此时 \(a_0>b_0\)，且在归纳法假定 \(a_k>b_k\) 之下
\[
\frac{a_{k+1}}{b_{k+1}}=\sqrt{\frac{1+a_k/b_k}{2}}>\sqrt{\frac{1+1}{2}}>1~\Rightarrow~\fbox{\(a_n>b_n,~~n=0,1,2,\cdots\)}
\]
上述结果允许俺们引进实参量 \(\tau_n>0\) 使得 \(a_n/b_n=\cosh\tau_n\)，特别 \(\tau_0={\rm arccosh}(A/B)=\ln[A/B+\sqrt{(A/B)^2-1}]\)。这些参量满足递推式
\[
\cosh\tau_{n+1}=\sqrt{\frac{1+\cosh\tau_n}{2}}=\cosh\frac{\tau_n}{2}~\Rightarrow~\fbox{\(\displaystyle\tau_n=\frac{\tau_{n-1}}{2}=\cdots=\frac{\tau_0}{2^n}\)}
\]
另一方面
\begin{align}
&a^2_{n+1}-b^2_{n+1}=a_{n+1}(a_{n+1}-b_n)=\frac{a^2_n-b^2_n}{4}~\Rightarrow~\fbox{\(\displaystyle\sqrt{a_n^2-b_n^2}=\frac{\sqrt{a^2_0-b^2_0}}{2^n}\)}\\
\therefore~ & b_n=\frac{\sqrt{a^2_0-b^2_0}}{2^n\sqrt{\cosh^2\tau_n-1}}=\frac{\sqrt{A^2-B^2}}{2^n\sinh(\tau_0/2^n)}~\stackrel{n\to\infty}{\longrightarrow}~\frac{\sqrt{A^2-B^2}}{\tau_0}=\frac{\sqrt{A^2-B^2}}{{\rm arccosh}(A/B)}
\end{align}