# Prove, that sequence $a_{n+1}=a_n^2-1$ is limited

I have problems with:
Let $a_1\in\mathbb{R}$ and $a_{n+1}=a_n^2-1$ for $\forall n\in\mathbb{N}$.

Prove, that if $|a_1|\leqslant \dfrac{1+\sqrt{5}}{2}$ then $a_n$ is limited, and otherwise,
if $|a_1|>\dfrac{1+\sqrt{5}}{2}$ then $a_n$ diverges to $+\infty$.

So I assumed that $a_{n+1}-a_n>0$. Then for sure $a_{n}\geqslant a_n^2-1$.

If we want $a_n$ to be increasing then it must lie between $\dfrac{1-\sqrt{5}}{2}$ and $\dfrac{1+\sqrt{5}}{2}$.

This is a place where problems appears. I am able to calculate it (to that point of course), but I don't fully understand what am I doing. Can somebody help me? Thanks in advance!

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$a_{n+1} - a_n > 0$ is equivalent to $a_n < a_n^2 - 1$. –  WimC Nov 14 '12 at 20:39

The recurrence is $$a_{n+1}=a_n^2-1$$ but, because $a_{n+1}$ does not care about the sign of $a_n$, we can instead analyse the boundedness/divergence of the non-negative sequence $|a_{n+1}|=|a_n|^2-1$.
Let us consider the $z$ for which $z^2-1 \le z$, rewriting this and solving the quadratic we get $(z - \varphi)(z - \bar \varphi) \le 0$ which can only happen when $\bar \varphi \le z \le \varphi$. This implies about our sequence that if $|a_n| \le \varphi$, $|a_{n+1}| \le \varphi$ (note, we can't say that $|a_{n+1}| \le |a_n|$).
We have found an interval $[0,\varphi]$ on which the recurrence is bounded. Let us consider $a_n > \varphi$, since $\varphi$ satisfies $z^2 - 1 = z$ let $a_1 = \varphi + r$ (some positive $r$) and observe $a_{n+1} = \varphi + (2 \varphi r + r^2)$ which is clearly bigger than $a_n$ (since both $r$ and $\varphi$ are positive). Therefore if $a_1 \in (\varphi,\infty)$ the sequence diverges.
Hint: Have a look at the function $f(x)=x^2-x-1$ and try to find its zero using the iteration method which is basicaly is your problem $x_{n+1}=x^2_n - 1.$