<s>[INST] Prove that the sequence $x_{n+1} = {\frac{x_n+1}{x_n+2}}, x_0 = 1$ converges to $\frac{1}{\varphi}$ I've established that $(x_n)$ is monotonically decreasing, but don't know if I should attempt to show that $\frac{1}{\varphi}$ is the infimum of $(x_n)$ or use another method.
 [/INST] Consider $a$ is the positive root of the equation $x^2 + x - 1 = 0$
We have $a^2 + a - 1 = 0, \ \dfrac{1-2a}{1-a} = - a$.
firstly, we proof $x_n > a, \forall n \ge 1$ by induct:
for n = 1, $x_1 = 1 > a$ is right.
If $x_n > a,$ then $ x_{n+1} -a  = \dfrac{x_n + 1}{x_n + 2} - a = \dfrac{(1-a)x_n + 1 - 2a}{x_n + 2}$
$x_{n+1} - a = (1-a) \dfrac{x_n + \dfrac{1-2a}{1-a}}{x_n + 2} = (1-a) \dfrac{x_n - a}{x_n + 2} > 0$
Implying $x_{n+1} > a$
Now we proof sequence ${x_n}$ decreasing:
Consider $x_{n+1} - x_n = \dfrac{x_n + 1}{x_n + 2} - x_n = \dfrac{1 - x_n - x_n^2}{x_n + 2}$
$x_{n+1} - x_n = \dfrac{a + a^2 - x_n - x_n^2}{x_n + 2} = \dfrac{(a - x_n)(1 + a + x_n)}{x_n + 2} < 0 $
Sequance ${x_n}$ decreasing, bounded below so that it tends to $a$ which is the positive root of the equation $ a = \dfrac{a + 1}{a+2}$. 
$a = \dfrac{-1 + \sqrt{5}}{2} = \dfrac{1}{\varphi}$
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