方企勤 第四章 级 数 第6题
📝 题目
例 6 讨论下列级数的收敛性:
(1) $\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n - 1}\frac{\ln n}{\sqrt{n}}$ ; (2) $\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n - 1}\frac{\left( {{2n} - 1}\right) !!}{\left( {2n}\right) !!}$ .
💡 答案解析
解 (1) 设 $f\left( x\right)$ 是 $\ln x/\sqrt{x}$ ,则
$$ {f}^{\prime }\left( x\right) = \frac{2 - \ln x}{{2x}\sqrt{x}} < 0\;\left( {x > {\mathrm{e}}^{2}}\right) . $$
因此当 $x > 9$ 时, $f\left( x\right)$ 单调递减,即得序列 $\left\{ {\ln n/\sqrt{n}}\right\}$ 单调递减. 又由洛必达法则, 有
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\ln n}{\sqrt{n}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{\sqrt{x}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1/x}{1/2\sqrt{x}} = 0. $$
由莱布尼茨判别法知原级数收敛.
(2)为证明此级数收敛, 根据莱布尼茨判别法, 只要证序列
$$ {a}_{n}\overset{\text{ 记为 }}{ = }\frac{\left( {{2n} - 1}\right) !!}{\left( {2n}\right) !!}\;\left( {n = 1,2,3,\cdots }\right) $$
单调下降并且趋于零. 由
$$ {a}_{n + 1} = \frac{\left( {{2n} + 1}\right) !!}{\left( {{2n} + 2}\right) !!} = \frac{\left( {{2n} + 1}\right) \left( {{2n} - 1}\right) !!}{\left( {{2n} + 2}\right) \left( {2n}\right) !!} $$
$$ < \frac{\left( {{2n} - 1}\right) !!}{\left( {2n}\right) !!} = {a}_{n}, $$
即可看出 ${a}_{n}$ 单调下降. 为了证明 ${a}_{n} \rightarrow 0\left( {n \rightarrow \infty }\right)$ ,有下面几种证法.
证法 1 根据基本不等式
$$ \frac{a}{b} < \frac{a + 1}{b + 1}\;\left( {b > a > 0}\right) , $$
显然
$$ {a}_{n} = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \cdots \cdot \frac{{2n} - 1}{2n} $$
$$ \leq \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \cdots \cdot \frac{2n}{{2n} + 1} = \frac{1}{{a}_{n}\left( {{2n} + 1}\right) }. $$
由此 $0 < {a}_{n} \leq 1/\sqrt{{2n} + 1}$ ,即得 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0}$ .
证法 $2\;{a}_{n} = \left( {1 - \frac{1}{2}}\right) \left( {1 - \frac{1}{4}}\right) \cdots \left( {1 - \frac{1}{2n}}\right) = {\mathrm{e}}^{\mathop{\sum }\limits_{{k = 1}}^{\infty }\ln \left( {1 - \frac{1}{2k}}\right) }$ . 又
$$ \ln \left( {1 - \frac{1}{2k}}\right) \sim - \frac{1}{2k}\;\left( {k \rightarrow + \infty }\right) , $$
而级数 $\displaystyle{- \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{2k}}$ 发散,故
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{k = 1}}^{n}\ln \left( {1 - \frac{1}{2k}}\right) = - \infty $$
从而
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mathrm{e}}^{\mathop{\sum }\limits_{{k = 1}}^{\infty }\ln \left( {1 - \frac{1}{2k}}\right) } = 0. $$
证法 3 由瓦里斯公式:
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left\lbrack \frac{\left( {2n}\right) !!}{\left( {{2n} - 1}\right) !!}\right\rbrack }^{2}\frac{1}{{2n} + 1} = \frac{\pi }{2}, $$
可知 ${a}_{n} \sim \sqrt{\frac{2}{\pi }} \cdot \frac{1}{\sqrt{{2n} + 1}}\left( {n \rightarrow \infty }\right)$ ,即得 ${a}_{n} \rightarrow 0\left( {n \rightarrow \infty }\right)$ .