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Question : Let $a_n=n\alpha-\lfloor n\alpha\rfloor\ (n=1,2,\cdots)$ where $\alpha$ is an irrational number. Then, does the limit $n\to\infty$ of $(a_n)^n$ exist?

I know that $\lim_{n\to\infty}(a_n)^n=0$ for a rational number $\alpha$. However, I don't have any good idea to solve the question. Can anyone help?

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1 Answer 1

up vote 5 down vote accepted

Clearly, we have $\liminf\limits_{n\to\infty} a_n^n = 0$. For irrational $\alpha$, the (simple) continued fraction expansion of $\alpha$ is infinite, and for every convergent $\frac{p_k}{q_k}$ of the continued fraction we have

$$\left\lvert \alpha - \frac{p_k}{q_k}\right\rvert < \frac{1}{q_k^2}.$$

The convergents with odd indices are larger than $\alpha$, so for odd $k$, we have

$$\alpha < \frac{p_k}{q_k} < \alpha + \frac{1}{q_k^2},$$

from which we deduce

$$q_k\alpha < p_k < q_k\alpha + \frac{1}{q_k} \iff p_k - \frac{1}{q_k} < q_k\alpha < p_k$$

and further

$$1-\frac{1}{q_k} < q_k\alpha - \lfloor q_k\alpha\rfloor = a_{q_k} < 1,$$

and hence

$$a_{q_k}^{q_k} > \left(1-\frac{1}{q_k}\right)^{q_k} \xrightarrow{q_k\to \infty} e^{-1},$$

thus

$$\limsup_{n\to\infty} a_n^n \geqslant e^{-1}.$$

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Thank you so much! –  mathlove Jul 6 at 17:05
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