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?

This problem can be solved in a more generic manner.
Let $g(x) = x - \sin(x)$. The key observation is 


*

*$g'(x) = 1 - \cos x \ge 0 $ for all $x$, so $g(x)$ is increasing over whole $\mathbb{R}$. 

*$g'(x) = 0$ at and only at isolated points $2n\pi$ for $n \in \mathbb{Z}$,
this implies $g(x)$ is strictly increasing.


In other words, $g$ is order preserving!
Start from any real number $b$, one can construct a sequence $( b_n )_{n\in\mathbb{Z}}$ by 
$$b_n = \begin{cases} b, & n = 1\\ g(b_{n-1}), & n > 1\end{cases}$$
There are three possibilities:


*

*$b_2 = g(b_1) = b$ : we will have $b_n = b$ for all $n$.

*$b_2 = g(b_1) > b$ : we will have 
$$b_2 > b_1 \implies b_3 = g(b_2) > g(b_1) = b_2 \implies b_4 > b_3 \implies \ldots$$
i.e. $b_n$ will be an increasing sequence.

*$b_2 = g(b_1) < b$ : we will have
$$b_2 < b_1 \implies b_3 = g(b_2) < g(b_1) = b_2 \implies b_4 < b_3 \implies 
\ldots$$
i.e. $b_n$ will be an decreasing sequence.
Furthermore, if $m, M$ are any pair of successive fixed points of $g(x)$, then
for any $b \in (m, M)$, 
$$ m < b_1 < M \implies m = g(m) < b_2 = g_(b_2) < M = g(M)\\ \implies
m < b_3 < M \implies m < b_4 < M \implies \ldots$$
i.e. the whole sequences $b_n$ will falls inside $(m,M)$. Being a bounded monotonic sequence, $b_n$ will converge to some numbers $b_\infty \in [m,M]$. Since $g(x)$ is continuous, $g(b_\infty) = b_\infty$. This implies $b_\infty \ne (m,M)$ and hence $b_\infty = m$ or $M$. 
For those cases where $g(x)$


*

*has a largest fixed points $m'$ and $b$ falls between $(m',\infty)$

*or a smallest fixed points $M'$ and $b$ falls between $(-\infty,M')$, 


the analysis will be similar. In general, we will have
$$b_{\infty} \stackrel{def}{=} \lim_{n\to\infty} b_n = \begin{cases} 
+\infty,& \text{ if } g(b) > b, b \in (m', \infty)\\
m',& \text{ if } g(b) < b, b \in (m', \infty)\\
M,& \text{ if } g(b) > b, b \in (m, M)\\
m,& \text{ if } g(b) < b, b \in (m, M)\\
M',& \text{ if } g(b) > b, b \in (-\infty,M')\\
-\infty,& \text{ if } g(b) < b, b \in (-\infty, M')
\end{cases}
$$
Apply this to our function $g(x) = x - \sin(x)$. Notice the fixed points of our $g(x)$ are located at $n\pi$ with $n \in \mathbb{Z}$. Over the two successive fixed points $0$ and $\pi$,
$$g(x) = x - \sin(x) < x, \quad\text{ for } x \in (0,\pi)$$
By above argument, we find $\;\lim\limits_{n\to\infty} a_n = 0$.
As one can see, what we really need from $g(x)$ is the properties that it is continuous and strictly increasing. If you generate a sequence by functional iteration of a strictly increasing continuous function, the sequences converges
to either one of its nearby fixed points or $\pm \infty$. Which fixed point
to converge is controlled by the sign of $b_2 - b_1$.
