[Blank lines to avoid giving things away] On Sunday 03 December 2006 23:08, Fred lunnon wrote:
I have to admit I've still no idea why
\sum \mu(k) / k^2 = 6 / \pi^2
I'm not sure whether "no idea why" means "no idea how to prove" or "no brief argument making it obvious that". I'll assume the former, because if you mean the latter then I "have no idea" either :-). I'd get there via sum 1/k^2 = pi^2/6, and I guess you would too. I'm not sure which portion is troubling you. There's no really short self-contained proof for sum 1/k^2 = pi^2/6, but if you don't mind assuming a bit of the theory of Fourier series then you can get there quite quickly via Parseval's theorem. To get from there to sum mu(k)/k^2 = 6/pi^2, note that it's enough to prove [sum mu(k)/k^2][sum 1/k^2] = 1. This is standard Dirichlet series stuff (LHS is the case s=2 of the product of the Dirichlet series for mu(k) and for 1, which is the "Dirichlet convolution" of those series, etc.), but it's easy enough from first principles: expand the product in the LHS to get sum {k,l} of mu(k)/(k^2.l^2); change variable via n=kl to get sum {n} of 1/n^2 . sum {k|n} mu(k); the inner sum is well known (and easily proved) to be 1 when n=1 and 0 otherwise. -- g