发布时间 : 星期三 文章陈文灯主编《微积分》第八章作业解答更新完毕开始阅读03f01e68011ca300a6c390c7
?(4)?(?1)n?1?nn?13n?1
??解:?(?1)n?1?nnnn?1??n?1 记un?n?13n?133n?1
n?1limun?1u?limn?13?3n??nn??nn?13<1
所以所给级数绝对收敛
9.求下列级数的和函数:
?(1)求级数?nxn的和函数
n?1解:liman?1?limn?1?1 R=1 收敛区间(-1,1) n??ann??n?当x??1,x?1时,级数?nxn均发散 收敛域是(-1,1)
n?1?设S(x)??nxn
n?1 当x?0S(x)??x??nxn?1?(?xn)??(1
n?0n?01?x)??1?1?x?2 所以 S(x)?x??1,1?x?2x???0??0?, 1 当x?0,S(0)?0
?所以和函数为
?nxn?xn?1?1?x?1 1?x?2??(2)求级数?n2xn?1的和函数
n?1解:liman?1?lim(n?1)2n??ann2?1n?? R=1 收敛区间(-1,1) 5
??当x?1时,?n2xn?1发散;当x??1时,?n2xn?1发散收敛域是(-1,1)
n?1n?1? 设S(x)??n2xn?1
n?1??? S(x)??n(n?1?1)xn?1?x?n(n?1)xn?2??nxn?1
n?1n?1n?1???又因为
?nxn?1??(xn)?(x1 n?1n?11?x)??1?1?x?2?1?x???所以
?n(n?1)xn?2??(nxn?1)??[1n?1n?1(1?x)]??22?
1?x?3 ??n2xn?1?1x?xn?1?
?1?x?1
1?x?2?2(1?x)3?1?1?x?3
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