Show that the sequence defined as $x_{n+1}=\frac{1}{1+x_n}$ converge Show that the sequence $(x_n)$ defined as $x_{n+1}=\frac{1}{1+x_n},x_0=0$ converge. My attempt is to show that $(x_n)$ is Cauchy so i would like to have a feedback on my proof,please.
First of all we show by induction that $2>x_n\ge0$. The previous inequality is true for $n=0$ so we can suppose that it works up to a certain $n$. We want to proof right now that it holds fot $n+1$ and so show that $2>x_{n+1}\ge0$:
$2>x_{n+1}\ge0 \iff 2>\frac{1}{1+x_n}\ge0$.
It is trivial that $x_{n+1}\ge0$ by induction hypothesis (because $x_n\ge0$). Moreover, as $2>x_n\ge0$, we have that
$x_{n+1}=\frac{1}{1+x_n}\le1<2$. So we can conclude that $2>x_n\ge0 \ \forall n\ge0$
Then, $\forall p\ge2$ we have:
$|x_p-x_{p-1}|=\Big|\frac{x_{p-2}-x_{p-1}}{(1+x_{p-1})(1+x_{p-2})}\Big|\le 1 \cdot|x_{p-2}-x_{p-1}|=...=1^{p-1}|x_0-x_1|=1^{p-1}$
We show now that $(x_n)$ is Cauchy. So, by définition of Cauchy sequence, we have to show that $\forall \epsilon>0 \ \exists N \ \forall m,n>N$: $|x_m-x_n|<\epsilon$. But,
$|x_m-x_n|\le|x_m-x_{m-1}|+|x_{m-1}+x_{m-2}|+...+|x_{n+1}-x_n|\le(m-n-1)$
So $(x_n)$ is Cauchy
I am not reall sure if this proof holds, because i think that in case where $m=n-1$ it doesn't work,or...?
 A: Let $x_{1}>0$ and $\displaystyle x_{n+1}=\frac{1}{1+x_{n}}$. Notice that $x_{n}>0$ (by induction) and that :
$$
|x_{n+2}-x_{n+1}|=\left|\frac{1}{1+x_{n+1}}-\frac{1}{1+x_{n}}\right|=\frac{|x_{n}-x_{n+1}|}{(1+x_{n+1})(1+x_{n})}\leq\frac{4}{9}|x_{n}-x_{n+1}|
$$
Hence, $x_{n}$ is a contraction and hence it is Cauchy, and so converges by completeness of $(\mathbb{R},|.|)$. Let $L$ be its limit. We have that $\displaystyle L=\frac{1}{1+L}$ that is :
\begin{align*}
0&=L^{2}+L-1\\&=(L+0.5)^{2}-0.25-1\\&=\left(L-\frac{-1+\sqrt{5}}{2}\right)\left(L-\frac{-1-\sqrt{5}}{2}\right)
\end{align*}
We obtain that either $\displaystyle L=\frac{-1+\sqrt{5}}{2}$, or $\displaystyle L=\frac{-1-\sqrt{5}}{2}$. However, since $L\geq0$ and since $x_{n}>0$, we conclude that :
$$
\lim_{n\to\infty} x_{n}=\frac{-1+\sqrt{5}}{2}
$$
where $\displaystyle\varphi:=\frac{-1+\sqrt{5}}{2}$ is the golden ratio.

As $\text{Theo Bendit}$ mentioned, What's wrong with your proof is that it  fails to make $\left|x_{m}-x_{n}\right|$ small.
