I learnt a definition: Let $X$ be a normed linear space and $J$ be a non-empty set. A family $x:J\rightarrow X$ is summable with sum $\overline{x}$ if for all $\epsilon>0$, there exists a finite subset $M_0$ of $J$ such that whenever $M_0\subset M\subset J$ with $M$ finite, $$\|\overline{x}-\sum_{j\in M}x_j\|<\epsilon.$$
I wondered whether this definition extended the usual notion of the convergence of a series. This is what I thought:
Let $J=\mathbb{N}$. If $x_1,x_2,\dots$ is summable with sum $\overline{x}$ as in the above definition, then $\sum_{n=1}^\infty x_n=\overline{x}$ in the usual sense. However, $\sum_{n=1}^\infty x_n$ being convergent does not imply that $x_1,x_2,\dots$ is summable as in the above definition. For example, take $X=\mathbb{R}$ and take $x_n=(-1)^nn^{-1}$. However, if $X$ is complete and $\sum_{n=1}^\infty x_n$ is absolutely convergent i.e. $\sum_{n=1}^\infty \|x_n\|<\infty$, then $x_1,x_2,\dots$ is summable as in the above definition.
So $x_1,x_2,\dots$ being summable is weaker than $\sum_{n=1}^\infty \|x_n\|<\infty$, but is "strictly stronger" than $\sum_{n=1}^\infty x_n$ being convergent.
Does anyone know if $x_1,x_2,\dots$ being summable is equivalent to $\sum_{n=1}^\infty \|x_n\|<\infty$?
Best Answer
Consider the sequence $x_n = e_n/n$ in $\ell^\infty$, where $e_n$ is the $n$'th standard unit vector ($e_n(n) = 1$, $e_n(i) = 0$ otherwise). This is summable with sum $x$ where $x(j) = 1/j$ for all $j$, but $\sum_n \|x_n\| = \infty$.