Proposition 9

The series defined by the sequence $(x_n)$ is commutatively convergent if and only if the sequence $(x_n)$ is summable; and then, for each permutation $\sigma$ of $\mathbf{N}$, we have

If the sequence is summable, then clearly the series is commutatively convergent. To prove the reverse implication we shall argue by reductio ad absurdum and suppose that the series $(x_n)$ is commutatively convergent but that the sequence $(x_n)$ is not summable. The image of the filter $\Phi$ under the mapping $H \to s_H$ therefore cannot be a Cauchy filter base in $G$, otherwise this filter base would converge, since by hypothesis it has a cluster point (Chapter II, § 3, no. 2, Proposition 5, Corollary 2). Hence there is a neighbourhood $V$ of 0 in $G$ such that, for each finite subset $J$ of $\mathbf{N}$, there is a finite subset $H$ of $\mathbf{N}$ which does not meet $J$ and is such that $\sum_{n \in H} x_n \notin V$. Hence, by induction, we can define a partition of $\mathbf{N}$ into finite subsets $H_k$ ($k \in \mathbf{N}$) such that $\sum_{n \in H_k} x_n \notin V$ for an infinite number of indices $k$. Clearly there is a permutation $\sigma$ of $\mathbf{N}$ such that for each $k$, the values of $n$ for which $\sigma(n) \in H_k$ are consecutive. If $\sigma$ is such a permutation, then the series whose general term is $x_{\sigma(n)}$ cannot be convergent, and we have a contradiction.

If the group $G$ is written multiplicatively, the infinite product defined by a sequence $(x_n)$ of points of $G$ (or infinite product whose general term is $x_n$, or even product $x_n$ if there is no possibility of confusion) is defined to be the pair formed by the sequence $(x_n)$ and the sequence of partial products

The infinite product is said to converge if the sequence $(p_n)$ converges, and the limit of this sequence is denoted by $\prod_{n=0}^{\infty} x_n$ (or $\prod_{n=0}^{\infty} x_n$ by abuse of notation). We leave to the reader the task of transcribing into multiplicative notation the properties of series which we have established.