Is Absolute Convergence Required for Evaluating Sums over Rational Numbers?

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SUMMARY

This discussion centers on the evaluation of sums over the set of rational numbers, specifically the expression \(\sum_{q} f(q)\) where \(q = \frac{m}{n}\) and \(m, n\) are positive integers. It is established that for suitable functions, sums like \(\sum_{q} f(qx)\) can be evaluated, particularly when \(f(0) = 0\). The conversation highlights the importance of the ordering of rational numbers and references the fundamental theorem of arithmetic to express integers as products of primes, leading to the exploration of sums over prime powers. A key conclusion is that all rearrangements of a series converge to the same value if and only if the series is absolutely convergent, which can significantly impact the sum's value.

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it is possible to evaluate sums over the set of Rational

so \sum_{q} f(q) with q= \frac{m}{n} and m and n are POSITIVE integers different from 0 ??

in any case for a suitable function is possible to evaluate

\sum_{q} f(qx) with f(0)=0 ??
 
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I would think so, as the rationals are countable.
 
However, in some cases the sum will depend on the ordering of the rational numbers given by the one-to-one correspondence with the positive integers.
 
um.. if i use the fundamental theorem of the arithmetic to express m and n as a product of primes could i write or consider at least series over prime or prime powers ? i mean

\sum_{m=-\infty}^{\infty}\sum_{p}f(p^{m})

in both case this sum is over prime and prime powers is this more or less correct ??

using suitable products of primes we can reproduce every positive rational can't we ?

so we can study 'invariant-under-dilation' formulae as follows

\sum_{m=-\infty}^{\infty}\sum_{p}f(xp^{m})
 
HallsofIvy is correct: all rearrangements of a series converge to the same value if and only if the series is absolutely convergent. So that can affect the sum.
 

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