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.

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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 $z_n=n$.

The limit is zero. The expression $\sin(n)$ for a large integer $n$ is a “random” number between $-1$ and $+1$. You might be afraid that by applying $\sin$ many times, it will be even more random.

But it’s not the case because $\sin x$ for any $x$ may be Taylor-expanded
$$\sin x = x – \frac{x^3}{3!} + \frac {x^5}{5!} – \dots $$
and for small $|x|$, only the first few terms are important, so the effect of computing the sines is to do almost nothing, except for a small reduction of the absolute value of $|x|$.

It is enough to notice that $|\sin n| < 1$ for all positive integers $n$; the inequality is strict because $\pi$ is irrational. We may rewrite the Taylor expansion above as an inequality
$$|\sin x| \leq |x| – \frac{|x|^3}{7} $$
for any $|x|<1$. To prove the inequality above, check that it holds for $|x|=1$, and it consequently has to hold for $|x|<1$ because for ever smaller $|x|$, the adjustments from $x^5/5!$ are ever smaller. The inequality is sharp unless $x=0$.

So we’re led to compute the $(n-1)$ times “embedded sine” of a number between $-1$ and $+1$. Each embedding subtracts at least $|x|^3/7$ from the value of $|x|$. It is easy to see that this sequence is decreasing and positive, so it must have a non-negative limit. And the limit can’t be positive, $L$, because one more iteration brings $L$ to a strictly smaller number which cannot converge back to $L$ because of the monotonic character of the sequence.

So the limit has to be zero.

By the way, for large enough $n$, the member of the sequence becomes independent of small variations of $n$ (except for the sign). So for example, for all $n$ around $10,000$, the composite sine is very close to $\pm 0.017$.

The absolute value of the $n$-th element of the sequence actually goes like $1.7/\sqrt{n}$ where the value $1.7$ is approximate. One may actually show why this decrease of $a_n$ is right. It is because
$$\frac{1.7}{\sqrt{n}} – \frac{1}{3!} \left( \frac{1.7}{\sqrt{n}} \right)^3 = \frac{1.7}{\sqrt{n+1}} + O(n^{-5/2}) $$
which is true and easily provable if you Taylor-expand this expression:
$$ 1.7 (n+1)^{-1/2} = 1.7 n^{-1/2} + 1.7 (-1/2) n^{-3/2} – \dots $$
Well, you can see that
$$1.7 \times (1/2) \approx 1.7^3/6$$
and it is actually possible to get the exact value of the constant $1.7$:
$$1.7 \approx \sqrt{3}$$
So for very large $n$,
$$ |a_n| = \frac{\sqrt{3}}{\sqrt{n}} + O(n^{-3/2}).$$