I found this (supposedly) hard problem of the ENS concours online, but I didn't manage to solve it and found no solutions. I tried to use a resummation formula such as Abel's theorem, or compare this series with an integral but without any success.
Let $\sum_n a_n$ be an absolutely convergent (it means that $\sum_n \vert a_n \vert$ converges) series of complex numbers such that for any integer $k>0$,
$$\sum_{n\geq 0} (a_n)^k =0$$
Show that $a_i=0$ for every $i$.
Let $M:=\max_{i\geq 0} \vert a_i\vert $, which is finite since the series is convergent. Dividing the series by $M$, we can restrict ourselves to the case of a series with terms of modulus less than or equal to $1$.If $\vert a_n \vert <1$, then $a_n^{k} \rightarrow 0$ when $k\rightarrow \infty$. We therefore just need to get rid of the terms satisfying $\vert a_n \vert =1$. Let $$S:= \lbrace n\in \mathbb{N} ~:~\vert a_n\vert =1\rbrace.$$ Note that $S$ is finite since the series is convergent. Let $\varepsilon >0$ and $n_0 \in \mathbb{N}$ be such that $\sum_{n\geq n_0}\vert a_n\vert \leq \varepsilon$. Then, for any $n\geq n_0$ one gets
$$\vert \sum_{n\notin S} a_n^k\vert \leq \underbrace{\sum_{\substack{n\notin S \\ n<n_0}}\vert a_n^k\vert}_{\rightarrow 0~\text{when } k\rightarrow \infty \text{ since } n\notin S} + \varepsilon.$$ Therefore, $\vert \sum_{n\notin S} a_n^k\vert $ goes to $0$ as $k\rightarrow \infty$. Let $\lambda_1,...,\lambda_m$ be the listed values of the $a_i$ for which $i\in S$. We are left with distinct complex numbers satisfying $$\lim_{k\rightarrow \infty}\sum_{i=1}^{m}n_i \lambda_i =0$$ where the $n_i$'s are (strictly) positive integers. This is intuitively impossible, because without varying the $n_i$'s we are able to set this sum to $0$ when taking powers of the $\lambda_i$'s. The result is a consequence of the following result, which can be proved by induction on $m$.
Lemma : Let $m\in \mathbb{Z}_{>0}$, let $\lambda_1,...,\lambda_m$ be distinct complex numbers of modulus $1$, and let $z_1,...,z_m$ be complex numbers, such that $$\lim_{k\rightarrow \infty}\sum_{i=1}^{m}z_i \lambda_i^k=0.$$ Then, $z_1=z_2=...=z_m=0$.
\Edit : I am adding the proof of the lemma since it has been asked by the owner of the question. Let me know if more details are needed.
Proof of the Lemma : By induction. If $m=1$, the result is clear. Suppose that it holds for $m-1$, and that not all the $z_j$'s are zero. Without loss of generality, one can thus assume that $z_m \lambda_m=1$ (otherwise just divide by any non-zero term of the form $a_l\lambda_l$. Define $$f(k)=\sum_{j=1}^{m}z_j \lambda_j^k=1+\sum_{j=1}^{m-1}z_j \lambda_j^k$$ By hypothesis, $\lim_{k\rightarrow\infty} f(k) =0$. Therefore, the same holds for $g(k):=f(k)-f(k-1)$. But
$$g(k) = \sum_{j=1}^{m-1}z_{j}(1-\frac{1}{\lambda_j})\lambda_j^{k}\overset{k\rightarrow \infty}{\longrightarrow}0.$$ Since $\lambda_m=1$ and the $\lambda_i$'s are distinct, applying the induction hypothesis to $g(k)$ leads to $z_1=...=z_m=0$. This is a contradiction since $1\neq 0$, hence all the $z_i$'s must be zero.
