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Let d(x) denote the distance from x to the nearest integer.

Are there any non-integral numbers X for which the sequence d(X), d(X^2), d(X^3), etc. converges to 0?

EDIT: Sorry, I forgot to exclude the trivial case where X is between -1 and 1. I was looking for more interesting cases.

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    $\begingroup$ Not a research level question, voting to close. $\endgroup$
    – Igor Rivin
    Commented Dec 17, 2011 at 0:46
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    $\begingroup$ Look up Pisot numbers on Google. $\endgroup$
    – fedja
    Commented Dec 17, 2011 at 0:52
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    $\begingroup$ $X=1/2$ seems to work. $\endgroup$
    – Alain Valette
    Commented Dec 17, 2011 at 10:28
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    $\begingroup$ It's disappointing to see the trivial answers of magnitude less than $1$ repeated. $\phi = (\sqrt 5 + 1)/2$ has this property. It follows from the closed form expression for Lucas numbers $L(n) = \phi^n + (-\phi)^{-n}$, and a similar argument works for other Pisot numbers. Nevertheless, this is not research level. $\endgroup$
    – Douglas Zare
    Commented Dec 17, 2011 at 15:26
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    $\begingroup$ Hi yrudoy: I can't comment on the level or interest of the question, it being far from my area of expertise. But the style can be improved. A good mathoverflow question includes some background and motivation: Why is this a question you are interested in? Why should I be interested in it? How does it relate to your research? Please read carefully mathoverflow.net/howtoask and look over other successful MO questions. If you still think a version of this question is appropriate, edit it, and either flag for moderator attention or leave a comment at tea.mathoverflow.net . $\endgroup$
    – Theo Johnson-Freyd
    Commented Dec 18, 2011 at 5:39

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Even assuming that $|x| > 1$, there are some counter-examples, for example, as noted in the comments (exercise!) $$x = \frac{1 + \sqrt{5}}{2}.$$ Let $\|\alpha\|$ denote the distance from a real number $\alpha$ to the closest integer. It is an interesting problem to classify the set $S$ of real numbers $|x| > 1$ such that $$\lim_{\rightarrow} \|x^n\| = 0.$$ A special subset of $S$ is given by the set of Pisot-Vijayaraghavan (or PV) numbers:

http://en.wikipedia.org/wiki/Pisot%2DVijayaraghavan_number

which are real algebraic integers $\theta$ all of whose conjugates have absolute value less than $1$. (The example above is of this class.) In this case, the convergence of $\|x^n\|$ to zero is exponential. Conversely, if $x \in S$ and the convergence is fast enough ($L^2$), then $x$ is a PV number (this was proved by Pisot). However, it is not known whether there are any other real numbers in $S$. Even worse, it's very hard to tell whether any given number (say $e$ or $\pi$) lies in $S$. On the other hand, a theorem of Koksma says that for almost all $x > 1$, the fractional parts of $x^n$ are uniformly distributed in $[0,1]$.

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