# Möbius sums and Eulers totient function

Let $\phi(a)$ be eulers totient function, and $\mu(k)$ be the Möbius function, how can I prove that for all a,

$$\phi(a)a=\sum_{\substack{\gcd(a,r)=1,\\1\leq r\leq a}}(\sum_{k=0}^\infty\frac{-\mu(ak+r)ln(ak+r)}{(ak+r)})$$

Also maybe take note on the special case $$1=\sum_{k=1}^\infty\frac{-ln(k)\mu(k)}{k}$$

I dont think this should be hard to prove with some tools from elementry number theory, so please if you can, try to avoid introducting dirichlet characters or other such things.

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An infinite sum, involving logarithms, and you want to do it by elementary number theory? Color me skeptical. –  Gerry Myerson Jan 11 '13 at 1:59
It seems to me that the special case has a lower order of complexity. Recall that $$\sum_{n\ge 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$$ so that we may suppose that (not a formal proof) $$\sum_{n\ge 1} \frac{-\log(n)\mu(n)}{n^s} = -\frac{\zeta'(s)}{\zeta(s)^2} \sim 1 - 2\gamma (s-1)$$ Setting $s=1$ then yields $$\sum_{n\ge 1} \frac{-\log(n)\mu(n)}{n}=1.$$ So this indicates that your conjecture for the special case is correct. –  Marko Riedel Jan 11 '13 at 2:11
Yes I know this derivation, you can get explict formula for all mobius function dirichlet series summed over a specific congruence in terms of the hurrwitz zeta function. –  Ethan Jan 11 '13 at 2:14