Partial sum $\sum_{d \leq x} \mu^2(d)/d$

Question

what is the value of the partial sum $ \sum_{d \leq x}\frac{\mu^2(d)}{d}$ Is there a function f(x) such that

$ \sum_{d \leq x}\frac{\mu^2(d)}{d} \ll f(x)$? Thanks

Answer 1

Using Abel's summation we have $$S=\sum_{d\leq x}\frac{\mu^{2}\left(d\right)}{d}=\frac{Q\left(x\right)}{x}+\int_{1}^{x}\frac{Q\left(t\right)}{t^{2}}dt $$ where $$Q\left(x\right)=\sum_{d\leq x}\mu^{2}\left(d\right) $$ is the counting function of the squarefree numbers up to $x$. It is well-known that $$Q\left(x\right)=\frac{6}{\pi^{2}}x+O\left(\sqrt{x}\right) $$ hence $$S=\frac{6}{\pi^{2}}+O\left(\frac{1}{\sqrt{x}}\right)+\frac{6}{\pi^{2}}\int_{1}^{x}\frac{1}{t}dt+O\left(\int_{1}^{x}\frac{1}{t^{3/2}}dt\right) $$ $$=\frac{6}{\pi^{2}}\log\left(x\right)+O\left(1\right).$$

Answer 2

• Find $f(n)$ such that $\mu(n)^2 = \sum_{d| n} f(d)$. Since $\sum_{n=1}^\infty \mu(n)^2 n^{-s} = \frac{\zeta(s)}{\zeta(2s)}$ you have $\sum_{n=1}^\infty f(n)n^{-s} = \frac{1}{\zeta(2s)}$ and $f(n) = \mu(n^{1/2})$

• $$\sum_{n < x} \mu(n)^2 =\sum_{n < x} \sum_{d | n}f(d)=\sum_{n < x} f(n) \lfloor x/n \rfloor = x \sum_{n < x} \frac{f(n)}{n} - \sum_{n < x} f(n) \{x/n\}$$

  where $\{x\} = x-\lfloor x\rfloor \in [0,1)$. Note that $\sum_{n=1}^\infty \frac{f(n)}{n} = \sum_{n=1}^\infty \frac{\mu(n)}{n^2} = \frac{1}{\zeta(2)} $ converges absolutely and $\sum_{n< x} \frac{f(n)}{n} = \frac{1}{\zeta(2)}-\sum_{n > x^{1/2}}\frac{\mu(n)}{n^2}=\frac{1}{\zeta(2)}+\mathcal{O}(x^{-1/2})$

  and $\sum_{n < x} f(n) \{x/n\} = \sum_{n^2< x}\mu(n) \{x/n^2\}=\mathcal{O}(x^{1/2})$

  overall since $\zeta(2) =\frac{\pi^2}{6} $ you have $$\sum_{n < x} \mu(n)^2 = \frac{6}{\pi^2}x+ \mathcal{O}(x^{1/2})$$

• and by summation by parts $$\sum_{n =1}^N \frac{\mu(n)^2}{n} = \frac{1}{N}\sum_{n=1}^N \mu(n)^2+\sum_{n=1}^{N-1} (\frac{1}{n}-\frac{1}{n+1})\sum_{n=1}^m \mu(m)^2$$ $$=\frac{6}{\pi^2}+ \mathcal{O}(N^{-1/2})+\sum_{n=1}^{N-1}\frac{\frac{6}{\pi^2}+ \mathcal{O}(n^{-1/2})}{n+1}= \frac{6}{\pi^2}(\log(N)+\gamma)+\mathcal{O}(1)$$ where I used $\sum_{n=1}^N \frac{1}{n} = \log(N) + \gamma+ \mathcal{O}(1/N)$ and it should be possible to improve the error term $\mathcal{O}(1)$ to $C+\mathcal{O}(N^{-1/2}\log(N))$