Compute $ \lim\limits_{n \to \infty }\sin \sin \dots\sin n$

I need your help with evaluating this limit:

$$ \lim_{n \to \infty }\underbrace{\sin \sin \dots\sin}_{\text{$n$ compositions}}\,n,$$

i.e. we apply the $\sin$ function $n$ times.

Thank you.

Solution 1:

The first sine is in $I_1=[-1,1]$ hence the $n$th term of the sequence is in the interval $I_n$ defined recursively by $I_1=[-1,1]$ and $I_{n+1}=\sin(I_n)$. One sees that $I_n=[-x_n,x_n]$ where $x_1=1$ and $x_{n+1}=\sin(x_n)$. The sine function is such that $0\le\sin(x)\le x$ for every nonnegative $x$ hence $(x_n)$ is nonincreasing and bounded below by zero hence it converges to a limit $\ell$. The sine function is continuous hence $\ell=\sin(\ell)$. The only fixed point of the sine function is zero hence $\ell=0$. This proves that $x_n\to0$, that the sequence $(I_n)$ is nonincreasing and that its intersection is reduced to the point zero and finally, that the sequence considered in the post converges to zero.

Edit: The argument above shows that for every sequence $(z_n)$, the sequence $(s_n)$ defined by $s_n=\sin\sin\cdots\sin(z_n)$ (the sine function being iterated $n$ times to define $s_n$) converges to zero. In other words, there is nothing particular about the choice $z_n=n$.