$a_n =0$ for all $n \in \mathbb N$ if $\sum a_n^k =0$ for all integers $k \ge 1$

Suppose that $\sum a_n$ is an absolutely converging complex series and that for all integers $k \ge 1$, $\sum a_n^k =0$.

How to prove that the $a_n$ are all equal to $0$?

I was looking to extend Newton's identities to infinite number of variables, but I'm not able to write a simple proper proof!

Assume that the conclusion is wrong, and let $m$ be the smallest index such that $a_m \ne 0$. Without loss of generality we can assume that $m=1$, i.e. $$ \tag 1 a_1 \ne 0 \, . $$

Let $0 < \varepsilon < \frac 12|a_1|$. $\sum_{n=1}^\infty a_n$ is absolutely convergent, therefore there is an integer $N$ such that
$$ \sum_{n=N+1}^\infty |a_n| < \varepsilon \, . $$ and in particular $|a_n| < \varepsilon$ for $n > N$. It follows that for all integers $k$, $$ \sum_{n=N+1}^\infty |a_n|^k \le \sum_{n=N+1}^\infty \varepsilon^{k-1}|a_n| = \varepsilon^{k-1} \sum_{n=N+1}^\infty |a_n| < \varepsilon^k \, . $$

Since $\sum a_n^k =0$, the $k$-th power sums $$ p_k := p_k(a_1, \ldots, a_N) = \sum_{n=1}^N a_n^k $$ satisfy $$ \tag 2 |p_k| = \bigl| \sum_{n=N+1}^\infty a_n^k \, \bigr| \le \sum_{n=N+1}^\infty |a_n|^k < \varepsilon^k $$ for $1 \le k \le N$.

Now let $$ e_k := e_k(a_1, \ldots, a_N) $$ be the $k$-th elementary symmetric polynomial in the variables $a_1, \ldots, a_N$.

Then $e_0 = 1$, and Newton's identities state that $$ \begin{aligned} e_1 &= p_1 \\ 2e_2 &= e_1 p_1 - p_2 \\ 3e_3 &= e_2 p_1 - e_1 p_2 + p_3 \end{aligned} $$ and generally $$ \tag 3 k e_k = \sum_{i=1}^k (-1)^{i-1} e_{k-i} p_i \, \text{ for } k \ge 0 \, . $$

From $(2)$ and $(3)$ it follows easily by induction that the elementary symmetric polynomials satisfy $$ |e_k| \le \varepsilon^k \text{ for } 0 \le k \le N \, . $$

Now define $$ P(x) = (x-a_1)(x-a_2) \cdots (x-a_N) \\ = x^N - e_1 x^{N-1} + e_2 x^{N-2} \cdots \pm e_N \, . $$ Then $P(a_1) = 0$ and therefore $r := |a_1|$ satisfies $$ r^N \le |e_1| r^{N-1} + |e_2| r^{N-2} + \cdots + |e_N| \\ \le \varepsilon r^{N-1} + \varepsilon^2 r^{N-2} + \cdots + \varepsilon^N \\ = \varepsilon r^{N-1} \bigl( 1 + \frac{\varepsilon}{r } + \cdots + (\frac{\varepsilon}{r })^{N-1} \bigr) $$ or $$ 1 \le \frac{\varepsilon}{r } \bigl( 1 + \frac{\varepsilon}{r } + \cdots + (\frac{\varepsilon}{r })^{N-1} \bigr) \, . $$ $\varepsilon$ was chosen such that $0 < \frac \varepsilon r < \frac 12$, therefore $$ 1 < \frac 12 \bigl( 1 + \frac 12 + \cdots + (\frac 12)^{N-1} \bigr) < 1 $$ which is a contradiction.

So the initial assumption is wrong, and it is proven that all $a_n$ are zero.

I am not sure if you also want to look at other proofs, but here is a complex-analytic proof anyway.

Define $f(z)$ by the following series

$$ f(z) = \sum_{n=1}^{\infty} \frac{a_n}{1 - a_n z}. $$

Now we make the following claim.

Claim. For $f(z)$ the followings are true:

• $f(z)$ is holomorphic near $z = 0$.
• $f(z)$ is meromoprhic on $\Bbb{C}$ and the termwise differentiation is applicable.

Assume first that this claim is true. Then notice that for any $k \geq 0$ we have

$$ f^{(k)}(0) = k! \sum_{n=1}^{\infty} a_n^{k+1} = 0. $$

This shows that $f(z) \equiv 0$ and in particular, $f(z)$ has no poles. On the other hand, if $a_m \neq 0$ for some $m$, then $f(z)$ should have a simple pole at $z = 1/a_m$. Indeed, we should have

$$ \underset{z=1/a_m}{\operatorname{Res}} f(z) = -(\# \{ n : a_n = a_m \}) \neq 0. $$

This contradiction implies that $a_n = 0$ for all $n$. ////

Proof of Claim. Both claims are easily verified if we prove the following claim: for any $R > 0$, there exists $N$ such that the tail sum

$$ T_{N}(z) := \sum_{n \geq N} \frac{a_n}{1-a_n z} $$

converges uniformly on $|z| \leq R$. Indeed, let $R > 0$ be arbitrary. Since $\sum |a_n| < \infty$, we can choose $N$ such that $|a_n|R \leq \frac{1}{2}$ for all $n \geq N$. Then on $|z| \leq R$

$$ \left| \frac{a_n}{1 - a_n z} \right| \leq 2|a_n|. $$

This proves that $T_N(z)$ converges uniformly by the Weierstrass M-test. ////

WLOG assume that $r=\max_{n\ge 1}|a_n|<1$ (otherwise do scaling). Define $$ f(z)=\sum_{n=1}^{\infty}\frac{a_n}{z-a_n}. $$

1. It converges uniformly absolutely to a bounded analytical function in $|z|\ge 1$ due to the estimate $$ \sum_{n=1}^{\infty}\frac{|a_n|}{|z-a_n|}\le \frac{1}{1-r}\sum_{n=1}^{\infty}|a_n|<+\infty. $$
2. Inside the unit disc the Cauchy integral gives $$ \frac{1}{2\pi i}\int_{|z|=1}z^{k-1}f(z)\,dz=\sum_{n=1}^\infty a_n^k=0,\quad\forall k\ge 1. $$
3. Hence (a slight modification to include $n=0$ of) F. and M. Riesz theorem implies that $f\in H_0^1$, that is, $f$ is analytical inside the disc and $f(0)=0$.
4. Finally, Liouville theorem makes $f\equiv 0$ and, thus, all $a_n=0$.