I'm doing the following exercise from Apostol's number theory book, a solution, or better a Hint would be welcome.
(Chapter 2 ex 14) Let $f(x)$ be defined for all rationals $x$ in $0 \leq x \leq 1$ and let
$$ F(n) = \sum_{k = 1}^n f(k/n), \quad F^*(n) = \sum_{\substack k = 1 \\ (k, \,n) = 1}^nf(k/n) $$
Prove that $F^* = \mu * F$, where $\mu$ is the Möbius function and $*$ is the Dirichlet product.
My attempt :
$$ \mu * F = \sum_{d | n} \mu(d) \sum_{k = 1}^{n/d} f(k/(n/d)) $$
But the summation in the inner part is equal to $$ \sum_{k \leq n/d} f(k/(n/d)) = \sum_{dk \leq n} f(dk/n) = \sum_{\substack t \leq n \\ d|t} f(t/n) $$
Using the last expresion in the sum of $\mu * F$ we get
$$ \mu *F = \sum_{d|n}\mu(d) \sum_{\substack t \leq n \\ d|t} f(t/n) $$
Struggling here (Trying to interchange the order of summation to make the $\mu(d)$ "go away").
Now for each $d|n$ we're summing all the $f(t/n)$ where $d|t$, So I can change the order of summation into all the $t$ less than $n$, and check all the $d$ that divide both $t$ and $n$
$$ \sum_{t \leq n} f(t/n) \sum_{\substack d|n \\ d|t} \mu(d) = \sum_{t \leq n} f(t/n) \sum_{ d|(t, n)} \mu(d) = \sum_{t \leq n} f(t/n) \lfloor 1/(t, n) \rfloor$$
And this last sum is the one over the $t$ relatively prime to $n$ which is what we were looking for.