Let $\alpha$ be an irational number. Consider sequence $x_n=\{n^p\alpha\}$, $n\in\mathbb{N}$ (it's the fractional part of $n^p\alpha$), where $p$ is a nonzero real number.

Question. For which values of $p$ the sequence $\{x_n\}_{n\in\mathbb{N}}$ is equidistributed on $[0,1)$? The other qusetion is when $\{x_n\}_{n\in\mathbb{N}}$ is dense on $[0,1)$.

It's known that if $p\in\mathbb{N}$, then it's true (it's a consequence of Weyl's equdsitribution criterion and van der Corput's difference theorem). However, it's not clear how to apply Weyl's criterion in case when $p\notin\mathbb{N}$. I encountered similar problem when I was working on this question Convergence of the product $\prod_{n=1}^{\infty}\left(1+\frac{x^n}{n^p}\right)\cos\frac{x^n}{n^q}$ (in order to study the behaviour of $\cos\frac{1}{n^q}$, so $\alpha=1/\pi$ in this case).

Update. It would be also intersting to investigate the distbution of $\left\{\frac{x^n}{n^p}\right\}$.

If the result is known, please give a link or reference. Any help would be appreciated.

  • 1
    $\begingroup$ One can, I believe, get an (oscillating) asymptotic formula for $\sum x_n$. I recommend checking Montgomery's Ten Lectures on the Interface between Analytic Number Theory and Harmonic Analysis. $\endgroup$ May 27, 2020 at 17:34

1 Answer 1


The following is Theorem 3.4 of 'Uniform Distribution of Sequences' by Kuipers and Niederreiter.


Let $(f(n))$, $n=1,2,\ldots,$ be a sequence of real numbers, and let $k$ be a positive integer. If $\Delta^k f(n)$ is monotone in $n$, if $\Delta^k f(n) \rightarrow 0$ and $n|\Delta^kf(n)|\rightarrow\infty$ as $n\rightarrow\infty$, then the sequence $(f(n))$ is uniformly distributed modulo $1$.

Here $\Delta f(n)=f(n+1)-f(n)$ and $\Delta^k = \Delta \circ \Delta^{k-1}$. When $p>0$ is real and not an integer, then we can find $k$ that satisfies conditions of the Theorem. Further, it is easy to see that $p\leq 0$ does not give uniform distribution. Hence, your sequence is uniformly distributed if and only if $p>0$.

  • $\begingroup$ Thanks. Am I right that we should choose $k=[p]$? $\endgroup$
    – richrow
    May 28, 2020 at 7:28
  • $\begingroup$ Actually $\lfloor p \rfloor +1$. $\endgroup$ May 28, 2020 at 16:11
  • $\begingroup$ By the way, it's easier to apply here the next theorem (3.5) (just the same but in differential form, i. e. if $f^{(k)}(x)\to 0$ monotonically and $x|f^{(k)}(x)|\to+\infty$, when $x\to\infty$, then we have uniform distribution of $(f(n))$). $\endgroup$
    – richrow
    Dec 1, 2020 at 8:00

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