# Sum over Rational numbers m/n

1. Mar 30, 2012

### zetafunction

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 ??

2. Mar 30, 2012

### A. Bahat

I would think so, as the rationals are countable.

3. Mar 31, 2012

### HallsofIvy

Staff Emeritus
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.

4. Mar 31, 2012

### 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

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

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

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

5. Mar 31, 2012

### 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.