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?
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$\begingroup$ Hint : Try to prove that $a_n \sim \sqrt{3/n}$. $\endgroup$– TheSilverDoeMay 5, 2021 at 11:44
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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
$$\frac{1}{a_{n}^2} - \frac{1}{a_{0}^2} \sim \frac{n}{3}.$$
answered May 5, 2021 at 11:45
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$\begingroup$ Dear downvoter, a comment would be appreciated! $\endgroup$ May 5, 2021 at 11:45
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$\begingroup$ Welcome to the club of victims of mysterious downvotes ! $\to +1$ $\endgroup$ May 5, 2021 at 12:13