TITLE: If $a_n$ goes to zero, can we find signs $s_n$ such that $\sum s_n a_n$ converges?
QUESTION [25 upvotes]: Let $a_n$ be a sequence of complex numbers that converge to zero. Can we always find $s_n \in \{-1,1\}$ such that $\sum_{n=1}^{\infty} s_n a_n$ converges?
If the $a_n$ are real numbers, we can find such a sequence $s_n$. If the partial sum of the first $N$ terms is positive we make sure the following terms are negative until the sum becomes less than zero. Then we switch to making the terms positive until the partial sum becomes greater than zero, and so on. It is easy to see that the partial sums will either tend monotonically towards zero, or oscillate around zero with decreasing amplitude. 
Edit: Actually there is no reason that the partials sums should go to zero in the monotone case. They will still converge however.

REPLY [11 votes]: Note: I had written this proof before the other answers were posted. I was about to discard it because the idea is actually the same as mixedmath's answer, but his answer has a small issue because it only proves the theorem for $k=n$, which can only prove that a subsequence of $b_n$ converges, not the whole sequence. So here is a hopefully correct proof.

First we prove the following theorem:

Theorem: For any finite sequence $a$ bounded by $M$, we can find signs $s$ so that $\left|\sum_{i=1}^k s_i a_i\right| \le 2M$ for all
  $k\le n$.

We say that $a$ and $b$ on the unit disk are compatible when either $a+b$ or $a-b$ lies on the unit disk.
Lemma: If $a,b,c$ lie on the unit disk, at least one of $(a,b)$, $(b,c)$ or $(a,c)$ is a compatible pair.
Proof. When $|a+b|>1$ with $a$ and $b$ in the unit disk, we also have $\left|\frac{a}{|a|}+\frac{b}{|b|}\right|>1$ and therefore the arguments of $a$ and $b$ differ by less than $2\pi/3$ (mod $2\pi$). Thus $a$ and $b$ are incompatible iff $\arg a - \arg b\in (-2\pi/3,-\pi/3)\cup(\pi/3,2\pi/3)$ (mod $2\pi$), which shows that $a,b,c$ cannot all be pairwise incompatible.
Lemma: For any finite sequence $a_1,\dots a_n$ with $a_i$ in the unit disk, we can find signs $s_1,\dots s_n$ and $t_1,\dots t_n$ so that for all $k\le n$, $\left|\sum_{i=1}^k s_i a_i\right|\le 2$, $\left|\sum_{i=1}^k s_i t_i a_i\right|\le 2$ and for each $\epsilon\in\{-1,1\}$, $u_\epsilon = \sum\limits_{\substack{i\le n\\t_i=\epsilon}} s_i a_i$ lies on the unit disk.
Proof. This is of course true for $n\le 2$. We continue by induction: assume we have built $s_1,\dots s_n$ so that the lemma applies. Then if $u_1$ and $u_{-1}$ are incompatible, $a_{n+1}$ is compatible with some $u_\epsilon$ so that we can find $s_{n+1}$ such that $u_\epsilon+s_{n+1}a_{n+1}$ lies on the unit disk, and letting $t_{n+1}=\epsilon$ finishes the proof. If $u_1$ and $u_{-1}$ are compatible so that $u_1+\epsilon~u_{-1}$ lies on the unit disk, we can pick any $s'_{n+1}$ and $t'_{n+1}$, and for $i\le n$ define $t'_i=-t'_{n+1}$ and $s'_i=s_i$ if $\epsilon=1$ and $s'_i=s_i t_i$ if $\epsilon=-1$.
We obtain the theorem as a corollary.

Now take an infinite sequence $a_n$ converging to 0. Let $n_i$ be an increasing sequence of integers such that $|a_n|<2^{-i}$ when $n\ge n_i$. We simply apply the theorem to each subpart $a_{n_i+1},\dots,a_{n_i}$ to define the sequence $s$. We can let $s_n=1$ for $n<n_1$.
Then we can prove that $b_n=\sum_{k=1}^n s_k a_k$ is a Cauchy sequence: for any $\varepsilon>0$, pick $i_0$ so that $4\cdot 2^{-i_0}\le \varepsilon$. Then for $n\ge N=n_{i_0}$, $|b_n-b_N|=|\sum_{k=N+1}^n s_k a_k|\le \sum_{i\ge i_0} 2\cdot 2^{-i}\le \varepsilon$.
So $b_n$ converges. $\square$

Can we improve the bound in the theorem?
Someone asked what the sharpest bound for the theorem was. If we restrict to the $k=n$ case, this is $\sqrt 2 M$. We have $\min(|a+b|^2,|a-b|^2) = |a|^2+|b|^2-2|a\cdot b|\le 2$ for $a,b$ in the unit disk ($\cdot$ is the scalar product). Since the lemma happens to prove that the sum can be indifferently written as $u_1+u_{-1}$ or $u_1-u_{-1}$ with $u_\epsilon$ in the unit disk, this proves the bound. The bound is sharp for $n\ge 2$: take $(a_1,\dots a_n)=(1,i,0,0,\dots)$.
However $\sqrt 2 M$ cannot be used to bound partial sums ($k<n$) when the set of signs is not allowed to vary: indeed for $a=1$, $b=\exp(i\pi/3+\varepsilon)$, $c=-\exp(i\pi/6)$, $|a+b+c|\le \sqrt 2$ is the only set of signs ensuring the bound, but $|a+b|=\sqrt 3-\varepsilon'>\sqrt 2$.
Does the theorem work in higher dimensions?
Yes! Call the dimension $p$. When $a$ and $b$ are incompatible with $a,b$ in the unit $(p-1)$-sphere, the distance between $a$ and $b$ is bounded from below (by 1). This means there is an upper bound $N$ to the size of pairwise incompatible sets on the unit $(p-1)$-sphere, and therefore also on the unit ball because $a\mapsto a/\|a\|$ preserves incompatible pairs (a consequence of the fact that $\|ta-b\|$ is a convex function of $t$ with value at most 1 for $t=0$). Then the proof of the second lemma still holds, using a partition of $1,\dots n$ into $N$ subsets and requiring that for all assignments of a sign to each subset, the sum be bounded by $N$.