How to show that $a_{n+2}=\frac{a_{n+1} ^2 -1}{a_n}$ . is bounded?

Question

Let $a_{n+2}=\frac{a_{n+1} ^2 -1}{a_n}$ be a sequence of real numbers where $a_n>0$ for all $n \in \mathbb{Z}_{+}$. It is given that $a_1=1 , a_2=b>0$. It is given that $1<b<2$ .Is it possible to show that $a_n$ is bounded from this given information?

Answer 1

It seems that for all integers $n\geq 1$, $a_n$ will be a rational fraction in $b$ of degree $n-1$, with leading coefficient $1$ (by that I mean that for all fixed $n\geq 1$, you can write that $a_n\sim b^{n-1}$ as $b$ goes to infinity.

So my intuition is that if $b>1$, then $a_n$ will behave like a diverging geometric sequence.

Answer 2

Well, bounded or not, it depends on $b$. I won't go in details but will give the outline in the following.

Firstly, you can observe that $ a_{n+2}a_n -a_{n+1}^2= a_{n+1}a_{n-1} -a_n^2$, and this leads to : $$ \dfrac{a_{n+2}+a_n}{a_{n+1}}= \dfrac{a_{n+1}+a_{n-1}}{a_{n}}= \dots = \dfrac{a_3+a_1}{a_2} =b$$

So now you have the following recursive relation: $$ a_{n+2} -ba_{n+1}+a_n=0$$

Solving that, we have some formula of $(a_n)$ as: $$ a_n= a \phi_1^n+b\phi_2^n$$

And rest is simple, normally, the first necessary condition we would get for the boundedness is $|b| \le 2$.