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?

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $enter image description here


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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.