Proof
(related to Lemma: Upper Bound of Harmonic Series Times Möbius Function)
- Let $x \ge 1$ be a real number, for which we want to estimate the partial sum.
$$\left|\sum_{n=1}^x\frac{\mu(n)}{n}\right|\le ?.$$
- By definition of the floor function, we have the estimation $$0\le\frac{x}{n}-\left\lfloor\frac{x}{n}\right\rfloor\begin{cases}
< 1&\text{ for }1\le n < x\\
=0&\text{ for }n = x.\end{cases}$$
- Multiplying these terms by the Möbius function $\mu(n)$ and building the sum for $n\le x$ gives us with the lemma about Möbius and floor functions combined:
$$\sum_{n=1}^x\mu(n)\left(\frac{x}{n}-\left\lfloor\frac{x}{n}\right\rfloor\right)=x\sum_{n=1}^x\frac{\mu(n)}{n}-\sum_{n=1}^x\mu(n)\left\lfloor\frac{x}{n}\right\rfloor=x\sum_{n=1}^x\frac{\mu(n)}{n}-1.$$
- With the triangle inequality and because $\mu(n)\le 1$ for all $n\ge 1,$ we get the estimation
$$\begin{array}{rcl}
\left|x\sum_{n=1}^x\frac{\mu(n)}{n}-1\right|&=&\left|\sum_{n=1}^x\mu(n)\left(\frac{x}{n}-\left\lfloor\frac{x}{n}\right\rfloor\right)\right|\\
&\le&\sum_{n=1}^x\left(\frac{x}{n}-\left\lfloor\frac{x}{n}\right\rfloor\right)\\
&\le&x-1.
\end{array}$$
- Bringing the term $-1$ on the other side of the inequation results in
$$\begin{array}{rcl}
\left|x\sum_{n=1}^x\frac{\mu(n)}{n}\right|&\le&1+(x-1)=x.
\end{array}$$
- Dividing both sides of the inequation by $x$ results in the required upper bound.
$$\begin{array}{rcl}
\left|\sum_{n=1}^x\frac{\mu(n)}{n}\right|&\le&1.
\end{array}$$
- This upper bound holds for all partial sums depending on $x\ge 1,$ therefore also for the whole infinite series.
∎
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References
Bibliography
- Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition
- Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
Footnotes
