📝 题目
例 7 讨论下列级数的收敛性:
(1) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{{n}^{1 + \frac{1}{n}}}$ ; (2) $\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n}\frac{\sin n}{n}$ .
💡 答案与解析
解 (1) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n} \cdot \frac{1}{\sqrt[n]{n}}$ 中,因 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n - 1}}{n}$ 收敛, $\sqrt[n]{n} =$ ${\mathrm{e}}^{\frac{\ln n}{n}}$ 当 $n$ 充分大时单调递减,故 $\frac{1}{\sqrt[n]{n}}$ 单调递增且有界,所以由定理 10 知级数收敛.
(2) $\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n}\frac{\sin n}{n} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\cos \left( {n\pi }\right) \sin n}{n}$
$$ = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin n\left( {1 + \pi }\right) - \sin n\left( {\pi - 1}\right) }{2n}. $$
因级数 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin n\left( {1 + \pi }\right) }{2n}$ 与 $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin n\left( {\pi - 1}\right) }{2n}$ 收敛,所以级数收敛.
又解 由 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin {nx}}{n}}$ 收敛,令 $x = 2$ ,知 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin {2n}}{n}}$ ,即 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin {2n}}{2n}}$ 收敛,再令 $x = 1$ ,知 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin n}{n}}$ 收敛,由可组合性得
$$ \mathop{\sum }\limits_{{n = 1}}^{\infty }\left\lbrack {\frac{\sin \left( {{2n} - 1}\right) }{{2n} - 1} + \frac{\sin {2n}}{2n}}\right\rbrack \text{ 收敛 } \Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin \left( {{2n} - 1}\right) }{{2n} - 1}\text{ 收敛 } $$
$$ \Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }\left\lbrack {\frac{\sin {2n}}{2n} - \frac{\sin \left( {{2n} - 1}\right) }{{2n} - 1}}\right\rbrack \text{ 收敛 } $$
$$ \Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n}\frac{\sin n}{n}\text{ 收敛. } $$