Corollary

In a complete normed space E, every absolutely convergent series is commutatively convergent.

The converse of Proposition 11 is in general false.

Consider for example the space $\mathscr{B}(\mathbf{N}; \mathbf{R})$ of bounded sequences $x = (x_n)_{n \in \mathbf{N}}$ of real numbers, with the norm $||x|| = \sup_n |x_n|$. Let $x_m$ be the sequence $(x_{mn})_{n \in \mathbf{N}}$ such that $x_{mn} = 0$ if $m \neq n$ and $x_{mm} = 1/m$ for $m \geq 1$. It is immediately verified that the sequence $(x_m)_{m \in \mathbf{N}}$ is summable in $\mathscr{B}(\mathbf{N}; \mathbf{R})$ and that its sum is the element $y = (y_n)$ such that $y_0 = 0$ and $y_n = 1/n$ if $n \geq 1$; but since $||x_m|| = 1/m$, the sequence of norms of the $x_m$ is not summable in $\mathbf{R}$.

However, we have seen in Chapter VII, § 3, no. 1, that every summable family in $\mathbf{R}^n$ is absolutely summable.