If for any $k$, $\sum\limits_{n=0}^\infty a_n^k=\sum\limits_{n=0}^\infty b_n^k$, then$(a_n)=(b_{σ(n)}),\ σ \in{\mathfrak S}_{\mathbb N}$ Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \  a_n, b_n \neq 0$$
Show that $(\forall  k \in \mathbb N^*,  \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})$
The vector-space should be a finite-dimensional $\mathbb R$-vector-space. 
I do not see how to deal with the hypothesis. If you have ideas/hints...
It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases.
 A: Assume without loss that $|b_i|\leq |b_0|$ for all $i>0$. 
You have 
$$\sum \left(\frac{a_i}{b_0}\right)^{2k}=\sum \left(\frac{b_i}{b_0}\right)^{2k}\to n$$
Where $n$ is the number of $b_i^2$'s equal to $b_0^2$.
Now let $k\to \infty$. The right hand side limits to $n$. The left hand side can go to $n$ if and only if for there are $n$, $i$'s such that $a_i^2=b_0^2$.
Now, to get the signs 
$$\sum \left(\frac{a_i}{b_0}\right)^{2k+1}=\sum \left(\frac{b_i}{b_0}\right)^{2k+1}$$ and both sides can have the same limits as $k\to \infty$ iff there are the same number of $1$,s as $-1$$'s.
A: Here is an approach which works for sequences in $\Bbb C$, obtained
by a small variation of Sangchul Lee's answer at $a_n =0$ for all $n \in \mathbb N$ if $\sum a_n^k =0$ for all integers $k \ge 1$.
Assume that for some $w_0 \in \Bbb C$
$$ \tag{*}
 \# \{ j \mid a_j = w_0 \} \ne \# \{ j \mid b_j = w_0 \} \, .
$$ 
and set $z_0 = 1/w_0$.
Choose any $R > |z_0|$.
We have that $ |a_n|R \leq \frac{1}{2}$
and $ |b_n|R \leq \frac{1}{2}$ for all sufficiently large $n \ge N$,
and therefore
$$
 \left| \frac{a_n}{1 - a_n z} \right| \leq 2|a_n| \, , \quad \left| \frac{b_n}{1 - b_n z} \right| \leq 2|b_n| \, .
$$
for $|z| < R$ and $n \ge N$.
Then
$$
 f(z) = \sum_{n=1}^{N-1} \left( \frac{a_n}{1 - a_n z} - \frac{b_n}{1 - b_n z}\right) 
 + \sum_{n=N}^{\infty} \left( \frac{a_n}{1 - a_n z} - \frac{b_n}{1 - b_n z}\right) 
$$
is well-defined in $B_R(0)$ as the sum of a rational function and a uniformly converging series of analytic functions, and a meromorphic
function in that disk (with at most finitely many poles).
Now $(*)$ implies that $f$ has a pole at $z_0$.
On the other hand,
$$
 \frac{f^{(k)}(0)}{k!} = \sum_{n=1}^{\infty} (a_n^{k+1} - b_n^{k+1}) = 0
$$
for each $k$, so that $f$ is identically zero, which is a contradiction.
