How slow does sine iteration converges?

Question

It is no hard to prove that the real sequence $\{a_n\}_{n=1}^\infty$ decided by $$ \begin{cases} a_1 = 1\\ a_{n+1} = \sin a_n \end{cases} $$ converges to $0$ as $n\to\infty$. However, it seems that such a convergence is extremely slow (at least not exponentially), since $a_{n+1}-a_{n} = o(a_n^3)$. However, I still want to know just how slow it is. To be more precise, $$ \text{is }a_n \text{ of order } \frac{1}{n}, \frac{1}{n^2}, \text{or}\ \frac{1}{\ln n}, \text{or something else?} $$ And how should we analyze a problem like this?

Answer 1

You can prove that

$$a_n \sim \sqrt{\frac{3}{n}}$$ based on the fact that

$$\frac{1}{a_{n+1}^2} - \frac{1}{a_{n}^2} \sim \frac{1}{3}$$ using Taylor expansion of $\sin x$. From there, you get that the series $\sum \left(\frac{1}{a_{n+1}^2} - \frac{1}{a_{n+1}^2}\right)$ diverges and

$$\frac{1}{a_{n}^2} - \frac{1}{a_{0}^2} \sim \frac{n}{3}.$$