Corollary

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

There is only a finite number of integers $n$ such that $|u_n| > 1$. Let $J = [a, m]$ be an interval of $\mathbf{N}$ containing all these integers. The product of the sequence $(e + u_n)$ is the product of $p_J$ and the element $\prod_{n \notin J} (e - u_n)$, all of whose factors are units (Chapter IX, § 3, no. 7, Corollary to Proposition 12), and is therefore itself a unit; since $p_J$ is the product of a finite number of non-zero divisors, it is not a zero divisor, and hence $\prod_{n \in \mathbf{N}} (e + u_n)$ is not a zero divisor.

The sufficient condition for multipliability given by Theorem 2 is not in general necessary (cf. Exercise 6). However, it is necessary in the important case where $A$ is an algebra of finite rank over the field $\mathbf{R}$ (i.e., $A$ is finite-dimensional as a vector space over $\mathbf{R}$); in particular this is the case if $A$ is the division ring of quaternions $\mathbf{H}$, or a matrix algebra $\mathbf{M}_n(\mathbf{R})$: