<s>[INST] Prove that $a_n \in [0,2)$ 
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$

Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0)  < 2$.
Inductive step:
Suppose that $0 \leq a_n  < 2$
So $$\begin {split} 0 \leq a_n  < 2 &\iff 0 \geq -a_n > -2 \\ &\iff 6 \geq 6-a_n > 6-2 \\ &\iff \frac{1}{6} \geq \frac{1}{6-a_n} > \frac{1}{4} \\ &\iff \frac{a_n}{6}+1 \geq \frac{6+a_n}{6-a_n} > \frac{3}{2} + \frac{a_n}{4} \end{split}$$ 
To be fair I have no idea if this is going somewhere.
 [/INST] Let's do it by induction:
$0 \le a_n < 2$ 
So $6= 6- 0 \ge 6- a_n > 6-2 = 4$ and $\frac 16 \le \frac 1{6-a_n} < \frac 14$
$6 + a_0 \ge 6 > 0$ so
$\frac {6 + a_n}6 \le \frac {6+a_n}{6-a_n} < \frac{6+a_n}4$.
And $6+a_n \ge 6+0 = 6$ and $6+a_n < 6+2 = 8$ so 
$0 \le 1 = \frac 66 \le \frac {6+a_n}6 \le \frac {6+a_n}{6-a_n} < \frac {6+a_n}4 < \frac 84 = 2$
So $a_{n+1} = \frac {6+a_n}{6+a_n} \in [0, 2)$.  (If fact $a_{n+1} \in [1,2)$).
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