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In their article "Calculations Related to Riemann's Prime Number Formula", Riesel and Gohl claim a simplification of Riemann's evaluation of the modified prime counting function \begin{align} \pi_{0}(x) & = \tfrac{1}{2} \lim_{\epsilon \to 0} \pi(x + \epsilon) + \pi(x - \epsilon) \\ & = \lim_{N \to \infty} R_{N}(x) - F_{N}(x) \end{align} where $F_{N}(x)$ is an infinite sum which depends only on the complex zeros of the zeta function in the critical strip and accounts for the discontinuities of $\pi_{0}$ at the primes, while \begin{align} R_{N}(x) = \sum_{n \geq 1} \frac{\mu(n)}{n} \mathsf{li}(\sqrt[n]{x}) + \frac{1}{2 \log x} \sum_{m = 1}^{N} \mu(n) + \frac{1}{\pi} \arctan \frac{\pi}{\log x} + \epsilon_{N}(x) \end{align} is a smooth function of $x$, and $\epsilon_{N} \to 0$ as $N \to \infty$. The first term is Riemann's well-known approximation to $\pi_{0}$, and the remaining terms depend only on the trivial zeros of the zeta function (at the negative even integers). One often finds the approximation, \begin{align} \pi_{0}(x) \approx \sum_{n \geq 1} \frac{\mu(n)}{n} \mathsf{li}(\sqrt[n]{x}) - \frac{1}{ \log x} + \frac{1}{\pi} \arctan \frac{\pi}{\log x} \end{align} usually attributed to them. What of the factor of $\frac{1}{2}$ and sum involving the Möbius function; doesn't this sum diverge?

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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').

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Thank you for your answer. How is $1.$ conjectural? One can replace the summation with the inverse zeta function and evaluate it at $s = 0$. The estimate of Merten's function precludes the conjecture anyway, no? – user02138 May 29 '12 at 20:28
@user02138: Well if you look at the picture of the Mertens function you'll notice that the function is (randomly) oscillating with increasing amplitude so that it is very probably divergent (I am not saying that a kind of 'pseudo-limit' isn't $-2$ but a limit requires convergence that seems unlikely here...) – Raymond Manzoni May 29 '12 at 20:39

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