Prove that $\sum_{d=1}^{n} M(\lfloor n/d \rfloor) = 1$

Question

In Wikipedia entry for Mertens function it says that

From [Lehman, R.S. (1960). "On Liouville's Function". Math. Comput. 14: 311–320.] we have that $$\sum_{d=1}^{n} M(\lfloor n/d \rfloor) = 1$$

I have looked at the cited paper and found that the identity $$\sum_{n\leq x} M \left(\frac{x}{n}\right)=1 \quad for \space x \geq 1$$ is cited at the end of Section 3 without giving neither a proof of its validity nor a reference.

I would appreciate a valid proof or reference of a proof of the identity.

Thanks!

Answer 1

$$\begin{eqnarray} & & \sum_{d=1}^{n}M(\lfloor n/d\rfloor) \\ & = & \sum_{d=1}^{n}\sum_{k=1}^{\lfloor n/d\rfloor}\mu(k) \\ & = & \sum_{d=1}^{n}\sum_{dk\leq n}\mu(k) \\ & = & \sum_{k=1}^{n}\sum_{dk\leq n}\mu(k) \\ & = & \sum_{r=1}^{n}\sum_{k|r}\mu(k)\quad\textrm{ where $r=dk$} \\ & = & \sum_{r=1}^{n}[\textrm{$1$ if $r=1$, else 0]} \\ & = & 1. \end{eqnarray}$$