📝 题目
例 4 求证:
(1) $\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}\left( {1 - x}\right) {x}^{n}$ 在 $\left\lbrack {0,1}\right\rbrack$ 上一致收敛;
(2) $\mathop{\sum }\limits_{{n = 0}}^{\infty }\left( {1 - x}\right) {x}^{n}$ 在 $\left\lbrack {0,1}\right\rbrack$ 上收敛但不一致收敛.
💡 答案与解析
证(1)固定 $x \in \left\lbrack {0,1}\right\rbrack$ ,序列 $\left( {1 - x}\right) {x}^{n}$ 单调下降且趋于零,由交错级数的余项估计式得
$$ \left| {S\left( x\right) - {S}_{n}\left( x\right) }\right| = \left| {{r}_{n}\left( x\right) }\right| = \left| {\mathop{\sum }\limits_{{k = n}}^{\infty }{\left( -1\right) }^{k}\left( {1 - x}\right) {x}^{k}}\right| $$
$$ \leq \left( {1 - x}\right) {x}^{n}\text{ . } $$
再求函数 ${u}_{n}\left( x\right) = \left( {1 - x}\right) {x}^{n}$ 的最大值.
$$ {u}_{n}^{\prime }\left( x\right) = \left( {n + 1}\right) {x}^{n - 1}\left( {\frac{n}{n + 1} - x}\right) , $$
$$ {u}_{n}^{\prime }\left( x\right) \overset{\text{ 令 }}{ \rightarrow }0 \Rightarrow x = \frac{n}{n + 1}. $$
又
$$ \left( {x - \frac{n}{n + 1}}\right) {u}_{n}^{\prime }\left( x\right) < 0\;\left( {x \neq \frac{n}{n + 1}}\right) , $$
所以 $\left| {S\left( x\right) - {S}_{n}\left( x\right) }\right| \leq \frac{1}{n + 1} \cdot {\left( \frac{n}{n + 1}\right) }^{n} \leq \frac{1}{n + 1}$ ,
故
$$ \mathop{\sup }\limits_{{0 \leq x \leq 1}}\left| {S\left( x\right) - {S}_{n}\left( x\right) }\right| \rightarrow 0\;\left( {n \rightarrow \infty }\right) , $$
即原级数在 $\left\lbrack {0,1}\right\rbrack$ 上一致收敛.
或由狄利克雷判别法: ${a}_{n}\left( x\right) \overset{\text{ 定义 }}{ = }\left( {1 - x}\right) {x}^{n}\xrightarrow[]{\left\lbrack 0,1\right\rbrack }0$ ,固定 $x$ , ${a}_{n}\left( x\right)$ 单调, ${b}_{n}\left( x\right) \frac{\text{ 定义 }}{}{\left( -1\right) }^{n},\left| {\mathop{\sum }\limits_{{k = 0}}^{n}{b}_{n}\left( x\right) }\right| \leq 1$ ,故原级数在 $\left\lbrack {0,1}\right\rbrack$ 上一致收敛.
(2)当 $\displaystyle{n \rightarrow \infty}$ 时,
$$ {S}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{{n - 1}}\left( {1 - x}\right) {x}^{k} \rightarrow S\left( x\right) = \left\{ \begin{array}{ll} 1, & 0 \leq x < 1, \\ 0, & x = 1, \end{array}\right. $$
这说明级数收敛. 由和函数不连续, 说明级数在 $\left\lbrack {0,1}\right\rbrack$ 上不一致收敛. 或因为
$$ \mathop{\sup }\limits_{{0 \leq x \leq 1}}\left| {S\left( x\right) - {S}_{n}\left( x\right) }\right| = \mathop{\sup }\limits_{{0 \leq x \leq 1}}{x}^{n} = 1 \rightarrow 0\;\left( {n \rightarrow \infty }\right) , $$
所以级数在 $\left\lbrack {0,1}\right\rbrack$ 上不一致收敛.