Convergence of the series $ \sum \dfrac{\sigma(n)}{n^{\alpha}} $

Question

Let $ \sigma $ be a permutation of $ \mathbb{N}^\star $. I must find for which real numbers $ \alpha $ the series $ \sum \dfrac{\sigma(n)}{n^{\alpha}} $ converges.

I showed that it does not converge when $ \alpha \leq 2$, but I don't know how to do in the other cases. I think it converges.

Is there a similar result for $\alpha$ complex ?

Answer 1

you seem to be ignoring the permutation. Define one where $$ \sigma( 2j) = j! $$ while all the other numbers are placed, in order, at the $n = 2j-1.$

With this permutation, no matter what $\alpha,$ the series diverges.

Let's see, $$ 3,1 , 4, 2, 5,6, 7, 24, 8, 120, 9, 720, 10, 5040, 11, 40230,12, 362880, 13, ... $$

Answer 2

For each $\alpha\in\mathbb{R}$, there is a permutation $\sigma$ of $N^*$ such that the series ${\large{\sum \!{\frac{\sigma(n)}{n^{\alpha}}}}}$ diverges.

Fix $\alpha\in\mathbb{R}$, and let $b=\max(2,\left\lceil{\alpha}\right\rceil)$.

Let $S=\{n^b\mid n\in N^*\}$, and let $T=N^*{\,\setminus\,}S$.

Let $p:N^*\to T$ be a bijection.

Define $\sigma:N^*\to N^*$ by $$ \sigma(n)= \begin{cases} \left(\frac{n}{2}\right)^b&\text{if $n$ is even}\\[4pt] p\bigl(\frac{n+1}{2}\bigr)&\text{if $n$ is odd}\\ \end{cases} $$ Then $\sigma$ is a permutation of $N^*$, but the terms of the series ${\large{\sum \!{\frac{\sigma(n)}{n^{\alpha}}}}}$ don't approach zero, hence the series diverges.