Is $c + \sin c + \sin\sin c + \sin\sin\sin c + \cdots$ convergent? I thought about sequences, and I had an idea.

Let's define a sequence $(a_i)$ such that 
  $$a_0 = c \qquad\text{and}\qquad a_{n+1}=\sin a_n$$
  where $c$ is arbitrary constant. Let's also define $$S_N := \sum_{i=0}^Na_i=c\;+\;\sin c\;+\;\sin\sin c\;+\;\cdots\;+\;\underbrace{\sin\sin\cdots\sin}_{N\text{ times}} c$$
Is $\lim_{N\to\infty} S_N$ convergent or divergent?

The problem is that I have no idea how I can, or if it is even possible to, determine $a_n$. Or if it is possible to show whether the sum converges.
 A: Let denote $\sin^{[n]}c$ to be the $n$-times iteration of $\sin$. It is quite clear we may well assume $c \in (0,1]$, since the first iteration $\sin c$ is in $[-1,1]$. If $\sin c = 0$ it is trivial, if $\sin c = [-1,0)$, just flip the sign in the subsequent arguments.
We first note that by concavity of $\sin$, $\sin x \geq x\sin 1$ for $x \in (0,1]$.
Now given any $\epsilon \in (0,\sin c)$, since $a_n$ is strictly decreasing, the set $\{n:a_n \geq \epsilon\}$ (non-empty) has a largest element $N$, in particular it holds that we may always find $N$ such that $\epsilon \sin 1 \leq \sin^{[N+1]} c < \epsilon$. 
By concavity of $\sin$ again $\sin^{[N+1+k]}(c) \geq \epsilon^{-k} (\sin \epsilon)^k \sin^{[N+1]} c \geq \epsilon^{-k} (\sin \epsilon)^k \epsilon \sin 1$ holds for all $k \in \mathbb{N}$, so that in particular $\sum_{k=0}^\infty a_{N+1+k} \geq \sin (1) \epsilon (1 - \epsilon^{-1} \sin \epsilon)^{-1}$.
Since $\epsilon (1 - \epsilon^{-1} \sin \epsilon)^{-1} \to \infty$ as $\epsilon \to 0$, we can conclude that for any $K > 0$, we may find $N$ such that $\sum_{n=N+1}^\infty a_n > K$, hence the sum diverges.
