I'm stuck with the prove of the following recurrence sequence which was part of and old exam.
$a_1:=\frac{1}{2}, a_{n+1}:=a_n(2-a_n)$ for $n \in \mathbb{N}$
I have to show that $0 <a_n < 1$ and as a second part (b), that its monotonic increasing (That's where I got stuck) and as a thid part (c) the convergence and limit of the sequence
a) My approach was to use induction to show that the sequence is upper and lower bounded:
Base case:
$ n=1, a_1 < 1 = \frac{1}{2} <1 $ ok
Inductive step: $n\rightarrow n+1$
$a_{n+2} = a_{n+1}(2-a_{n+1}) < 1$
Then my idea was to rewrite it as using the given definition from $a_{n+1}$:
$a_{n+2}=a_n(2-a_n)(2-(a_n(2-a_n))$ and I know that $a_n<1$
Then I used the base case that $a_n=\frac{1}{2}<1$
so that I've got $ 0 < \frac{3}{4}*(2-\frac{3}{4}) = 0.9375 <1$
Question: Is this right?
b)
For the second part of the task, to show montonic increasing I know that
$a_{n+1} \leq a_n$ or $a_{n+1}-a_n\leq 0$
$a_{n+1}=a_n(2-a_n) \geq 0 \Leftrightarrow a_n-a_n^2 \geq 0 \Rightarrow a_n^2-a_n \geq 0$
Here I got stuck:
Question: How do I make the final conclusion?
c)
Limes:
$a_n \rightarrow a$
$a=a(2-a) \Leftrightarrow a=2a-a^2 \Rightarrow a^2-a=0$
Solving gives $a_1=1$ and $a_2=0$
Because the sequence is monotonic increasing it converges and it must be $>0$ so the $\lim(a_n) =1$
Question: Is that conclusion "exact", I mean can I use this conlcusion or is there something missing?
Best Regards,
Christoph
Best Answer
Hint: If $0\lt a_n\lt1$, then $1\lt2-a_n\lt2$, therefore, $$ \begin{align} a_{n+1} &=a_n(2-a_n)\\ &\gt a_n \end{align} $$ Furthermore, $$ \begin{align} a_{n+1} &=a_n(2-a_n)\\ &=1-(a_n-1)^2\\ &\lt 1 \end{align} $$ After showing the limit exists, take limits: $$ \begin{align} \lim_{n\to\infty}a_n &=\lim_{n\to\infty}a_{n+1}\\ &=\lim_{n\to\infty}a_n(2-a_n)\\ &=\lim_{n\to\infty}a_n(2-\lim_{n\to\infty}a_n) \end{align} $$