方企勤 第一章 分析基础 第12题
📝 题目
例 12 设 $\left\{ {a}_{n}\right\}$ 单调下降,且 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0,{b}_{n}\frac{\text{ 定义 }{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{n}}$ . 求证:
(1) ${b}_{n}$ 单调下降; (2) ${b}_{2n} \leq \frac{1}{2}\left( {{a}_{n} + {b}_{n}}\right)$ ; (3) $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = 0}$ .
💡 答案解析
证(1)由已知条件有
$$ {a}_{n} \downarrow \Rightarrow {b}_{n} = \frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{n} \geq {a}_{n + 1} $$
$$ \Rightarrow {b}_{n + 1} = \frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n} + {a}_{n + 1}}{n + 1} $$
$$ = \frac{n{b}_{n} + {a}_{n + 1}}{n + 1} \leq \frac{n{b}_{n} + {b}_{n}}{n + 1} = {b}_{n} \Rightarrow {b}_{n} \downarrow . $$
(2) ${b}_{2n} = \frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n} + {a}_{n + 1} + \cdots + {a}_{2n}}{2n} = \frac{1}{2}{b}_{n} + \frac{{a}_{n + 1} + \cdots + {a}_{2n}}{2n}$
$$ \leq \frac{1}{2}\frac{1}{2}{b}_{n} + \frac{n{a}_{n}}{2n} = \frac{1}{2}\left( {{a}_{n} + {b}_{n}}\right) . $$
( 3 )由第 (1 ) 小题及 ${b}_{n} \geq 0,{b}_{n}$ 是单调下降有下界序列,因此极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b}$ 存在. 对第 (2) 小题的不等式两边取极限,得 $b \leq \frac{1}{2}b \Rightarrow b \leq$ 0; 又 $b \geq 0$ ,即得 $b = 0$ ,即 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = 0}$ .
引申
$$ \frac{1}{\sqrt[n]{n!}} = \sqrt[n]{\frac{1}{n!}} \leq \frac{1 + \frac{1}{2} + \cdots + \frac{1}{n}}{n}\underset{\frac{1}{n + \infty }n}{\underbrace{\frac{1}{\lim \frac{1}{n} = 0}}}\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{\sqrt[n]{n!}} = 0. $$