My task is to prove the well-known identity
Here all variables are positive integers
I only know I should use mobius inversion formula, but how to proceed I am getting confusion, please any one can help me in this, and is there any article which shows the applications of mobius inversion formula.
$$\begin{aligned}\sum_{m,n} \frac{(m,n)^r}{m^s n^t} &= \sum_{d=1}^\infty \sum_{(m,n) = d} \frac{(m,n)^r}{m^s n^t} \\ &=\sum_{d=1}^\infty \sum_{(m,n) = 1} \frac{d^r}{(dm)^s (dn)^t} \\ &=\zeta(s+t-r) \sum_{(m,n) = 1} \frac{1}{m^s n^t} \end{aligned}$$ Let $$A_d = \sum_{(m,n) = d} \frac{1}{m^s n^t} $$ Note that $$\sum_{r|d} A_d = \frac{1}{r^{s+t}}\zeta(s)\zeta(t)$$ Therefore, $$A_1 = \sum_{r=1}^\infty \mu(r) \sum_{r|d} A_d = \zeta(s)\zeta(t) \sum_{r=1}^\infty \frac{\mu(r)}{r^{s+t}} = \frac{\zeta(s)\zeta(t)}{\zeta(s+t)}$$ completing the proof.