Investigate convergence of the functional series $\sum\limits_{n>0}\sin^{\circ n}x$ When I was in college my professor challenged us to investigate the convergence of the following beauty 
$$\sum\limits_{n>0}\sin^{\circ (n)}x=\sin x+\sin(\sin x)+\sin(\sin(\sin x))+\ldots$$
He claimed that for many years in his carrier only once he saw a valid proof from one student.  The standard Cauchy and Ratio limits $$\lim_{n\to\infty}\left|\frac{\sin^{\circ (n+1)}x}{\sin^{\circ (n)}x}\right|=\lim_{n\to\infty}\left|\sin^{\circ (n)}x\right|^\frac{1}{n}$$ are challenging to compute. My failed attempt included Taylor expansion around $0$ but even the compositions of polynomials like $$\left(x-\frac{x^3}{6}+\frac{x^5}{120}+\ldots\right)^{\circ n}$$ are difficult to estimate since it is not straightforward how truncation for the expansion series will affect the original infinite sum. Clearly we could start by assuming that without losing generality that $0<x<<1$, and even the necessary convergence test is obvious only from the geometric argument.
I think it's a fun problem and hope someone will enjoy cracking it.

If $x$ is a multiple of $\pi$, the series is identically zero hence it converges. Otherwise, $0\lt|\sin x|\leqslant1$ hence $\sin x$ is not a multiple of $\pi$. Assume without loss of generality that $\sin x\gt0$, then the sequence $(\sin^{\circ n}x)_{n\geqslant1}$ converges to zero hence, either the series $\sum\limits_{n\geqslant1}\sin^{\circ n} x$ converges absolutely, or it diverges.
Now, consider $x_n=\frac1{\sqrt{n}}$ and compare $\sin x_n$ with $x_{n+1}$. Limited expansions are 
$$\sin x_n=x_n-\tfrac16x_n^3+o(x_n^3)=\tfrac1{\sqrt{n}}-\tfrac16\tfrac1{n\sqrt{n}}+o\left(\tfrac1{n\sqrt{n}}\right),
$$ 
and 
$$
x_{n+1}=\tfrac1{\sqrt{n}}\left(1+\tfrac1n\right)^{-1/2}=\tfrac1{\sqrt{n}}-\tfrac12\tfrac1{n\sqrt{n}}+o\left(\tfrac1{n\sqrt{n}}\right).
$$
In particular, for $n$ large enough, say for every $n\geqslant k$, $\sin x_n\geqslant x_{n+1}$.
If $\sin x\gt0$, there exists some $i\geqslant k$ such that $\sin x\geqslant x_{i+1}$. Then $\sin^{\circ n} x\geqslant x_{i+n}$ for every $n\geqslant1$. For every $i$, the series $\sum\limits_{n\geqslant1}x_{i+n}$ diverges. This proves that, for every $x$ not a multiple of $\pi$, the series $\sum\limits_{n\geqslant1}\sin^{\circ n} x$ diverges.
