no. 2, Theorem 1).
Conversely, suppose that (ui) is summable in R. For each ε such that 0<ε<1, there is a finite subset J of I such that, for each finite subset H of I disjoint from J, we have 0≤∑i∈Hui≤ε. By (3), we have therefore ∏i∈H(1−ui)≥1−ε; but since 1+u≤1−u1 for every number u such that 0≤u<1, it follows that
1≤i∈H∏(1+ui)≤1−ε1,
and this shows that (1+ui) is multipliable (Cauchy’s criterion).
The proof is similar when all the ui are ≤0. To show that (ui) is summable if (1+ui) is multipliable, use formula (2) and the inequality 1−u≤1+u1 (0≤u<1); to show that (1+ui) is multipliable if (ui) is summable, use formula (3).
In Chapter V (§ 4), the topological study of the group R∗ will enable us to give another criterion for multipliability of a family in R∗ with the help of the logarithmic function; in a later volume we shall establish the equivalence of this criterion with the one above, using the differential properties of the logarithm.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.