<s>[INST] 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! 
 [/INST] 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
$$ \prod_{n\geq 0}(1+a_n^2) = \prod_{k\geq 0}\sec^2\frac{\varphi}{2^n} = \frac{\varphi^2}{\sin^2\varphi\cos^2\varphi}=\left(a_0+\frac{1}{a_0}\right)^2\arctan^2 a_0.$$
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