I view the orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ as the sequence $(x_{n})$. So
$x_n = \frac{1}{2}, f\bigg(\frac{1}{2}\bigg), f\bigg(f\bigg(\frac{1}{2}\bigg)\bigg), \dots$ $(x_n)$ appears to dance around the fixed point $\frac{2}{3}$ while slowly converging to it.
The odd-numbered iterates are increasing and the even-number iterates are decreasing. Let $(a_{k})$ be the sequence of odd terms of $(x_{n})$ where $k = 2n-1$ and $(b_{k})$ be the sequence of even terms of $(x_{n})$ where $k = 2n$.
Show $(a_k)$ is monotonic & bounded. 1. Increasing:
$a_{k}
$x_{2n+1} - x_{2n-1}>0$ $(-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+9x_{2n-1})-(x_{2n-1}) > 0$ $-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+8x_{2n-1}> 0$ $x_{2n-1}(2-3x_{2n-1})^3 > 0$ $0
Hence, the subsequence $(a_k)$ is strictly increasing and $(a_k)$ is bounded below by the first point of the subsequence $a_1=\frac{1}{2}$.
Question: How do I show $a_{k}$ is bounded above by the fixed point $\frac{2}{3}$?
Maybe, use epsilon delta proof to show convergence, which implies $(x_n)$. Or by induction, assume $x_{2n-1}< \frac{2}{3}$ show $x_{2n+1}=f(f(x_{2n-1}))<\frac{2}{3}$