Given four conditionally convergent series, is there a single sequence of naturals such that each corresponding subseries sums to $\pm\infty$?

Question

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.

Answer 1

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.