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Section 11.4 More properties of convergents

By induction. Note that

\begin{equation*} q_0 = 1\leq a_1 = q_1. \end{equation*}

Now assume that the lemma holds for \(k = m,\) where \(1\leq m \lt n.\) Then

\begin{equation*} q_{m+1} = a_{m+1}q_m + q_{m-1} \gt a_{m+1}q_m \geq q_m. \end{equation*}

This proves the lemma.

We compute the difference \(C_{k+2}-C_k\) to give

\begin{align*} C_{k+2}-C_k \amp = (C_{k+2}-C_{k+1}) + (C_{k+1}-C_k)\\ \amp =\left(\frac{p_{k+2}}{q_{k+2}}-\frac{p_{k+1}}{q_{k+1}}\right)+\left(\frac{p_{k+1}}{q_{k+1}}-\frac{p_{k}}{q_{k}}\right)\\ \amp =\frac{p_{k+2}q_{k+1}-p_{k+1}q_{k+2}}{q_{k+2}q_{k+1}}+\frac{p_{k+1}q_{k}-p_kq_{k+1}}{q_{k+1}q_k}\\ \amp =\frac{(-1)^{k+1}}{q_{k+2}q_{k+1}}+\frac{(-1)^k}{q_{k+1}q_k}\\ \amp =\frac{(-1)^{k}(q_{k+2}-q_{k})}{q_kq_{k+1}q_{k+2}}. \end{align*}

Recall that \(q_i\gt 0\) for all \(i\geq 0\) and that \(q_{k+2}-q_k\gt 0\) by the previous lemma, we have that \(C_{k+2}-C_k\) has the same sign as \((-1)^k.\) Considering the even and odd cases proves the first part of the theorem.

For the second part of the theorem, we note that since

\begin{equation*} p_kq_{k-1}-q_kp_{k-1} = (-1)^{k-1}, \end{equation*}

when we divide by \(q_kq_{k-1}\) we have

\begin{equation*} \frac{p_k}{q_k}-\frac{p_{k-1}}{q_{k-1}}=\frac{(-1)^{k-1}}{q_kq_{k-1}}; \end{equation*}

that is

\begin{equation*} C_k-C_{k-1} = \frac{(-1)^{k-1}}{q_kq_{k-1}}. \end{equation*}

Note that \(q_kq_{k-1}\gt 0.\) So if \(k = 2j\) then we have

\begin{equation*} C_{2j}-C_{2j-1} =\frac{(-1)}{q_{2j}q_{2j-1}}\lt 0. \end{equation*}

That is \(C_{2j}\lt C_{2j-1}.\) Now if \(2u\) and \(2v-1\) are arbitrary even and odd integers, then using the first part of the theorem we have

\begin{equation*} C_{2u} \lt C_{2u+2v} \lt C_{2u+2v-1} \lt C_{2v-1}. \end{equation*}

This proves the theorem.