A possible upper bound on the fractional part of exponential functions

Question

Does there exist a real number $a$ bigger than $1$ that's not a rational power of an integer such that $\displaystyle\lim_{n\to\infty}${$a^n$}$=0$ where values for $n$ are positive integers?

(P.S. $\{ x \} = x - \lfloor x \rfloor$ or namely the fractional part function)

I just saw this problem today. I've just started studying about the fractional part approximations so unfortunately I've made no progress solving this.
*I made a slight change in the problem to avoid any confusions about small values

Answer 1

It is hard to find such $a$. Indeed, according to Wikipedia, the set of such $a$ is countable. (Remark that Corollary 3 of Zeldich’s attraction theorem states that a set $\{a\in (1,\infty): \{a^n\}\mbox{ is not dense in }[0,1]\}$ is meager in $(1,\infty)$.)

Moreover, if the answer to a longstanding Pisot-Vijayaraghavan problem is affirmative then there is no such $a$. Indeed, for a contradiction pick such $a$. The affirmative answer to Pisot-Vijayaraghavan problem implies that $a$ is a Pisot–Vijayaraghavan number. Let $a_1=a$ and $a_2, \dots, a_m$ be the conjugates of $a$. Then for each natural $n$, $\sum_{k=1}^m a_k^n$ is integer.

For each $k=1,\dots, m$ let $a_k=r_ke^{\varphi_ki}$ for some positive $r_k$ and $\varphi_k$. Let $G=\Bbb T^m=\{z\in\Bbb C:|z|=1\}$ be a (multiplicative) topological group and $g=(e^{\varphi_1i}, e^{\varphi_2i},\dots, e^{\varphi_m i})$ be an element of $G$. Since the group $G$ is compact, it is well-known and easy to show that $G$ topologically periodic, so for any neighborhood $U$ of the identity of $G$ there exists arbitrarily big natural number $n$ such that $g^n\in U$. Pick $U_0=\{(x_1,\dots,x_m)\in\Bbb T^m: \forall i (\operatorname{Re} x_i\ge 0)\}$.

Let $N>0$ be any number. Since $a$ is a Pisot-Vijayaraghavan number, we have $r_k<1$ for each $k=2,\dots, m$. So there exists $M>0$ such that $\sum_{k=2}^m r_k^M<1/2$. Pick $n>N,M$ such that $g^n\in U_0$. Then for some integer $K$ we have

$$K=\sum_{k=1}^m a_k^n=\sum_{k=1}^m r_k^ne^{n\varphi_ki} =\sum_{k=1}^m \operatorname{Re} r_k^n (e^{n\varphi_ki})= a^n+\sum_{k=2}^m r_k^n \operatorname{Re} e^{n\varphi_ki}.$$

So $a^n\le K\le a^n+\sum_{k=2}^m r_k^n\le a^n+\sum_{k=2}^m r_k^M<a^n+1/2$. Thus $\{a^n\}>1/2$, a contradiction.

Answer 2

I don't have a positive answer to OP's question; but note that it's easy to find solutions to the seemingly similar question where you would like to have $\lim_{n\rightarrow\infty}\{a^n\}=1$. For instance, $$a_m = \frac{1}{2}\left(m + \sqrt{m^2-4}\right) \in (1,\infty)$$ has this property for any integer $m > 2$. This works because the linear recurrence defined by $f_0=2,$ $f_1=m,$ and $f_{n+2}=mf_{n+1} - f_n$ can be explicitly solved to yield $$ f_n = a_m^n+b_m^n, $$ where $b_m=\frac{1}{2}\left(m - \sqrt{m^2-4}\right)\in(0,1)$. Hence $$ \{a_m^n\}=\{f_n - b^n_m\}=\{-b^n_m\}\rightarrow 1. $$ It seems as if it should be possible, perhaps with a higher-range linear recurrence, to do the same thing with the opposite sign: we would need, say, $f_n=a^n-Bb^n+...$ for some $a>1$ and $B> 0$ and $b\in(0,1)$, and for the remaining terms to be real and decaying faster than $b^n$. Some constraints are placed on this by the fact that the sum of the eigenvalues of an integer matrix is integral.