Prove $x_n$ is convergent and $\lim_{n\to\infty}x_n=0$ $$f:\mathbb R\to \mathbb R, f(x)=x-\arctan(x)$$
Let series $(x_n)_{n\in\mathbb N}$, $x_0>0$, and $x_{n+1}=f(x_{n})$.
Prove $x_n$ is convergent and $\lim_{n\to\infty}x_n=0$ 
I tried calculating $f'(x)$ and it comes out to:
$f'(x)=\frac{x^2}{x^2+1}$, which is positive for any $x\in\mathbb R$.
This means that series $x_n$ is increasing, so it is monotone. It is also bounded by $0$ but that's all I know to say. From my experience, this kind of exercises are solved by using squeeze theorem, but I do not know what an upper bound would be for $x_n$.  
I would greatly appreciate any help.
 A: From $f(x)=x-\arctan(x)$, we get
$$f'(x)=\frac{x^2}{1+x^2}$$
Since $f(0)=0$,$\;$and$\;f'(x) > 0\;$for all$\;x > 0$,$\;$it follows that$\;f(x) > 0\;$for all$\;x > 0$.

Hence$\;x_n > 0\;$for all$\;n$.

Identically, we have
$$x_n - x_{n+1}=x_n-f(x_n)=x_n-(x_n-\arctan(x_n))=\arctan(x_n) > 0$$
hence the sequence $(x_n)$ is strictly decreasing.

Since the sequence $(x_n)$ is positive and decreasing, it follows that $(x_n)$ approaches a limit, $L$ say, with $L \ge 0$.
\begin{align*}
\text{Then}\;\;&x_{n+1}-f(x_n)=0\\[4pt]
\implies\;&\lim_{n\to\infty}x_{n+1}-f(x_n)=0\\[4pt]
\implies\;&L-f(L)=0\;\;\;\;\;\text{[$f(x_n)\;$approaches$\;f(L)\;$since $f$ is continuous]}\\[4pt]
\implies\;&L=f(L)\\[4pt]
\implies\;&L=L-\arctan(L)\\[4pt]
\implies\;&\arctan(L)=0\\[4pt]
\implies\;&L=0\\[4pt]
\end{align*}
A: Let $x_{0}>0$ be given. Fix $M>0$ such that $0<x_{0}<M$. Let $X=[0,M]$,
which is a complete metric space with respect to the usual metric.
Let $f:X\rightarrow X$ be defined by $f(x)=x-\arctan x$. We need
to show that $f$ is well-defined. Let $g:[0,M]\rightarrow\mathbb{R}$
be defined by $g(x)=x-\arctan x$, then $g'(x)=\frac{x^{2}}{1+x^{2}}>0$
for all $x\in(0,M)$ and it follows that $g$ is strictly increasing
on $[0,M]$. In particular, $f(x)=g(x)\geq g(0)=0$. On the other
hand, $f(x)\leq x\leq M$, so $f(x)\in[0,M]$.
Next, we show that $f$ is a contractive map. Let $x,y\in X$ with
$x\neq y$. Without loss of generality, we may assume that $x<y$.
By mean-value theorem, there exists $\xi\in(x,y)$ such that $|f(y)-f(x)|=|f'(\xi)(y-x)|\leq c|y-x|$,
where $c=\frac{M^{2}}{1+M^{2}}<1$. (Here, observe that $x\mapsto\frac{x^{2}}{1+x^{2}}$,
$x\in[0,M]$, is strictly increasing and hence $|f'(\xi)|\leq\frac{M^{2}}{1+M^{2}}$).
By Banach Fixed Point Theorem, the sequence $(x_{n})$ constructed
by $x_{n}=f(x_{n-1})$ is convergent. Moreover, if $x=\lim_{n\rightarrow\infty}x_{n}$,
$x$ is the unique fixed point of $f$. That is, $f(x)=x$ and hence
$x=0$.
