How to prove this sequence is uniformly distributed mod 1? I will first define everything to make my question clear.
For a real number $x$, let $[x]$ denote the integral part of $x$ and let $\{x\}=x-[x]$ denote the fractional part of $x$.
Let $\omega = (x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers; let $N \in \mathbb{N}$ and let $E\subseteq [0,1)$.
Let $A=(E;N;\omega)$ be defined as the number of terms $x_n$ up to the $N$th term of $\omega$ for which $\{x_n\} \in E$.
We say $\omega$ is uniformly distributed modulo $1$ if for every pair $a,b$ of real numbers with $0\leqslant a <b \leqslant 1$ we have:
$$\lim_{N\rightarrow \infty}\frac{A([a,b);N;\omega)}{N} = b-a$$
I am just a beginner with this concept. I seek help in proving - by the definition - that the sequence $$0/1,0/2,1/2,0/3,1/3,2/3,\ldots,0/k,1/k,2/k,\ldots,(k-1)/k,\ldots$$ is uniformly distributed modulo 1
 A: A first thing to notice is that the first $k(k+1)/2$ terms of the sequence consists exactly of all rational numbers in $[0,1)$ with denominator $\leq k$.  Let's first prove the desired limit
$$
\lim_{N \to \infty} \frac{A([a,b), N, \omega)}{N} = b-a
$$
only along the subsequence $N_k = k(k+1)/2$, and we'll upgrade it to the full result at the end.  We have
\begin{align*}
A([a,b), k(k+1)/2, \omega) &= \#\{\text{rational numbers in $[a,b)$ with denominator $\leq k$} \} \\
&= \sum_{q = 1}^{k} \#\{ p: p/q \in [a,b) \}.
\end{align*}
For a given $a,b,q$, the number of such $p$ is $q(b-a) \pm 1$ (the $\pm 1$ is because you don't know exactly how the endpoints line up with the rational points).  So you have
$$
A([a,b), k(k+1)/2, \omega) = \sum_{q=1}^{k} q(b-a) + O(1) = (b-a) \frac{k(k+1)}{2} + O(k).
$$
This proves that
$$
\lim_{k \to \infty} \frac{A([a,b), k(k+1)/2, \omega)}{k(k+1)/2} = b-a.
$$
Now we can use a very common trick to upgrade the convergence along a subsequence to full convergence.  Fix $N$, and define $k_N$ to be such that $k_N(k_N+1)/2 \leq N < (k_N+1)(k_N+2)/2$.  Then we have
$$
\frac{\frac{k_N(k_N+1)}{2}}{N} \frac{A([a,b), \frac{k_N(k_N+1)}{2}, \omega)}{\frac{k_N(k_N+1)}{2}} \leq \frac{A([a,b), N, \omega)}{N} \leq \frac{A([a,b), \frac{(k_N+1)(k_N+2)}{2}, \omega)}{\frac{(k_N+1)(k_N+2)}{2}} \frac{\frac{(k_N+1)(k_N+2)}{2}}{N}.
$$
As $N \to \infty$, both the leftmost and rightmost terms converge to $b-a$ by the subsequential convergence already established and the fact that $N \approx \frac12 k_N^2$.  Thus the middle term also converges to $b-a$.
