$ \ \ $ I want to ask an estimation of $\sum_{1 \leq n\leq x} \mu(n)n^{-1}$. According to a paper: http://arxiv.org/pdf/0908.4323v5.pdf of Terry Tao (See the theorem 1.3 on page 4 if you want), for an arbitrary set $\mathcal{P}$ which consists of primes (finite of infinite),
$$ \sum_{n \in \mathcal{P} \atop n\leq x}\frac{\mu(n)}{n}=\bigg(\prod_{p \in \mathcal{P}}1-\frac{1}{p}\bigg)+o(1),$$ where the part of small-o depends on the set $\mathcal{P}$. But the part of product be $0$ when $card (\mathcal{P})=\infty$, so the equation depends on the part of small-o.
$ \ \ $ Then, if we choose the set as the set of the primes,
\begin{align} \sum_{n \in \mathcal{P} \atop n\leq x}\frac{\mu(n)}{n}&=c_1\bigg(\prod_{p \leq x}1-\frac{1}{p}\bigg) \ \ \ ? \\ \text{i,e.}\\ &=\frac{c_2}{\log x}+o\Big(\frac{1}{\log x} \Big) \ \ ? \ \ (\text{for some constant $c_1, c_2$.}) \end{align}