If $a_n\to0$, there exists $\pm$ such that $\sum\limits_n\pm a_n$ converges Our Analysis I lecturer in his last lecture for the course gave us a problem to think about. I've been thinking about it for a while and has been bothering me for some time.
It looks like a generalisation of a theorem( or lemma) in the real 1D case. 
The problem is as follow:

Let $(a_n)$ be a complex sequence tending towards $0$. Show that there exists a sequence $\delta_n\in\{-1,1\}$ for all $n$ such that $\sum\limits_n\delta_na_n$ converges.

This is true in the case for real sequence.
My only vaguely hopeful line of attacking the problem is to reduce it down to a finite case problem as follows.
Suppose we have $A=\{a_i:n\le i\le m\}$ such that $a_i\le\epsilon$ for some $\epsilon$. 
Then I wish to show that I can choose $\delta_n$ such that the partial sum is always bounded by some multiple of $\epsilon$ (say $\sqrt2$ or possibly $2$).
I suspect this is true but I'm not entirely sure how one goes about doing this since the size of our set A is arbitrary.
 A: I already pointed to a related question in the comment above, but as I thought about this question a while ago, I'll give a slightly different proof here:
For brevity, I'll call $\sum_{i\in I} \alpha_iz_i$ an additive
  combination of the sequence of numbers $(z_i \mid i \in I)$ if
$\alpha_i\in\{1,-1\}$ for all $i\in I$.
Claim1:
If $z_1\in\mathbb{C}$ has an absolute value not larger than $\sqrt2$ and if
$z_2,z_3\in\mathbb C$ are two numbers on the unit circle, then 
one can find an additive combination $z$ of $z_2$ and $z_3$ such that
$|z_1+z|\leq\sqrt2$.
Proof: We'll argue partly geometrically by identifying complex numbers with
the corresponding points or vectors in the Argand plane.
It is clear (e.g. by taking the scalar product) that $z_2+z_3$ and
$z_2-z_3$ are orthogonal, so the four additive combinations of $z_2$
and $z_3$ form a rhombus with sides of length $2$:

That implies that in whichever direction $z_1$ points it will fall
between two of the four combinations (which we'll call $a$ and $b$)
and we'll have something like this (where the Thales circle
corresponding to the right angle at the origin $\cal O$ is included in
the picture):

We want to prove that at least one of the lines $l_a$ and $l_b$ is not
longer than $\sqrt2$ (because they correspond to $z_1-a$ and $z_1-b$
respectively) and we'll consider three cases.
We start with the case that $z_1$ is within the circle but not on the
same side of the diameter through $a$ and $b$ as $\cal O$ is - as
pictured above.  Let's assume for a contradiction that we can mark
points $a'$ and $b'$ on $l_a$ and $l_b$ respectively that both have a
distance of $\sqrt2$ from $z_1$ and both lie in the interior of the
circle.  Then, because the angle at $z_1$ from $a$ to $b$ is clearly
less than $\pi$, the segment of the circle of radius $\sqrt2$ around
$z_1$ from $a'$ to $b'$ lies completely within the Thales circle which
implies that some point between $z_1$ and the origin $\cal O$ must be
on this segment too.  This contradicts $|z_1|\leq \sqrt 2$.
For the next case let $z_1$ be on the other side of the diameter:

Here we have an obtuse triangle the sides of which are $l_a$, $l_b$
and the circle's diameter.  This implies $|l_a|^2+|l_b|^2\leq 2^2=4$
which in turn means that $l_a$ and $l_b$ cannot both be longer than
$\sqrt2$.
In the last case, $z_1$ lies outside of the circle:

In the quadrilateral with the corners $\cal O$, $b$, $z_1$ and $a$,
the angle at $\cal O$ is $\frac\pi2$ and the one at $z_1$ is less than
$\frac\pi2$ (otherwise $z_1$ would lie in or on the circle) and thus
at least one of the other two angles must be greater than
$\frac\pi2$, w.l.o.g. let it be the angle at $a$ as in the picture.
It follows that in the triangle with the corners $\cal O$, $z_1$ and
$a$ the line from $\cal O$ to $z_1$ (which has length at most
$\sqrt2$) is the longest side and thus $l_a$ must be shorter.
Claim 2: If $(z_n)_{n\in\mathbb N}$ is a sequence of complex numbers lying on the unit
circle, then at least one additive combination of this sequence is
bounded.
Proof: We'll use the first claim to recursively construct a sequence
$(\alpha_n)_{n\in\mathbb N}$ of elements of $\{1,-1\}$ such that
$|s_{2n}|\leq\sqrt2$ for all $n\in\mathbb N$ where
$s_n=\sum_{k=0}^n\alpha_kz_k$.  We start with $\alpha_0=1$ which implies
$s_0=z_0$ and thus $|s_0|=|z_0|=1$.
Now, if $|s_{2n}|\leq\sqrt2$, then by the lemma we can find
$\alpha_{2n+1}$ and $\alpha_{2n+2}$ such that
$s_{2n}+\alpha_{2n+1}z_{2n+1}+\alpha_{2n+2}z_{2n+2}=s_{2n+2}$ has an
absolute value not greater than $\sqrt 2$ which is what we wanted.
For the odd values we now have
$|s_{2n+1}|\leq|s_{2n}|+|z_{2n+1}|\leq\sqrt2+1$ by the triangle
inequality and thus the whole series is bounded.
Claim 3: If $(z_n)_{n\in\mathbb N}$ is a sequence of complex numbers with
$\lim_{n\rightarrow\infty}z_n=0$, then at least one additive
combination of this sequence converges.
Proof: W.l.o.g. $z_n\neq0$ for $n\in\mathbb N$.  We can now find by the claim above a sequence $(\alpha_n)_{n\in\mathbb N}$ such that
$\sum_{n=0}^\infty \alpha_n|z_n|^{-1}z_n$ is bounded, say by
$C>0$.
We want to show that $\sum_{n=0}^\infty \alpha_nz_n$ converges using
Cauchy's convergence test, so let $\varepsilon>0$ be given.  Because
of $\lim_{n\rightarrow\infty}z_n=0$ we can find an $N\in\mathbb N$ with
$|z_n|\leq\frac{\varepsilon}{2C}$ for all $n\geq N$.
Now for $m,n\in\mathbb N$ with $n>m\geq N$ and with $z^\ast=\max\{|z_k| \mid m \leq k \leq n \}$ we have
$$
\left|\sum_{k=m}^n \alpha_kz_k\right|
= |z^\ast| \cdot \left|\sum_{k=m}^n \alpha_k{z^\ast}^{-1}z_k\right|
\leq |z^\ast| \cdot \left|\sum_{k=m}^n \alpha_k|z_k|^{-1}z_k\right|
\leq |z^\ast| \cdot 2C \leq \frac{\varepsilon}{2C} \cdot 2C = \varepsilon
$$
which concludes the proof.
