Rate of convergence of ${x_n}=\arctan ({x_{n-1}})$ Find the rate of convergence of : 
${x_n}=\arctan ({x_{n-1}})$. Well, 
$$\lim\limits_{n\to\infty}\frac{{x_{n+1}}}{{x_n}}=\lim\limits_{n\to\infty}\frac{\arctan({x_n})}{\arctan({x_{n-1}})}=l,$$
where $l$ is the point where the series converges to. I can't calculate this limit so I can not find $l$.
The next step would be to calculate: $$\lim\limits_{n\to\infty}\frac{{x_{n+1}}-l}{({x_n}-l)^p}$$
Basically this question has to do with the $l$ calculation which will help me define the rate of converge for ${x_n}$. Any hint about the limit would be appreciated. Thanks in advance!
 A: As pointed out in comments, both the limits $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}$ and $\lim_{n\to\infty} \frac{x_{n+1}}{x_n^p}$ are easy to compute given the observation that $x_n \to 0$. So let me demonstrate a method of finding an asymptotic expansion of $(x_n)$ instead.

A hint is that you can estimate the growth of $x_n^{-2}$ more easily. For instance, assuming that $x_1 > 0$,
Step 1. Since $(x_n)$ is strictly decreasing and bounded, $(x_n)$ converges. The limit must be a fixed point of $\arctan$, which is exactly $0$.
Step 2. Notice that $\frac{1}{\arctan^2 x} = \frac{1}{x^2} + \frac{2}{3} + \mathcal{O}(x^2)$. From this,
$$ \frac{1}{x_{n+1}^2} - \frac{1}{x_n^2} = \frac{2}{3} + \mathcal{O}(x_n^2). $$
In view of Stolz–Cesàro theorem, we have
$$ \lim_{n\to\infty} \frac{x_n^{-2}}{n} = \lim_{n\to\infty} \frac{\frac{2}{3} + \mathcal{O}(x_n^2)}{(n+1) - n} = \frac{2}{3} $$
and hence $x_n^{-2} \sim \frac{2}{3}n$.
Step 3. Now using $\frac{1}{\arctan^2 x} = \frac{1}{x^2} + \frac{2}{3} - \frac{1}{15}x^2 + \mathcal{O}(x^4)$ and the previous step,
$$ \lim_{n\to\infty} \frac{x_n^{-2} - \frac{2}{3}n}{\log n} = \lim_{n\to\infty}\frac{-\frac{1}{15}x_{n+1}^2 + \mathcal{O}(x_n^4)}{\log(n+1) - \log n} = -\frac{1}{10} $$
and hence $x_n^{-2} = \frac{2}{3}n - \frac{1}{10}\log n + o(\log n)$.
Step 4. Now write
$$ \frac{1}{x_n^2} - \left( \frac{2}{3}n - \frac{1}{10}\log n \right)
= \frac{1}{x_1^2} - \frac{2}{3} + \sum_{k=1}^{n-1} \left( \frac{1}{x_{k+1}^2} - \frac{1}{x_k^2} - \frac{2}{3} + \frac{1}{10}\log\left(1+\frac{1}{k}\right) \right). $$
Using the previous step, it is easy to check that the right-hand side converges as $n\to\infty$ with the error term decaying at most as fast as $\mathcal{O}(\frac{\log n}{n})$. So there exists a constant $C$ such that
$$ x_n^{-2} = \frac{2}{3}n - \frac{1}{10}\log n + C + \mathcal{O}\left(\frac{\log n}{n}\right). $$

Using this estimation above, we can easily check that
$$ x_n = \sqrt{\frac{3}{2n}} \left( 1 + \frac{3}{40}\frac{\log n}{n}\right) + \mathcal{O}\left(\frac{1}{n^{3/2}}\right). $$
A: It seem to me for recursive schemes it is convinient to use the following. If we are already at the limit, then $a= arctan(a)$. It remains to find a. That is $tg(a)= a$, one solution is $a=0$, there is no other solutions as $tg(a)$ grows faster than $a$ for $a \in (0, \frac{\pi}{2}]$, and decreases faster for $a \in [-\frac{\pi}{2},0)$.
The rate at some point $x$ is $\lim_{n\to \infty} \frac{arctan(x_{n})}{x_{n}}$. Which is the same as $\lim_{x\to 0} \frac{arctan(x)}{x}$, which is the same as $\lim_{x\to0}\frac{x}{tgx}$, which is one (as $\lim_{x\to 0 }\frac{x}{sinx}=1$ and $cos(0)=1$).
