Prove that this iteration cuts a rational number in two irrationals $\sum_{n=0}^\infty \frac{1}{q_n^2-p_n q_n+1}+\lim_{n \to \infty} \frac{p_n}{q_n}$

As an answer for the question in the title I propose the following (using the results from the OP):

$$A=\sum_{n=0}^\infty \frac{q_n}{q_{n+1}} \tag{1}$$

We have:

$$\frac{q_{n+2}}{q_{n+1}}=q_{n+1}q_n+1+\frac{q_{n+1}^2}{q_n^2} \left(\frac{q_{n+1}}{q_n}-1 \right) \tag{2}$$

Set $a_n=\frac{q_n}{q_{n-1}}$ and $b_n=q_{n-1}q_{n-2}+1$, then we have:

$$a_n=q_{n-1}q_{n-2}+1+a_{n-1}(a_{n-1}-1):=a_{n-1}(a_{n-1}-1)+b_{n-1}$$

Thus, according to this paper: The Approximation of Numbers as Sums of Reciprocals, the sum in $(1)$ is the greedy expansion of the number $A$:

$$A=\sum_{n=1}^\infty \frac{1}{a_n}$$

According to the paper, every such expansion for a real number has the form:

$$x=\frac{1}{a_1}+\frac{1}{a_2}+\dots$$

$$a_{k+1}=a_k(a_k-1)+b_k,~~~a_1 \geq 2,~~b_k > 1,~~~~a_k,b_k \in \mathbb{N}$$

All of the requirements are met. (To prove that $a_n$ are all integers we only need to look at the initial definition of $q_n$).

And since the greedy expansion for a rational number is finite, but the sequence $a_n$ is not, we have proved that $A$ is irrational.