[proofplan]
The proof uses the standard recurrence for [continued fraction](/page/Continued%20Fraction) convergents and the determinant identity for adjacent convergents. We first prove the alternating determinant formula $p_{n+1}q_n-p_nq_{n+1}=(-1)^n$, then substitute the recurrence two indices later to determine the sign of
\begin{align*}
\frac{p_{n+2}}{q_{n+2}}-\frac{p_n}{q_n}.
\end{align*}
This gives monotonicity of the even and odd subsequences, and the adjacent determinant identity then places every even convergent below every odd convergent.
[/proofplan]
[step:Establish positivity of the convergent denominators]
We use the standard recurrence convention for the numerators and denominators of the convergents. Define $p_{-2}:=0$, $p_{-1}:=1$, $q_{-2}:=1$, and $q_{-1}:=0$, and for every $n \geq 0$ define
\begin{align*}
p_n:=a_np_{n-1}+p_{n-2},\qquad q_n:=a_nq_{n-1}+q_{n-2}.
\end{align*}
Thus $p_n/q_n$ is the $n$th convergent under this convention. We first record that $q_n>0$ for every $n \geq 0$. Since $q_0=a_0q_{-1}+q_{-2}=1$, the base case holds. Also
\begin{align*}
q_1=a_1q_0+q_{-1}=a_1>0.
\end{align*}
For $n \geq 2$, the recurrence gives
\begin{align*}
q_n=a_nq_{n-1}+q_{n-2},
\end{align*}
where $a_n>0$ and, inductively, $q_{n-1},q_{n-2}>0$. Hence $q_n>0$. Thus every quotient $p_n/q_n$ is well-defined, and multiplying or dividing by products $q_mq_n$ preserves signs.
[/step]
[step:Compute the alternating determinant for adjacent convergents]
Define the adjacent determinant sequence $\Delta_n \in \mathbb{Z}$ for $n \geq 0$ by
\begin{align*}
\Delta_n := p_{n+1}q_n-p_nq_{n+1}.
\end{align*}
For $n=0$, using $p_0=a_0$, $q_0=1$, $p_1=a_1a_0+1$, and $q_1=a_1$, we obtain
\begin{align*}
\Delta_0=(p_1q_0-p_0q_1)=(a_1a_0+1)\cdot 1-a_0a_1=1.
\end{align*}
For $n \geq 1$, substituting the recurrence
\begin{align*}
p_{n+1}=a_{n+1}p_n+p_{n-1}
\end{align*}
and the recurrence
\begin{align*}
q_{n+1}=a_{n+1}q_n+q_{n-1}
\end{align*}
into the definition of $\Delta_n$ gives
\begin{align*}
\Delta_n=(a_{n+1}p_n+p_{n-1})q_n-p_n(a_{n+1}q_n+q_{n-1})=p_{n-1}q_n-p_nq_{n-1}=-\Delta_{n-1}.
\end{align*}
By induction,
\begin{align*}
\Delta_n=(-1)^n
\end{align*}
for every $n \geq 0$.
[guided]
The purpose of this step is to track the sign of the difference between neighbouring convergents. Since
\begin{align*}
\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}=\frac{p_{n+1}q_n-p_nq_{n+1}}{q_{n+1}q_n},
\end{align*}
and the denominator is positive, the sign is controlled entirely by the numerator. We therefore define, for every $n \geq 0$,
\begin{align*}
\Delta_n := p_{n+1}q_n-p_nq_{n+1}.
\end{align*}
We compute the first value directly. From the recurrence,
\begin{align*}
p_0=a_0,\qquad q_0=1.
\end{align*}
Also
\begin{align*}
p_1=a_1a_0+1,\qquad q_1=a_1.
\end{align*}
Therefore
\begin{align*}
\Delta_0=(p_1q_0-p_0q_1)=(a_1a_0+1)\cdot 1-a_0a_1=1.
\end{align*}
Now we show that the determinants alternate sign. For $n \geq 1$, substitute the recurrence relation
\begin{align*}
p_{n+1}=a_{n+1}p_n+p_{n-1}
\end{align*}
and the recurrence relation
\begin{align*}
q_{n+1}=a_{n+1}q_n+q_{n-1}
\end{align*}
into $\Delta_n$. This gives
\begin{align*}
\Delta_n=(a_{n+1}p_n+p_{n-1})q_n-p_n(a_{n+1}q_n+q_{n-1}).
\end{align*}
Expanding and cancelling the two identical terms $a_{n+1}p_nq_n$ gives
\begin{align*}
\Delta_n=p_{n-1}q_n-p_nq_{n-1}=-\bigl(p_nq_{n-1}-p_{n-1}q_n\bigr)=-\Delta_{n-1}.
\end{align*}
Starting from $\Delta_0=1$, this recurrence gives
\begin{align*}
\Delta_n=(-1)^n
\end{align*}
for every $n \geq 0$. This alternating determinant identity is the algebraic source of the alternating behaviour of the convergents.
[/guided]
[/step]
[step:Compare convergents whose indices differ by two]
For every $n \geq 0$, define the two-apart determinant $E_n \in \mathbb{Z}$ by
\begin{align*}
E_n:=p_{n+2}q_n-p_nq_{n+2}.
\end{align*}
Using
\begin{align*}
p_{n+2}=a_{n+2}p_{n+1}+p_n.
\end{align*}
Also
\begin{align*}
q_{n+2}=a_{n+2}q_{n+1}+q_n,
\end{align*}
we compute
\begin{align*}
E_n=(a_{n+2}p_{n+1}+p_n)q_n-p_n(a_{n+2}q_{n+1}+q_n)=a_{n+2}(p_{n+1}q_n-p_nq_{n+1})=a_{n+2}(-1)^n.
\end{align*}
Since $a_{n+2}>0$ and $q_{n+2}q_n>0$, we have
\begin{align*}
\frac{p_{n+2}}{q_{n+2}}-\frac{p_n}{q_n}=\frac{p_{n+2}q_n-p_nq_{n+2}}{q_{n+2}q_n}=\frac{a_{n+2}(-1)^n}{q_{n+2}q_n}.
\end{align*}
Thus this difference is positive when $n$ is even and negative when $n$ is odd.
[/step]
[step:Deduce monotonicity of the even and odd subsequences]
Taking $n=2r$ in the preceding identity, where $r \geq 0$, gives
\begin{align*}
\frac{p_{2r+2}}{q_{2r+2}}-\frac{p_{2r}}{q_{2r}}
=
\frac{a_{2r+2}}{q_{2r+2}q_{2r}}
>0.
\end{align*}
Hence
\begin{align*}
\frac{p_{2r}}{q_{2r}} < \frac{p_{2r+2}}{q_{2r+2}}
\end{align*}
for every $r \geq 0$, so the even convergents form a strictly increasing sequence.
Taking $n=2r+1$ gives
\begin{align*}
\frac{p_{2r+3}}{q_{2r+3}}-\frac{p_{2r+1}}{q_{2r+1}}
=
-\frac{a_{2r+3}}{q_{2r+3}q_{2r+1}}
<0.
\end{align*}
Hence
\begin{align*}
\frac{p_{2r+1}}{q_{2r+1}} > \frac{p_{2r+3}}{q_{2r+3}}
\end{align*}
for every $r \geq 0$, so the odd convergents form a strictly decreasing sequence.
[/step]
[step:Place every even convergent below every odd convergent]
The adjacent determinant identity gives, for every $n \geq 0$,
\begin{align*}
\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}=\frac{p_{n+1}q_n-p_nq_{n+1}}{q_{n+1}q_n}=\frac{(-1)^n}{q_{n+1}q_n}.
\end{align*}
In particular, for every $r \geq 0$,
\begin{align*}
\frac{p_{2r}}{q_{2r}} < \frac{p_{2r+1}}{q_{2r+1}},
\end{align*}
and for every $r \geq 1$,
\begin{align*}
\frac{p_{2r}}{q_{2r}} < \frac{p_{2r-1}}{q_{2r-1}}.
\end{align*}
Now fix arbitrary $r,s \geq 0$. If $r \leq s$, then monotonicity of the even subsequence and the adjacent even-to-odd inequality give
\begin{align*}
\frac{p_{2r}}{q_{2r}}
\leq
\frac{p_{2s}}{q_{2s}}
<
\frac{p_{2s+1}}{q_{2s+1}}.
\end{align*}
If $r \geq s+1$, then monotonicity of the odd subsequence gives
\begin{align*}
\frac{p_{2s+1}}{q_{2s+1}}
\geq
\frac{p_{2r-1}}{q_{2r-1}},
\end{align*}
and the adjacent odd-to-even inequality gives
\begin{align*}
\frac{p_{2r}}{q_{2r}}
<
\frac{p_{2r-1}}{q_{2r-1}}.
\end{align*}
Combining these inequalities yields
\begin{align*}
\frac{p_{2r}}{q_{2r}}
<
\frac{p_{2s+1}}{q_{2s+1}}.
\end{align*}
Thus every even convergent is strictly less than every odd convergent, completing the proof.
[/step]
