The problem
Given $h_0=0.6$ I want to prove that the following iterative sequence converges. I also want to find the value it converges to.
$$h_{t+1}= \sigma(3h_t-1)=\frac{1}{1+e^{-(3h_t-1)}}$$
My attempt
We can show that $h_{t+1}$ is increasing with respect to the previous value:
$$\frac{dh_{t+1}}{dh_t}=\frac{3e^{-(3h_t-1)}}{(1+e^{-(3h_t-1)})^2} > 0$$
Although I've shown this, I cannot prove that $h_{t+1}$ > $h_t$. I can only show that $h_1$>$h_0$ by actually finding $h_1$.
$$h_1 = \frac{1}{1+e^{-(1.8-1)}}=0.69 > h_0$$
Then, it can iteratively be proven that each next h is greater than the previous, since: $$h_1 > h_0 => \sigma(3h_1-1) > σ(3h_0-1)=>h_2 > h_1 \\ h_2>h_1=>\sigma(3h_2 - 1)>\sigma(3h_1-1)=>h_3>h_2$$
But I'm not sure if that's correct and there has to be a cleaner way to show this.
And I'm still far from proving convergence. I thought I had a clue about the converging value, since this is a sigmoid and has an upper bound of $1$. But upon thinking of it, the convergence value cannot be $1$ since the next $h$ would be smaller.
Another idea
We could just impose that $h_{t+1}=h_t$, but this did not lead me anywhere either. Or to be more precise, I could not solve for $h_t$.
$$h_{t}=\sigma(3h_t-1)=\frac{1}{1+e^{-(3h_t+1)}}$$
We can show: $\{h_n\}$ increasing sequence by induction on $n \ge 0$. This means we show: $h_{n+1} > h_n, n \ge 0$. The base case $n = 0$, we have: $h_1 = 0.69 > 0.6 = h_0$. Assume $h_n > h_{n-1}, n \ge 1$, we show: $h_{n+1} > h_n$. We have: $h_n > h_{n-1} \implies 3h_n - 1 > 3h_{n-1} - 1\implies e^{3h_n-1} > e^{3h_{n-1}-1}\implies e^{-(3h_n-1)} < e^{-(3h_{n-1}-1)}\implies 1+e^{-(3h_n-1)} < 1+e^{-(3h_{n-1}-1)}\implies \dfrac{1}{1+e^{-(3h_n-1)}}> \dfrac{1}{1+3^{-(3h_{n-1}-1)}}\implies h_{n+1} > h_n. $. Thus the sequence $\{h_n\}$ is strictly increasing. Since it is bounded above by $1$, it converges to a value $L$ which satisfies the equation: $L = \dfrac{1}{1+e^{-(3L-1)}}$.We can verify $L$ with wolfgram alpha and get $L \approx 0.8041$.