<s>[INST] Fractional part of rational power arbitrary small I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).
 [/INST] I will show that
if $a = 1+\sqrt{2}$
then the limit points of
$\{a^n\}$
are $0$ and $1$.
I know that this doesn't tell anything
about rational $a$,
but this might be of use.
Note that this method can show this
for $a$ and $b$ roots of
$x^2-2ux-v = 0$
where $u$ and $v$ are
positive integers such that
$v < 2u+1$.
This case is $u=v=1$;
$u=1, v=2$ also works.
If $a = \sqrt{2}+1$
and
$b = 1-\sqrt{2}$
then
$ab = -1$
and
$a+b = 2$.
Therefore $a$ and $b$
are the roots of
$x^2-2x-1 = 0$.
If $u_n = a^n+b^n$,
then
$u_0 = 2,
u_1 = 2$.
Since
$a^{n+2}
=a^n(a^2)
=a^n(2a+1)
=2a^{n+1}+a^n$
and similarly for $b$,
$\begin{array}\\
u_{n+2}
&=a^{n+2}+b^{n+2}\\
&=2a^{n+1}+a^n+2b^{n+1}+b^n\\
&=2u_{n+1}+u_n\\
\end{array}
$
Therefore
$u_n$ is a
positive integer for all $n$.
Since
$|b| < 1$,
$|b^n| \to 0$.
Since
$b < 0$,
$b^{2n} > 0$
and
$b^{2n+1} < 0$.
Therefore,
since
$a^n = u_n - b^n$,
$\{a^{2n}\}
=\{u_{2n}-b^{2n}\}
=1-b^{2n}
$
so
$\{a^{2n}\}
\to 1
$
and
$\{a^{2n+1}\}
=\{u_{2n+1}-b^{2n+1}\}
=\{u_{2n+1}+|b^{2n+1}\}
=|b^{2n+1}|
$
so
$\{a^{2n+1}\}
\to 0
$.
Therefore
the limits points of
$\{a^n\}$
are $0$ and $1$.
</s>