<s>[INST] Showing convergence of recursive sequence $A_{n+1}=\frac 1 {1+A_n}$ 
Given : $\forall n\in\Bbb N,\quad A_{n+1} = \frac 1 {1+A_n}$ and $A_1 = 0$
Show the sequence converges and find its limit.

Briefly what I did was to create two sub-sequences with an index difference of 2. Doing that I got two recursive formulas that depend on the initial A1, be it 0 for the odds or 1 for the evens.
After showing the sequences are bounded, I got stuck showing the sequence is monotonic.
Thanks,
 [/INST] The recurrence is of the form
$$x_{n+1} = \phi(x_n)$$
where $\phi(x) = \frac{1}{1+x}$. The map $\phi$ has two fixed points $x_{1,2}$,  roots of the equation $x^2 + x -1=0$, that is $\frac{-1\pm \sqrt{5}}{2}$. Now, we can write $\phi(x)$ in the form
$$\frac{\phi(x) -x_1}{\phi(x) - x_2} = k \cdot \frac{x-x_1}{x-x_2}$$
$k$ can be determined by givind a particular value for $x$, say $x=0$. We get
$$\frac{1-x_1}{1-x_2} = k \frac{x_1}{x_2}$$
and so
$$k = \frac{x_2(1-x_1)}{x_1(1-x_2)} = \frac{x_2 + 1}{x_1 + 1}=\frac{1-\sqrt{5}}{1+\sqrt{5}}$$
We get
$$\frac{\phi(x) -x_1}{\phi(x) - x_2} = \frac{1-\sqrt{5}}{1+\sqrt{5}} \cdot \frac{x-x_1}{x-x_2}$$
In other words, $\phi$ is conjugate to the map $t \mapsto \frac{1-\sqrt{5}}{1+ \sqrt{5}}\cdot t$.  The monotony and convergence should be easy now.
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