Let ϱ1,...,ϱn be real numbers >0 and let U⊂Cn be the polydisc product of the open discs with center 0 and radius ϱi. Let
∑c(k1,...,kn)X1k1⋯Xnkn∈A[[X1,...,Xn]]
(k1,...,kn)∈Nn
be a formal series with coefficients in A. Suppose that this series converges in U, and let f denote the holomorphic function in U of which it is the sum.
Let a=(a1,...,an)∈An be such that ϱ(ai)<ϱi for 1⩽i⩽n. Then Spn(a)⊂U, the family (c(k1,...,kn)a1k1⋯ankn) of elements of A is absolutely summable, and
Θa(f)=∑c(k1,...,kn)a1k1⋯ankn
(k1,...,kn)∈Nn
For every character χ of A and every integer i such that 1⩽i⩽n, one has ∣χ(ai)∣⩽ϱ(ai)<ϱi, hence Spn(a)⊂U by definition of the simultaneous spectrum. Let z1,...,zn be the restrictions to U of the coordinate functions on Cn. Then the family (c(k1,...,kn)z1k1⋯znkn) is summable in O(U;A) and has sum f. In view of lemma 10 and the continuity of the mapping ΘaU, the family (c(k1,...,kn)a1k1...ankn) is therefore summable in A and has sum Θa(f). For 1⩽i⩽n, let λi be a real number such that ϱ(ai)<λi<ϱi. There exists Mi<+∞ such that ∥aik∥⩽Miλik for every integer k⩾0. One then has
∑∥c(k1,...,kn)∥∥a1k1⋯ankn∥⩽
(k1,...,kn)∈Nn
M1⋯Mn∑∥c(k1,...,kn)∥λ1k1⋯λnkn
(k1,...,kn)∈Nn which is finite by hypothesis, hence the family (c(k1,...,kn)a1k1⋯ankn) is absolutely summable.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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