Asymptotic expansion of $u_{n + 1} = \frac12 \arctan(u_n)$ (I'm aware of Asymptotic expansion of $v_n = 2^nu_n$ where $u_{n+1} = \dfrac{1}{2}\arctan(u_n)$ but it has no answers…)
Let be $u_0 \in \mathbb{R}$ and the sequence $(u_n)_n$ defined by: $u_{n + 1} = \frac12 \arctan(u_n)$.
I define also: $v_n = 2^n u_n$, so I can show that: $\lim (u_n)_n = 0$ (by studying $x \mapsto \frac12 \arctan(x)$), thus, I can show that $(v_n)_n$ is monotone and converges because it is bound.
Now, I conclude: $u_n \sim \dfrac{l}{2^n}$, I'd like to determine $l$ more precisely.
Here is what I tried, I suspect $l$ to be something like $f(\pi)$ for some $f$ :


*

*push the asymptotic expansion of $\arctan$ to the 2nd order and reinject it ;

*use $\arctan(u_n) + \arctan(1/u_n) = \dfrac{\pi}{2}$ ;

*use series techniques to look for $\sum v_{n + 1} - v_n$, maybe conclude using Cesaro summation


For convenience make a slight generalization of the problem. Let $\,u_0\,$ and $\,y\,$ be two given numbers and suppose
$\,u_{n+1} = y \arctan(u_n)\,$ for all $\,n\ge 0\,$ where $\ y=1/2\ $
in the original recursion. Define with power series the function
$$ F(x,y,z)\! :=\! z\!\left(\!x \!-\! \frac{1\!-\!z^2}{1\!-\!y^2}
\frac{x^3}3 \!+\!\frac{(1\!-\!z^2)((3\!-\!2z^2)\!+\!y^2(2\!-\!3z^2)}
{(1-y^2)(1-y^4)}\frac{x^5}{15}
   \!+\! O(x^7)\! \right) \tag{1}$$
which is required to satisfy the equation
$\,F(x,y,yz) = \arctan(F(x,y,z))\,y.\,$
Use iteration to get the equation
$\, u_n = F(x,y,y^n)\,$ where $\, x = \lim_{n\to\infty} u_n/y^n.\,$
More terms in the power series expansion can be easily found.
Equation $(1)$ implies
$\, u_n \approx y^n(x - (1-y^{2n})x^3/(3(1-y^2))).\,$
