TITLE: How to evaluate $\lim_{ n\to \infty }\frac{a_n}{2^{n-1}}$, if $a_0=0$ and $a_{n+1}=a_n+\sqrt{a_n^2+1}$?
QUESTION [7 upvotes]: Let $a_1,a_2,..,a_n$ be sequence of real numbers such that $a_{n+1}=a_{n}+\sqrt{1+a_n^2}$ and $a_0=0$.
How to evaluate $\lim_{ n\to \infty  }\frac{a_n}{2^{n-1}}$ ?

REPLY [15 votes]: Since:
$$ \cot\frac{x}{2}=\cot x+\sqrt{1+\cot^2 x} $$
and $a_0=\cot\frac{\pi}{2}$ we have, by induction:
$$ a_n = \cot\frac{\pi}{2^{n+1}} $$
and the wanted limit equals $\displaystyle{\color{red}{\frac{4}{\pi}}}$.