📝 题目
例 3 求证: $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{nx}{1 + {n}^{5}{x}^{2}}}$ 在 $\displaystyle{\left| x\right| < + \infty}$ 上一致收敛.
💡 答案与解析
证法 1 由 ${a}^{2} + {b}^{2} \geq {2ab}$ ,可得
$$ \left| \frac{nx}{1 + {n}^{5}{x}^{2}}\right| = \left| {\frac{1}{2{n}^{3/2}} \cdot \frac{2{n}^{5/2}x}{1 + {n}^{5}{x}^{2}}}\right| \leq \frac{1}{2{n}^{3/2}}\;\left( {\left| x\right| < + \infty }\right) . $$
又 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{2{n}^{3/2}}}$ 收敛,由 $M$ 判别法即得原级数在 $\displaystyle{\left| x\right| < + \infty}$ 上一致收敛.
证法 2 记 ${u}_{n}\left( x\right) = \frac{nx}{1 + {n}^{5}{x}^{2}}$ ,先求函数 ${u}_{n}\left( x\right)$ 的最大值,由于 ${u}_{n}\left( x\right)$ 为奇函数,只需讨论 $x \geq 0$ 的情形.
$$ {u}_{n}^{\prime }\left( x\right) = {u}_{n}\left( x\right) \left\lbrack {\frac{1}{x} - \frac{2{n}^{5}x}{1 + {n}^{5}{x}^{2}}}\right\rbrack = \frac{n\left( {1 - {n}^{5}{x}^{2}}\right) }{{\left( 1 + {n}^{5}{x}^{2}\right) }^{2}}, $$
$$ {u}_{n}^{\prime }\left( x\right) \overset{\text{ 令 }}{ \Rightarrow }0 \Rightarrow x = {n}^{-5/2}. $$
又 $\left( {x - {n}^{-5/2}}\right) {u}_{n}^{\prime }\left( x\right) < 0\left( {x > 0,x \neq {n}^{-5/2}}\right)$ ,故 $x = {n}^{-5/2}$ 是函数 ${u}_{n}\left( x\right)$ 的最大值点. 因此
$$ \left| \frac{nx}{1 + {n}^{5}{x}^{2}}\right| \leq \frac{n \cdot {n}^{-5/2}}{1 + {n}^{5}{\left( {n}^{-5/2}\right) }^{2}} = \frac{1}{2{n}^{3/2}}\;\left( {\left| x\right| < + \infty }\right) . $$
下同证法 1.