📝 题目
例 12 求证: 将级数 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n} = \ln 2$ 重排后的级数
$$ 1 + \frac{1}{3} - \frac{1}{2} + \cdots + \frac{1}{{4k} - 3} + \frac{1}{{4k} - 1} - \frac{1}{2k} + \cdots $$
的和为 $\frac{3}{2}\ln 2$ .
💡 答案与解析
证 记级数 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}$ 的部分和为 ${\sigma }_{n}$ ,重排后级数的部分和为 ${S}_{n}$ ,则
$$ {S}_{3n} = \mathop{\sum }\limits_{{k = 1}}^{n}\left( {\frac{1}{{4k} - 3} + \frac{1}{{4k} - 1} - \frac{1}{2k}}\right) $$
$$ = {\sigma }_{4n} + \frac{1}{{2n} + 2} + \frac{1}{{2n} + 4} + \cdots + \frac{1}{4n} $$
$$ = {\sigma }_{4n} + \frac{1}{2}\left( {\frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2n}}\right) $$
$$ = {\sigma }_{4n} + \frac{1}{2}\left\lbrack {c + \ln {2n} + {r}_{2n} - c - \ln n - {r}_{n}}\right\rbrack $$
$$ = {\sigma }_{4n} + \frac{1}{2}\left( {\ln 2 + {r}_{2n} - {r}_{n}}\right) \rightarrow \ln 2 + \frac{1}{2}\ln 2 = \frac{3}{2}\ln 2, $$
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{{3n} + 1} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{3n} + \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{4n} + 1} = \frac{3}{2}\ln 2, $$
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{{3n} + 2} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{{3n} + 1} + \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{4n} + 3} = \frac{3}{2}\ln 2 $$
$$ \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} = \frac{3}{2}\ln 2, $$
即重排后级数和为 $\frac{3}{2}\ln 2$ .