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Lemma: Möbius and Floor Functions Combined

For every real number $x \ge 1$ we have:
$$\sum_{n=1}^{\lfloor x\rfloor}\mu(n)\left\lfloor\frac xn\right\rfloor=1.$$

| | | | | created: 2019-04-06 18:02:18 | modified: 2019-04-06 19:33:27 | by: bookofproofs | references: [701], [1272]

1.Proof: (related to "Möbius and Floor Functions Combined")

2.Explanation: Changing the Index Of Sums in Number Theory


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Bibliography (further reading)

[1272] Landau, Edmund: “Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie”, S. Hirzel, Leipzig, 1927

[701] Scheid Harald: “Zahlentheorie”, Spektrum Akademischer Verlag, 2003, 3. Auflage

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