Prove that $a_n \in [0,2)$

Question

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.

Answer 1

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 $[1, 2) \subset [0, 2)$, $a_{n+1} \in [0, 2)$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $a_{n+1} \in [1,2)$).