Ufyukyu

3 1 $begingroup$ The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$ We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above. I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem. elementary-number-theory mobius-inversion share | cite | improve this question asked Jan 27 at 10:39 windircurse windircurse ...