$(\log n)_{n \in \mathbf{N}}$ not uniformly distributed mod 1.

Question

Let $(x_n)_{n \in \mathbf{N}}$ be a sequence of real numbers, we say that $(x_n)$ is uniformly distributed mod 1 (u.d. mod 1) if

$$\lim_{N \to \infty} \frac{|\{1\leq n \leq N : (x_n -\lfloor x_n \rfloor) \in (a,b) \}}{N} = b-a.$$

for any $a,b \in \mathbf{R}$ such that $0\leq a\leq b\leq 1$. Many easily defined sequences are u.d. mod 1; for instance, $(n^c)_{n \in \mathbf{N}}$ for any $c \in (0,1)$ (by Fejer's criterion). Obviously, $(n)_n$ is not u.d. mod 1. So the next natural questions would be:

Is $(\log{n})_{n \in \mathbf{N}}$ u.d. mod 1?

Surprisingly, (or at least to me at first) it is not. I found a proof in Kuipers & Niederreiter, pages 8-9, but I don't like it. I've tried to prove the result myself, using the direct definition and also Weyl's criterion for uniform distribution. But, I haven't been able to succeed. As someone once told me: "Any good theorem ought to be proved more than once". So, does anyone know of a different proof of this claim, or care to give me some hints on how to proceed?

Answer 1

Intuitively, this doesn't work because the spans of fractional parts of logs in an interval get longer and longer as we go up. To be more explicit, let us consider $a=0, b=0.1$ and I claim that the limit does not exist. For natural $k$, let $m=\lceil e^k \rceil$. Then let $m'=\lfloor e^{k+0.1} \rfloor$ For large enough $k$ the fraction in your limit is $\frac {m'-m+1}m \gt \exp(0.1-\epsilon) \gt 0.105$ because all the numbers $m$ through $m'$ have the fractional part between $0$ and $0.1$

Answer 2

I ran across this myself recently in a book, and was surprised. So, I did a numerical experiment. In Mathematica, run these:

t = Table[N[FractionalPart[Log[n]]], {n, 2, 10^4}];

Histogram[t]

The data clearly show that the fractional parts are not falling equally into the bins.