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 $a_1 > 2$?
 A: 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
$$
a_1 \in (0,2) \Longrightarrow a_n \to 1.
$$
A: You can show that if you ever get a negative number, the sequence will be monotonically decreasing from that point onwards. If it had a limit, that would therefore be negative, and you have proved that the only possible limits are $0$ and $1$. Therefore it cannot have a limit.
You can show that for $a_n\in(0,1)$ the next element in the sequence is still in $(0,1)$ but larger. Since the sequence is bounded and monotonic it must converge to something, and you have proved that $1$ is the only possible limit that's larger than $a_n$.
$a_0=0$ obviously converges to itself, and for symmetry reasons a sequence starting at $a_0>1$ joins with that starting at $2-a_0$ after one step, and therefore has the same limit.
Together, these cases cover everything.
