Equidistribution mod $1$ of sequence of rational numbers from interval $(0,1]$. Define a sequence of rational numbers $\frac{a}{b}$ (gcd$(a,b)=1$) from interval $(0,1]$ as follows: $\frac{a_1}{b_1}$ comes before $\frac{a_2}{b_2}$ if $b_1 < b_2$ and $\frac{a_1}{b}$ before $\frac{a_2}{b}$ if $a_1 < a_2$.
Show that such sequence is equidistributed mod $1$.
I think this should be done using Weyl's criterion but I cant figure a way to do it.
 A: Let $\mathcal{F}_b$ be the set of rationals in $(0,1]$ such that the denominator in the reduced fraction is at most $b$. For $q\leq b$, let $\mathcal{G}_q\subseteq \mathcal{F}_b$ be the ones with denominator exactly $q$. 
We use Weil's criterion. Let $n\neq 0$ be an integer. First, we consider the expontial sum for all members of $\mathcal{F}_b$: 
$$
\sum_{r\in\mathcal{F}_b} e^{2\pi i n r}=\sum_{q\leq b} \sum_{r\in\mathcal{G}_q} e^{2\pi i nr}.
$$
We recognize that the inner sum is Ramanjuan sum $c_q(n)$ when $n>0$. For negative $n$, use $c_q(-n)=\overline{c_q(n)}$. This satisfies $|c_q(n)|\leq n$ for a fixed $n>0$ and uniformly for $q$. Thus, for any fixed $n\neq 0$, we have $|c_q(n)|\leq |n|$ uniformly for $q$. 
Then the exponential sum has bound:
$$
\left|\sum_{r\in\mathcal{F}_b} e^{2\pi i nr}\right|\leq \sum_{q\leq b} |c_q(n)|\leq |n|b.
$$
On the other hand, 
$$|\mathcal{F}_b|=\sum_{q\leq b}\phi(q)=\frac3{\pi^2} b^2 + O(b\log b).$$
Therefore, 
$$
\left|\frac1{|\mathcal{F}_b|}\sum_{r\in\mathcal{F}_b} e^{2\pi i n r}\right|\leq \frac{|n|b}{\frac3{\pi^2} b^2 + O(b\log b)}\rightarrow 0 \ \ \mathrm{as} \ \ b\rightarrow\infty.$$
The case with incomplete $\mathcal{G}_b$ can be handled with an error of $O(b)$ on the numerator and the denominator. Hence, the sequence of rationals with this order is equidistributed.
