How to evaluate $\lim_{ n\to \infty }\frac{a_n}{2^{n-1}}$, if $a_0=0$ and $a_{n+1}=a_n+\sqrt{a_n^2+1}$? [duplicate]

Solution 1:

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 $\displaystyle{\color{red}{\frac{4}{\pi}}}$.

Solution 2:

Write $$ a_{n+1} = 2a_n + [\sqrt{1+a_n^2} - a_n], $$ and note that $\sqrt{1+a_n^2} - a_n \approx \frac{1}{2a_n}$. So roughly speaking, $a_{n+1} \approx 2a_n$, so that $a_n/2^{n-1}$ does approach a limit; this needs to be argued more formally, but I leave you the details. There is no particular reason that the limit have any nice expression; calculation shows that it is about 1.27323954473516.