Let us say that a set $A \subseteq \mathbb N$ sends a series $\sum_{n \in \mathbb N}a_n$ of real numbers to infinity if the subseries $\sum_{n \in A}a_n$ sums either to $\infty$ or to $-\infty$.

Given four or more conditionally convergent series, is there a single $A \subseteq \mathbb N$ that sends them all to infinity?

Notice that I do not require that all of the subseries sum to $\infty$, or that all of them sum to $-\infty$. This may well be impossible, even for just two series $\sum_{n \in \mathbb N}a_n$ and $\sum_{n \in \mathbb N}b_n$, since we may have $a_n = -b_n$ for all $n$. However, we do require that each subseries is made to diverge to either $\infty$ or $-\infty$, and not merely to diverge by oscillation. (If divergence by oscillation is allowed, then the answer to my question becomes a relatively easy yes.)

For one series this is trivial: let $A$ be the set of indices of the positive terms.

For two series, the question requires thought, but it is doable. [The idea is that two conditionally convergent series naturally partition $\mathbb N$ into four sets, the partition being determined by where each of the series is positive or non-positive. With a little work, you can prove that either one of these four sets works, or else the union of two of them will.]

For three series, I was able to prove that the answer is yes, but my proof is long, complicated, and frankly . . . ugly. I can't help but think that I'm going about it the wrong way, and that there must be a better approach leading to an easier proof.

For four or more series, I'm stumped.

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    $\begingroup$ Perhaps a quick summary of my now deleted attempt at an answer for those who can't see it: It's tempting to try to deduce this from the Levy-Steinitz theorem on rearrangements of vector valued series (see the linked question), but apparently this only gives the following (without further analysis, that is): we can find an $A$ and a $v\in\mathbb R^k$, with $v_n\not=0$ for all $n$, so that the partial sums of $\sum_A x_k$ will be eventually be in the half space $\langle x, v\rangle \ge C$ for any $C>0$. $\endgroup$ – Christian Remling May 2 '17 at 21:12
  • $\begingroup$ Will, I would be interested in a sketch of the result for three series, even if you feel your current version is ugly. $\endgroup$ – Andrés E. Caicedo May 3 '17 at 20:13
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    $\begingroup$ @AndrésE.Caicedo: I have already typed out most of it (I wanted to write it down anyway in order to make sure I hadn't missed any cases or muddled any details). I'll clean it up a bit and make it available soon. $\endgroup$ – Will Brian May 4 '17 at 15:17

There is a counterexample with 4 series. Notice that the problem is equivalent to asking if for every sequence of vectors $X_j$ in $\mathbb R^4$ with lengths tending to $0$ and the infinite sum of absolute values of the projections to each coordinate axis, you can find a sequence of signs $\varepsilon_j=\pm 1$ such that $\sum_j \varepsilon_j X_j$ diverges to plus or minus infinity in each coordinate (in one way, it is trivial: just view taking or not taking a vector as deviating from one half of it up or down). Now build the following blocks of fast increasing lengths $N_k$: all vectors are $(\frac 1{k},\frac 1{k},1/N_k,0)$ if $k$ is odd and $(\frac 1{k},-\frac 1{k},0,1/N_k)$ if $k$ is even. Suppose we try to diverge to $(+\infty,+\infty)$ or $(-\infty,-\infty)$ in the first two coordinates. Then every even block of length $N_k$ will create a terrible setback unless it is balanced up to $(N_1+\dots+N_{k-1})k\approx kN_{k-1}$. But then all we can gain in the fourth position is $\sum_{k\text{ even}}\frac{kN_{k-1}}{N_k}$ and that series hopelessly converges. If we try to get infinities of different signs, the odd blocks and the third position will create a problem. Now, to avoid firing any big cannons to show that the equivalence of the problems holds in both directions, just take the blocks of even length and get the 4 series requested as alternating sums of the specified vectors.

  • $\begingroup$ Elegant and simple (afterwards, that is) example! One can actually start reading at the end, and then the business with the signs is unnecessary: as you stated, the example now becomes $(-1)^n (1/k,1/k, 1/N_k, 0)$ etc., and your argument as stated can also be read as being about making selections within each block rather than imposing signs (what we look at is now called $\sum_{A_k}(-1)^n$ in this version). $\endgroup$ – Christian Remling May 3 '17 at 22:54
  • $\begingroup$ @ChristianRemling Thanks! I preserved the $\pm$ reformulation because it helps a bit when thinking about 3 series too. In fact, you have even more freedom: you can skip vectors as well due to the sign version of Levy-Steinitz. Once you realize that, the three series case becomes a relatively simple casework. $\endgroup$ – fedja May 4 '17 at 0:47
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    $\begingroup$ @ChristianRemling BTW, Christian, once you like such problems, take a look at math.stackexchange.com/questions/2256846/… To my shame, I do not know how to handle 3 or more permutations... $\endgroup$ – fedja May 4 '17 at 0:54
  • $\begingroup$ I'm with Christian -- this is a lovely answer. I am both ashamed of how long I tried proving this for four series . . . and thankful that I can now stop. Thanks! $\endgroup$ – Will Brian May 4 '17 at 15:15

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