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We consider numbers $\alpha\in \mathbb{R}$ with $|\alpha|>1$. Is there any result about a characterization of those $\alpha$ so that $\{\alpha^n\}_{n\in \mathbb{N}}$ is convergent modulo 1?

I already found that characterizing those $\alpha$ such that $\{\alpha^n\}_{n\in \mathbb{N}}$ is uniform distributed modulo 1, is a hard problem.

Motivation : Let $\varphi=\frac{1+\sqrt{5}}{2}$. Calculate the value $\lim_{n\to \infty} \cos(\frac{\pi}{\sqrt{5}}\varphi^{3n})$.

Since $\varphi^{3n}-\varphi^{-3n}=\sqrt{5}F_{3n}$ where $F_n$ is a $n$-th Fibonacci number, we can calculate the value $\lim_{n\to \infty} \cos(\frac{\pi}{\sqrt{5}}\varphi^{3n})$.

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    $\begingroup$ en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number $\endgroup$
    – Puddles the turtle
    Aug 7, 2019 at 2:29
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    $\begingroup$ @Puddlestheturtle: I think your comment pretty much answers the question. I recommend that you turn it into an answer! $\endgroup$
    – GH from MO
    Aug 7, 2019 at 2:39
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    $\begingroup$ @GHf, I don't think the Wikipedia page says anything about non-algebraic $\alpha$ (in the absence of square-summability). Then again, I don't think anything much is known, other than that uniform distribution holds outside a set of measure zero (we're tacitly assuming $|\alpha|>1$). $\endgroup$
    – Gerry Myerson
    Aug 7, 2019 at 6:27
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    $\begingroup$ As far as I'm aware, it's not even known that $\{e^n\}$ doesn't converge to $0$. $\endgroup$
    – Greg Martin
    Aug 7, 2019 at 7:29
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    $\begingroup$ @GHfromMO: I think your comment pretty much answers the question. I recommend that you turn it into an answer! $\endgroup$
    – Puddles the turtle
    Aug 7, 2019 at 12:06

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