Sums involving square of Moebius function

Question

I try to estimate the following sum: $$ \sum_{n \leq x}\mu(n)^2 f(n) $$ where $\mu(n)$ is a Moebius function and $f(n)$ is some multiplicative arithmetic function. If I understand it correctly it is the sum of values of some function $f$ for all square-free numbers upto $x$.

I found the identity involving such sums in the comment to this answer to the question about Mertens function. But I can't grasp what $k$ means. Can someone explain it?

Thanks!

Answer 1

It's just changing the order of summation ($k$ there equals $n/d^2$): $$\sum_{n\leq x}\mu^2(n)f(n)=\sum_{n\leq x}f(n)\sum_{d^2|n}\mu(d)=\sum_{n\leq x}f(n)\sum_{n=d^2k}\mu(d)=$$ $$=\sum_{d\leq \sqrt{x}}\mu(d)\sum_{\substack{n\leq x\\ n=d^2k}}f(n)=\sum_{d\leq \sqrt{x}}\mu(d)\sum_{k\leq x/d^2}f(d^2k).$$