I think $\sum_{\alpha\in A}x_\alpha=\sum_{\beta\in B}x_{\phi(\beta)}$ always holds even if $(x_\alpha)_{\alpha\in A}$ is not absolutely convergent.

Question

The following is from "An introduction to measure theory" by Terence Tao.

Motivated by this, given any collection $(x_\alpha)_{\alpha\in A}$ of numbers $x_\alpha\in [0,+\infty]$ indexed by an arbitrary set $A$ (finite or infinite, countable or uncountable), we can define the sum $\sum_{\alpha\in A} x_\alpha$ by the formula $$\sum_{\alpha\in A} x_\alpha=\sup_{F\subset A,F\text{ finite}}\sum_{\alpha\in F} x_\alpha.$$

Note from this definition that one can relabel the collection in an arbitrary fashion without affecting the sum; more precisely, given any bijection $\phi:B\to A$, one has the change of variables formula $$\sum_{\alpha\in A}x_\alpha=\sum_{\beta\in B}x_{\phi(\beta)}.$$
Note that when dealing with signed sums, the above rearrangement identity can fail when the series is not absolutely convergent (cf. the Riemann rearrangement theorem).

I think the above rearrangement identity always holds.

Example:
Let $A:=\{1,2,\dots\}.$
Let $x_\alpha:=(-1)^{\alpha-1}1/\alpha.$
$$\sum_{\alpha\in A}x_\alpha=\sum_{\beta\in B}x_{\phi(\beta)}=\infty.$$

Am I wrong?

Note that $$\{\sum_{\alpha\in F} x_\alpha:\text{ }F\text{ is a finite subset of }A\}=\{\sum_{\beta\in G} x_{\phi(\beta)}:\text{ }G\text{ is a finite subset of }B\}$$ holds.
And if $S=T$ holds, then $\sup S=\sup T$ holds.

Answer 1

You are wrong, sort of. It depends on what you meant and on what Tao meant. See this, as Tao says.

The essential problem is that a conditionally but not absolutely convergent series of reals will necessarily have that both its negative and positive terms diverge when taken as individual series. So you have a lot of money in the bank: you can keep adding negative terms to get to an arbitrary negative point, or keep adding positive terms to get to an arbitrary positive point. More precisely the Wikipedia page shows that for any extended real number $M$, you can reorder the terms of the series so that it converges to $M$. In this sense, reindexing with some $\phi$ miserably fails!

You took $1-1/2+1/3-1/4+\cdots$ as an example. In the sense of: $\sum_A=\sup_{F\subset A}\sum_F$ (which is actually a Lebesgue integral!!) you’re right that the sum of this series is $\infty$. This will always be true for a conditionally-but-not-absolutely convergent series by the Riemann series theorem. So in that sense I suppose Tao means that allowing negative terms means you get pathological behaviour. Series you think are convergent end up not being ‘integrable’. Moreover it’s usual to define a Lebesgue integral (series) in that way for positive functions (sequences) only and then extend it to generic functions (sequences) by taking the difference of the series of positive and negative terms. That would be undefined since we’d have an $\infty-\infty$ situation.

If you take a conventional series we have: $$\sum_{n\ge1}\frac{(-1)^{n-1}}{n}=\ln2$$

Consider $1-1+1/2-1/2+1/3-1/3+\cdots$ which, as written, converges to zero as a conventional series. Rearranging as follows: $1+(1/2-1)+1/3+(1/4-1/2)+1/5+(1/6-1/3)+\cdots$ yields $\ln2\neq0$. Whether or not rearrangement is well defined depends on whether or not you want this Lebesgue-style definition of series or the more conventional one (limit of partial sums).

Answer 2

This is called Riemann's Rearrangement Theorem.

I call this problem the creative accountant's task .

Given an infinite collection of declining daily paper losses and paper profits, both of which add up to infinite losses and profits, the accountant is tasked with recording these values in an order so that the sequence of reported net proceeds converges to any pre-assigned final target value.

Hypotheses of the RRT. The individual positive terms, when sorted in non-increasing order, are assumed to tend to zero, as are the magnitudes of the individual negative terms.

Pictorial Proof. Sort the positive terms in non-increasing order as $a_1 \geq a_2 \geq a_3\ldots$ and likewise for the negative terms $\{-b_k\}$, sort their magnitudes as $b_1\geq b_2, \ldots$. Then construct the sequence of partial sums $A_k =\sum_1^k a_j$ and likewise $B_k$. If both partial sum sequences converge, the series is absolutely convergent. If exactly one converges, then the other diverges, and their sum diverges.

Focus therefore on the interesting case where both sequences of partial sums diverge. Plot these partial sums on two orthogonal axes. The gridlines on this diagram intersect in ordered pairs of "lattice points" that travel to infinity. The rectangular pixels on this grid have their two dimensions decreasing toward zero as we travel rightward and/or upward.

The goal is to show that we can obtain any preassigned limit value $L =\lim _{k\to \infty} A_{m_k} - B_{n_{k}}$ using some subsequence of lattice points. Here $(m_k, n_k)$ label the lattice points.

Draw the target line $A-B=L$ in the $(A,B)$ plane. To construct a sequence of partial sums that tends to $L$ from above, consider all pixels that lie strictly above the target line. Since the lattice pixels get tiny, this region gets arbitrarily close to the target line as we travel outward. As we slide forward along the target line and look behind us, mark the pixel corners that come closest to the target line. This sequence of lattice points is what we want. We use them to define a zig-zag path on the lattice that always lies above the target line but gets arbitrarily close to it as we move rightward. Note: The zigs and zags are horizontal and vertical distances that represent a sequence of incremental selections of positive and negative terms from our original list $a_k$ and $b_j$. Because our zig-zag path travels outward to infinity, all terms $a_k$ and $b_j$ eventually get used. QED.