关于函数项级数一致收敛的专题讨论
$\bf命题:$设$f(x)$在[-1,1]上有连续的导函数,且$f(0)=0$
$(1)$证明:$\sum\limits_{n = 1}^\infty {\frac{1}{n}f\left( {\frac{x}{n}} \right)} $在[-1,1]上一致收敛
$(2)$设$S\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{1}{n}f\left( {\frac{x}{n}} \right)} $,证明:$S\left( x \right)$在[-1,1]上连续可导
$\bf命题:$$\bf(14武大七)$设函数项级数$\sum\limits_{n = 1}^\infty {\frac{{{n^{n + 2}}}}{{{{\left( {1 + nx} \right)}^n}}}} $,则
$(1)$级数在$\left( {1, + \infty } \right)$上收敛
$(2)$级数在$\left( {1, + \infty } \right)$上非一致收敛,但在$\left( {1, + \infty } \right)$上连续
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$\bf命题:$$\bf(04大连理工)$设${u_n}\left( x \right)\left( {n = 1,2, \cdots } \right)$在$[a,b]$上可微,$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $在${x_0} \in \left[ {a,b} \right]$处收敛,$\sum\limits_{n = 1}^\infty {{u_n}^\prime \left( x \right)} $在$[a,b]$上一致收敛,证明:$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $在$[a,b]$上一致收敛
1
$\bf命题:$当$|x|<r$时,$\sum\limits_{n = 0}^\infty {{a_n}{x^n}} $收敛于$f(x)$,证明:当${\frac{{{a_n}}}{{n + 1}}{r^{n + 1}}}$收敛时,$\int_0^r {f\left( x \right)dx} = \sum\limits_{n = 0}^\infty {\frac{{{a_n}}}{{n + 1}}{r^{n + 1}}} $
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$\bf命题:$$\bf(09武大九)$设${u_n}\left( x \right) = \frac{1}{{{n^3}}}\ln \left( {1 + {n^3}x} \right),n = 1,2, \cdots $,记$S\left( x \right) = \sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $
(1)证明:$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $在$[0,b]$上一致收敛,而在$\left( {0, + \infty } \right)$上非一致收敛
(2)讨论$S(x)$的可微性
$\bf命题:$
附录
$\bf命题:$$(\bf{Bendixon判别法})$设$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $为$\left[ {a,b} \right]$上的可微函数项级数,且$\sum\limits_{n = 1}^\infty {{u_n}^\prime \left( x \right)} $的部分和函数列在$\left[ {a,b} \right]$上一致有界
证明:如果$\sum\limits_{n = 1}^\infty {{u_n}\left( x \right)} $在$\left[ {a,b} \right]$上收敛,则必在$\left[ {a,b} \right]$上一致收敛
$\bf命题:$$(\bf{Dini定理})$设函数项级数$\sum\limits_{n = 0}^\infty {{u_n}\left( x \right)} $的每一项及其和函数均在$\left[ {a,b} \right]$上连续,且对每个$x \in \left[ {a,b} \right]$,
有$\sum\limits_{n = 0}^\infty {{u_n}\left( x \right)} $是正项级数或负项级数,则$\sum\limits_{n = 0}^\infty {{u_n}\left( x \right)} $在$x \in \left[ {a,b}\right]$上一致收敛
参考答案