📝 题目
例 8 (1) 求 ${\ln }^{2}\left( {1 + x}\right)$ 在 $x = 0$ 点的幂级数展开式;
(2)求 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n + 1}\left\{ {1 + \frac{1}{2} + \cdots + \frac{1}{n}}\right\}$ 的和；
(3) 求 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}\left\{ {1 + \frac{1}{2} + \cdots + \frac{1}{n}}\right\}$ 的和.
💡 答案与解析
解 (1) $\ln \left( {1 + x}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}\frac{{x}^{n + 1}}{n + 1}\left( {\left| x\right| < 1}\right)$ 是一绝对收敛的级数. 由于绝对收敛级数可以任意相乘,记
$$ {a}_{n} = \frac{{\left( -1\right) }^{n}}{n + 1} $$
则有
$$ {\ln }^{2}\left( {1 + x}\right) = {x}^{2}{\left( \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{x}^{n}\right) }^{2} = {x}^{2}\mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}{x}^{n}, $$
其中
$$ {c}_{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}{a}_{n - k} = {\left( -1\right) }^{n}\mathop{\sum }\limits_{{k = 0}}^{n}\frac{1}{\left( {k + 1}\right) \left( {n - k + 1}\right) } $$
$$ = \frac{{\left( -1\right) }^{n}}{n + 2}\mathop{\sum }\limits_{{k = 0}}^{n}\frac{\left( {k + 1}\right) + \left( {n - k + 1}\right) }{\left( {k + 1}\right) \left( {n - k + 1}\right) } $$
$$ = \frac{{\left( -1\right) }^{n}}{n + 2}\mathop{\sum }\limits_{{k = 0}}^{n}\left\{ {\frac{1}{k + 1} + \frac{1}{n - k + 1}}\right\} $$
$$ = \frac{2{\left( -1\right) }^{n}}{n + 2}\mathop{\sum }\limits_{{k = 0}}^{n}\frac{1}{k + 1} $$
即得
$$ {\ln }^{2}\left( {1 + x}\right) = {x}^{2}\mathop{\sum }\limits_{{n = 0}}^{\infty }\left\{ {\frac{2{\left( -1\right) }^{n}}{n + 2}\mathop{\sum }\limits_{{k = 0}}^{n}\frac{1}{k + 1}}\right\} {x}^{n} $$
$$ = 2\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n + 1}\left\{ {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}\right\} {x}^{n - 1} $$
$$ \left( {\left| x\right| < 1}\right) \text{ . } $$
(2)对 ${\ln }^{2}\left( {1 + x}\right)$ 展开的幂级数,用阿贝尔引理得
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n + 1}\left\{ {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}\right\} = \frac{{\ln }^{2}2}{2}. $$
(3) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}\left\{ {1 + \frac{1}{2} + \cdots + \frac{1}{n}}\right\}$
$$ = 1 + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}\left\{ {1 + \frac{1}{2} + \cdots + \frac{1}{n - 1}}\right\} $$
$$ + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{n}^{2}} $$
$$ = - \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\left( -1\right) }^{k - 1}}{k + 1}\left\{ {1 + \frac{1}{2} + \cdots + \frac{1}{k}}\right\} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{n}^{2}} $$
$$ = \frac{{\pi }^{2}}{12} - \frac{1}{2}{\ln }^{2}2. $$