📝 题目
1.3.7 设 $0 < {a}_{1} < {b}_{1}$ ,令
$$ {a}_{n + 1} = \sqrt{{a}_{n} \cdot {b}_{n}},\;{b}_{n + 1} = \frac{{a}_{n} + {b}_{n}}{2}\;\left( {n = 1,2,\cdots }\right) . $$
求证: 序列 $\left\{ {a}_{n}\right\} ,\left\{ {b}_{n}\right\}$ 的极限存在.
💡 答案与解析
1.3.7 $0 < {a}_{1} < {b}_{1}$ \_\_\_\_\_ 数学归纳法 $0 < {a}_{n} < {b}_{n} \Rightarrow {a}_{n} \uparrow ;{b}_{n} \downarrow ;{b}_{2} - {a}_{2} =$ $\frac{{b}_{1} - {a}_{1} + 2\left( {{a}_{1} - {a}_{2}}\right) }{2} < \frac{{b}_{1} - {a}_{1}}{2},\cdots ,{b}_{n} - {a}_{n} < \frac{{b}_{1} - {a}_{1}}{{2}^{n}}$ . 用区间套定理肯定序列 $\left\{ {a}_{n}\right\}$ , $\left\{ {b}_{n}\right\}$ 的极限存在,并趋于同一极限.