Let $(x_n)_{n\geq0}$ be a sequence such that $x_0>0$ and $$x_{n+1}=\frac{x_n}{1+\sqrt{1+x_n^2}}$$ Find the limit of $(x_n)_{n\geq0}$ and $$\lim_{n\to+\infty} (1+x_0^2)(1+x_1^2)...(1+x_n^2)$$
One can see that the sequence is decreasing and positive, so it has la limit and when we replace $x_{n+1}$ and $x_n$ with that limit in the given relation we obtain that limit of $(x_n)_{n\geq0}$ is equal to $0$. I can't find, however, the second limit. My guess is that it is $+\infty$.
What I tried: I considered $(y_n)_{n\geq0}$ such that $y_n=1+x_n^2$. From the given relation, if we write it as $$x_{n+1}^2=(\frac{x_n}{1+\sqrt{1+x_n^2}})^2$$ we obtain that $$y_{n+1}=\frac{2\sqrt{y_n}}{1+\sqrt{y_n}}$$which is the harmonic mean of $1$ and $\sqrt{y_n}$, but I couldn't do anything from this point. Any suggestion?
$x_{n+1}-x_n=\frac{x_n}{1+\sqrt{1+x_n^2}}-x_n=-\frac{x_n\sqrt{1+x_n^2}}{1+\sqrt{1+x_n^2}}$. Hence it is decreasing and bounded by $0$ from below. It is convergent and the limit satisfies the equation $l=\frac{l}{1+\sqrt{1+l^2}}$ whose unique solution is $l=0$.