I found the following exercise in a real analysis book:
If $r$ is a positive rational number, it can be uniquely expressed as $r=\frac pq$ with $p,q$ comprime, non-zero natural numbers. Let $d(r)=q$. Is the sum $$\sum_{r\in \mathbb Q \cap ]0,1[} \frac{1}{(d(r))^3}$$ finite?
It easy to prove that the answer is yes by observing that
$$\sum_{r\in \mathbb Q \cap ]0,1[} \frac{1}{(d(r))^3}=\sum_{p Question: Is it actually possible to explicitly compute the value of this generalized series?
Recall that $\#\left\{m\mid\gcd(m,n)= 1, 1 \le m \le n\right\} = \phi(n)$. Also note that $\phi(n)$ is multiplicative and that $\phi(p^k) = (1-p^{-1})p^{k}$ so that $$\sum_{r\in \mathbb Q \cap ]0,1[} \frac{1}{d(r)^3} = \sum_{n=1}^\infty \frac{\phi(n)}{n^3} = \prod_p \left(1+\sum_{k=1}^\infty \frac{\phi(p^k)}{p^{3k}}\right)=\prod_p \left(1+\sum_{k=1}^\infty (1-p^{-1}) p^{-2k}\right)$$ $$ = \prod_p \left(1+\frac{(1-p^{-1})p^{-2}}{1-p^{-2}}\right)= \prod_p \frac{1-p^{-3}}{1-p^{-2}}= \frac{\zeta(2)}{\zeta(3)}$$