<s>[INST] Does the maxima of $\sin x + \sin (\sin x ) + \sin(\sin (\sin x )) + \sin(\sin(\sin (\sin x ))) + ...$ converge? Does the maxima of $\sin x + \sin (\sin x ) + \sin(\sin (\sin x )) + \sin(\sin(\sin (\sin x ))) + ...$ converge?
Given that $f^{n+1}(x) = f(f^n(x))$, where $f(x) = \sin x$, each "next" term grows by $f^{n+1}(x)$. Thus, if there's a convergence point, it would mean that $f^{\infty}(\pi/2) = 0$. Is there any way to prove this?
Another form of this expression would be the repeated integral bounds of $\cos x$, something like $\int_0^{\int^{pi/2}_0\cos xdx}\cos x dx$ repeating.
 [/INST] Given that, when $0<x<1$ you have $$x>\sin x >x\left(1-\frac{x^2}6\right)$$ you get, when $x=1/n$ for $n\geq 2,$ $$\begin{align}\sin\frac1n&>\frac{n^2-1/6}{n^3}\\&>\frac{n^2-1}{n^3+1}\\&=\frac{n-1}{n^2-n+1}\\&=\dfrac{1}{n+\frac{1}{n-1}}\\&\geq \frac1{n+1}
\end{align}$$
So $\sin(1/n)\geq \frac{1}{n+1}.$
So let $x_n=f^n(x)$ then when $\frac{1}{n}<x_k<1$ then $\frac{1}{n+1}<x_{k+1}<1$ and thus by induction, $\frac{1}{n+m}<x_{k+m}.$
So $\sum_{i=1}^{\infty} x_i$ diverges because $$\sum_{i=0}^N x_{k+i}>\sum_{i=0}^N \frac1{n+i}$$
The same happens when $-1<x_k\leq -\frac1{n}.$ We know $|\sin(\sin x))|<1.$ If it is non-zero, then $|\sin(\sin x))|>\frac{1}{n}$ for some $n,$ and thus the sum diverges.
So the only time you can get convergence is when $\sin(\sin x)=0.$ but that is only possible if $\sin(x)=0.$
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