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What is the value of

$$\sum_{x\in\mathbb{Q}\setminus\{0\}, |x|<1}(\mathrm{denominator}\;\; \mathrm{of}\;\; x)^{-2}?$$

(the denominator of a nonzero rational number $x$ is defined to be $b$ where $x=a/b$ with $a\in\mathbb{Z}\setminus\{0\}$, $b\in\mathbb{N}_{>0}$, and $a,b$ are relatively prime).

Or, is there a "nice" expression for the above sum?

Edit: Forgot to write a hypothesis. I'm considering only those rationals with $|x|<1$.

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Does this series even converge in some sense? –  DonAntonio Oct 8 '12 at 12:16

3 Answers 3

up vote 1 down vote accepted

Note that if you replace the exponent -2 by -3, then one can show your sum, over nonzero rationals of absoloute value <= 1, ends up converging, to a number bounded above by

$$2 \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{3}.$$

To see this note that each denominator $n$ appears in $2\phi(n)$ of the terms, and use that $\phi(n) \le n$ for $n \ge 1$. So the contribution from denominator n is at most $2n/(n^3)=2/n^2$. The extra 2 is from also counting the negative numbers in [-1,0).

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Great. Thank you! –  Simplicius Oct 26 '12 at 12:27

The series diverges. Consider when $x$ ranges over just those rational numbers whose denominator is 1, which is strictly less than your sum here.

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Of course. Wait... I'm going to edit my question, as I forgot writing the complete hypothesis (I was actually considering only rationals with $|x|<1$ but I forgot writing it!) –  Simplicius Oct 8 '12 at 12:26

The sum of the reciprocals of the primes diverges, and you can use that to prove that even if you only take the rationals with prime denominator, you get divergence.

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Fair enough. Accepted! (I was to hasty in writing the question. In fact, I was trying to invent a nice "exercise" that involved a sum over the rationals and their denominators. Appearently, I failed. :) ) –  Simplicius Oct 8 '12 at 12:39
    
It is funny that you, Simplicius, got a simple answer! :P –  000 Oct 8 '12 at 22:10
    
Even if this answer is ok, I think I should accept coffeemath's answer instead, because it's more complete. –  Simplicius Oct 26 '12 at 12:28

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