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Consider the following problem:

Let $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ be real numbers such that $$a_1^k+\cdots a_n^k= b_1^k+ \cdots+b_n^k$$ for all $1\leq k\leq n$. Then there is a bijection $\varphi:\{1, 2, \ldots, n\}\to \{1, 2, \ldots, n\}$ such that $a_i=b_{\varphi(i)}$ for all $i$.

This can be proved using Newton's identities, which gives that the polynomials $p(x)=(x-a_1)\cdots(x-a_n)$ and $q(x)=(x-b_1)\cdots(x-b_n)$ are identical.

My question is, can this be said about countable sequences. More precisely, is the following true?

Let $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3,\ldots$ be real numbers such that $$a_1^k+a_2^k+a_3^k+\cdots = b_1^k+ b_2^k+b_3^k+ \cdots$$ are equal for all $k\geq 1$ (Here we insist that both sums are convergent for each $k$). Then there is a bijection $\varphi:\mathbb N \to \mathbb N$ such that $a_i=b_{\varphi(i)}$ for all $i$.

caffeinemachine
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1 Answers1

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Lemma. If $c_n\in\Bbb R$, $|c_n|<1$ for all $n$ and $s_k:=\sum_{n=1}^\infty c_n^k$ converges absolutely for all $k\ge 1$, then $\lim_{k\to\infty}s_k=0$.

Proof. The series with $k=2$ converges absolutely, hence so do all series with $k\ge 2$. Thus for $k>2$, we have $$ \left|\sum_n c_n^k\right|\le \sum_n|c_n|^k\le |c_1|^{k-2}\sum_nc_n^2\to 0.$$ $\square$

For $c>0$ and $m\in\Bbb N_0$, consider $$\tag1 \lim_{k\to\infty}\left(c^{-k}\sum_na_n^k-(-1)^km\right).$$ Using the lemma, we see that

  • For $c=\max |a_n|$, this limit exists iff and $m=\#\{\,n\mid a_n=-c\,\}$ and then it equals $\#\{\,n\mid a_n=c\,\}$.
  • For $c>\max|a_n|$, the limit exists only for $m=0$ and then it equals $0$.
  • For $c<\max|a_n|$, the limit does not exist for any choice of $m$.

Thus from merely considering the sequence $\{\sum_na_n^k\}_k$, we can read off the following properties of $\{a_n\}_n$: The maximal absolute value of its terms as well as the multiplicity of the positive and negative of this maximum. We conclude that $\{b_n\}_n$ has the same maximal absolute value with the same multiplicities by sign. Thus we can put these $\pm$maximal values of both sequences in bijection. After removing these $\pm$maximal terms, we repeat the process, thereby bijecting the terms with second-largest absolute value, and so on. In the limit, this gives us a permutation of $\Bbb N$ as desired.

Hagen von Eitzen
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