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Proof

(related to Lemma: Möbius and Floor Functions Combined)

• Let $x \ge 1$ be a real number.
• Applying the proposition about the sum of Möbius function over divisors for $n=1,2,\ldots,\lfloor x\rfloor$ and adding up all results gives us $$1=\sum_{n=1}^{\lfloor x\rfloor}\sum_{d\mid n}\mu(d)=\sum_{d=1}^{\lfloor x\rfloor}\left\lfloor \frac xd\right\rfloor\mu(d).$$
• In the last step we have used that there are exactly $\left\lfloor \frac xd\right\rfloor$ multiples of $d$ less than $x$ and therefore the inner sum can be replaced[^1] by weighting the Möbius function $\mu(d)$ by $\left\lfloor \frac xd\right\rfloor.$
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References

Bibliography

1. Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition
2. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927

Footnotes