新讲 第18章 数项级数 第8题
📝 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
$$
\mathop{\lim }\limits_{{n \rightarrow + \infty }}\frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}k{b}_{k} = 0.
$$