Show that the sequence $\frac{0}{1}, \frac{0}{2}, \frac{1}{2}, \frac{0}{3}, \frac{1}{3}, \frac{2}{3}, \dots \frac{k-1}{k}$ is equidistributed mod 1 I need to show that the sequence $\frac{0}{1}, \frac{0}{2}, \frac{1}{2}, \frac{0}{3}, \frac{1}{3}, \frac{2}{3}, \dots , \frac{0}{k}, \frac{1}{k}, \dots , \frac{k-1}{k}$ is equidistributed in the interval $[0,1]$ (and so equidistributed mod $1$).
Ah, of course I can't use any further theorem developped in the theory of equidistibution of sequences like, for example, Weyl 's criterion. I must deduce it from the definition.
This is taken from Kuipers and Niederreiter, exercize 1.13.
I found this old question but as you can see noone answered it.
Thank you in advance
 A: Let $(x_n)_{n \geq 0}$ be this sequence. We want to show that, for any function $f \in \mathcal{C} ([0,1], \mathbb{R})$,
$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} f(x_k) = \int_0^1 f(t) \ \text{d}t.$$
The sequence is a concatenation of finite subsequence $(k/n)_{0 \leq k < n}$. We shall proceed in three steps: first by proving convergence for these subsequences, the for the initial sequence $(x_n)_{n \geq 0}$ along a nice subsequence, and then the equidistribution.
Let $f \in \mathcal{C} ([0,1], \mathbb{R})$.
First step
We have 
$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} f(k/n) = \int_0^1 f(t) \ \text{d}t,$$
since the left ahnd-side is a Riemann sum for $f$. We write $u_n$ for the $n$th Riemann sum $\frac{1}{n} \sum_{k=0}^{n-1} f(k/n)$.
Second step
Let $N \geq 0$, and set $n(N) := \frac{N(N+1)}{2}$. Then
$$\frac{1}{n(N)} \sum_{k=0}^{n(N)-1} f(x_k) = \frac{2}{N(N+1)} \sum_{\ell = 1}^N \sum_{k=0}^{\ell-1} f(k/\ell) = \sum_{\ell = 1}^N \frac{2\ell}{N(N+1)} u_\ell.$$
Set $v_\ell^{(N)} := \frac{2\ell}{N(N+1)}$ for $\ell \leq N$, and $0$ for $\ell \geq N+1$. Then
$$\frac{1}{n(N)} \sum_{k=0}^{n(N)-1} f(x_k) = \sum_{\ell = 1}^{+ \infty} v_\ell^{(N)} u_\ell.$$
In addition :


*

*$\sum_{\ell=1}^{+\infty} v_\ell^{(N)} = 1$ for all $N$, 

*$\lim_{M \to + \infty} \sum_{\ell=1}^{M} v_\ell^{(N)} = 0$,

*$\lim_{\ell \to + \infty} u_\ell = \int_0^1 f(t) \ \text{d}t$.
Hence, by a lemma I'll leave aside (but can be proved with some $\epsilon$-$\delta$ arguments),
$$\lim_{N \to + \infty} \frac{1}{n(N)} \sum_{k=0}^{n(N)-1} f(x_k) = \lim_{N \to + \infty} \sum_{\ell = 1}^{+ \infty} v_\ell^{(N)} u_\ell = \int_0^1 f(t) \ \text{d}t.$$
Third step
Let $n \geq 0$. Let $N$ be the largest integer such that $\frac{N(N+1)}{2} \leq n$. Then $n-n(N) \leq n(N+1)-n(N) = N+1 \leq \sqrt{2(n+1)}$, so that
$$\frac{1}{n} \sum_{k=0}^{n-1} f(x_k) = \frac{n(N)}{n} \frac{1}{n(N)} \sum_{k=0}^{n(N)-1} f(x_k) + \frac{1}{n} \sum_{k=n(N)}^{n-1} f(x_k) 
= (1+o(1)) \frac{1}{n(N)} \sum_{k=0}^{n(N)-1} f(x_k) + O\left(\frac{\|f\|_{\infty}}{\sqrt{n}} \right),$$
and thus the full sequence also converges to the integral of $f$.
