Are There Sophisticated Methods for Proving Uniform Distribution of Sequences?

[snipez90]

Let $$\left(a_n\right)$$ be the sequence

$$\frac{1}{2},\, \frac{1}{3},\, \frac{2}{3},\, \frac{1}{4},\, \frac{2}{4},\, \frac{3}{4},\, \frac{1}{5},\, \frac{2}{5},\, \frac{3}{5},\, \frac{4}{5},\, \frac{1}{6},\, \frac{2}{6},\,\mbox{...}$$

Suppose that $$0\leq a<b \leq 1.$$ Let $$N(n;a,b)$$ be the number of integers $$j \leq n$$ such that $$a_j \in \left[a,b\right].$$ Prove that
$$\lim_{n\rightarrow \infty}\frac{N(n;a,b)}{n} = b-a.$$

I already know how to do this based on the definition of a sequence. The basic idea is to take the set of rational numbers {1/n, 2/n, ... , (n-1)/n} for an arbitrary n and consider the smallest member of the set which is also in [a,b], giving us a bound on a, and a similar consideration for the largest member of the set in [a,b] gives a bound on b. This allows us to estimate the number of elements of the set (for that particular n) that are also in [a,b].

My proof of this was rather long, but entirely elementary (the source is Spivak). I was wondering if there are more sophisticated methods of dealing with this type of problem. Thanks in advance.

 

[fresh_42]

It is combinatorics by nature ending with the law of large numbers. Any proof will have to bridge this gap, so the answer is probably not, although it is hard to tell without knowing your proof.