Given the following recursive sequence $x_{n+1}=\frac{x_n}{2}(1-x_n)$ , $n\geq1$ , $x_1=a$ and $a\in(0,\frac{1}{2})$, prove by induction that it is monotonic, bounded, convergent and find its limit.
I've tried already to solve this problem in my introduction to calculus class and i'm don't getting it. It would be very helpful for me if someone could give me an insight on it. Thanks.
Best Answer
$x_{n+1}= \frac{x_n(1- x_n)}{2}= x_n\left(\frac{1- x_n}{2}\right)$
In particular, $x_1= x_0\left(\frac{1- x_0}{2}\right)$. Since $0< x_0< \frac{1}{2}$, $-\frac{1}{2}< x_0< 0$, $1- \frac{1}{2}= \frac{1}{2}< 1- x_0< 1$, $\frac{1}{4}< \frac{1- x_0}{2}< \frac{1}{2}$ so $0< \frac{x_0}{4}< x_{n+1}< \frac{x_0}{2}< x_0< \frac{1}{2}$.
That is the first step in showing that the sequence is decreasing. For the induction step, assume that, for some k, $0< x_{k+1}< x_k$. Now look at $x_{k+2}= x_k\left(\frac{1- x_k}{2}\right)$. Since $0< x_{k+1}< x_k< \frac{1}{2}$, as above, $\frac{1- x_k}{2}< \frac{1}{2}$