?limn??n?1?1 n 所以 收敛半径R?1 当x?1时，级数成为
?n；当x??1时，级数成为???1?
nn?1n?1?? 两个级数都发散，故收敛域为??1,1?
xx2x3xn 2．???????
22?42?4?62?4?6??2n? 解：因为??limn??an?1 an ?limn??12?4?6??2n??2n?2?
12?4?6??2n? ?limn??1
2n?2 ?0
所以 收敛半径R???，收敛域为???,???
2n 3．?2xn
n?1n?1? 解：因为??limn??an?1 an2n?1 ?limn???n?1?2?1
2nn2?11 ?2
??112n?1?1?。当x??时，级数成为?2??????1n2 所以 收敛半径R?
2?22?n?1n?1n?1?n?1n??x2n?1 4．???1?
2n?1n?1?n 解：因为
limn??un?1?x??limun?x?n??x2n?3??1?2n?3 2n?1nx??1?2n?1n?1 ? 故 当x2limn??x2?x
22?1即x?1时级数收敛；当x?1即x?1时级数发散，故收敛半径R?1；当
x??1时，级数成为收敛，故收敛域为[?1,1]
5．
2n?12n?2 x?n2n?1un?1?x??limun?x?n??2n_12nx2n?1
2n?12n?2x2n? 解：因为
limn?? ?limn??x22?x22
故当
x22?1即x?2时级数收敛，当
?x22?1时级数发散，则收敛半径为2。
2n?12n?2?2n?1 当x??2时级数成为? 2??n22n?1n?12n?1 当x?2时级数成为??2n2n?1????2n?2????1?n?1?n2n?1 2 以上级数均发散。故级数
2n?12n?2的收敛域为?2,2 x?n2n?1?? 6．
?n?1??x?5?nn
? 解：令t?x?5，则上述级数变为
?n?11ntn
1 因为 ??limn??an?1?limann??n?1?1 1n 所以收敛半径R?1。收敛区间为t?1，即?1?1，也即?1?x?5?1。则4?x?6
当x?4时，级数成为
?n?1????1?n，此交错级数收敛
n1n，此级数发散。
当x?6时，级数成为
?n?1 故级数
?n?1??x?5?nn的收敛域为?4,6?
三、利用逐项求导或逐项积分，求下列级数的和函数：
x3x5x7???? 1．x?357解：显然，所给幂级数的收敛区间为??1,1?。收敛域为??1,1?。 设 s?x???2n?1n?1???1?n?1x2n?1?x?x3?x53x7??? 57?????1?n?12n?1?? s??x?????n?12n?1x?
???n?1????1?x2n?1?? ?????n?1?2n?1? ?n?12n?2 ???1x?n?1?? ????x?n?12n?1
?1 21??x?? s'(x)? s?x??11?x2 注意到s(0)=0
?x01dx?arctaxnx???1,1? 21?x357 2. 2x?4x?6x?8x??
解：显然，所给幂级数收敛域为(?1,1)
s?x??2x?4x?6x?8x????2nx2n?1
357n?1???2n?1?2nx????dx sxdx??0???0n?1?xx ???n?1?x02nx2n?1dx
??xn?1?2nx2 ?21?xx???1,1?
??x2?2x? s?x?????1?x2?1?x2????2x4n?1 3．?
n?14n?1?解：显然，所给幂级数的收敛域为??1,1?
x4n?1 s?x???
4n?1n?1????x4n?1? s??x?????4n?1??
?n?1???x4n?1? ????4n?1??
n?1??? ??xn?1?4n
x4 ?1?x4xx???1,1?x注意到s?0??0
x411?x1dx?ln?arctanx?x s?x???s??x?dx??01?x4041?x2习题十
一、`（1）C,（2）A，（3）A，（4）C（5）A
x?[?1,1)
??11?11?1二、解：? ?????lims?lim(1?)?1???n??nn??n?1?n?1n=1n(n+1)n?1?nn=1n(n+1)??1三、
（1）解：nnn3?2?2n3?12?11n1,而?1发散，故n2n?1?发散 n2n?1n3?2??（2）解：nn5?2?nn5?1n3,而?13收敛，故?收敛
n2n?1n2n?1n5?2四、
（）1unn?n3?2?n?112n321, n2??而?1发散，故???1?nn?1但交错级数n?1n2n?1n3非绝对收敛。?2???1?nnn?1n3?2满足：limnn??n3?2?0,（2）??n?n3?2?单调递减. ??故由莱布尼兹判别法知原级数条件收敛.
2）unn1?1n?收敛，故原级数绝对收敛。
n5?2?n5?3,而?3n2n?1n2a1n?五、证明：
aan?bnanbn?n22，nn?2，
?????而正项级数?a1都收敛，故n?1n?1n?1n?aan，?bn，?2nbn与?n都收敛.
n?1n?1n六、见教材P467 例4
??七、解：设?b??11??2?1nn?2?n?2??n?1?n?1?????2n?2n?1?, n?2n?1?而?1是调和级数,故发散。 n?2n?1八、?1?解：由题设条件知：lima??n?1?1,令t?x?1,则?nannn??an?x?1???nant n3，n?1n?1lim?n?1?an?1an?11n??na?limnn??a?,故收敛半径为3。 n3?2）解：幂级数?an?（nx在点x?2收敛??x?2,?anxn都绝对收敛.
n?1n?1?故?an(?1)n绝对收敛.
n?11）
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