Finding the limit of $x_1=0 , x_{n+1}=\frac{1}{1+x_n}$ I have had big problems finding the limit of the sequence $x_1=0 , x_{n+1}=\frac{1}{1+x_n}$. So far I've only succeeded in proving that for $n\geq2$: $x_n>0\Rightarrow x_{n+1}>0$
(Hopefully that much is correct: It is true for $n=2$, and for $x_{n+1}>0$ exactly when $\frac{1}{1+x_n}>0$, which leads to the inequality $x_n>-1$ which is true by the induction assumption that $x_n>0$.)
On everything else I failed to come up with answers that make sense (such as proving that $x_{n+1}>x_n \forall n\geq1$). I'm new to recursive sequences, so it all seems like a world behind the mirror right now. I'd appreciate any help, thanks!

It is obvious that $f:x\mapsto\frac1{1+x}$ is a monotonically decreasing continuous function $\mathbf R_{\geq0}\to\mathbf R_{\geq0}$, and it is easily computed that $\alpha=\frac{-1+\sqrt5}2\approx0.618$ is its only fixed point (solution of $f(x)=x$). So $f^2:x\mapsto f(f(x))$ is a monotonically increasing function that maps the interval $[0,\alpha)$ into itself. Since $x_3=f^2(x_1)=\frac12>0=x_1$ one now sees by induction that $(x_1,x_3,x_5,...)$ is an increasing sequence bounded by $\alpha$. It then has a limit, which must be a fixed point of $f^2$ (the function mapping each term of the sequence to the next term). One checks that on $ \mathbf R_{\geq0}$ the function $f^2$ has no other fixed point than the one of $f$, which is $\alpha$, so that must be value of the limit. The sequence $(x_2,x_4,x_6,...)$ is obtained by applying $f$ to $(x_1,x_3,x_5,...)$, so by continuity of $f$ it is also convergent, with limit $f(\alpha)=\alpha$. Then $\lim_{n\to\infty}x_n=\alpha$.
