📝 题目
例 17 求证: 若 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = + \infty}$ ,则有 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{n} = + \infty}$ .
💡 答案与解析
证 因为 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = + \infty}$ ,所以对 $\forall M > 0,\exists m$ ,当 $n > m$ 时, ${a}_{n} >$ ${3M}$ . 令 ${b}_{n} = {a}_{1} + {a}_{2} + \cdots + {a}_{n}$ ,改写
$$ \frac{{b}_{n}}{n} = \frac{{b}_{m}}{n} + \frac{{b}_{n} - {b}_{m}}{n} = \frac{{b}_{m}}{n} + \frac{{b}_{n} - {b}_{m}}{n - m}\left( {1 - \frac{m}{n}}\right) $$
$$ > {3M}\left( {1 - \frac{m}{n}}\right) - \left| \frac{{b}_{m}}{n}\right| \text{ . } \tag{4.5} $$
又因为 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {1 - \frac{m}{n}}\right) = 1,\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{b}_{m}}{n} = 0$ ,所以存在 $N > m$ ,使得当 $n > N$ 时, 有
$$ 1 - \frac{m}{n} > \frac{1}{2},\;\left| \frac{{b}_{m}}{n}\right| < \frac{M}{2}\overset{\text{ 由 }\left( {4.5}\right) }{ \Rightarrow }\frac{{b}_{n}}{n} > M, $$
即
$$ \frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{n} > M\;\left( {n > N}\right) . $$
故有
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{n} = + \infty . $$