A recurrent sequence and its fixed point

1. $x=\frac{x+2}{x+1}\iff x^2+x=x+2\iff x^2=2$. So $\alpha=\sqrt2$.

2. $f(x)=1+\frac1{x+1}>1$ for $x\ge0$, so $u_n\ge1$ for $n\ge1$ and $u_0=1$. On $[1,+\infty)$, $|f'(x)|=\frac1{(x+1)^2}\le\frac14$. By the mean value theorem,
$$|u_{n+1}-\alpha|=|f(u_n)-f(\alpha)|\le\frac14|u_n-\alpha|,$$
so $|u_n-\alpha|\le4^{-n}|u_0-\alpha|\to0$.

3. $f'(\sqrt2)=-\frac1{(\sqrt2+1)^2}=-(\sqrt2-1)^2\approx-0.1716$. Since $0<|f'(\alpha)|<1$, convergence is geometric with ratio $\approx0.17$. The errors alternate in sign: $u_1=1.5$, $u_2=1.4$, $u_3=1.41667$, $u_4=1.41379$, ...

Remark. This is the continued fraction $\sqrt2=1+\cfrac1{2+\cfrac1{2+\cdots}}$.