Examine the convergence of a sequence ${a_{n}}$: $a_{n+1}=a_{n}-\sin(a_{n}),0\le a_{1}<\pi$ ${a_{n}}$: $a_{n+1}=a_{n}-\sin(a_{n}),0\le a_{1}<\pi$
One way to do it is to show that the sequence is bounded and monotonous.
How to show that it is bounded? 
If
$$-1\le \sin(a_{n})\le 1$$
then
$$0\le a_{n}<2\pi$$
Is it right that then
$$0\le a_{n+1}\le 2\pi$$
and the sequence is bounded (below and above)?
From
$$a_{n+1}-a_{n}=a_{n}-\sin(a_{n})-a_{n}=-\sin(a_{n}),a_{n}\in[0,2\pi]$$
the sequence is monotonous and non-increasing.
Can someone check this problem?

Your first implication does not hold. A counterexample would be $a_n=100$. Its sine is certainly between $-1$ and $1$, but $100$ itself is not between $0$ and $2\pi$.
I think it would be easier to say that if $a_n\in[0,\pi]$ then $a_{n+1}\in[0,\pi]$ because $0\le\sin x\le x$ when $x\in[0,\pi]$. Then, by induction, every $a_n$ is in this interval.
It is also true that if $a_n\in[0,2\pi)$ then $a_{n+1}\in[0,2\pi)$, but on one hand it is more work to argue that (because $\sin a_n$ can then be both positive or negative), and on the other hand it doesn't tell you enough that you can argue that the sequence is monotonic. For that you need to know that the $a_n$s stay within the range where the sine is nonnegative.
