TITLE: Convergence of series of nested sequence
QUESTION [6 upvotes]: Let $a_n =\underbrace{\sin \left ( \sin \left ( \sin \cdots (\sin x) \cdots \right ) \right )}_{n \; \rm {times}}, \; \; x \in (0, \pi/2)$. Examine if the series:
$$S=\sum_{n=1}^{\infty} a_n$$
converges.
I don't know how to tackle this.It seems very difficult to me!

REPLY [5 votes]: The sequence $(a_n)$ is given by :$$\left\{ \begin{array}{l} a_0=x \\  \forall n\geq 0, \ a_{n+1}=\sin(a_n) \end{array}\right.$$
Hence $(a_n)$ belongs to $[0,\pi/2]$ and is decreasing since $$\sin(x)\leq x$$ and thus $(a_n)$ converges to $\ell\in [0,\pi/2]$.
Moreover, by continuity of the $\sin$ function, $\ell$ must satisfies $$\sin(\ell)=\ell$$ and so $\ell=0$.
Since $a_n \to 0$, one has $$a_{n+1}=\sin(a_n) = a_n -\dfrac{1}6 a_n^3 + o(a_n^4)$$
and 
$$a_{n+1}^2 = a_n^2-\dfrac{1}{3}a_n^4+o(a_n^4).$$
Now, one can use the classical Cesaro mean theorem:  since $$\dfrac{1}{a_{n+1}^2}-\dfrac{1}{a_n^2}=\dfrac{a_n^2-a_{n+1}^2}{a_n^2a_{n+1}^2} \underset{n\to +\infty}{\sim} \dfrac{1}{3}$$
we get $$\dfrac{1}{n}\left(\dfrac{1}{a_n^2}-\dfrac{1}{a_0^2}\right)= \dfrac{1}{n} \sum_{k=0}^{n-1} \dfrac{1}{a_{k+1}^2}-\dfrac{1}{a_k^2} \underset{n\to +\infty}{\longrightarrow} \dfrac{1}{3}.$$
Hence, $$\dfrac{1}{na_n^2}\underset{n\to +\infty}{\sim} \dfrac{1}3$$ and $$a_n\underset{n\to +\infty}{\sim} \sqrt{\dfrac{3}n}.$$
From this equivalent, you get the divergence of the series.