A question on convergence of series Suppose $(z_i)$ is a sequence of complex numbers such that $|z_i|\to 0$ strictly decreasing. If $(a_i)$ is a sequence of complex numbers that has the property that for any $n\in\mathbb{N}$
$$
 \sum_{i}a_iz_i^{n}=0  
$$ 
does this imply that $a_i=0$ for any $i$?
Edit: For the "finite dimensional" case, when we have $n$ distinct $(z_i)$, then $(a_i)$ must be $0$. This amounts to solving a homogeneous system of $n$ equations with $n$ unknowns, which only has the trivial solution in the case of distinct $(z_i)$. I am really curious what happens in the infinite dimensional case. My intuition tells me the same must be true, but I don't have a proof for it. 
Edit 2: Very interesting, looking were this question originated, the fact that all $a_i=0$ when $(a_i)\in l_1$ is "expected". I was hoping to get a counterexample otherwise. However, if a non-trivial sequence $a_i$ exists (at least for some sequences $z_i$), I would "expect" to be able to choose it in $l_2$. Looking at Davide and Julien answers below, it seems $(a_i)\in l_1$ is an essential assumption in their argument. 
 A: Under the assumption that $\sum_{i=0}^\infty |a_i| < \infty$, we can use power series. This assumption can be weakened: it is enough to assume that there exists $n \in \mathbb{N}$ such that $(a_i z_i^n) \in \ell^1$ (since we can work with $a'_i = a_i z_i^n$).
For $|\lambda| < |z_0|^{-1}$, define
$$
f(\lambda) =  \sum_{i=0}^\infty \frac{a_i}{1 - \lambda z_i} = \sum_{n=0}^\infty \sum_{i=0}^\infty (\lambda z_i)^na_i =0.
$$
If $k = \min\{i \in \mathbb{N} \mid a_i \neq 0\}$ exists, then $f(\lambda) \sim \dfrac{a_k}{1-\lambda z_k}$ when $\lambda \to z_k^{-1}$, wich is absurd.
A: It's a partial answer to @Tomas' comment which ask what happens if we assume the sequence $\{a_j\}$ in $\ell^1$, and it would be too long for a comment.
I interpret the hypothesis $\sum_{j=0}^{+\infty}a_jz_j^n=0$ for all $n$ as "the sequence of partial sums converges and its limit is $0$". In particular, it implies that $\{a_j\}$ is convergent to $0$, hence $\{|a_j|\}$ is in particular a bounded sequence, say by $M$. We show by induction that $a_j=0$ for all $j$. First, we have for all integer $p$ that 
$$|a_0z_0^p|\leqslant\sum_{j\geqslant 1}|a_j||z_j|^p$$
so 
$$|a_0|\leqslant \sum_{j\geqslant 1}|a_j|\left(\frac{|z_j|}{|z_0|}\right)^p.$$
Let $b_{j,p}:=|a_j|\left(\frac{|z_j|}{|z_0|}\right)^p$. We have $|b_{j,p}|\leqslant |a_j|\left(\frac{|z_j|}{|z_0|}\right)^p$, and as $\{a_j\}\in\ell^1$, we can conclude by monotone convergence theorem that $a_0=0$. 
Now assume that $a_0=\dots=a_n=0$. Then 
$$|a_{n+1}|\leqslant \sum_{j\geqslant n+2}|a_j|\left(\frac{|z_j|}{|z_{n+1}|}\right)^p,$$
and an other application of monotone convergence theorem yields $a_{n+1}=0$. 
A: Suppose that the statement is not true, and let $N$ be the smallest index such that $a_N \neq 0$. 
Then $-a_N z_N^n= \sum_{i > N} a_i z_i^n \neq 0$ for all $n$. Then
$0 < |a_N| \leq |\sum_{i \geq N} a_i \frac{z_i^n}{z_N^n}| 
 \leq \sum_{i > N}|a_i|(|z_i|/|z_N|)^n$.
Taking $n \to \infty$ in this strict inequality would yield $0 < 0$, a contradiction.
