Convergence of series of nested sequence 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!
 A: First we note that the sequence $(a_n)$ is (strictly) monotonically decreasing, and $\lim\limits_{n\to\infty} a_n = 0$, whatever the choice of $x$. Thus, choosing an (almost) arbitrary large $k \in \mathbb{N}$, there is an $n_k$ with
$$\frac{1}{k} \leqslant a_{n_k} < \frac{1}{k-1}.\tag{1}$$
Now we use that for $0 < y < 1$ we have
$$0 < y - \sin y < \frac{y^3}{6},\tag{2}$$
and therefore
$$a_{n_k + m} > a_{n_k} - \frac{m}{6(k-1)^3}.$$
For $0 \leqslant m < k+1$ we thus have
$$a_{n_k+m} > a_{n_k} - \frac{k+1}{6(k-1)^3} \geqslant \frac{1}{k} - \frac{k+1}{6(k-1)^3} \geqslant \frac{1}{k+1},$$
provided $k$ is so large that
$$\frac{k+1}{6(k-1)^3} < \frac{1}{k(k+1)},$$
or, equivalently $\frac{k(k+1)^2}{(k-1)^3} < 6$, which means $k > 3$.
Thus
$$\sum_{n = n_k}^{n_k + k} a_n > \sum_{n = n_k}^{n_k+k} \frac{1}{k+1} = 1,$$
which shows that the series is divergent.
A: 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.
