$\lim \{r^n\}$ exists, Is $r$ an integer?

Question

$r\in\Bbb R$, $|r|\gt1$ and $\lim\limits_{n\to\infty}\{r^n\}$ exists. Can one conclude that $r$ is an integer? Here, $\{x\}=x-[x] $ is the fractional part of $x\in\Bbb R$

If $r\in\Bbb Q$, the answer is yes!

Thanks a lot!

Answer 1

Nope!

For example, the number

$$(1 - \sqrt{2})^n + (1 + \sqrt{2})^n $$

is always an integer, which can be seen in various ways, such as using the binomial theorem to expand the two binomials and seeing that the odd powers of $\sqrt{2}$ cancel out.

However, $|1 - \sqrt{2}| < 1$, and so the first term converges to $0$ as $n \to \infty$.

Thus, the difference between $(1 + \sqrt{2})^n$ and the nearest integer decreases to $0$ as $n \to \infty$.

This sequence actually alternates between being slightly above and slightly below an integer: to get the fractional part to converge, we can just take every other term. Thus

$$r = (1 + \sqrt{2})^{2} = 3 + 2 \sqrt{2}$$

is a counterexample to your conjecture.