📝 题目
例 13 求证:
(1) $\frac{1}{n + 1} < \ln \left( {1 + \frac{1}{n}}\right) < \frac{1}{n}$ ;
(2) $\displaystyle{\exists \mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n}\right\rbrack}$ .
💡 答案与解析
证(1)已知序列 ${\left( 1 + \frac{1}{n}\right) }^{n}$ 严格 $\uparrow$ ,且
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( 1 + \frac{1}{n}\right) }^{n} = \mathrm{e} \Rightarrow {\left( 1 + \frac{1}{n}\right) }^{n} < \mathrm{e}. \tag{3.2} $$
又设 ${y}_{n}\overset{\text{ 定义 }}{ = }{\left( 1 + \frac{1}{n}\right) }^{n + 1}$ ,显然 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n} = \mathrm{e}}$ . 再根据 $n + 2$ 项的平均值不
等式, 有
$$ \frac{1}{{y}_{n}} = {\left( \frac{n}{n + 1}\right) }^{n + 1} \cdot 1 \leq {\left( \frac{\left( {n + 1}\right) \frac{n}{n + 1} + 1}{n + 2}\right) }^{n + 2} = \frac{1}{{y}_{n + 1}} $$
$$ \Rightarrow {y}_{n}\text{ 严格 } \downarrow \Rightarrow {y}_{n} > \mathrm{e}\text{ . } \tag{3.3} $$
联合 (3.2) 与 (3.3) 式即得
$$ {\left( 1 + \frac{1}{n}\right) }^{n} < \mathrm{e} < {\left( 1 + \frac{1}{n}\right) }^{n + 1} $$
$$ \overset{\text{ 两边取对数 }}{ = }\frac{1}{n + 1} < \ln \left( {1 + \frac{1}{n}}\right) < \frac{1}{n}. $$
(2)记 ${x}_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n$ ,由第 (1) 小题结论,有
$$ {x}_{n + 1} - {x}_{n} = \frac{1}{n + 1} - \ln \left( {n + 1}\right) + \ln n $$
$$ = \frac{1}{n + 1} - \ln \left( {1 + \frac{1}{n}}\right) < 0 \Rightarrow {x}_{n}\text{ 严格 } \downarrow \text{ . } $$
再由第 (1) 小题结论, 有
$$ {x}_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n $$
$$ > \ln \left( {1 + \frac{1}{1}}\right) + \ln \left( {1 + \frac{1}{2}}\right) + \cdots + \ln \left( {1 + \frac{1}{n}}\right) - \ln n $$
$$ = \ln \left( {n + 1}\right) - \ln n > 0, $$
即 ${x}_{n}$ 有下界. 从而极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}}$ 存在.
评注 (1) 极限值 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n}\right\rbrack}$ 称为欧拉常数,它等于 0.577216 ... .
(2)用第(1)小题的不等式推导第(2)小题结论也可以用下一个例题的结果.