第一章 分析基础 · 第20题

例题

📝 题目

例 20 设 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{a}_{n + 1} - {a}_{n}}\right) = a$ ,求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{n}}{n} = a}$ .

💡 答案与解析

证 注意到 ${a}_{n} = \left( {{a}_{n} - {a}_{n - 1}}\right) + \left( {{a}_{n - 1} - {a}_{n - 2}}\right) + \cdots + \left( {{a}_{2} - {a}_{1}}\right) + {a}_{1}$ ,

$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{n}}{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\left( {{a}_{n} - {a}_{n - 1}}\right) + \left( {{a}_{n - 1} - {a}_{n - 2}}\right) + \cdots + \left( {{a}_{2} - {a}_{1}}\right) }{n} + \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{1}}{n} $$

$$ = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\left( {{a}_{n} - {a}_{n - 1}}\right) + \left( {{a}_{n - 1} - {a}_{n - 2}}\right) + \cdots + \left( {{a}_{2} - {a}_{1}}\right) }{n} $$

$$ \text{ 用