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## Proof: (related to "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.1

1 The existence of such an upper bound means that the above series either converges or oscillates in the real interval $[-1,1].$ Which of these two cases is true, can be proven but requires more sophisticated methods we will introduce in analytic number theory. Anticipating the right answer, the series converges and its limit is $0.$

q.e.d

(none)