Limit $\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$ for $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$

I found the following question in a book:-

$Q:$Let $a_1, a_2, ... , a_n$ be a sequence of real numbers with $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$. Prove that $$\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$$ I tried many things none of which seemed fruitful. First thing I did was to define $a_n=2^{n-1}b_n$. Substituting this into the condition, we get $$2^{n}b_n=2^{n-1}b_n+\sqrt{1+2^{2n-2}b_n^2}\implies b_{n+1}=\frac {b_n}2+\sqrt{\frac1{2^{2n}}+\left(\frac {b_n}2\right)^2}$$ Simplifying this gives $$b_{n+1}^2-b_{n+1}b_n=\frac1{2^{2n}}\implies b_{n+1}(b_{n+1}-b_n)=\frac1{2^{2n}}$$ This doesn't lead anywhere. One lead that I got was by substituting $a_n=\tan \theta_n$. This gives $$\begin{align}a_{n+1}&=\tan \theta_n+\sec \theta_n\\&=\frac{\sin\theta_n+1}{\cos\theta_n}\\&=\frac{1+\tan\frac{\theta_n}2}{1-\tan\frac{\theta_n}2}\\&=\tan\left(\frac {\theta_n}2+\frac{\pi}4\right)\end{align}$$ Hence $$\tan\theta_{n+1}=\tan\left(\frac {\theta_n}2+\frac{\pi}4\right)\implies \theta_{n+1}=\frac {\theta_n}2+\frac{\pi}4\implies \theta_n=\frac {\pi}2+c\left(\frac 12\right)^n$$From initial conditions we get $\theta_0=0\implies c=-\frac{\pi}2$. Therefore $$\lim_{n \to \infty} \frac{a_n}{2^{n-1}}=\lim_{n\to\infty}\frac{\tan\left(\frac {\pi}2-\frac{\pi}2 \left( \frac 12 \right)^n\right)}{2^{n-1}}=\lim_{n\to\infty}\frac{2\left(\frac12\right)^n}{\tan\left(\frac {\pi}2\left(\frac12\right)^n\right)}=\lim_{n\to\infty}\frac{2\left(\frac12\right)^n}{\frac{\pi}2\left(\frac12\right)^n}=\frac4{\pi}$$ Is this proof valid and are there any other ways of doing it?

Solution 1:

Make use of :

$$ \cot\frac{x}{2}=\cot x+\sqrt{1+\cot^2 x} $$

Since $a_0=\cot\frac{\pi}{2}$, one can notice ( use an inductive argument): $$ a_n = \cot\frac{\pi}{2^{n+1}} $$ So the limit obviously becomes: $$\frac{4}{\pi} $$