📝 题目
例 14 设 ${a}_{n}$ 单调增加, ${b}_{n}$ 单调下降,且 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{b}_{n} - {a}_{n}}\right) = 0$ ,求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}}$ 和 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n}}$ 都存在,且 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n}}$ .
💡 答案与解析
证 用反证法. 假定极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}}$ 不存在. 因为 ${a}_{n}$ 单调增加,所以 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = + \infty}$ ,这时由条件 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{b}_{n} - {a}_{n}}\right) = 0$ 推知 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = + \infty}$ ,这与 ${b}_{n}$ 单调下降矛盾,故极限 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}}$ 存在. 又
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {\left( {{b}_{n} - {a}_{n}}\right) + {a}_{n}}\right\rbrack = \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}. $$
引申 设
$$ {b}_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n, $$
$$ {a}_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n - 1} - \ln n. $$
则 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{b}_{n} - {a}_{n}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n} = 0$ ,根据