📝 题目
例 3 判别下列级数的收敛性:
(1) $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{3}^{\ln n}}}$ (2) $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{3}^{\sqrt{n}}}}$ .
💡 答案与解析
解 (1) 改写 ${3}^{\ln n} = {\mathrm{e}}^{\ln n \cdot \ln 3} = {n}^{\ln 3}$ ,并记 $p = \ln 3$ ,则原级数 $=$ $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{p}}\left( {p > 1}\right)$ ,从而原级数收敛.
(2)因为 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\ln n/\sqrt{n} = 0}$ ,所以 $\exists {n}_{0}$ ,当 $n \geq {n}_{0}$ 时, $\ln n < \sqrt{n}$ , 从而
$$ \frac{1}{{3}^{\sqrt{n}}} < \frac{1}{{3}^{\ln n}}\;\left( {n \geq {n}_{0}}\right) . $$
由比较判别法 (利用 (1) 的结果), 知原级数收敛.
又解 用比较判别法的极限形式. 设 ${a}_{n} = 1/{3}^{\sqrt{n}},{b}_{n} = 1/{n}^{2}$ . 已知级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}}$ 收敛,又由洛必达法则,
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{n}}{{b}_{n}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{n}^{2}}{{3}^{\sqrt{n}}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{2}}{{3}^{\sqrt{x}}}\overset{\text{ 令 }x = {t}^{2}}{ = }\mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{{t}^{4}}{{3}^{t}} $$
$$ = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{4{t}^{3}}{{3}^{t} \cdot \ln 3} = \cdots = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{4!}{{3}^{t}{\left( \ln 3\right) }^{4}} = 0 < + \infty , $$
从而原级数收敛.