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).
 A: I think that the wolfram link cited the first result (Vijayaraghavan 1941) incorrectly where it reports that 

$\{ (3/2)^n \}$ has infinitely many accumulation points in both $[0, 1/2)$ and $[1/2, 1]$.

The first result actually is 

$\{ (3/2)^n \}$ has infinitely many accumulation points in $[0,1]$. 

The second result cited by wolfram is by Flatto, Lagarias, Pollington (1995) which is available here: http://matwbn.icm.edu.pl/ksiazki/aa/aa70/aa7023.pdf. In the introduction of this paper by them, it reports that

Vijayaraghavan later remarked that it was striking that
  one could not even decide whether or not $(3/2)^n$ mod $1$ has infinitely
  many limit points in $[0, 1/2)$ or in $[1/2, 1)$.

This paper proved that 

$$\limsup\{ (3/2)^n \} - \liminf \{(3/2)^n\} \geq \frac13. $$

A few later results are mentioned (Dubickas 2006, 2008) in Yann Bugeaud's book "Distribution Modulo One and Diophantine Approximation" (for example, p. xi, p. 67 and p. 68). 

$\{ (3/2)^n \}$ has a limit point in $[0.238117 . . . , 0.761882 . . .]$ which has length $0.523764 . . .$.
$\{(3/2)^n\}$ has a limit point in $[0, 8/39] \cup
[18/39, 21/39] \cup [31/39, 1]$, of total length $19/39$. 

A: 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 $1$.
A: Edit: sorry, I now realize that the question asks whether 0 is a limit point of $\{a^n\}$. My answer below is not a paraphrase of the question but is something weaker. But I still hope it helps.

Let me rephrase the question:
Let $a$ be a rational greater than 1, is it true that $\lim_{n \to \infty}\{a^n\} = 0$?
When $a$ is an integer, yes, $\lim_{n \to \infty}\{a^n\} = \lim_{n \to \infty} 0 = 0$.
However, When $a$ is a noninteger rational, $\lim_{n \to \infty}\{a^n\} \ne 0$.
Proof by contradiction, assume $\lim_{n \to \infty}\{a^n\} = 0$. let $$ be the denominator of $\{a\}$, we can find an $$ such that $∀ n > $, $\{x^n\} < \frac{1}{aq}$.
Let $ > $, we have
$$\{a^m\} < \frac{1}{aq} \tag{1}\label{1}$$ and
$$\{a^{m+1}\} < \frac{1}{aq} < \frac{1}{q} \tag{2}\label{2} $$
where
\begin{align}
\{a^{m+1}\} &=  \{ (⌊a^m⌋ + \{a^m\}) (⌊a⌋ + \{a\}) \} \\  & = \{ ⌊a^m⌋⌊a⌋ + ⌊a^m⌋\{a\} +  \{a^m\}(⌊a⌋ + \{a\}) \} \\  
&= \{⌊a^m⌋\{a\} +  \{a^m\}(⌊a⌋ + \{a\})\} \\
&= \{⌊a^m⌋\{a\} +  \{a^m\}a\} \\
& = \{\{⌊a^m⌋\{a\}\} +  \{a^m\}a\}
\end{align}
where the last step comes from the observation that { + } = {{} + }.
Note $\{⌊a^m⌋\{a\}\}$ is one of $1/, 2/, 3/... (-1) / $, in particular
$$ \frac{1}{q} \le \{⌊a^m⌋\{a\}\} \le \frac{q-1}{q}  \tag{3}\label{3}$$
We have $1/ < \{⌊a^m⌋\{a\}\} +  \{a^m\}a < 1$ by \eqref{1} \eqref{3}, which implies $\{a^{m+1}\} = \{\{⌊a^m⌋\{a\}\} +  \{a^m\}a\} > 1/$ which contradicts \eqref{2}.
