📝 题目
例 21 (1) 设 $0 < {x}_{1} < 1,{x}_{n + 1} = {x}_{n}\left( {1 - {x}_{n}}\right)$ ,求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}n{x}_{n} = 1}$ .
(2)设 $0 < q < 1,0 < {x}_{1} < \frac{1}{q},{x}_{n + 1} = {x}_{n}\left( {1 - q{x}_{n}}\right)$ ,求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}n{x}_{n} = \frac{1}{q}}$ .
💡 答案与解析
证(1)用数学归纳法容易证明
$$ {x}_{1} \in \left( {0,1}\right) \Rightarrow {x}_{n} \in \left( {0,1}\right) \;\left( {n = 1,2,\cdots }\right) . $$
从而
$$ 0 < \frac{{x}_{n + 1}}{{x}_{n}} = 1 - {x}_{n} < 1\;\left( {n = 1,2,\cdots }\right) $$
$$ \Rightarrow {x}_{n} \downarrow \Rightarrow \exists \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = x\text{ . } $$
$$ {x}_{n + 1} = {x}_{n}\left( {1 - {x}_{n}}\right) \overset{\text{ 令 }n \rightarrow \infty }{ = }\frac{}{}x\left( {1 - x}\right) = x $$
$$ \overset{{x}_{n} \downarrow \Rightarrow {x}_{n} < 1}{ \Rightarrow }x = 0. $$
令 ${a}_{n}\frac{\text{ 定义 }1}{{x}_{n}}\left( {n = 1,2,\cdots }\right)$ ,则当 $\displaystyle{n \rightarrow \infty}$ 时,
$$ {a}_{n + 1} - {a}_{n} = \frac{1}{{x}_{n + 1}} - \frac{1}{{x}_{n}} = \frac{1}{{x}_{n}\left( {1 - {x}_{n}}\right) } - \frac{1}{{x}_{n}} $$
$$ = \frac{1}{1 - {x}_{n}} \rightarrow 1\overset{\text{ 由