📝 题目
例 9 讨论级数 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\ln \left( {1 + \frac{{\left( -1\right) }^{n}}{{n}^{p}}}\right)$ 的收敛性与绝对收敛性 $(p >$ 0).
💡 答案与解析
解 记 ${a}_{n} = {\left( -1\right) }^{n}/{n}^{p},{b}_{n} = \ln \left( {1 + {a}_{n}}\right) ,{c}_{n} = {a}_{n} - {b}_{n}$ ,则有
$$ {b}_{n} = \ln \left( {1 + \frac{{\left( -1\right) }^{n}}{{n}^{p}}}\right) = \frac{{\left( -1\right) }^{n}}{{n}^{p}} - \frac{1}{2{n}^{2p}} + o\left( \frac{1}{2{n}^{2p}}\right) , $$
$$ {c}_{n} = \frac{1}{2{n}^{2p}} + o\left( \frac{1}{{n}^{2p}}\right) \sim \frac{1}{2{n}^{2p}}. $$
当 $0 < p \leq \frac{1}{2}$ 时,级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}}$ 条件收敛, $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}$ 发散 $\displaystyle{\Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}}$ 发散.
当 $\frac{1}{2} < p \leq 1$ 时,级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}}$ 条件收敛,级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}$ 绝对收敛 $\Rightarrow$ $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}}$ 条件收敛,要不然, $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}$ 绝对收敛,由
$$ {a}_{n} = {c}_{n} + {b}_{n} \Rightarrow \left| {a}_{n}\right| \leq \left| {b}_{n}\right| + \left| {c}_{n}\right| \Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {a}_{n}\right| \text{ 收敛. } $$
矛盾！
当 $p > 1$ 时,级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n},\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}$ 皆绝对收敛 $\displaystyle{\Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}}$ 绝对收敛.
评注 级数 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\ln \left( {1 + \frac{{\left( -1\right) }^{n}}{{n}^{p}}}\right)$ 为交错级数,当 $0 < p \leq \frac{1}{2}$ 时,由于一般项 $\left| {b}_{n}\right|$ 不单调,级数可以发散；还说明对于非正项级数,由
$$ \ln \left( {1 + \frac{{\left( -1\right) }^{n}}{{n}^{p}}}\right) \sim \frac{{\left( -1\right) }^{n}}{{n}^{p}} $$
及 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n}}{{n}^{p}}$ 收敛,得不出 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\ln \left( {1 + \frac{{\left( -1\right) }^{n}}{{n}^{p}}}\right)$ 收敛.