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Sum over Rational numbers m/n

  1. Mar 30, 2012 #1
    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 ??
  2. jcsd
  3. Mar 30, 2012 #2
    I would think so, as the rationals are countable.
  4. Mar 31, 2012 #3


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    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.
  5. Mar 31, 2012 #4
    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]
  6. Mar 31, 2012 #5
    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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