Riesel and Gohl's Approximation of the Modified Prime Counting Function, $\pi_{0}$

Solution 1:

They used the classical formula $$\frac 1{\zeta(s)}=\sum_{n\ge1} \frac{\mu(n)}{n^s}$$ and applied it with different values of $s$ to get :

1. $\sum_{n\ge1} \mu(n)=-2\ $ ($s=0$ and using $\zeta(0)=-\frac 12$) (warning : conjectured only!)
2. $\sum_{n\ge1} \frac{\mu(n)}{n}=0\ $ ($s=1$ and since $\frac 1{\zeta(1)}=0$)

3. $\sum_{n\ge1} \frac{\mu(n)\log(n)}{n}=-1\ $ (at $s=1$ with $\left(\frac 1{\zeta(s)}\right)'_{s=1}=1$ and $\left(\frac 1{n^s}\right)'_{s=1}=-\frac{\log(n)}n$)

   The transformation used was the first one in your case (with $+$ in front of the initial $\frac 1{\log(x)} \sum \mu(n)$ I think).

   But this doesn't seem satisfying for a proof since Hardy wrote about (2) that it was as "deep" as the prime number theorem (in "Ramanujan" page 24, I think that (2) was proved by Landau, for a recent review see Terence Tao's "A remark on partial sums involving the Möbius function").

   Concerning your first sum this wikipedia article about the Mertens function should be helpful.
   If we define the Mertens function as $M(n)= \sum_{k=1}^n \mu(k)$ then $M(n)=\operatorname{O}\left(n^{\frac 12+\epsilon}\right)$ for every $\epsilon >0$ is equivalent to the Riemann Hypothesis (Anderson 'On the Möbius function').