设$\frac{p_0}{q_0},\frac{p_1}{q_1},\frac{p_2}{q_2},\cdots$为连分数$[a_0,a_1,a_2,\cdots]$的收敛项，则
\begin{equation}
\label{eq:2.07}
\frac{p_{n-1}}{q_{n-1}}-\frac{p_n}{q_n}=\frac{(-1)^n}{q_{n-1}q_n}
\end{equation}
证明:当$n=1$时，
\begin{equation}
\frac{p_0}{q_0}-\frac{p_1}{q_1}=\frac{-1}{a_1}
\end{equation}此时成立.假设当$n=k(k\geq 1)$时有
\begin{equation}
\frac{p_{k-1}}{q_{k-1}}-\frac{p_k}{q_k}=\frac{(-1)^k}{q_{k-1}q_k}
\end{equation}
则
\begin{equation}
\frac{p_k}{q_k}-\frac{p_{k+1}}{q_{k+1}}=\frac{p_k}{q_k}-\frac{a_{k+1}p_k+p_{k-1}}{a_{k+1}q_k+q_{k-1}}=\frac{\frac{p_k}{q_k}-\frac{p_{k-1}}{q_{k-1}}}{a_{k+1}\frac{q_k}{q_{k-1}}+1}
\end{equation}
而
\begin{equation}
\frac{\frac{p_k}{q_k}-\frac{p_{k-1}}{q_{k-1}}}{a_{k+1}\frac{q_k}{q_{k-1}}+1}=\frac{\frac{(-1)^{k+1}}{q_{k-1}q_k}}{a_{k+1}\frac{q_k}{q_{k-1}}+1}=\frac{\frac{(-1)^{k+1}}{q_k}}{a_{k+1}q_k+q_{k-1}}=\frac{(-1)^{k+1}}{q_kq_{k+1}}
\end{equation}(根据数论概论(Joseph H.Silverman) 定理39.1 连分数的递归公式)综上所述，根据数学归纳法，命题得证.