Finding limit of a recursive sequence Let $a_1 = \alpha$. Define a sequence $\{a_n\}$ recursively by 
$$a_{n+1} = 3 a_n^2, \quad \forall n \in  \Bbb{N}$$
(i) Show that if $\lim_{n\to\infty}a_n = a$, then $a = \frac{1}{3}$ or $a =0$
(ii) Does limit exist? 
(iii) Discuss the contradiction in the above two parts.
My attempt:
(i) $a= 3a^2$
$3a^2 - a =0$
$a(3a - 1) =0$
$a=0$ or $(3a-1) = 0 \Rightarrow a = 1/3$.
(ii) How do I show this? I know first I must show whether the function is increasing or decreasing. But I am having trouble doing this in terms of $\alpha$.
 A: (i) Correct
(ii) Consider the cases:
Case 1: $\alpha >\frac{1}{3}$. Then $a_1 > \frac{1}{3}$ and if $a_n > \frac{1}{3}$, then by multiplying by $3a_n$, we get that $a_{n+1} = 3a_n^2>a_n$. By induction, the sequence is strictly increasing. if $a_n$ converges, then it must converge to a value greater than $a_1$. This is a contradiction.
Case 2: $0 \leq \alpha \leq \frac{1}{3}$. Then the sequence is bounded below by $0$ and by a similar induction argument, we get that the sequence is decreasing. So it is convergent.
Case 3: $\alpha < -\frac{1}{3}$. Notice that $a_2 = 3\alpha^2 = 3(-\alpha)^2$. So except for the first term, this sequence is the same as the one given by $a_1=-\alpha$. So the convergence is the same as in the first case. 
Case 4: $-\frac{1}{3} \leq \alpha \leq 0$. By a similar argument, the convergence is the same as in the second case.
(iii) The contradiction is as discussed in the first case of (ii).
A: 
Fix $\alpha\in\mathbb{R}$. Let $a_n$ be a sequence defined by
  $$
a_1 = \alpha, \quad a_{n+1} = 3a_n^2 \quad\text{for $n\geq1$}
$$

Note that
$$
\{a_n\} = \{ \alpha, 3\alpha^2=3^{2^1-1}\alpha^{2^1}, 3(3^{2^1-1}\alpha^{2^1})^2=3^{2^2-1}\alpha^{2^2}, 3(3^{2^2-1}\alpha^{2^2})^2=3^{2^3-1}\alpha^{2^3}, \dotsc \}
$$
Thus $a_{n+1}=3^{2^n-1}\alpha^{2^n}$ for $n\geq0$.

(i) Show that if $\lim_{n\to\infty}a_n=a$, then $a=\frac{1}{3}$ or $a=0$.

Your attempt is right.

(ii) Does limit exist?

Since $a_{n+1} = 3^{2^n-1}\alpha^{2^n} = \tfrac{1}{3}(3^{2^n}\alpha^{2^n}) = \tfrac{1}{3}(3\alpha)^{2^n}$, the sequence $\{a_n\}$ converges only if $|3\alpha|\leq1$. That is,
$$
\lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \tfrac{1}{3}(3\alpha)^{2^n}
= \begin{cases} \tfrac{1}{3} & \text{if $|3\alpha|=1$} \\ 0 & \text{if $|3\alpha|<1$} \end{cases}
$$

(iii) Discuss the contradiction in the above two parts.

It is trivial that $\{a_n\}$ diverges when $|3\alpha|>1$.
