Does a irrational number $x > 1$ exist such that $\{x^n \} \le \frac{1}{2}$ for
all positive integers $n$ ?


$x=1+ \sqrt 2$ holds for $n$ odd, but not in even



Fractional part power
The OP asks for an instance of what Dubickas [1] has called a ${\cal Z}$-number: A real number $x>1$ for which there exists a real $\xi\neq 0$ such that $\{\xi x^n\}<1/2$ for every integer $n$.


An example [2] of an irrational ${\cal Z}$-number is $x=\tfrac{1}{2}(7+\sqrt{41})$, when $\{\xi x^n\}$ with $\xi=\tfrac{1}{4}(2+3x)$, converges to $1/4$ for $n\rightarrow\infty$.


The OP asks for an irrational ${\cal Z}$-number with $\xi=1$. It is known [3] that there exists no algebraic non-integer $x>1$ such that $\{x^n\}$ converges to 0. A stronger result (theorem 4 of [2]) indicates that a non-integer ${\cal Z}$-number with $\xi=1$ cannot be a [Pisot number](https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number). I am not aware of any results for non-algebraic numbers.


1. [Even and odd
integral parts of powers of a real number](https://doi.org/10.1017/S0017089506003090), A. Dubickas (2006).
2. [On the
limit points of the fractional parts of powers of Pisot numbers](https://www.emis.de/journals/AM/06-2/am1306.pdf),
A. Dubickas (2006).
3. [On a
question of G. Kuba](https://link.springer.com/article/10.1007/s000130050442), F. Luca (2000).


