What is the best way to show the monotonicity of this recursive sequence?

Question

This is the sequence:

$a_{n+1}$ = $\sqrt{3a_{n} - 2}$
$a_{0}$ = 3

Before evaluating monotonicity I know that the sequence is converting towards either 1 or 2.

My approach is to assume it is monotonously decreasing and then:

1. (observed) $a_{0} = 3 \geq \sqrt{7} = a_{1}$
2. (assumed) $a_{n} \geq a_{n+1}$
3. (hyptothetised) $a_{n+1} \geq a_{n+2}$

from 2. i can say

$\Rightarrow3a_{n} \geq 3a_{n+1}$

$\Rightarrow3a_{n} - 2 \geq 3a_{n+1} - 2$

$\Rightarrow\sqrt{3a_{n} - 2} \geq \sqrt{3a_{n+1} - 2}$

(i think i can take the square root if i know that it converges to either 1 or 2, correct?)

$\Rightarrow a_{n+1} \geq a_{n+2}$

Can i do it like this? If yes, is it conclusive/ is there a better way or a good alternative? If no, what is the right approach?

Thanks in advance

EDIT: Since most of the answers address the boundedness of the sequence, i should add this (sorry if it comes late): I made the assumption that for $lim(n\rightarrow\infty)$ that $a_{n} = a_{n+1}$ and arrive at

$(a_{n=\infty} - 2)(a_{n=\infty} - 1) = 0$

So i know the limit is either 1 or 2, right? Since in both cases $3a_{n} -2 > 0$, would this be enough to make my proof of monotonicity (above) valid? Or is this argument again circular?

Answer 1

If $a_0 =3$, then by induction $a_n > 2$ for all $n$:

$$a_{n-1} > 2 \implies 3a_{n-1}-2 > 4 \implies a_{n} =\sqrt{3a_{n-1}-2} > 2$$

Hence, your proof of monotonicity is valid (you can take the square root).

Now you have a decreasing sequence bounded below, which must converge.

Answer 2

You get a typo there...It is $a_{0} = 3 \geq \sqrt{7} = a_{1}$. ;)

You can not take square root directly, since you haven't proven it converges. Yes, you know the final answer... but a rigor proof can't use anything you didn't prove, or it will be circular argument.

So you may prove that $a_n>0$ first. But here is a tricky thing: $a_n>0$ is not enough to say $a_{n+1}>0$. So as you observed, you may prove $a_n \geq 1$, by induction.

Then you can use the square root idea.