How do you find the value of $\displaystyle \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2}$
I'm having trouble getting the desired error term, but I'm close to solving the problem. Maybe you could polish my method to get what you want...
$\displaystyle \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2} = \sum\limits_{n \leq x} \mu(n)\left(\frac{x}{n}-\left\{\frac{x}{n}\right\}\right)^{2} $
$\displaystyle = \sum\limits_{n \leq x} \mu(n)\frac{x^2}{n^2}-2\sum\limits_{n \leq x} \mu(n)\frac{x}{n}\left\{\frac{x}{n}\right\}+\sum\l imits_{n \leq x} \mu(n)\left\{\frac{x}{n}\right\}^{2} $
$\displaystyle = x^2\sum\limits_{n \leq x}\frac{\mu(n)}{n^2}-2x\cdot O\left(\sum\limits_{n \leq x}\frac{\mu(n)}{n}\right)+O\left(\sum\limits_{n \leq x} \mu(n)\right) $
First sum: $\displaystyle x^2\sum\limits_{n \leq x}\frac{\mu(n)}{n^2} = x^2\sum\limits_{n=1}^{\infty} \frac{\mu(n)}{n^2} - x^2\sum\limits_{n > x}\frac{\mu(n)}{n^2} = \frac{x^2}{\zeta(2)} + O\left(x^2\sum_{n>x}\frac{1}{n^2}\right) \longleftarrow $ Here's where you need to improve the error term.
Second sum: $\displaystyle 2x\cdot O\left(\sum\limits_{n \leq x}\frac{\mu(n)}{n}\right) = o(x) $ since $\displaystyle \sum\limits_{n \leq x}\frac{\mu(n)}{n} = o(1) $ (which is equivalent to the prime number theorem!).
Third sum: $\displaystyle O\left(\sum\limits_{n \leq x} \mu(n)\right) = O(x) $ since $\displaystyle |\mu(n)|\leq 1 $.
Enumerating we get $\displaystyle \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2} = \frac{x^2}{\zeta(2)} + O(f(x)) + o(x) + O(x) = \frac{x^2}{\zeta(2)} + O(f(x)) $. Where $\displaystyle f(x) $ is the error term that you need to improve upon.
Dear
Chiph, sorry for the delay in writing. I hope you are aware of the Eulers summation formula which says:
If $\displaystyle f$ has a continuous derivative on the interval $\displaystyle [y,x]$ then
$\displaystyle \sum\limits_{y<n \leq x}f(n) = \int\limits_{y}^{x} f(t) \ dt + \int\limits_{y}^{x} (t-[t]) f'(t) \ dt + f(x)([x]-x) -f(y)([y]-y)$
Please apply $\displaystyle f(n)= \frac{1}{n^s} $ in the Eulers Summation formula to get the desired result.
All of these can be found in Tom Apostol's Analytic Number Theory book.
Just to point out a combinatorial interpretation of the sum $\displaystyle \sum\limits_{k \leq n} \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^{2}$
It's equal to the cardinal of $\displaystyle \left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that gcd}(x,y)=1\right\}
$ (1)
The explanation is simple:
Fix $\displaystyle n$, let $\displaystyle A_p$ be $\displaystyle \left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that } p|x \wedge p|y\right\}$ (for each prime $\displaystyle p\leq n$ ).
Now (1) is the set of pairs of integers in $\displaystyle [n]=\left\{1,2,...,n\right\}$, such that they - the pairs- do not belong to the union $\displaystyle \bigcup_{p\leq n}A_p$.
Note that $\displaystyle A_{i_1}\cup...\cup A_{i_m}=\left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that }(i_1...i_m)|x\wedge (i_1...i_m)|y\right\}$ -since we are working with primes- which cardinal is $\displaystyle \left\lfloor\frac{n}{i_1\cdot ...\cdot i_m}\right\rfloor^2$
The rest is inclusion-exclusion. Evidently you can substitute that $\displaystyle 2$ for any $\displaystyle m\in\left\{2,3,...\right\}$