Evaluate $\sum\limits_{(m,n) \in D,m < n} \frac{1}{n^2 m^2} $ where $\gcd(m,n)=1$ i have no clue on how to evaluate:
$$\sum\limits_{(m,n) \in D,m < n} \frac{1}{n^2 m^2} \text{ where }D = \{ (m,n) \in (\mathbb{N}^*)^2 \mid \gcd(m,n) = 1\} $$
If someone is able to give me a hint...
Thanks much in advance.
 A: Call your sum as $S$. Then $S= \frac{1}2(\displaystyle\sum_{(m,n)=1}\frac{1}{m^2n^2}-1)$$$2S+1=\displaystyle\sum_{m,n}\displaystyle\sum_{d|(m,n)}\frac{\mu(d)}{m^2n^2}=\displaystyle\sum_{n=1}^\infty \frac{\mu(n)}{n^4}\displaystyle\sum_{r,s}\frac{1}{r^2s^2}=\frac{\zeta(2)^2}{\zeta(4)}$$
Thus, $S=\frac{3}{4}$
A: There is an inclusion-exclusion approach to this. First, note that if $$S_d:=\sum_{m,n=1}^\infty \frac{1}{(dm)^2(dn)^2} = \frac{1}{d^4} \zeta^2(2)$$
Then a sum related to your sum:
$$S:=\sum_{\;m,n=1\\(m,n)=1}^\infty \frac{1}{m^2n^2} 
= \sum_{d=1}^\infty \mu(d) S_d = \zeta(2)^2\sum_{d=1}^\infty \dfrac{\mu(d)}{d^4}=\dfrac{\zeta^2(2)}{\zeta(4)}$$
Your sum is $\frac{S-1}{2}$.
The key is showing that $S=\sum \mu(d)S_d$.
The answer is rational, btw, so there might be an approach which is much easier.
If you replace the terms with $\dfrac{1}{(mn)^s}$ with $\mathcal{Re}\; s>1$, then $S$ is $\dfrac{\zeta^2(s)}{\zeta(2s)}$.
The product formula for $\zeta$ gives us the product form:
$$\prod_p \frac{p^s+1}{p^s-1}$$
Perhaps there is an easier way to see that the above sum is this product
More generally, if $s,t>1$ then $\sum_{(m,n)=1} \dfrac{1}{m^sn^t} = \dfrac{\zeta(s)\zeta(t)}{\zeta(s+t)}$. But that equation isn't symmetric, so you couldn't use it to get the case of $m<n$.
A: Just an observation which is far from an answer.
$$\frac{\pi^2}{6}-1=\sum_{n> 1}\frac{1}{n^2}<\sum_{(m,n):\gcd(m,n)=1,\ m<n}\frac{1}{m^2n^2}\le \left(\sum_{n\ge 1}\frac{1}{n^2}\right)^2=\frac{\pi^4}{36}$$
