Proposition 3

Let $(k_n)$ be a strictly increasing sequence of integers $\geqslant 0$ ($k_0 = 0$) : if the infinite product of general factor $x_n$ converges, and if one sets $u_n = \prod_{p=k_n}^{k_{n+1}-1} x_n$, the infinite product of general factor $u_n$ is convergent, and one has $\prod_{n=0}^{\infty} u_n = \prod_{n=0}^{\infty} x_n$.

Indeed, the sequence of partial products of the sequence $(u_n)$ is extracted from the sequence of partial products of the sequence $(x_n)$.

Finally, by the same reasoning as for commutative groups (III, p. 44), one sees that if, in a normed algebra $A$, a sequence $(x_n)$ is multipliable, the product of general factor $x_n$ is convergent, and one has $\prod_{n=0}^{\infty} x_n = \prod_{n \in \mathbf{N}} x_n$ (which is also written $\prod_{n=0}^{\infty} x_n$); the converse is of course false (cf. IX, p. 126, exerc. 7).

Exercises