Theorem 2

Let $A$ be a complete normed algebra. If $(u_n)$ is an absolutely convergent series of elements of $A$, then the sequence $(e + u_n)$ is multipliable in $A$; and if all the elements $e + u_n$ are units in $A$, then so is $\prod_{n \in \mathbf{N}} (e + u_n)$.

Let us apply the criterion of Theorem 1. For every finite subset $L$ of $\mathbf{N}$, we have $p_L - e = \prod_{n \in L} (e + u_n) - e = \sum_M \left( \prod_{n \in M} u_n \right)$, where $M$ runs through the set of all non-empty subsets of $L$ (linearly ordered by the induced ordering). Since $\prod_{n \in M} u_n \leq \prod_{n \in M} |u_n|$, we may write

Now since the series whose general term is $|u_n|$ is convergent by hypothesis, the sequence $1 - |u_n|$ is multipliable in $\mathbf{R}_+^*$ (Chapter IV, § 7, no. 4, Theorem 4). Hence for each $\varepsilon > 0$ there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $L$ of $\mathbf{N}$ which does not meet $J_0$, we have $\left| \prod_{n \in L} (1 + ||u_n||) - 1 \right| \leq \varepsilon$; hence the result.