# Prove that $x_n$ has infinity limit

$$x_1 = a^2$$ and $$x_n = (x_{n-1} - a)^2$$ where a>2. I proved that it's increasing sequence, but I don't know how to prove that it's not bounded above.

• Have you tried to suppose that it's bounded, and taking a limit. Odds are, the delta will come negative (I suspect it will, but I didn't check) – Jakobian Oct 14 '18 at 17:02

Hint:

Assume on the contrary that the sequence $$x_n$$ is bounded above. You have already proved that this is a increasing sequence. This imply that sequence is convergent and has limit. Let us call the limit $$l$$

Then taking limits on both sides of the given expression we get

We have $$l=(l-a)^2$$

Can you arrive at some contradiction from here?

Write $$x_k=ay_k$$, where $$y_k$$ is defined by $$y_1=a$$, and $$y_{n+1}=a(y_n-1)^2,\ \forall n=1,2,\cdots$$. Then $$y_{n+1}-1=a(y_n-1)^2-1.$$ Suppose $$y_n-1=1+\varepsilon$$. Then $$y_{n+1}-1>1+4\varepsilon+2\varepsilon^2>1+4\varepsilon$$.

This shows that $$\displaystyle\lim_{n\rightarrow\infty}y_n=\infty$$, and therefore $$x_n$$ is unbounded.

Hope this helps.