# How slow does sine iteration converges? [duplicate]

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?

• Hint : Try to prove that $a_n \sim \sqrt{3/n}$. May 5, 2021 at 11:44

$$a_n \sim \sqrt{\frac{3}{n}}$$ 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
$$\frac{1}{a_{n}^2} - \frac{1}{a_{0}^2} \sim \frac{n}{3}.$$
• Welcome to the club of victims of mysterious downvotes ! $\to +1$ May 5, 2021 at 12:13