Sequence and two limits to calculate Let $(a_n)_{n \ge 0}$ with $a_0>0$ and $a_{n+1}=\frac{a_n}{1+\sqrt{1+a_n^2}}$, for every $n \ge 0$. Find $\lim_{n\to \infty} a_n$ and $\lim_{n\to \infty} (1+a_0^2)(1+a_1^2)...(1+a_n^2)$.
I proved $(a_n)_n$ is convergent and I found $\lim_{n\to \infty} a_n=0$ and for the second limit I write $1+a_n^2=(\frac{a_n-a_{n+1}}{a_{n+1}})^2$ and I obtained $(1+a_0^2)(1+a_1^2)...(1+a_n^2)=(\frac{(a_0-a_1)(a_1-a_2)...(a_n-a_{n+1})}{a_1a_2...a_{n+1}})^2$, but I don't know how to continue. Please help me! 
 A: If $a_n=\tan\theta$ for some $\theta\in\left(0,\frac{\pi}{2}\right)$, then $a_{n+1}=\tan\frac{\theta}{2}$ (familiarity with the bisection formulas).
By induction $a_n = \tan\frac{\arctan a_0}{2^n}=\tan\frac{\varphi}{2^n}$ and $1+a_n^2=\sec^2\frac{\varphi}{2^n}$.
By the duplication formula for the sine function
$$ \sin(\theta)=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=4\sin\frac{\theta}{4}\cos\frac{\theta}{4}\cos\frac{\theta}{2}=\ldots =2^n\sin\frac{\theta}{2^n}\prod_{k=1}^{n}\cos\frac{\theta}{2^k} $$
hence
$$ \prod_{k=1}^{n}\sec\frac{\theta}{2^k} = \frac{2^n\sin\frac{\theta}{2^n}}{\sin\theta}\to\frac{\theta}{\sin\theta}\text{ as }n\to+\infty $$
and
$$ \prod_{n\geq 0}(1+a_n^2) = \prod_{k\geq 0}\sec^2\frac{\varphi}{2^n} = \frac{\varphi^2}{\sin^2\varphi\cos^2\varphi}=\left(a_0+\frac{1}{a_0}\right)^2\arctan^2 a_0.$$
A: The crucial remark here, using only the duplication formula of the sine, is that, if $a=\tan\theta$ with $\theta$ in $(0,\pi/2)$ then $$\frac{a}{1+\sqrt{1+a^2}}=\tan\left(\frac\theta 2\right)$$ Thus, picking $\theta_0$ in $(0,\pi/2)$ such that $$a_0=\tan\theta_0$$ one gets, for every $n$, 

$$a_n=\tan(\theta_0/2^n)$$ 

Now, $$1+a_n^2=\frac1{\cos^2(\theta_0/2^n)}$$ hence $$\frac1{\sin^2(\theta_0/2^n)}\prod_{k=0}^n(1+a_k^2)=\frac1{\sin^2(\theta_0/2^n)}\prod_{k=0}^n\frac1{\cos^2(\theta_0/2^k)}=\frac4{\sin^2(\theta_0/2^{n-1})}\prod_{k=0}^{n-1}\frac1{\cos^2(\theta_0/2^k)}$$ which implies that, for every $n$,

$$\prod_{k=0}^n(1+a_k^2)=\frac{4^n\sin^2(\theta_0/2^n)}{\sin^2\theta_0\cos^2\theta_0}$$ 

Taking the limit of the numerator when $n\to\infty$ yields $$\prod_{k=0}^\infty(1+a_k^2)=\frac{\theta_0^2}{\sin^2\theta_0\cos^2\theta_0}$$ or, equivalently, 

$$\prod_{k=0}^\infty(1+a_k^2)=\frac{(1+a_0^2)^2}{a_0^2}\arctan^2a_0$$

