Prove Logistic sequence is convergent when $2<k\le3$

Question

Let $0<x_1<1$, $x_{n+1}=kx_n(1-x_n)$ be Logistic sequence. If $2<k\le3$, how to prove the sequence always converges to $1-1/k$?

Edit: I need a direct proof in calculus without using any result from dynamic system.

Answer 1

The fixed points are solutions to $$x=kx(1-x)$$ which are $x=0$ and $x=1-1/k$

We need to show that the fixed point $1-1/k$ is an attractor for $2<k<3$

We take derivative of $f(x)=kx(1-x)$ and show that for the given value of the fixed point $|f'(x)|<1$

Note that $|f'(1-1/k)| = |k-2|$ which is less than $1$ for $2<k<3$