$x_1 = a^2$ 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.

  • $\begingroup$ 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) $\endgroup$ – Jakobian Oct 14 '18 at 17:02


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy