I want to determine the convergence of $$\begin{cases}x_0=1 & \\ x_n=\sin(x_{n-1}) \end{cases}$$ I can see that $x_n=\sin(x_{n-1}) \geq -1$. Which means the sequence is bounded. However, it isn't strictly decreasing in $[-1,1]$, so I can't prove that it is convergent. Is there a way to get a better bound and show that it goes to $0$?
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3The answer to your question is here Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$. – Somos Oct 13 '20 at 21:06
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2It is stricktly decreasing. The inequality $\sin(x) \leq x$ is useful. – Winther Oct 13 '20 at 21:07
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First you can easily prove that $]0,\frac{\pi}{2}]$ is stable by $\sin$ therefore your sequence $(x_n)_n$ is valued in $]0,\frac{\pi}{2}]$ since $1 \in ]0,\frac{\pi}{2}]$ .
You can use the inequality :
$$\sin{x} <x, \quad \forall x >0$$
to prove that $x_{n+1} \leq x_n$ and your sequence is strictly decreasing and bounded by $0$ so it converges.
Then, by the continuity of $x \mapsto \sin(x)$, you know the sequence converges toward the solution of $\sin{l}=l$, i.e $l=0$.
Velobos
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