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 $\sum_{n=1}^\infty \|x_n\|<\infty$?

  • $\begingroup$ Here is a related question. $\endgroup$ – David Mitra Dec 7 '14 at 8:09
  • $\begingroup$ One should maybe note that in finite dimensional spaces, absolute convergence is equivalent to unconditional convergence (i.e. the notion of convergence you describe). Basically, this reduces to the case of the reals and there you know that a conditionally convergent series could be reordered to converge to a different value, contradiction. $\endgroup$ – PhoemueX Dec 7 '14 at 8:39

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 $\sum_n \|x_n\| = \infty$.


If $\sum_{j\in J} ||x_j||<\infty$, then $x_j=0$ for all but countably many $j$, so we can suppose $J=\mathbb{N}$. Then for any $\epsilon,$ let $N$ be such that $\sum_{N}^\infty||x_j||<\epsilon$. Then for every $k$, $||\sum_N^{N+k} x_j||<\epsilon$ by the triangle inequality, which shows the sequence of partial sums of the $x_j$ is Cauchy and thus converges to a unique limit $\bar x$. The converse is actually false, as the example of $\sum 1/n e_n$ in $\ell^2$ shows.


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