81
$\begingroup$

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.

  • 2
    @Arybhata: As I said, I usually write what I thi$n$k or what I did try to do, here I didn't have anything smart to say, So I just asked. So sure, there are rules about the site, and again, cause you having a hard time to understand that, I don't care AT ALL about the votedowns, Just wanted to know if I wrote something wrong. Please check my previous questions and see that I did write what I think and I tried. this is my last comment about this, Have agreat night\day.2011-06-14

1 Answers 1

164

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 $z_n=n$.

  • 1
    Okay thanks for the clarification. You didn't mention it explicitly, so I thought I was doing something wrong when I had to use the fact $\sin$ is non-decreasing on $[-1,1]$.2014-12-23