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.
 A: First, let's rewrite the recurrence formula:
$$a_{n+1}=\frac{6+a_n}{6-a_n}=\frac{a_n-6+12}{6-a_n}=-1+\frac{12}{6-a_n}$$
Now, we have:
$$0\leq a_n<2$$
$$6\geq 6-a_n > 4$$
$$2\leq \frac{12}{6-a_n} < 3$$
$$1\leq -1+\frac{12}{6-a_n}=a_{n+1} < 2$$
Thus, $a_{n+1} \in [1, 2)$ and since $[1, 2) \subset [0, 2)$, $a_{n+1} \in [0, 2)$.
A: 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.
A: Let:
$$f(x)=\frac{6+x}{6-x}$$
We have that $f$ is increasing, because;
$$f'(x)=\frac{12}{(6-x)^2}>0$$
where $x\neq 6$
Hence we have that $a_{k+1}>a_k$.
It's clear that $a_1=1$, so now we let $0\leq \epsilon \leq 1$, where $a_k=(2-\epsilon)$ for an arbitrary $k>1$, and we evaluate:
$$\frac{6+(2-\epsilon)}{6-(2-\epsilon)}=\frac{8-\epsilon}{4+\epsilon}=2-\frac{3\epsilon}{4+\epsilon}<2$$
This proves that $a_k<2$ for any $k$ and so $a_k \in [0,2)$ as required.
A: 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 $a_{n+1} \in [1,2)$).
