# ${α⋅ \log(n)}$ is not uniformly distributed mod1 in $[0,1]$

### $$\qquad \qquad \bbox[15px,border:2px solid red] { x_n:=\text{\{α\cdot log(n)\}}_{n\in \mathbb N}}$$

I want to show that the sequence $$x_n$$ is not uniformly distributed mod1 in $$[0, 1]$$ for any $$α\in \mathbb R$$.

Note:

### 1)$$\qquad \qquad \qquad \qquad$$Euler summation Formula:

$$\qquad \qquad \bbox[15px,border:2px solid red] { \sum_{n=1}^Nf(n)=\int_1^Nf(t)dt+\frac{1}{2}(f(1)+f(N)) +\int_1^N(\text{\{t\}-\frac{1}{2}) }f'(t)dt }$$

### 2) $$\qquad \qquad \qquad \qquad$$ Weyl's equidistributed criterion:

$$\qquad \qquad \qquad \qquad \quad \qquad$$ The following are equivalent
$$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {x_n \quad \text{is equivalent modulo 1} }$$ $$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {\forall \text{continuous & 1-peridic f:}\quad \frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f }$$ $$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {\forall k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 }$$

I've already proved it by using (1) & (2) , is there any other way to approach this problem?

• How did you use the Euler summation formula to show $x_n$ is not uniformly distributed? Aug 4, 2020 at 19:11
• Do you mean $\{x_n\}$ is equidistributed in $[0,1]$? The adjective Uniformly distributed means a different thing, that is $X:(\Omega,\mathscr{A},\mathbb{P})\rightarrow[0,1]$ is uniformly distributed if $\mathbb{P}[X\in A]=\lambda(A)$ for all measurable set $A\subset[0,1]$, where $\lambda$ is the Lebesgue measure. Aug 4, 2020 at 19:13
• @Verun Vejalla : math.rice.edu/~michael/teaching/426_Spr14/UDmod1A.pdf page 8 EXAMPLE 2.4. Aug 4, 2020 at 19:15
• @JohnMars: Weyl's equidistribute theorem is the natural thing to use. Whether one uses Euler summation, Abel summation or Sonin's summation is just a tool of calculation. Aug 4, 2020 at 20:06
• Another way would be directly from dynamics. If this sequence is uniformly distributed, it is recurrent around 0 (namely has syndetic set of return times). Pick M large so that $\alpha\cdot\log(M)$ is near $0$, one can now estimate that for many consecutive numbers n after $M$, $\alpha\cdot\log n$ is small, by some Taylor expansion or so. Hence having large gaps.
– Asaf
Aug 5, 2020 at 17:32

Yea, just think about what's going on. For $$n \approx e^k$$, $$\alpha\log(n) \approx k\alpha$$ and for $$n \approx e^{k+\frac{1}{2\alpha}}$$, $$\alpha\log n \approx k\alpha+\frac{1}{2}$$. So $$\{\alpha \log n\}$$ is in a particular interval of size $$\frac{1}{2}$$ for $$n$$ between $$e^k$$ and $$e^{k+\frac{1}{2\alpha}}$$. And for large $$k$$ ($$\alpha$$ is fixed), nearly all positive integers less than $$e^{k+\frac{1}{2\alpha}}$$ are greater than $$e^k$$. We conclude that nearly all $$n \le e^{k+\frac{1}{2\alpha}}$$ (you can put a floor function if you want) have $$\{\alpha \log(n)\} \in (k\alpha,k\alpha+\frac{1}{2}) \pmod{1}$$, clearly violating uniform distribution.

• I think the last part is missing something: the proportion of integers between $e^k$ and $e^{k+u}$ (with $u=1/2\alpha$) for large $k$ among integers less than $e^{k+u}$ is $1-e^{-u}$. So you only get a contradiction if $1-e^{-u} > 1/2$ ie $e^{-u} < 1/2$, ie if $\alpha$ is small… by taking variable length intervals I think we can reach a contradiction if $\alpha < 1$. In general we should be able compute the exact limit densities and deduce a contradiction for (almost?) every $\alpha$. Nov 25, 2021 at 19:57