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Thinking about Riemann's rearrangement theorem, I asked myself the following question: for any real number $\alpha >1$ fixed, are there two integers $1 \leqslant M < N$ and a sequence $(\varepsilon_n)_n \in \{-1,0,1\}^{\mathbb{N}}$ such that $$ \sum_{k=M+1}^N \frac{\varepsilon_k}{k^{\alpha}} = \frac{1}{M^{\alpha}} \, ? $$

I have tried to bound the sum as in the proof of the series-integral theorem, but it doesn't seem to be enough. For example, by taking $\alpha=2$, $M=5$ and $N=12$, the sum exceeds the right member ... But is it possible to have equality?

It seemed obvious to me that it is not, but I can't prove it. Do you have an idea please? Thank you in advance! :)

Kermatoni
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Have a look at this (beautiful) paper, it gives some interesting conditions. For example, corollary $A$ shows that for $\alpha = 2$, there are actually loads of solutions.

TheSilverDoe
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  • Thank you for this article! Corollary 1 is indeed very close to answering my question. I'm just trying to convince myself that (trivially) starting the sum at 1 in the article or starting at M+1 (with $M\geqslant 1$) as in my question don't change anything :) – Kermatoni Nov 12 '20 at 22:24
  • In addition, my sequence can also take the value -1 – Kermatoni Nov 12 '20 at 22:36
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    Actually, your question is equivalent to : does there exists $M<N$ and $\epsilon_n$ such that $$\sum_{n=M}^N \frac{\epsilon_n}{n^{\alpha}}=0$$ You just want to know if there is a "block" of the sum which sums to $0$, eventually changing the signs and omitting some terms. The given article gives stronger results (in some cases) because it answers to the same question as yours, except that it allows only one $-1$ sign (equivalently, it expresses one of the terms with all the others having a non-negative sign). – TheSilverDoe Nov 13 '20 at 07:50
  • Okay, thank you very much ! :) – Kermatoni Nov 13 '20 at 11:07