关于函数项级数一致收敛的专题讨论

$f命题：$设$f(x)$在[-1,1]上有连续的导函数，且$f(0)=0$

   $(1)$证明：$sumlimits_{n = 1}^infty  {frac{1}{n}fleft( {frac{x}{n}} ight)} $在[-1,1]上一致收敛

   $(2)$设$Sleft( x ight) = sumlimits_{n = 1}^infty  {frac{1}{n}fleft( {frac{x}{n}} ight)} $，证明：$Sleft( x ight)$在[-1,1]上连续可导

1

$f命题：$$f(14武大七)$设函数项级数$sumlimits_{n = 1}^infty  {frac{{{n^{n + 2}}}}{{{{left( {1 + nx} ight)}^n}}}} $，则

   $(1)$级数在$left( {1, + infty } ight)$上收敛

   $(2)$级数在$left( {1, + infty } ight)$上非一致收敛，但在$left( {1, + infty } ight)$上连续

1

$f命题：$$f(04大连理工)$设${u_n}left( x ight)left( {n = 1,2, cdots } ight)$在$[a,b]$上可微，$sumlimits_{n = 1}^infty  {{u_n}left( x ight)} $在${x_0} in left[ {a,b} ight]$处收敛，$sumlimits_{n = 1}^infty  {{u_n}^prime left( x ight)} $在$[a,b]$上一致收敛，证明：$sumlimits_{n = 1}^infty  {{u_n}left( x ight)} $在$[a,b]$上一致收敛

1

$f命题：$当$|x|<r$时，$sumlimits_{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 {fleft( x ight)dx}  = sumlimits_{n = 0}^infty  {frac{{{a_n}}}{{n + 1}}{r^{n + 1}}} $

1

$f命题：$$f(09武大九)$设${u_n}left( x ight) = frac{1}{{{n^3}}}ln left( {1 + {n^3}x} ight),n = 1,2, cdots $，记$Sleft( x ight) = sumlimits_{n = 1}^infty  {{u_n}left( x ight)} $

(1)证明：$sumlimits_{n = 1}^infty  {{u_n}left( x ight)} $在$[0,b]$上一致收敛，而在$left( {0, + infty } ight)$上非一致收敛

(2)讨论$S(x)$的可微性

$f命题：$

附录

$f命题：$$（f{Bendixon判别法}）$设$sumlimits_{n = 1}^infty {{u_n}left( x ight)} $为$left[ {a,b} ight]$上的可微函数项级数，且$sumlimits_{n = 1}^infty {{u_n}^prime left( x ight)} $的部分和函数列在$left[ {a,b} ight]$上一致有界

证明：如果$sumlimits_{n = 1}^infty {{u_n}left( x ight)} $在$left[ {a,b} ight]$上收敛，则必在$left[ {a,b} ight]$上一致收敛

1

$f命题：$$（f{Dini定理}）$设函数项级数$sumlimits_{n = 0}^infty {{u_n}left( x ight)} $的每一项及其和函数均在$left[ {a,b} ight]$上连续，且对每个$x in left[ {a,b} ight]$，

有$sumlimits_{n = 0}^infty {{u_n}left( x ight)} $是正项级数或负项级数，则$sumlimits_{n = 0}^infty {{u_n}left( x ight)} $在$x in left[ {a,b} ight]$上一致收敛

参考答案