<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] The function $f(x) = \frac{6+x}{6-x}$ is strictly increasing, which you can check by taking the derivative:
$$f'(x) = \frac{(6-x) + (6+x) }{(6-x)^2} = \frac{12}{(6-x)^2} > 0$$
Hence if you assume $0 \le a_n < 2$, you get
$$a_{n+1} = \frac{6+a_n}{6-a_n} \ge \frac{6+0}{6-0} = 1 \ge 0$$
$$a_{n+1} = \frac{6+a_n}{6-a_n} < \frac{6+2}{6-2} = 2$$
Therefore $0 \le a_{n+1} < 2$ as well.
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