综上所述，原级数的收敛区间为[-4,0). （6）此级数（x-1）的幂级数
p?limn??an?1a?limn??2nn?1?2
n故收敛半径r?12. 于是当|x?1|?1?32即
1?x22时，原级数绝对收敛.
当|x?1|?1即x?1或x?3222时，原级数发散.
3? 当x?时，原级数变为?1是调和级数，发散.
2n?0n? 当x?1(?1)n1，是收敛的交错级数.
2时，原级数变为?n?1n综上所述，原级数的收敛区间为?13???2,2?. ?2． 求下列幂级数的和函数：
??(1)
?(?1)nxn； (2)
?2nx2n?1；
n?1nn?1??(3)
?1n； (4)
?(2n?1)xn．
n?1n(n?1)xn?0解：（1）可求得所给幂级数的收敛半径r=1.
?n设S(x)??(?1)nxn?1n,则
?? S?(x)???n?(?1)nx??(?1)nxn?1??1n?1?n????n?11?x
xx?S(x)??S?(x)dx???ln(1?x) (|x|?1)0??1dx01?x
又当x=1时，原级数收敛，且S(x)在x=1处连续.
?n ??(?1n)x(1x)? ? x(?1
1)n?1n??ln??（2）所给级数的收敛半经r=1，设S(x)??2nx2n?1,当|x|?1时，有n?1 41
x?S(x)dx?0??0?x??2nx2n?1dx?n?1??n?1x2nx02n?1dx
??n?1x2n?x221?x
?2?x?2x于是s(x)?? ?2?22(1?x)?1?x?又当x??1时，原级数发散.
?故
?n?12nx2n?1?2x(1?x)22 (|x|?1)
（3）可求所给级数的收敛半径为1.
? 令s(x)??n?1?xnn(n?1)xn?1?1??xxn?1n?1n(n?1)?(x?0)
令g(x)?x?n?1n(n?1),则g??(x)?x?n?1xn?1?11?x
?g??(x)dx?g?(x)?g?(0)?0?11?x0dx
g?(0)?0,g?(x)??ln(1?x)
x?0g?(x)dx?g(x)?g(0)???ln(1?x)dx,g(0)?0
0xx所以g(x)???ln(1?x)dx?x?ln(1?x)?xln(1?x);
0所以S(x)?1???1??1?ln(1?x),|x|?1,且x?0. ?x??当x?1时，级数为?n?11n(n?1)?和?(?1)n?1n1n(n?1),它们都收敛.且显然有S(0)?0.
??1?1??1???ln(1?x)故S(x)???x??0?x?(?1,0)?(0,1)x?0,x?1.
?（4）可求得所给级数的收敛半径为r=1且x?1时，级数发散，设S(x)??n?0nxn?1,
则?s(x)dx?0x??n?0xn?11?x.
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于是S(x)?(11?x)??1(1?x)??2,即?nxn?1?n?1n?1?1(1?x)2.
?n所以?(2n?1)x?2x?nxn?1n?0??n?1xn
?2x?3． 求下列级数的和：
?1(1?x)2?11?x?1?x(1?x)2 (|x?|1 )(1)
?n?1n52n?； (2)
?n?1?1(2n?1)2n(n?1)2nn；
?(3)
?n?12n?122n?1； (4)
?n?1．
?n22n解：（1）考察幂级数?nx，可求得其收敛半径r?1 ，且当x?1时，级数的通项un?nx，
n?1?lim|un|?limnn??n??2???，因而limun?0，故当x?1时，级数?nxn??n?12n发散，故幂级数
??n?1nx的收敛区间为（-1，1）.
2n??设S(x)??n?1?nx (|x|?1)，则S(x)?x?nxn?12n2n?1
令S1(x)??n?1nx2n?1,则?S1(x)dx?0xx???n?1nxn?x?nxn?1n?1.
??再令S2(x)??n?1nxn?1,则?S2(x)dx?0?n?1xn?x1?x.
?x故S2(x)???1?x?xx1?S(x)dx?，从而有. ?(|x|?1)1?22?0(1?x)(1?x)????x1?xS1(x)??? (|x|?1) 2?3(1?x)(1?x)??于是 S(x)?xS1(x)?x?x23(1?x) (|x|?1)
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112取x?1(1?)?n25?(5)5，则S5?n??3?15n?15?1??32.
1?5???（2）考察幂级数?1
2n?x2n，可求得收敛半径r=1，设n?11?S(x)??1?x2n?x?1n?1 (|x|?1)n?12n?1n?12n?1x2??令S11(x)??12n?11?(x)??x2n?2?n?12n?1x,则Sn?11?x2.
xx?S?(x)dx?11?x01?dx1-x2?ln021?x
即 S11(x)?S1(0)?ln1?x (s,(0)?0)21?x.
于是 S1(x)?1ln1?x,(|x|<1)，从而 21?xS(x)?xSx1(x)?ln1?x (|x|?1)21?x ?111?1取x?1,则S(1)?ln222?n?1(2n?1)2n?221?1 2 ?1ln(1?2)
2?（3）考察幂级数?(2n?1)x2n?1，可求得其级数半经为r=1,因为
n?1????(2n?1)x2n?1??2nx2n?1??x2n?1
n?1n?1n?1? 令x?2S1(x)??2nx2n?1，则?S1(x)dx?x2n?xn?10?.
n?11?x2所以S?x2??2x1(x)??? (|x|?1)?1?x2??(1?x2)2,于是 ???
?(2n?1x)2n?1??2nx2n?1??xn2?
n?1n?1n?1 44
2xxx?x3 ?(1?x2)2?1?x2?(1?x2)2 (|x|?1)
取x?12，得
1?(13S?1?2n?1???2?2)?2??2n?1?n?12?1?109.
?1?()2??2??? （4）考察幂级数?n(n?1)xn，可求得其收敛半径r=1.
n?1? 设S(x)??n(n?1)xn (|x|?1)
n?1 则x???S(x)dx?nxn?1?x20??nxn?1.
n?1n?1?x?又设S1(x)??nxn?1则?S1(x)dx?xn?xn?10?n?11?x.
从而Sx)??x??11(??1?x??, ?(1?x)2x2?S(x)dx?x2S(x)?x01(1?x)2
2?S(x)??x?2x?? |?(1?x)2?x|?1?(1?x)3 取x?12，则
1?S?1?n(n?1)2?????2?8
?2??3n?12n?1??1??2??习题9-5
1． 将下列函数展开成x的幂级数： (1) cos2xx22; (2) sinx； (3) xe?2; (4)
1 (5)1?x2； cos(x?π)4． 45