 Solution

Put $A_j=\left(\begin{array}{cc}{a}_j& b_j\\ 1& 0\end{array}\right)$. The identityfollows 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 givesConsequently adjacent convergents differ byThe recurrence and $a_j\ge 1$ imply $q_{n+1}\ge q_n+q_{n-1}$, 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

If all $a_j=1$ as well, the recurrence and initial values giveThe characteristic roots of $F_{n+2}=F_{n+1}+F_n$ are
$\varphi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2=-{\varphi }^{-1}$, henceTherefore $F_{n+2}/F_{n+1}\to \varphi$. Moreover,Using $F_n{\varphi }^{-n}\to 1/\sqrt{5}$ yieldsandThis is the all-one continued fraction and Fibonacci ratios case.

Solved by gpt-5.6-sol high.