Proof Help: Showing Divergence and Convergence For the question below:
Let $x_1 = \alpha$ and $x_{n+1} = 3x^2_n$
for $n ≥ 1$.
(b) Does $\lim x_n$ exist? Explain.
I proved in the first part that if the limit of $x_n =x$ then $x=0$ or $x=1/3$
In this part, I'm required to show if limit exists or not. I know that for $\alpha=1/3$, limit is 1/3. For $0<\alpha<1/3$, the sequence is decreasing and converges to $0$. And that for $\alpha>1/3$, the sequence diverges
Can anyone please help me formally prove the divergence for $\alpha>1/3$. Any suggestions/hints?
Thank you.
 A: If $x_1 = \alpha, x_{n+1} = 3x_n^2$ then first you form the geometric picture in your mind, which you did correctly. Now, to see what exactly is happening, you must first inspect $x_{n+1} = 3x_n^2$ as a growth pattern, and then ram home the argument formally.
This lemma is our key tool.

If $0 < \alpha < \frac 13$, then $x_{n+1} < x_n$ for all $n$.

Proof : By induction. $x_1 = \alpha$, and $x_2= 3\alpha^2$, so $x_1 - x_2 = \alpha(1- 3\alpha) > 0$ since both the terms are greater than $0$.
For general $n$, start with $x_{n+1} = 3x_n^2$. By induction, $x_n < x_{n-1} ... < x_1 < \frac 13$. So, $x_n - x_{n+1} $ $= x_n(1-3x_n) > 0$ since $x_n > 0$ (since it's three times the square of something positive) and the other term is positive as $x_n < \frac 13$. Therefore, $x_n > x_{n+1}$ , completing the proof.

A bounded decreasing sequence converges.

This is easy to see. The candidate for the limit is the infimum of the sequence (as a set). Use $\epsilon - \delta$ and the definition of infimum.
Now, you showed that if there is a limit, it is $0$ or $\frac 13$. But the limit cannot be $\frac 13$ here : we saw that $x_n < \alpha$ for all $n$, so $|x_n - \frac 13| > |\alpha - \frac 13|$ for all $n$.
So, the sequence must go to $0$.

If $\alpha > \frac 13$, then a similar argument shows that $x_{n+1} > x_n$ for all $n$. However, here something different happens : if $x_n$ did converge, then it would go to either zero or $\frac 13$, but here neither is possible. Therefore, by my second lemma's contrapositive, $x_n$ must be divergent.

If $\alpha = \frac 13$ then the sequence is constant.

I leave you to see the nature of the sequence for $\alpha \leq 0$. In particular, you will notice that sign doesn't matter, since the square appears in $x_2$ itself, so this is only a question of the absolute value of $\alpha$. 
