# Möbius inversion formula and $\sum_{k\leq n} \frac{\mu(k)}{k}$

I have tried to apply what is stated at the Generalizations of Möbius inversion formula section of Wikipedia to bound $$\sum_{k\leq n} \frac{\mu(k)}{k}$$ The application seems simple and straightforward. Let us consider the function $$F\left(\frac{n}{k}\right)=\frac{n}{k\cdot H_n}$$ Where $$H_n$$ is the $$n_{th}$$ harmonic number. Then, we can set that, for all $$n\geq 1$$, $$G(n)=\sum_{k\leq n} F\left(\frac{n}{k}\right)=n$$ And thus, applying the generalization of Möbius inversion formula, we have that $$F(n)=\sum_{k\leq n} \mu(k)G\left(\frac{n}{k}\right)=\frac{n}{H_n}$$ As we have that $$G\left(\frac{n}{k}\right)=\frac{n}{k}$$ for all $$n,k\geq 1$$, we have that $$F(n)=\sum_{k\leq n} \mu(k)\left(\frac{n}{k}\right)=\frac{n}{H_n}$$ And from this result, that $$\sum_{k\leq n} \frac{\mu(k)}{k}=\frac{1}{H_n} \quad (1)$$

I have several doubts relating this process and result:

(1) Is the process of application correct? If incorrect, where is the error?

(2) If correct, which is the interpretation of the result? It is known that the Riemann Hypothesis is equivalent to the statement $$\sum_{k\leq n} \frac{\mu(k)}{k}=O(n^{-\frac{1}{2}+\epsilon})$$. As $$H_n \leq \log(n) + \gamma + \frac{1}{2n}$$, where $$\gamma$$ is the Euler-Mascheroni constant, would the result imply the falsehood of the Riemann Hypothesis?

I am 99,9% sure that there is some error, basically because it suffices to compute the first values of $$\sum_{k\leq n} \frac{\mu(k)}{k}$$ to check that the equality concluded in (1) does not hold. However, I am not able to spot where.

• You gave a formula for $F(n/k)$ but then what is $F(x)$? You can express a rational $x$ in many different ways giving different values for $F(x)$.
• $F$ is not well defined at rational numbers and it's not defined for irrational numbers. Commented Nov 13, 2023 at 0:50
• I think the problem is you defined $F$ in terms of $k$ as well as $n$. The definition $F(x)=1$ is consistent with $$G(x)=\sum_{k\leq x} F\left(\frac{x}{k}\right)=\lfloor x\rfloor$$ (note that $G(n)=n$ for $n\in\mathbb{N}$) and $$F(x)=\sum_{k\leq x} \frac{\mu(k)}{k}\, G\left(\frac{x}{k}\right)=\sum_{k\leq x} \frac{\mu(k)}{k}\, \left\lfloor\frac{x}{k}\right\rfloor=1$$ is consistent with the preceding. Commented Nov 13, 2023 at 3:42