Forcing series convergence I am trying to figure this out:

$\mathscr{S}=\big\{(a_n),(b_n),\dots \big\}$ is a finite set of real, null sequences. Does there exist a sequence $(\epsilon_n)$, where $\epsilon_k=\pm 1$ for each $k$, such that:
$$\forall\;(x_n)\in\mathscr{S}:\quad \sum_n \epsilon_n x_n<\infty\;?$$

A special case of this problem was posed by one of my lecturers: does every null sequence in $\mathbb{C}$ admit a sequence $(\epsilon_n)$ of signs such that $\sum\epsilon_nz_n$ converges?

 The answer is yes. We can always choose signs so that $|\epsilon_1z_1+\cdots\epsilon_nz_n|\leq \sqrt{3}$ for all $n$, with some assumptions on $|z_n|$.

The geometric nature of the proof prevents me from generalising though. Any ideas how to deal with the general case?
 A: The original problem was given to this year's $\mathsf{IA}$ students by Prof. W. T. Gowers, and oddly enough I remember discussing this generalisation with a fellow classmate.

The more general result is made less obscure by collapsing the sequences into a vector sequence. Two simple observations are then enough to solve the problem:

Lemma Given $q$ vectors $\mathbf{v}_1\dots\mathbf{v}_q$ in $\mathbb{R}^n$ with $\lVert \mathbf{v}_k\rVert\leqslant 1$ there exists $\epsilon_1\dots \epsilon_q\;[\epsilon_k=\pm 1]\;\text{s.t.:}$
  $$\left\lVert \sum_{k=1}^q \epsilon_k\mathbf{v}_k\right\rVert\leqslant \sqrt{q}$$

$\text{Pf}.:$ for vectors $\mathbf{x},\mathbf{y}$ with $\lVert\mathbf{x}\rVert\leqslant a,\lVert\mathbf{y}\rVert\leqslant b,\;\min_{\pm}\lVert\mathbf{x}\pm\mathbf{y}\rVert\leqslant (a^2+b^2)^{1/2}\;\;(\star)$. Suppose we have chosen $\epsilon_1\dots\epsilon_{m}$ such that $\lVert\sum_1^{m}\epsilon_k\mathbf{v}_k\rVert\leqslant m^{1/2}$ then set $\mathbf{x}=\sum_1^m\epsilon_k\mathbf{v}_k,\;\mathbf{y}=\mathbf{v}_{m+1}$ and apply $(\star)$ to get $\lVert\sum_1^{m+1}\epsilon_k\mathbf{v}_k\rVert\leqslant (m+1)^{1/2}\;\text{etc.}$
Remark For $q\leqslant n,\;\sqrt{q}$ is the best bound, as observed by taking $\mathbf{v}_1\dots\mathbf{v}_q$ an orthonormal basis.
Without loss of generality $\lVert\mathbf{v}_k\rVert\leqslant 1$ for each $k$. The crux of the problem is now showing:

Lemma Given sufficiently many vectors $($say $\geqslant N$ for appropriate $N)$ it is always possible to choose $\epsilon_1\dots\epsilon_q$ such that $\sum_1^q\epsilon_k\mathbf{v}_k$ lies inside the $n$-ball of radius $\sqrt{N}$, centered at the origin.

$\text{Pf.:}$ let $N$ be such that given any $N$ lines in $\mathbb{R}^n$, at least two will share an angle $\leqslant \pi/3$. Let $m\geqslant N$. Suppose we can always find $\epsilon_1\dots \epsilon_m$ such that:
$$\left\lVert \sum_{k=1}^m \epsilon_k\mathbf{v}_k\right\rVert\leqslant \sqrt{N}$$
Pick $\mathbf{v}_i,\mathbf{v}_j\in\{\mathbf{v}_1\dots\mathbf{v}_m\}$ such that the corresponding lines share an angle $\leqslant\pi/3$. Then there are $\epsilon_i,\epsilon_j$ such that $\lVert\epsilon_i\mathbf{v}_i+\epsilon_j\mathbf{v}_j\rVert\leqslant 1$. Set $\mathbf{w}_1=\epsilon_i\mathbf{v}_i+\epsilon_j\mathbf{v}_j$, label $\{\mathbf{v}_k\;|\;1\leqslant k\leqslant m+1,\;k\neq i,j\}$ as $\{\mathbf{w}_2\dots\mathbf{w}_m\}$ and apply the inductive hypothesis to $\mathbf{w}_1\dots\mathbf{w}_m$ to get:
$$\left\lVert \sum_{k=1}^{m+1} \epsilon_k\mathbf{v}_k\right\rVert\leqslant \sqrt{N}$$

We are done; for since $\mathbf{v}_n\to\mathbf{0}$ we can enclose the tail of the series into smaller and smaller $n$-balls. More concretely, let $\epsilon>0$ and $M_{\epsilon}$ such that $\lVert\mathbf{v}_{n\geqslant M_{\epsilon}}\rVert\leqslant \epsilon$ then for some $P\geqslant M_{\epsilon}$ we have:
$$\left\lVert \sum_{k=M_{\epsilon}}^{Q} \epsilon_k\mathbf{v}_k\right\rVert\leqslant \epsilon\sqrt{N},\quad \forall\,Q\geqslant P\quad \text{i.e.}\quad \sum_{k=1}^n\epsilon_k\mathbf{v}_k\;\;\text{converges}$$

Addendum
Just noticed this through a linked question: the statement can rather satisfyingly be pushed to $\mathscr{S}$ countable - see a proof here.
