📝 题目
1.3.6 求证:
(1) $\frac{1}{2\sqrt{n + 1}} < \sqrt{n + 1} - \sqrt{n} < \frac{1}{2\sqrt{n}}$ ;
(2)序列 ${x}_{n} = 1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} - 2\sqrt{n}$ 的极限存在.
💡 答案与解析
1.3.6 (1) 将 $\sqrt{n + 1} - \sqrt{n}$ 改写成 $\frac{1}{\sqrt{n + 1} + \sqrt{n}}$ .
(2) 用第 (1) 小题,
$$ {x}_{n + 1} - {x}_{n} = \frac{1}{\sqrt{n + 1}} - 2\left( {\sqrt{n + 1} - \sqrt{n}}\right) < 0 \Rightarrow {x}_{n} \downarrow , $$
以及
$$ {x}_{n} > \mathop{\sum }\limits_{{k = 1}}^{n}2\left( {\sqrt{k + 1} - \sqrt{k}}\right) - 2\sqrt{n} $$
$$ = 2\left( {\sqrt{n + 1} - 1}\right) - 2\sqrt{n} > - 2\text{ . } $$