Yesterday Prof. Feng gave a seminar which was about Furstenberg’s x2, x3 theorem. In the talk, he used the following fact:

Let : , . Arrange in an increasing order . Then

.

He said the proof was easy and it was left as an exercise, with the hint that is an irrational number.

However, after I discussed this question with some undergraduates and postgraduates, we couldn’t figure out a rigorous proof in short time as there is no good pattern in this sequence. (Yuen Chi Ho suggested that one could approach this question by using Hurwitz’s theorem, but I don’t know how to use it.)

Can anyone solve this “easy” exercise?

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ASSUME the limit exists, then it can be shown that , as follows.

First of all, clearly . On the other hand, since is irrational, is uniformly distributed on . Thus for any small , there is such that for some natural , which is equivalent to . So if , . Since is arbitrary, must be .

Oh, I know I should use equidristribution property somewhere, but could not work out the inequality..thanks for the “half proof”!

It turns out that we can modify the above to prove that the limit is 1 directly.

Define . Let , choose so large such that by equidistribution (of ), for all , there exist for some . In particular, for all , there exists for some . Suppose , then there is some natural such that . Letting , we thus have and . The result follows.

How can you guarantee that is a natural number (i.e., a non-negative integer)? Is it possible that it is negative and thus may be negative.

This is because and thus .

Good job KKK.

Here we give an alternate proof using the theory of continued fraction. The calculations here are constructive. Let . Let . Let , then

if and only if

Suppose are the -th convergent of the continued fraction of . It is known that (See Hardy, Wright, Theorem 163)

We need a slightly stronger result, in fact by the same theorem

and

Let be sufficiently large such that . Comparing with (2), we have both

and

satisfy (2). However, given above may not be positive. To ensure that at least one of the set gives positive solutions, we need

There are only finitely many such not satisfying the above. Thus for sufficiently large, we can find positive integer satisfy (1).

Thank for the nice proofs. Feel better as it seems not so easy (but somehow elementary, though).