方企勤 第四章 级 数 第4题
📝 题目
例 4 求级数 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{\left( {{2n} - 1}\right) \left( {{2n} + 1}\right) }$ 的和.
💡 答案解析
解法 1 令 $f\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{\left( {{2n} - 1}\right) \left( {{2n} + 1}\right) }{x}^{{2n} + 1}$ ,容易求出此幂级数的收敛半径 $R = 1$ ,且 $f\left( 0\right) = 0$ . 由逐项积分定理得
$$ {f}^{\prime }\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{2n} - 1}{x}^{2n}\;\left( {\left| x\right| < 1}\right) . \tag{3.1} $$
令 $g\left( x\right) \overset{\text{ 定义 }}{ = }{f}^{\prime }\left( x\right) /x\left( {x \neq 0}\right) ,g\left( 0\right) = 0$ ,则由 (3.1) 式得
$$ {g}^{\prime }\left( x\right) = {\left( \frac{{f}^{\prime }\left( x\right) }{x}\right) }^{\prime } = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n - 1}{x}^{{2n} - 2} $$
$$ = \frac{1}{1 + {x}^{2}}\;\left( {\left| x\right| < 1}\right) , $$
从而 $g\left( x\right) = {\int }_{0}^{x}{g}^{\prime }\left( t\right) \mathrm{d}t = {\int }_{0}^{x}\frac{\mathrm{d}t}{1 + {t}^{2}} = \arctan x\;\left( {\left| x\right| < 1}\right)$ ,
即得 ${f}^{\prime }\left( x\right) = x\arctan x$ ,于是
$$ f\left( x\right) = {\int }_{0}^{x}{f}^{\prime }\left( t\right) \mathrm{d}t = {\int }_{0}^{x}t \cdot {\tan }^{-1}t\mathrm{\;d}t $$
$$ = \frac{{x}^{2}}{2}\arctan x - \frac{x}{2} + \frac{1}{2}\arctan x\;\left( {\left| x\right| < 1}\right) . $$
容易证明 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{\left( {{2n} - 1}\right) \left( {{2n} + 1}\right) }{x}^{n}$ 在 $x = 1$ 收敛,再根据阿贝尔引理得
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{4{n}^{2} - 1} = \mathop{\lim }\limits_{{x \rightarrow 1}}f\left( x\right) = \frac{1}{2}\arctan 1 - \frac{1}{2} + \frac{1}{2}\arctan 1 = \frac{\pi - 2}{4}. $$
解法 2 先对原级数进行如下分解:
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{4{n}^{2} - 1} = \frac{1}{2}\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n - 1}\left\lbrack {\frac{1}{{2n} - 1} - \frac{1}{{2n} + 1}}\right\rbrack $$
$$ = \frac{1}{2}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{2n} - 1} - \frac{1}{2}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{2n} + 1} $$
$$ = \frac{1}{2}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{2n} - 1} + \frac{1}{2}\mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\left( -1\right) }^{k - 1}}{{2k} - 1}. \tag{3.2} $$
又由逐项积分定理,对 $\forall x \in \left( {-1,1}\right)$ ,有
$$ \arctan x = {\int }_{0}^{x}\frac{\mathrm{d}t}{1 + {t}^{2}} = {\int }_{0}^{x}\mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( -1\right) }^{k - 1}{t}^{{2k} - 2}\mathrm{\;d}t $$
$$ = \mathop{\sum }\limits_{{k = 1}}^{\infty }{\left( -1\right) }^{k - 1}\frac{{x}^{{2k} - 1}}{{2k} - 1}. $$
再由阿贝尔引理得
$$ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\left( -1\right) }^{k - 1}}{{2k} - 1} = \mathop{\lim }\limits_{{x \rightarrow 1}}\arctan x = \frac{\pi }{4}. \tag{3.3} $$
联合 (3.2),(3.3) 式得
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{4{n}^{2} - 1} = \frac{1}{2} \cdot \frac{\pi }{4} + \frac{1}{2}\left\lbrack {\frac{\pi }{4} - 1}\right\rbrack = \frac{\pi - 2}{4}. $$