He encontrado lo siguiente en un libro:-
$Q:$ % Que $a_1, a_2, ... , an$ser una secuencia de números verdaderos con $a{n+1}=a_n+\sqrt{1+a_n^2}$ y $a_0=0$. Demostrar que \thetan$$\lim{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$$ I tried many things none of which seemed fruitful. First thing I did was to define $$. Substituting this into the condition, we get $$2^{n}b_n=2^{n-1}b_n+\sqrt{1+2^{2n-2}bn^2}\implies b{n+1}=\frac {b_n}2+\sqrt{\frac1{2^{2n}}+\left(\frac {b_n}2\right)^2}$$ Simplifying this gives $\theta0=0\implies c=-\frac{\pi}2$b{n+1}^2-b_{n+1}bn=\frac1{2^{2n}}\implies b{n+1}(b_{n+1}-bn)=\frac1{2^{2n}}$$ This doesn't lead anywhere. One lead that I got was by substituting $$ $. This gives $$\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}$a_n=2^{n-1}bn$ Hence $$\tan\theta{n+1}=\tan\left(\frac {\thetan}2+\frac{\pi}4\right)\implies \theta{n+1}=\frac {\theta_n}2+\frac{\pi}4\implies \thetan=\frac {\pi}2+c\left(\frac 12\right)^n$$From initial conditions we get $$. Therefore $$\lim{n \to \infty} \frac{an}{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}$a_n=\tan es esta prueba válida y hay algún otro modo de hacerlo?
