= Solution
Put $A_j=\left(\begin{smallmatrix}a_j&b_j\\1&0\end{smallmatrix}\right)$. The identity
$$
A_0\cdots A_n=
\begin{pmatrix}p_n&b_np_{n-1}\\q_n&b_nq_{n-1}\end{pmatrix}
$$
follows by induction: multiplying the formula for $n-1$ by $A_n$ gives the two stated recurrences in the first column and $b_n(p_{n-1},q_{n-1})^T$ in the second. The associated Möbius transformations $z\mapsto a_j+b_j/z$, applied from the right to $a_n$, show that the first-column ratio is the displayed <positive generalized continued fraction>.
Now set every $b_j=1$. Taking <determinants> in the <matrix> identity gives
$$
p_nq_{n-1}-q_np_{n-1}=(-1)^{n+1}.
$$
Consequently adjacent convergents differ by
$$
\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}
=\frac{(-1)^{n+2}}{q_nq_{n+1}}.
$$
The recurrence and $a_j\geq1$ imply $q_{n+1}\geq q_n+q_{n-1}$, so $q_n\to\infty$. The <determinant> signs show that the even convergents increase, the odd convergents decrease, and every even one is below every odd one. Their adjacent separation tends to zero, so both subsequences have a common <limit> $x$. Since $x$ lies between each adjacent pair, the <convergence of a positive simple continued fraction> gives exactly
$$
\left|\frac{p_n}{q_n}-x\right|
+\left|\frac{p_{n+1}}{q_{n+1}}-x\right|
=\frac1{q_nq_{n+1}}.
$$
If all $a_j=1$ as well, the recurrence and initial values give
$$
p_n=F_{n+2},\qquad q_n=F_{n+1}.
$$
The characteristic roots of $F_{n+2}=F_{n+1}+F_n$ are
$\phi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2=-\phi^{-1}$, hence
$$
F_n=\frac{\phi^n-\psi^n}{\sqrt5}.
$$
Therefore $F_{n+2}/F_{n+1}\to\phi$. Moreover,
$$
F_{n+1}-\phi F_n=\psi^n,\qquad
F_{n+2}-\phi F_{n+1}=\psi^{n+1}.
$$
Using $F_n\phi^{-n}\to1/\sqrt5$ yields
$$
F_nF_{n+1}\left|\frac{F_{n+1}}{F_n}-\phi\right|
=F_{n+1}\phi^{-n}\longrightarrow\frac{\phi}{\sqrt5},
$$
and
$$
F_nF_{n+1}\left|\frac{F_{n+2}}{F_{n+1}}-\phi\right|
=F_n\phi^{-(n+1)}\longrightarrow\frac1{\phi\sqrt5}.
$$
This is the <all-one continued fraction and Fibonacci ratios> case.
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