方企勤 第一章 分析基础 第3题
📝 题目
例 3 求证:
(1) $\frac{1}{2} \cdot \frac{3}{4} \cdot \cdots \cdot \frac{{2n} - 1}{2n} < \frac{1}{\sqrt{{2n} + 1}}$ ;
(2) $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{2} \cdot \frac{3}{4} \cdot \cdots \cdot \frac{{2n} - 1}{2n} = 0}$ .
💡 答案解析
证(1)令 ${x}_{n} = \frac{1}{2} \cdot \frac{3}{4} \cdot \cdots \cdot \frac{{2n} - 1}{2n}$ . 注意到
$$ 0 < a < b \Rightarrow \frac{a}{b} < \frac{a + 1}{b + 1}, $$
有
$$ {x}_{n} < \frac{2}{3} \cdot \frac{4}{5} \cdot \cdots \cdot \frac{2n}{{2n} + 1} = \frac{1}{{x}_{n}} \cdot \frac{1}{{2n} + 1} $$
$$ \Rightarrow {x}_{n}^{2} < \frac{1}{{2n} + 1} \Rightarrow {x}_{n} < \frac{1}{\sqrt{{2n} + 1}}. $$
(2)由第(1)小题, $0 < {x}_{n} < \frac{1}{\sqrt{{2n} + 1}} < \frac{1}{\sqrt{n}}$ . 于是,对任给定 $\varepsilon > 0$ , 取 $N = \left\lbrack \frac{1}{{\varepsilon }^{2}}\right\rbrack + 1$ ,当 $n > N$ 时,便有 $0 < {x}_{n} < \frac{1}{\sqrt{N}} < \varepsilon$ . 所以 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = 0}$ .
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