(This is a cross-posted simplification of this question posted in MSE which did not have a complete answer.)
I am trying to get the asymptotic behaviour when $n$ grows to infinity of a partial sum of the form $$S(n)=\sum_{k=1}^{\alpha} \mu(k) G\left(\frac{n}{k}\right)$$
After some back and forth, here is more or less what I have as a possible "set up" for obtaining the asymptotic behaviour of $S(n)$ when $n$ grows to infinity:
Given facts
We have that $\alpha= C_n \cdot \sqrt{n}$ where $1<C_n<2$ is some real number that is different for each $n$ but approaches $1$ as $n$ grows to infinity.
We have that $G(n)=D_n \cdot \frac{n}{\log(n)}$ where $1<D_n<2$ is some real number that is different for each $n$ but approaches $1$ as $n$ grows to infinity. .
Other hand, we have that $D_n \cdot \frac{n}{\log(n)} = K_n \cdot \sum_{k=1}^{n} \left(\frac{\log\left(\frac{n}{k}\right)}{\log\log\left(\frac{n}{k}\right)}\right)$ where $1<K_n<2$ is some real number that is different for each $n$ but approaches $1$ as $n$ grows to infinity.
Applicable method
An application of the generalization of Möbius inversion formula states that, if $F(x)$ and $G(x)$ are complex-valued functions such that $G(x) = \sum_{k=1}^{x} F\left(\frac{x}{k}\right)$, we have that $\sum_{k=1}^{x} \mu(k) G\left(\frac{x}{k}\right)=F(x)$.
As we have that $G(x) = K_x \cdot \sum_{k=1}^{x} F\left(\frac{x}{k}\right)$, where $1<K_{x}<2$ is some real number that is different for each $x$ but approaches $1$ as $x$ grows to infinity, one can apply the generalization of Möbius inversion formula stated to obtain that $$\sum_{k=1}^{n} \mu(k) G\left(\frac{n}{k}\right)=K_n \cdot \frac{\log (n)}{\log\log (n)}$$
Problem: This application is not correct for some values of $n$. The functions $G(x)$ and $F(x)$ are not defined on the interval $[1, \infty)$; concretely, they are not defined for $1$. Can one dismiss this fact, as we are looking for an asymptotic bound when $n$ grows to infinity? If not, can we overcome it somehow?
Conclusion
- We have that $G\left(\frac{n}{k}\right)=H_n \cdot G\left(\frac{\alpha}{k}\right)$, where $2\sqrt{n}<H_{n}<4\sqrt{n}$ is some real number that is different for each $n$ but approaches $2\sqrt{n}$ as $n$ grows to infinity. Therefore, we have that $$\sum_{k=1}^{\alpha} \mu(k) G\left(\frac{n}{k}\right)=H_{\alpha} \cdot \sum_{k=1}^{\alpha} \mu(k) G\left(\frac{\alpha}{k}\right)= H_{\alpha} \cdot K_{\alpha} \space \cdot \space \frac{\log \alpha}{\log\log \alpha}$$ where $H_{\alpha} \cdot K_{\alpha}$ approaches $2\sqrt{n}$ and $\frac{\log \alpha}{\log\log \alpha}$ approaches $\frac{\log \sqrt{n}}{\log\log \sqrt{n}}$ as $n$ grows to infinity. Therefore, one could state that, as $n$ grows to infinity, $$S(n)\sim 2\sqrt{n} \cdot \frac{\log \sqrt{n}}{\log\log \sqrt{n}}$$
Questions:
The ones stated at the "Problem" note.
Is the "set up" correct? Is there any missing step, flaw or mistake I should take into account?
Can "sharp" upper and lower bounds be obtained for $S(n)$? How? (please detail your answer)
Thanks in advance for your time!
