I like this problem very much. To be honest, I was first trying to show that it is not true by coming up with a clever counterexample, designed around fooling Henry's answer. But to cut that vein short, couldn't.
If I may boil down the entire proof into 2 statements:
- Given a finite sequence $z_1, \dots, z_k$ of $k$ complex numbers with $|z_i| < M$ for all $i$, then there exists a constant $K(M) = K \in \mathbb{R}$ independent of the length of the sequence $k$ and a set of signs $s_i \in \{-1, 1\}$ s.t. $|s_1z_1 + \dots s_kz_k| < K$
- Break a given sequence $\{z_i\}_{i \in \mathbb{N}}$ with $z_i \to 0$ into finite subsequences $\{z_{n_i}\}_{i \in I}$ with $|I|$ finite and with $|z_{n_j}| < 2^{-n}$ for each $j \in I$, and so that each $z_i$ is included in exactly one of the finite subsequences. Then applying the bound from above, we get the result.
So we now have a plan to follow.
Proof of [1.]: ($\spadesuit$)
Note that if $z_1, z_2$ are two complex units, then there are signs $s_i$ s.t. $|s_1z_1 + s_2z_2| \leq \sqrt 2 \leq 2$ (where $\sqrt 2$ is optimal, but where I only use $2$ because I can state it relying on trivial proof). Similarly, if $z_1, z_2$ have $|z_1|, |z_2| \leq M$, then there are signs $s_i$ s.t. $|s_1z_1 + s_2z_2| \leq 2M$.
Lemma: ($\diamondsuit$) If $z_1, z_2, z_3$ are three complex units, then we can always choose two of them (say $z_1, z_2$) s.t. $|z_1 + z_2|\leq 1$ or $|z_1 - z_2|\leq 1$
Proof: Assume not. Then $z_1$ must have an angle of between $60^\circ$ and $120^\circ$ with each of $z_2$ and $z_3$. But then it's easy to show that $z_2$ or $-z_2$ will have a relative angle less than $60^\circ$ with $z_3$. Thus $|z_3 + z_2| \leq 1$ or $|z_3 - z_2| \leq 1$. Contradiction. $\diamondsuit$
So given a finite sequence of complex numbers $z_1, \dots, z_k$, with $k \geq 3$ and $|z_i| \leq 1$, we may inductively choose signs to bound pairs $z_\alpha, z_\beta$ by a single element $z_\gamma$ satisfying $|z_\gamma| \leq 1$, until we are left with only 2 elements. But then we may apply the trivial bound first mentioned. So we have that there are signs so that $|s_1 z_1 + \dots + s_k z_k| \leq 2$. Scaling, we get the general result ($\leq 2M$ where $M$ is the bound on the $|z_i|$). $\spadesuit$
So given a finite sequence, we can choose signs to bound the sum irrespective of the (finite) length of the sequence. Now we use this on our infinite sequence.
Proof of the main result: ($\clubsuit$)
So we have a sequence $z_1, z_2 , \dots$ with $|z_i| \to 0$. I'm going to abuse a bit of notation here. Then for each $n \in \mathbb{N}$, there is a lowest index which I denote by $n_1$ so that $|z_i| > 2^{-n}$ for some $i < n_1$ and $|z_j| \leq 2^{-n}$ for all $j \geq n_1$. Then the elements sequentially between $z_{n_1}$ and $z_{(n+1)_1}$ can be relabelled as $z_{n_2}, z_{n_3}, \dots, z_{n_l}$, and there are finitely many numbers in each. For ease, denote the set of elements with index $1_j$ for some $j$ by $I_j$, so that $z_{1_1}, z_{1_2} \in I_1, z_{3_5} \in I_3$, etc.
So now, we are essentially done. Tossing aside the (finitely many) initial terms before $z_{1_1}$, we know that there are signs that can be given to $z_i \in I_j$ so that $|\sum_{I_j} z_i| < 2\cdot 2^{-j}$ for every $j$, and that the $I_j$ evenly cover the entire sequence $\{z_i\}$ in order. As $2\sum 2^{-j}$ converges, the triangle inequality gives us that our sequence converges as well. $\clubsuit$
EDIT
As Johan points out below, there is a detail missing. I was in the process of writing a completed solution when I noticed that my missing detail is included in Generic Human's post, and in the exact way I was going to do it. So I defer the missing detail.
Best Answer
$|a_n| \leq |a_1|+|a_2-a_1|+|a_2-a_3|+\cdots+|a_n-a_{n-1}|$. From this and the hypothesis conclude that $(a_n)$ is bounded. If $|a_n| \leq M$ then $|a_n x^{n}| \leq M|x|^{n}$ and $\sum M|x|^{n}$ is a convergent geometric series. By Comparison Test the series $\sum a_nx^{n}$ is absolutely convergent.