I want to show that $\displaystyle\sum_{i=0}^n\left(\mu(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\{jx\}\right)=O(n)$ for $x\in (0,1)$. I have tried to use the result that $\displaystyle\sum_{i=0}^n\left(\mu(i)\frac{n}{i}\right)\le n$ but haven't got far. I have even tried substituting $\displaystyle \sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\{jx\}=\frac{n}{i} c_i$, $(c_i\in(0,1))$.
Also note that the sum can switched as $\displaystyle\sum_{i=0}^n\left(\mu(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\{jx\}\right)=\sum_{i=0}^n\left(\{jx\}\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\mu(i)\right)$.