Corollary

If the series of general term $u_n$ is absolutely convergent, and if none of the elements $e + u_n$ is a divisor of 0 in $A$, the product $\prod_{n \in \mathbf{N}} (e + u_n)$ is not a divisor of 0 in $A$.

Indeed, there are only a finite number of integers $n$ such that $\|u_n\| > 1$. Let $J = \{0, m\}$ be an interval of $\mathbf{N}$ containing all these integers. The product of the sequence $(\mathbf{e} + u_n)$ is the product of $p_J$ and of the element $\prod_{n > m} (\mathbf{e} + u_n)$, all of whose factors are invertible (IX, p. 39, corollary), and which is consequently itself invertible; since $p_J$ is the product of a finite number of elements which are not divisors of 0, it is not a divisor of 0, and the same is true of $\prod_{n \in \mathbf{N}} (\mathbf{e} + u_n)$.

The sufficient condition for multipliability given in th. 2 is not necessary in general (cf. IX, p. 126, exerc. 6). It is, however, necessary in the important case where $A$ is an algebra of finite rank over the field $\mathbf{R}$ (in particular when $A$ is the field of quaternions $\mathbf{K}$, or a matrix algebra $\mathbf{M}_n(\mathbf{R})$):