证法二:仍然证楼上的式 (*),显然 `a_n` 恒为正有理数,故
\[\bigl(a_{n+1}-\sqrt7\bigr)\bigl(a_n-\sqrt7\bigr)=\left( \frac{a_n+7}{a_n+1}-\sqrt7 \right)\bigl(a_n-\sqrt7\bigr)=\frac{\bigl(1-\sqrt 7\bigr)\bigl(a_n-\sqrt7\bigr)^2}{a_n+1}<0,\]于是
\begin{align*}
(*)&\iff\led
&\ln a_{n+1}-\ln\sqrt7<\frac12\bigl(\ln\sqrt7-\ln a_n\bigr),&&\text{if }a_n<\sqrt7,\\
&\ln\sqrt7-\ln a_{n+1}<\frac12\bigl(\ln a_n-\ln\sqrt7\bigr),&&\text{if }a_n>\sqrt7,
\endled\\
&\iff\led
&\frac{a_{n+1}^2}7<\frac{\sqrt7}{a_n},&&\text{if }a_n<\sqrt7,\\
&\frac7{a_{n+1}^2}<\frac{a_n}{\sqrt7},&&\text{if }a_n>\sqrt7,
\endled\\
&\iff\led
&a_n\left( \frac{a_n+7}{a_n+1} \right)^2<7\sqrt7,&&\text{if }a_n<\sqrt7,\\
&a_n\left( \frac{a_n+7}{a_n+1} \right)^2>7\sqrt7,&&\text{if }a_n>\sqrt7,
\endled
\end{align*}令
\[F(x)=x\left( \frac{x+7}{x+1} \right)^2,\quad x>0,\]则 `F\bigl(\sqrt7\bigr)=7\sqrt7` 且
\[F'(x)=\frac{(x+7)(x^2-4x+7)}{(x+1)^3}>0,\]所以式 (*) 成立。 |