📝 题目
例 8 设 $\displaystyle{\sum {b}_{k}}$ 是收敛级数,求证
$$ \mathop{\lim }\limits_{{n \rightarrow + \infty }}\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}k{b}_{k} = 0 $$
💡 答案与解析
证明 我们记
$$ {B}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{b}_{k},\;B = \mathop{\lim }\limits_{{n \rightarrow + \infty }}{B}_{n} = \mathop{\sum }\limits_{{k = 1}}^{{+\infty }}{b}_{k}. $$
对有限和
$$ \mathop{\sum }\limits_{{k = 1}}^{n}k{b}_{k} $$
应用分部求和公式得
$$ \mathop{\sum }\limits_{{k = 1}}^{n}k{b}_{k} = - \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}{B}_{k} + n{B}_{n}. $$
于是
$$ \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}k{b}_{k} = {B}_{n} - \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{{n - 1}}{B}_{k}. $$
因为
$$ \mathop{\lim }\limits_{{n \rightarrow + \infty }}\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{{n - 1}}{B}_{k} = \mathop{\lim }\limits_{{n \rightarrow + \infty }}{B}_{n} = B, $$
所以
$$ \mathop{\lim }\limits_{{n \rightarrow + \infty }}\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}k{b}_{k} = 0. $$