[proofplan]
The proof uses only the recurrence and the hypotheses that the sequence $(a_n)_{n=0}^{\infty}$ is infinite and satisfies $a_n \geq 1$ for every $n \geq 1$. We first record that the denominator recurrence keeps $q_n$ positive for every $n \geq 0$. Then we prove the alternating determinant invariant $p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1}$, which forces the relevant numerator-denominator pairs to be coprime. Taking absolute values gives exactly the determinant condition for Farey neighbours.
[/proofplan]
[step:Prove the denominators remain positive under the recurrence]
By hypothesis, the partial quotients $a_n$ are defined for every $n \geq 0$, and $a_n \geq 1$ for every $n \geq 1$. The initial denominator values are $q_0=1$ and $q_1=a_1$, so $q_0>0$ and $q_1>0$. If $n \geq 2$ and $q_{n-1}>0$ and $q_{n-2}>0$, then the denominator recurrence gives
\begin{align*}
q_n = a_n q_{n-1}+q_{n-2}>0.
\end{align*}
By induction on $n$, $q_n>0$ for every $n \geq 0$.
[/step]
[step:Derive the alternating determinant identity from the recurrence]
For each integer $n \geq 1$, define the integer $D_n$ by
\begin{align*}
D_n := p_n q_{n-1} - p_{n-1} q_n.
\end{align*}
Using the recurrence relations for $p_n$ and $q_n$, for every $n \geq 2$ we compute
\begin{align*}
D_n = (a_n p_{n-1} + p_{n-2})q_{n-1} - p_{n-1}(a_n q_{n-1} + q_{n-2}).
\end{align*}
Expanding and cancelling the two equal terms $a_n p_{n-1}q_{n-1}$ gives
\begin{align*}
D_n = p_{n-2}q_{n-1} - p_{n-1}q_{n-2}.
\end{align*}
By the definition of $D_{n-1}$, this last expression equals $-D_{n-1}$. Thus $D_n=-D_{n-1}$ for every $n \geq 2$.
For $n=1$, the initial values $p_0=a_0$, $q_0=1$, $p_{-1}=1$, and $q_{-1}=0$ give
\begin{align*}
D_1 = (a_1p_0+p_{-1})q_0-p_0(a_1q_0+q_{-1}) = (a_1a_0+1)\cdot 1-a_0a_1 = 1.
\end{align*}
Therefore, by induction on $n$,
\begin{align*}
p_n q_{n-1} - p_{n-1}q_n = D_n = (-1)^{n-1}
\end{align*}
for every $n \geq 1$.
[guided]
The recurrence relations for consecutive convergents are designed so that the coefficient $a_n$ cancels in the determinant. For each $n \geq 1$, define the integer $D_n$ by
\begin{align*}
D_n := p_n q_{n-1} - p_{n-1} q_n.
\end{align*}
This is the determinant of the matrix whose columns are the consecutive numerator-denominator vectors $(p_n,q_n)$ and $(p_{n-1},q_{n-1})$, up to the chosen order. To understand how $D_n$ changes with $n$, substitute the recurrence formulas $p_n=a_n p_{n-1}+p_{n-2}$ and $q_n=a_n q_{n-1}+q_{n-2}$. For every $n \geq 2$ this gives
\begin{align*}
D_n = (a_n p_{n-1}+p_{n-2})q_{n-1}-p_{n-1}(a_n q_{n-1}+q_{n-2}).
\end{align*}
Expanding the products, the two terms $a_n p_{n-1}q_{n-1}$ occur with opposite signs, so cancellation gives
\begin{align*}
D_n = p_{n-2}q_{n-1}-p_{n-1}q_{n-2}.
\end{align*}
The remaining expression is the negative of the previous determinant, because
\begin{align*}
D_{n-1}=p_{n-1}q_{n-2}-p_{n-2}q_{n-1}.
\end{align*}
Hence $D_n=-D_{n-1}$ for every $n \geq 2$.
It remains to identify the starting value. Since $p_0=a_0$, $q_0=1$, $p_{-1}=1$, and $q_{-1}=0$, the recurrence gives $p_1=a_1a_0+1$ and $q_1=a_1$. Therefore
\begin{align*}
D_1 = p_1q_0-p_0q_1 = (a_1a_0+1)\cdot 1-a_0a_1 = 1.
\end{align*}
Because each passage from $D_{n-1}$ to $D_n$ changes the sign and does not change the absolute value, induction gives
\begin{align*}
D_n = (-1)^{n-1}
\end{align*}
for every $n \geq 1$. Equivalently,
\begin{align*}
p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1}.
\end{align*}
[/guided]
[/step]
[step:Use the determinant identity to prove both fractions are reduced]
Fix $n \geq 1$. For integers $r$ and $s$, let $\gcd(r,s)$ denote the greatest non-negative common divisor of $r$ and $s$. If $d$ is a positive common divisor of $p_n$ and $q_n$, then $d$ divides $p_n q_{n-1}$ because $d$ divides $p_n$, and $d$ divides $p_{n-1}q_n$ because $d$ divides $q_n$. Hence $d$ divides their difference:
\begin{align*}
p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1}.
\end{align*}
Therefore $d$ divides $1$, so $d=1$. Hence $\gcd(p_n,q_n)=1$.
The same argument applied to the index $n-1$ when $n \geq 2$ gives $\gcd(p_{n-1},q_{n-1})=1$. When $n=1$, we have $p_0=a_0$ and $q_0=1$, so $\gcd(p_0,q_0)=1$. Thus both fractions are in lowest terms.
[/step]
[step:Identify the Farey-neighbour determinant condition]
From the determinant identity,
\begin{align*}
\left|p_n q_{n-1} - p_{n-1}q_n\right| = \left|(-1)^{n-1}\right| = 1.
\end{align*}
Together with
\begin{align*}
\gcd(p_{n-1},q_{n-1}) = \gcd(p_n,q_n) = 1,
\end{align*}
this is precisely the [Farey-neighbour](/page/Farey%20Neighbour) condition for the reduced rational numbers
\begin{align*}
\frac{p_{n-1}}{q_{n-1}} \quad \text{and} \quad \frac{p_n}{q_n}.
\end{align*}
Therefore consecutive recurrence fractions are Farey neighbours for every $n \geq 1$.
[/step]
