Hey all,

I'd like to know if anyone has a link to a proof of the following?

Take two infinite sequences $a_n$ and $b_n$ such that $$\sum_{n=1}^\infty a_n^s=\sum_{n=1}^\infty b_n^s=finite$$

for all $s\in\mathbb{N}$. Then, after some rearrangement, $a_i=b_i$ for all i? I have my own proof which I am pretty sure is correct and I'd very much like to know if such a proof has been given before (to see if it is any different)

Thanks in advanced for any help :)

  • 1
    $\begingroup$ Out of curiosity: is the question by chance inspired by one of the problems on the latest Putnam? $\endgroup$ – t3suji Dec 10 '10 at 15:05
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    $\begingroup$ @Joe : I think you need some additional assumption. Are the $a_n,b_n$ also supposed to be positive? Otherwise, you can start with, say, $a_n=1/n^2$, and add infinitely many zeros in any order you want, to form the $b_n$. $\endgroup$ – Andrés E. Caicedo Dec 10 '10 at 15:46
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    $\begingroup$ I don't know about prior art, but here is a simple proof (assuming they are positive as per Andres Caicedo's comment). Rearrange in descending order, find the first mismatch and remove the preceding terms. Now $a_1>b_1\ge b_i$ for all $i$. Multiply by a constant so that $a_1=1$. Now the first sum is at least 1 but the second goes to 0 as $s\to\infty$. $\endgroup$ – Sergei Ivanov Dec 10 '10 at 16:01
  • $\begingroup$ @t3suji: Not at all, it was of my own interesting @Andres and Sergi: Yes, I would have been better specifying the summation $\sum_{n}\frac{1}{a_n^s}$ and $\sum_{n}\frac{1}{b_n^s}$ and giving the condition that all $a_n,b_n$ must lie outside the unit disk in the complex plane so as to ensure that the summations are finite $\endgroup$ – backstoreality Dec 10 '10 at 16:51

If $\mathbb{N}$ is the set of positive integers, then it is formally false, we may add arbitrary number of zero terms to our series. But if all $a_i$ and $b_i$ are different from 0 complex numbers, and the series converge absolutely, then the answer is yes. Just rearrange both sequences so that $|a_1|\geq |a_2|\geq |a_3|\geq \dots$ and the same for $b_i$'s. Note that $\limsup |\sum a_i^s|/|a_1|^s$ equals $\max\{k: |a_k|=a_1\}$. It follows that $|b_1|=|a_1|$ and both sequences have the same number of terms with absolute value equal to $|a_1|$. Denote the number of such terms by $k$, let them be $a_1,\dots,a_k$ and $b_1,b_2,\dots,b_k$ respectively. I claim that $A:=\{a_1,\dots,a_k\}=\{b_1,\dots,b_k\}=:B$ in the sense of multi-set equalities. Indeed, without loss of generality we may suppose that $|a_1|=1$. Choose a sequence $\{s_i\}$ such that $a_m^{s_i}$ and $b_m^{s_i}$ tends to $1$ for all $m\in \{1,2,\dots,k\}$. Such sequence does exist by Kronecker lemma. Then for fixed postive integer $k$ one has $\lim_j \sum_i a_i^{s_j+p}=\sum_{i=1}^k a_i^p$, the same for $b$'s. That is, power sums of arrays $A$ and $B$ are equal, it follows that those arrays are equal. Then remove such arrays from both sequences and proceed the same way.

For real sequences, from convergence of $\sum a_i^2$ the absolute convergence of $\sum a_i^k$ for $k>2$ follows, and if we consider only $k$'s divisible by 3 we conclude that $\{a_i^3\}$ is a permutation of $\{b_i^3\}$, what desired.

I do not know the answer for conditionally convergent complex sequences.

  • $\begingroup$ @Fedor : Very nice. $\endgroup$ – Andrés E. Caicedo Dec 10 '10 at 16:03

Here is a link to a counterexample for conditionally convergent complex series.


Since for sequences $\|a\|_p\to \|a\| _\infty$, maximum term is determined by the values of $\sum_i a_i^s$. Arguing inductively on the powers sum of the remaining terms one concludes. (this assuming all $a_i$ are positive; and similarly if they have a sign).

  • $\begingroup$ That's quite slick. $\endgroup$ – Alex B. Dec 11 '10 at 4:46

This is just Sergei Ivanov's comment finished with Pietro Majer's answer. If all the terms are real and positive then rearrange both $a:=(a_n)$ and $b:=(b_n)$ so they are non-increasing. If $a \neq b$, then by removing the longest initial segment on which they agree we may assume that $a_1 \neq b_1$.

But now

$$ a_1 = \|a\|_\infty=\lim_{p \to \infty} \|a\|_p = \lim_{p \to \infty} \|b\|_p=\|b\|_\infty=b_1, $$

a contradiction.


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