Why is $\sum_{j^2\leq \sqrt x} \mu(j)[\frac x {j^2}]=x\sum_{j^2\leq \sqrt x}\frac {\mu(j)} {j^2}+ O(\sqrt x)$?

Question

I was studying neighbourhood methods from Overholt's book of Analytic Number theory(P No 42). There to estimate $Q(x)=\sum_{n \leq x}\mu^2(n)$ they have used a statement that

$$\sum_{j^2\leq x} \mu(j)\left[\frac x {j^2}\right]=x\sum_{j\leq \sqrt x}\frac {\mu(j)} {j^2}+ O(\sqrt x).$$

I am not getting this statement as $[\frac x {j^2}]=\frac x {j^2}-\{\frac x {j^2}\}$ and $\{\frac x {j^2}\}\leq 1$. Can I get something from here?

Answer 1

Since $[x]=x-\{x\}$, we have $[x]=x+O(1)$, and so $$ \sum_{j^2\leq x} \mu(j)\left[\frac x {j^2}\right]= \sum_{j^2\leq x} \mu(j) \left(\frac{x}{j^2} + O(1) \right) =\sum_{j^2\leq x} \mu(j) \frac{x}{j^2} + O(\sqrt{x}) = x \sum_{j \le \sqrt{x}} \frac{\mu(j)}{j^2} +O(\sqrt{x}). $$

Answer 2

Writing $M(x)=\sum_{n\leq x}\mu(n)$ and then applying Abel's summation formula gives us that:

$$Q(x)=\sum_{j^2\leq x} \mu(j)\left\lfloor\frac x {j^2}\right\rfloor=\sum_{n\leq \sqrt{x}} \mu(n)\left(\frac{x}{n^2}-\left\{\frac{x}{n^2}\right\}\right)=x\sum_{n\leq \sqrt{x}}\frac{\mu(n)}{n^2}-\sum_{n\leq \sqrt{x}}\mu(n)\left\{\frac{x}{n^2}\right\}\\=\frac{6}{\pi^2}x-\sum_{n\leq \sqrt{x}}\mu(n)\left(-1+\left\{\frac{x}{n^2}\right\}\right)-2x\int_{\sqrt{x}}^{\infty}\frac{M(t)}{t^3}dt$$

Therefore we can write:

$$Q(x)=\frac{6}{\pi^2}x-\sum_{n\leq \sqrt{x}}\mu(n)\left\{\frac{x-n^2}{n^2}\right\}-2x\int_{\sqrt{x}}^{\infty}\frac{M(t)}{t^3}dt\\\implies \left|Q(x)-\frac{6}{\pi^2}x\right|\leq \left|\sum_{n\leq \sqrt{x}}\mu(n)\left\{\frac{x-n^2}{n^2}\right\}\right|+2x\left|\int_{\sqrt{x}}^{\infty}\frac{M(t)}{t^3}dt\right|\\\leq \sum_{n\leq\sqrt{x}}\left|\mu(n)\right|+2x\int_{\sqrt{x}}^{\infty}\frac{\left|Q(t)\right|}{t^3}dt\leq Q(x^{1/2})+2x\int_{\sqrt{x}}^{\infty}\frac{1}{t^2}\frac{\left|Q(t)\right|}{t}dt\leq 3x^{1/2}\\\implies \left|\frac{6}{\pi^2}x-Q(x)\right|\leq 3\sqrt{x}\iff \left|\frac{6}{\pi^2}-\frac{Q(x)}{x}\right|\leq \frac{3}{\sqrt{x}}$$

Which finally proves:

$$Q(x)=\sum_{j^2\leq x} \mu(j)\left\lfloor\frac x {j^2}\right\rfloor=\sum_{n\leq x}\mu(n)^2=\sum_{n\leq x}|\mu(n)|=\sum_{\substack{n\leq x\\ n{\small \text{ is squarefree}}}}1=\frac{6}{\pi^2}x+\mathcal{O}(x^{1/2})$$

Where intuitively this should make sense from a probabilistic perspective because $Q(x)$ is counting the number of square-free natural numbers $n\leq x$ and the natural density of positive integers not divisible by the square of any prime $p$ is $\left(1-\frac{1}{p^2}\right)$ now since a natural number is not divisible by any square iff it is not divisible by the square of any prime, we see the natural density of squarefree positive integers should roughly approach $\prod_{p}\left(1-\frac{1}{p^2}\right)=\frac{6}{\pi^2}$ which aligns with our bounds on the asymptotic density of square free integers in the interval $\left[1,x\right]$ as we showed $Q(x)/x\sim \frac{6}{\pi^2}$. Also for future reference one typically writes $\lfloor x\rfloor$ instead of $\left[x\right]$ to express the floor function applied to $x$ as the notation $\left[x\right]$ is occasionally used to denote the ceiling function $\lceil x\rceil$ which can result in some ambiguities if you're not careful.