Let $f(x)$ be defined over all rationals $x$ in $[0,1]$ and let $F(n) = \sum_{i=1}^n f(\frac in)$ also define $$F^*(n) = \sum_{i=1\,\,(i,n)=1}^n f(\frac in)$$ then prove that $$F^* = \mu * F$$
where $\mu$ is the Möebius function and the $*$ means the Dirichlet convolution.
I tried the Bell series which gave me this:
$$F^*(p^k) = \sum_{i=1}^{p^k} f(\frac i {p^k}) - \sum_{i=1}^k f(\frac{p^i}{p^k})$$
$$F^*(p^k) = F(p^k) - \sum_{i=1}^k f(\frac{p^i}{p^k})$$
(assuming $k\geq1$ and $p$ an arbitrary prime)
$$x^kF^*(p^k) = x^kF(p^k) - x^k\sum_{i=1}^k f(\frac{p^i}{p^k})$$
$$\sum_{k=1}^{+\infty}x^kF^*(p^k) = \sum_{k=1}^{+\infty}x^kF(p^k) - \sum_{k=1}^{+\infty}(x^k\sum_{i=1}^k f(\frac{p^i}{p^k}))$$
because $$F(1) = F^*(1)$$ we have
$$h_p(F*) = h_p(F) - \sum_{k=1}^{+\infty}(x^k\sum_{i=1}^k f(\frac{p^i}{p^k}))$$
where $h_p(f)$ represents the bell series of $f$ with respect the prime $p$
all I need to prove is that
$$xh_p(F) = \sum_{k=1}^{+\infty}(x^k\sum_{i=1}^k f(\frac{p^i}{p^k})) $$
which is equivalent to 
$$h_p(F) = \sum_{k=1}^{+\infty}(x^{k-1}\sum_{i=1}^k f(\frac{p^i}{p^k})) $$
or if you will:
$$\sum_{k=0}^{+\infty} x^k \sum_{i=1}^{p^k}f(\frac i{p^k}) = \sum_{k=1}^{+\infty}(x^{k-1}\sum_{i=1}^k f(\frac{p^i}{p^k})) $$
$$\sum_{k=0}^{+\infty} x^k \sum_{i=1}^{p^k}f(\frac i{p^k}) = \sum_{k=0}^{+\infty}(x^{k}\sum_{i=1}^{k+1} f(\frac{p^i}{p^{k+1}})) $$
$$\sum_{i=1}^{k+1} f(\frac{p^i}{p^{k+1}})) =  \sum_{i=1}^{p^k}f(\frac i{p^k})$$
it seems to me that the sum in the right contains more terms than the sum in the left (and also contains the whole sum of the left)
 A: Here is a standard derivation of the result. To help us in the derivation we start by introducing the $\delta$-function $\delta_{m,n}:\mathbb{Q}\times \mathbb{Q} \to \{0,1\}$ by $$\delta_{m,n} = \left\{\matrix{1 & m=n\\0 & m\not= n}\right.$$  By construction, the $\delta$-function satisfy the following identity:

Identity 1: If $d\mid n$ and $i\leq n/d$ then $\sum_{j=1}^n\delta_{i,j/d}h(j) = h(id)$ for any function $h$.

Another useful identity is the Möbius function sum property in the following form:

Identity 2: $\sum_{d\mid n}\mu(d)\sum_{i=1}^{n/d}\delta_{i,j/d} = 1$ if $(n,j)= 1$ and $0$ otherwise. 

This follows by writing out
  $$\sum_{d\mid n}\mu(d)\sum_{i=1}^{n/d}\delta_{i,j/d} = \sum_{d\mid n,\,d\mid j}\mu(d) = \sum_{d\mid (n,j)}\mu(d) = \delta_{(n,j),\, 1}$$ 
Using the two identities above the derivation is pretty straight forward:
$$
\begin{align}(\mu * F)(n) &\equiv \sum_{d\mid n}\mu(d)\sum_{i=1}^{n/d}f(id/n) \\&= \sum_{d\mid n}\mu(d)\sum_{i=1}^{n/d}{\sum_{j=1}^{n}\delta_{i,j/d}\,f(j/n)}~~~~~~~\text{(using identity 1)}\\&= \sum_{j=1}^{n}f(j/n){\sum_{d\mid n}\mu(d)\sum_{i=1}^{n/d}\delta_{i,j/d}} ~~~~~~~\text{(switched the summation order)}\\&= \sum_{j=1,\,(n,j)=1}^nf(j/n)~~~~~~~~~~~~~~~~~~~~~~~~~\text{(using identity 2)}
\end{align}
$$
