Let $(u_n)$ be a real sequence defined $u_0>0$ and $u_{n+1} =\dfrac{1}{2} \arctan(u_n)$. We define $v_n = 2^n u_n$ and write $l = \lim v_n$.
Find an equivalent of $u_n - \dfrac{l}{2^n}$ as $n\to \infty$.
What I did:
I've shown that $(u_n)$ converges to 0, $(v_n)$ is convergent. So we get $u_n \sim \dfrac{l}{2^n}$. I don't know how to continue.
