方企勤 第四章 级 数 第4题
📝 题目
例 4 判别下列级数的收敛性:
(1) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{\ln \left( {n + 1}\right) }\sin \frac{1}{n}$ ;
(2) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {\sqrt[n]{a} - \sqrt{1 + \frac{1}{n}}}\right) \left( {a > 0}\right)$ .
💡 答案解析
解 (1) $\frac{1}{\ln \left( {n + 1}\right) }\sin \frac{1}{n} \sim \frac{1}{n\ln \left( {n + 1}\right) } \sim \frac{1}{n\ln n}\left( {n \rightarrow \infty }\right)$ ,因此,由定理 5 便知 $\displaystyle{\mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{1}{n\ln n}}$ 发散 $\Rightarrow \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{\ln \left( {n + 1}\right) }\sin \frac{1}{n}$ 发散.
(2) $\sqrt[n]{a} - \sqrt{1 + \frac{1}{n}} = {\mathrm{e}}^{\frac{1}{n}\ln a} - {\left( 1 + \frac{1}{n}\right) }^{\frac{1}{2}}$
$$ = 1 + \frac{1}{n}\ln a + \frac{{\left( \ln a\right) }^{2}}{2{n}^{2}} + o\left( \frac{1}{{n}^{2}}\right) - \left\lbrack {1 + \frac{1}{2n} - \frac{1}{8{n}^{2}} + o\left( \frac{1}{{n}^{2}}\right) }\right\rbrack $$
$$ = \left( {\ln a - \frac{1}{2}}\right) \frac{1}{n} + \left\lbrack {\frac{{\left( \ln a\right) }^{2}}{2} + \frac{1}{8}}\right\rbrack \cdot \frac{1}{{n}^{2}} + o\left( \frac{1}{{n}^{2}}\right) . $$
当 $a \neq {\mathrm{e}}^{\frac{1}{2}}$ 时, $\sqrt[n]{a} - \sqrt{1 + \frac{1}{n}} \sim \left( {\ln a - \frac{1}{2}}\right) \frac{1}{n} \Rightarrow$ 级数发散;
当 $a = {\mathrm{e}}^{\frac{1}{2}}$ 时, $\sqrt[n]{a} - \sqrt{1 + \frac{1}{n}} \sim \frac{1}{4{n}^{2}} \Rightarrow$ 级数收敛.