analysis - Convergence/Divergence of Sequence defined by a recurrence relation

Given the following sequence:
$$a_{n+1} = a_n(2 - a_n)$$

for which values $a_1 \in \mathbb{R}$ does this sequence converges or diverges.

By trial and error I found that for $a_1 \in (0, 2)$ it converges to $1$, for $a_1 \in \{ 0 , 2 \}$ it converges to $0$ and for all other values it goes to $-\infty$.

But how can we prove these facts? If a limit exists then I can show that it must be $1$ or $0$ by the following calculation. It holds
$$\begin{align*} \lim_{n \to \infty} a_{n+1} = \lim_{n\to \infty} a_n(2-a_n) \end{align*}$$
So if $\lim_{n\to \infty} a_n = a$, then
$$a = 2a - a^2 \Leftrightarrow a^2 = a \Leftrightarrow a = 1 \lor a = 0.$$
But how to show that a limit exists when $a_1 \in (0,2)$, and there is no limit when $a_1 < 0$ or $a_1 > 2$?

Answer

Now that you have a guess for what the limit should be, try making a substitution to reflect that. In this case, let

$$a_n = 1 + b_n$$

for all $n$. Substituting this in, the original recurrence $a_{n+1} = a_n(2 - a_n)$ becomes

$$1 + b_{n+1} = (1 + b_n)(1 - b_n) = 1 - b_n^2,$$

so that

$$b_{n+1} = -b_n^2.$$

By iterating this recurrence we can solve it explicitly as

$$b_{n+1} = -b_1^{2n}.$$

Thus if $|b_1| > 1$ we see that $b_n \to -\infty$. This is equivalent to the statement

$$a_1 \in (-\infty,0) \cup (2,\infty) \Longrightarrow a_n \to -\infty.$$

If $|b_1| = 1$ then $b_n = -1$ for all $n > 1$. This is equivalent to the statement

$$a_1 \in \{0,2\} \Longrightarrow a_n = 0 \text{ for all } n > 1.$$

Finally if $|b_1| < 1$ then $b_n \to 0$, and this is equivalent to

$$a_1 \in (0,2) \Longrightarrow a_n \to 1.$$