Order of summation of Moebius function with summations of fractional parts as coefficients

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 $\displaystyle\sum_{i=0}^n\left(\mu(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\{jx\}\right)=\sum_{i=0}^n\left(\{jx\}\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}\mu(i)\right)$.

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Welcome to MSE! Do you have any thoughts and can share what you have tried? Regards – Amzoti Jun 19 '13 at 14:28
Switch the order, split the sum into two ranges, and use the bound $\sum_{n\leq x} \mu (n)\ll \frac{x}{\log x}$. Try adding some more to the question, explaining what you have tried, and why you are asking this question. In its current form, it is the way a problem would appearing on an assignment, and it is unlikely that anyone will give a complete response. – Eric Naslund Jun 19 '13 at 16:38
Thank you for your replies. I have added more information to my question. @Eric Thats a new inequality for me. I am going to try that out :). – biswajitsc Jun 20 '13 at 4:47
Using the bound $\sum_{n\le x} \mu(n)\ll \frac{x}{\log x}$, i get it to be $o(\mbox{Ei} (2 \log n))$ which is of a greater order than $O(n)$. – biswajitsc Jun 22 '13 at 4:45