Theorem 2

Let $A$ be a complete normed algebra; if $(u_n)$ is an absolutely convergent series of points of $A$, the sequence $(e + u_n)$ is multipliable in $A$; moreover, if all the elements $e + u_n$ are invertible in $A$, the same is true of $\prod_{n \in \mathbf{N}} (e + u_n)$.

Apply the criterion of $IX$, p. 79, th. 1; for every finite subset $L$ of $N$, we have $p_L = e - \prod_{n \in L} (e + u_n) - e = \sum_M \left( \prod_{n \in M} u_n \right)$, $M$ ranging over the set of the non-empty subsets of $L$ (totally ordered by the induced order). Since $\left| \prod_{n \in M} u_n \right| \leq \prod_{n \in M} \| u_n \|$, one can write

Now, since the series of general term $\| u_n \|$ is convergent by assumption, the sequence $(1 + \| u_n \|)$ is multipliable in $\mathbf{R}_+^*$ ($IV$, p. 35, th. 4); for every $\varepsilon > 0$, there therefore exists a finite subset $J_0$ of $N$ such that, for every finite subset $L$ of $N$ not meeting $J_0$, we have $\left| \prod_{n \in L} (1 + \| u_n \|) - 1 \right| \leq \varepsilon$; whence the theorem.