Let [tex]\left(a_n\right)[/tex] be the sequence(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\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{....}[/tex]

Suppose that [tex]0\leq a<b \leq 1.[/tex] Let [tex]N(n;a,b)[/tex] be the number of integers [tex]j \leq n[/tex] such that [tex]a_j \in \left[a,b\right].[/tex] Prove that

[tex]\lim_{n\rightarrow \infty}\frac{N(n;a,b)}{n} = b-a.[/tex]

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.

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# Uniformly Distributed Sequence

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