Let
- $I\ne\emptyset$ be a set
- $E$ be a normed $\mathbb R$-vector space
$(x_i)_{i\in I}\subseteq E$ is called summable with sum $x\in E$ $:\Leftrightarrow$ $$\forall\varepsilon>0:\exists J\subseteq I\text{ with }|J|<\infty:\forall K\subseteq I\text{ with }|K|<\infty\text{ and }J\subseteq K:\left\|x-\sum_{k\in K}x_k\right\|_E<\varepsilon\;.\tag1$$ If $I=\mathbb N^2$, then $\sum_{m,\:n=1}^\infty x_{(m,\:n)}$ is called absolutely convergent $:\Leftrightarrow$ $$\sum_{m,\:n=1}^\infty\left\|x_{(m,\:n)}\right\|_E<\infty\;.\tag2$$
If $I=\mathbb N^2$, are we able to show that $(x_i)_{i\in I}$ is summable with sum $x$ iff $\sum_{m,\:n=1}^\infty x_{(m,\:n)}$ is absolutely convergent and $x=\sum_{m,\:n=1}^\infty x_{(m,\:n)}$?
Both statements are not equivalent: Set $E=l^2$ with the usual norm. Define $$ x_{n,m} = (0,\dots, 0, \frac1{nm^2}, 0\dots), $$ where the only non-zero element is at the $n$-th position. Then $$ \sum_{m=1}^\infty \|x_{n,m}\|_{l^2} = \frac{\pi^2}6 \frac1n, $$ hence $$ \sum_{n=1}^\infty\sum_{m=1}^\infty \|x_{n,m}\|_{l^2} = +\infty, $$ and $(x_{n,m})$ is not absolutely convergent.
However, $(x_{n,m})$ is summable with sum $$ x=(1,1/2,1/3,\dots)\in l^2. $$ To see this, take $\epsilon>0$ and fix $N$ such that $\sum_{N+1}^\infty \frac1{n^2}<\frac{\epsilon^2}4$. Then fix $M$ such that $$ \sum_{n=1}^N \sum_{M+1}^\infty \frac 1{(n m^2)^2} <\frac{\epsilon^2}4. $$ Then the index set $J= \{1\dots N\} \times \{1\dots M\}$ does the job.