Sequence $x_n$ such that $x_0 \in [0, 2]$ is defined by $x_{n+1} = \frac{1}{4}x_n(5-x_n)$. For what values of $x_0 \in [0, 2]$ does $x_n$ converges and to what limit?
For $x_0 = 0$, $\lim_{n\to\infty}x_n = 0$. I guess that in other cases $\lim_{n\to\infty}x_n = 1$. The sequence is bounded by $\max(2, \frac{25}{16})$ from AM-GM. How can I prove my assertion that $x_n$ converges to $1$?
I've tried to prove that for $x_0 \in (0, 1)$ sequence increases and for $x_0 \in (1, 2)$ decreases, but when proving first part I got stuck, because I don't know how to prove inductive step that $x_n < 1$ implies $\frac{x_n(5-x_n)}{4} < 1$
Answer
The function $f(x)=\frac {x(5-x)} 4$ maps $[0,2]$ into itself and it is increasing there. From this it follows by induction that $x_n \in [0,1]$ for all $n$ and that $x_{n+1} \geq x_n$. Hence $a\equiv \lim x_n$ exists and since $a=\frac {a(5-a)} 4$ we get $a=0$ or $a=1$. Since $(a_n)$ is increasing it follows that $a=0$ if $x_0=0$ and $a=1$ otherwise.
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