I have been given the following exercise:
Get a function $G(x)$ so that, being $x$ a positive integer,
$$x+\sqrt{x}= \sum^x_{n=1} \mu(n) G(\frac{x}{n})$$
The result does not have to be exact, but it must approach to the equation as $x \to \infty$. In other words, there can be an error term $E(x)$ if it does not diverge.
My try:
We know that $$F(x)=\sum \mu(n) G(\frac{x}{n})$$ iff $$G(x)=\sum F(\frac{x}{n})$$
Then, $$G(x)=\sum \frac{x}{n}+ \sqrt{\frac{x}{n}}=x\log x + \gamma x +1 +2x +\zeta (1/2) \sqrt{x} +E(x)$$
Where $E(x) \to 0$ as $x \to \infty$.
The problem is that, when I try to compute:
$$\sum^x_{n=1} \mu(n) \left ( \frac{x}{n}\log \frac{x}{n} + \gamma \frac{x}{n} +1 +2\frac{x}{n} +\zeta (1/2) \sqrt{\frac{x}{n}} \right ) $$
I always get a much larger result than what I need (i.e. $x+\sqrt{x}$).
Any help finding the flaw?
Thanks!
Edit: changed $+\frac{1}{2}$ to $+1$,