# Summability vs Unconditional Convergence

I'm reading that summability and unconditional convergence are the same. (At least if the index set is countable, so unconditional convergence makes sense.)

Then unconditional convergence is nothing else than convergence of the net of all rearranged partial sums. Thus uniqueness of the limit becomes fairly easy as the limit of every net in a Hausdorff space is unique.

But the usual proof deals with separating functionals on Banach spaces, which is a much more serious technique. So I think there must be a hook in my reasoning. Can somebody resolve this issue? Thanks a lot!

Best Regards, Freeze_S

• They are the same only if your space if of finite dimension (cf wikipedia) – Thomas Dec 19 '13 at 21:36
• can you just precise what you mean by Summability just to be sure we are talking about the same thing ? – Thomas Dec 19 '13 at 21:50
• @Thomas A family $\{x_\alpha : \alpha \in A\}$ is summable with sum $s$ if and only if for every $\varepsilon > 0$, there is a finite subset $A_\varepsilon \subset A$, such that for every finite $B \subset A$ with $A_\varepsilon \subset B$ you have $$\left\lVert \sum_{\alpha \in B} x_\alpha - s\right\rVert < \varepsilon.$$ That is a much weaker condition than absolute convergence. – Daniel Fischer Dec 19 '13 at 21:57
• Thats what i meant as summability. For sure it is much weaker than absolute convergence but isn't it the same unconditional convergence? – C-Star-Puppy Dec 19 '13 at 22:02
• @Alex It certainly is not weaker than unconditional convergence, and I'm almost sure that it is equivalent (restricting summable families to countably many members, which isn't much of a restriction, since in a normed space, a summable family can only have countably many nonzero terms). But I'll need to think about the one direction. – Daniel Fischer Dec 19 '13 at 22:23

For countable families $\{ x_n : n \in \mathbb{N}\}$ in a Banach (or Fréchet) space, summability and unconditional convergence are the same.

Let $\mathscr{F} = \{ F \subset \mathbb{N} : \operatorname{card}F < \infty\}$ the directed (by inclusion) set of finite subsets of $\mathbb{N}$, and for $F\in\mathscr{F}$, let $s_F = \sum\limits_{n\in F} x_n$.

The family is (by definition) summable if and only if the net $(s_F)_{F\in \mathscr{F}}$ converges (to $\lambda$, say).

Let the family be summable. Every bijection $\sigma \colon \mathbb{N}\to \mathbb{N}$ induces a map $\sigma^\ast \colon \mathbb{N}\to \mathscr{F}$ via $\sigma^\ast(n) = \{ \sigma(k) : 0 \leqslant k \leqslant n\}$, and $\{\sigma^\ast(n) : n \in \mathbb{N}\}$ is cofinal in $\mathscr{F}$, so $s^\sigma \colon n \mapsto s_{\sigma^\ast(n)}$ is a subnet of $(s_F)_{F\in \mathscr{F}}$ and hence also converges to $\lambda$. That is the unconditional convergence of $\sum\limits_{n=0}^\infty x_n$.

For the converse, we argue indirectly. Suppose the family is not summable. We have to show that it isn't unconditionally convergent.

Consider the sequence of partial sums

$$s_n = \sum_{k=0}^n x_k.$$

If that sequence doesn't converge, we're done, otherwise let $\mu$ be its limit. That the family isn't summable to $\mu$ means that there is an $\varepsilon > 0$, such that for every $F_1 \in \mathscr{F}$, there is a $c(F_1)\in \mathscr{F}$ with $F_1 \subset c(F_1)$ and

$$\left\lVert\sum_{n\in c(F_1)} x_n - \mu \right\rVert \geqslant \varepsilon.$$

We can now construct a bijection $\sigma \colon \mathbb{N}\to\mathbb{N}$ such that

$$s^\sigma_n = \sum_{k=0}^n x_{\sigma(k)}$$

does not converge to $\mu$, which shows that $\{x_n\}$ is not unconditionally convergent.

Let $n_1 = \min \{n\in \mathbb{N} : \lVert s_n - \mu\rVert \leqslant \varepsilon/2\}$. Whenever $n_k$ is defined, we let $F_k = \{0,\dotsc,n_k\}$. For $k \leqslant n_1$, let $\sigma(k) = k$. Let $m_1 = \operatorname{card} c(F_1)$, and $\sigma(n_1+1),\dotsc,\sigma(m_1)$ the elements of $c(F_1)\setminus F_1$ in ascending order. Let $r_1 = \max c(F_1)$ and $n_2 = \min \{n > r_1 : \lVert s_n-\mu\rVert \leqslant \varepsilon/2\}$, and $\sigma(m_1+1),\dotsc,\sigma(n_2)$ the elements of $F_2\setminus c(F_1)$ in ascending order.

We continue in this way, when $n_k$ and $\sigma\lvert_{F_k}$ are determined, we let $m_k = \operatorname{card} c(F_k)$, $\sigma(n_k+1),\dotsc,\sigma(m_k)$ the elements of $c(F_k)\setminus F_k$ in ascending order, $r_k = \max c(F_k)$, $n_{k+1} = \min \{ n > r_k : \lVert s_n - \mu\rVert \leqslant \varepsilon/2\}$, and $\sigma(m_k+1),\dotsc,\sigma(n_{k+1})$ the elements of $F_{k+1}\setminus c(F_k)$ in ascending order.

The construction yields $n_k < m_k \leqslant r_k < n_{k+1}$ for all $k$, and

$$\left\lVert \sum_{k=0}^{m_k} x_{\sigma(k)} - \mu\right\rVert \geqslant \varepsilon,$$

so $(s^\sigma_n)$ does not converge to $\mu$ (it does not converge at all, in fact).

• can you tell me which step you use the condition banach? – noname1014 Oct 2 '17 at 3:06
• @noname1014 For this equivalence, we in fact don't need completeness, only first countability. But one needs completeness to have other useful properties of summable families (like that every subfamily of a summable family is summable), so I'm used to always assume a complete space when dealing with summable families. – Daniel Fischer Oct 2 '17 at 11:57

I don't think that you need complicated tools.

Consider that you are working in R and that your elements are all positive. If your serie is summable, then it is pretty easy to show that any rearrangement leads to the same limit. If you have positive and negative terms, just separate them into 2 big sums and the same result applies.

For Banachs : Suppose $x_n$ summable. Then $x_{\sigma(n)}$ is summable too : let $\epsilon >0$. Let N be such as $\sum_{n>N} |x_n| < \epsilon$. Then $\forall n < N'$ big enough, $\sigma(n)$ < N so $\sum_{n>N'} |x_{\sigma(n)}| < \epsilon$

To show that they have the same limit for any $\sigma$ and $\sigma'$, do the subtraction and apply the same principle

Ok, so let me summarize.

In principal, they are two different kind of objects: Series vs. Sums

More precisely, a series is usually defined as the sequence of its partial sums while a sum is the net of all finite sums. Both give rise to some meaning of an infinite sum but in a slightly differenet fashion: ordered and unordered*. This ordering causes (what we already know from our first year analysis course) that a series might change its value if one changes its ordering, or it can even change from convergent to divergent. This does not happen if one considers sums. It is either convergent or divergent and if it is convergent it has a precise value unless the space is non-Hausdorff (which is different matter).

Now, how does this fit into our understanding of what a "sum" is?

The answer to this seems to be hidden in the way of indexing. A series can be redefined as the net of certain finite sums ordered by inclusion while some of them are simply left out. Thus the ordering tells you precisely which finite sums to take care of and which not! Following this idea, reordering gives rise to a net of different finite sums. However all of these have one thing in common: They all are a subnet of the net of all finite sums. This leads us directly to sums. A sum is by definition the net of all finite sums. This approach has the tremendous advantage that it can deal with arbitrary index sets not just countable ones. Luckily, it can be resolved for series which I will mention in a moment.

The final and crucial point is what value we'd like to assign to a "sum".

In any case it is taken as the limit. As we saw already, different reorderings of a series can change the value or even make it divergent. This corresponds to different subnets as different "branches" of the net of all finite sums. Similar to subsequences this can give convergent objects while the initial object was divergent, e.g. the alternating sequence (1,0,1,0,...). Here, we face a subtle: In the case of a sequence we know that it converges iff all of its subsequences converge and the limit being unique (which comes for free). So one might hope that a family is summable iff all of its corresponding series converge and the limit being unique. However, this fails since one can simply leave out a summand in a series. So the next idea would be to take care of every index while ordering them linearly. But this fails to work in principle since the index set can be uncountable. The resolution to this problem lies in the the fact that summable families necessarily have only countable many contributing summands. So the promising hypothesis is that a family is summable iff every series over all contributing summands converge and the limit being unique (which interestingly enough comes for free too). This, thanks to Daniel Fischer, proves to be true.

Concluding: A not necessary countable family is summable iff every series over all contributing summands converges.

*Unordered might be misleading rather ordered by inclusion (cf. https://www.math.ucdavis.edu/~hunter/m201b_old/sums.pdf).