# Möbius sums and Eulers totient function

Let $\phi(a)$ be Euler's 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 don't think this should be hard to prove with some tools from elementary number theory, so please if you can, try to avoid introducing Dirichlet characters or other such things.

• 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