Sum over Rational numbers m/n

zetafunction

it is possible to evaluate sums over the set of Rational

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

in any case for a suitable function is possible to evaluate

[tex] \sum_{q} f(qx) [/tex] with f(0)=0 ??

A. Bahat

I would think so, as the rationals are countable.

HallsofIvy

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.

zetafunction

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

[tex] \sum_{m=-\infty}^{\infty}\sum_{p}f(p^{m}) [/tex]

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

[tex] \sum_{m=-\infty}^{\infty}\sum_{p}f(xp^{m}) [/tex]

A. Bahat

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.