Basically the problem asks to show that $$\Sigma _{n \leq x} \mu(n) \lfloor \frac{x}{n} \rfloor ^2 = \frac{x^2}{\zeta(2)} + \mathcal{O}(x\log(x))$$ where $x$ is a real number and $n$ integer

I might be wrong but I showed that $\lfloor \frac{x}{n} \rfloor = \frac{\lfloor x \rfloor}{n} $, since if $x = x_0 +x_1$ where $x_0$ is the integer part of $x$ and $x_0 = s+t$ where $s$ is the largest number less than $x$ that is divisble by $n$ then $\lfloor \frac{x}{n} \rfloor = \lfloor \frac{x_0}{n}+\frac{x_1}{n} \rfloor=\frac{s}{n}+ \lfloor \frac{t}{n}+\frac{x_1}{n} \rfloor \lt \frac{s}{n}+\lfloor \frac{n-1}{n} + \frac{1}{n}\rfloor = \frac{s}{n}+1$ thus $\lfloor \frac{x}{n} \rfloor= \frac{s}{n}=\frac{\lfloor x \rfloor}{n}$

and so $\Sigma _{n \leq x} \mu(n) \lfloor \frac{x}{n} \rfloor ^2 = \lfloor x \rfloor ^2 \Sigma _{n \leq x} \frac{\mu(n)}{n^2}= \lfloor x \rfloor^2(\frac{1}{\zeta(2)}+\mathcal{O}(\frac{1}{x}))) = \frac{x^2}{\zeta(2)}+\mathcal{O}(x\log(x))$

I just want to check that my proof is alright and I thought there might be some better bound than $\mathcal{O}(x\log(x)$ it seems to me that it should be much smaller than this but I couldn't find any other expression that doesn't include $\lfloor x \rfloor$


You are on the right track, but some things are not quite right. For instance, it is not true in general that $\lfloor x/n \rfloor = \lfloor x \rfloor /n$.

We have $\lfloor x/n \rfloor = x/n + O(1)$, so for $n \leq x$ one has $$\lfloor x/n\rfloor^2 = \frac{x^2}{n^2} + O \left( \frac{x}{n} \right).$$ After this it should be clear what to do.


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