?|Un|n1?2?n?1，14．解：此为交错级数，∵(n??)而级数?发散，1n?1n?1nn?故?|Un|发散，即原级数非绝对收敛，显然
n?1n单调递减且趋向于零，故原n2?1级数条件收敛。
an?1113n?1nnR?x?15．解：∵lim，∴，当时，?lim?lim3??3nn??an??n??33n?1n?13n级数为?n?1??11n1发散，当x??时，级数为???1?收敛。故原级数的收敛区
3nnn?1?11?间为??,?。
?33?an?1nn1116．解：∵???0，?n???，∴R??，收敛n?1nann?1?n?1??1??1???n?区间为???,???。
an?111???0，?n???，∴R?0。 17．解：∵nann?1?1??1??n??an?12nn118．解：∵lim?limn?1?，∴R?2。故当|x?1|?2，即
n??an??2?n?1?2n?1?x?3时收敛，当x??1或x?3时发散，当x??1时，级数为???1?n?1?n1，收n1敛；当x?3时，级数为?，发散。故收敛区间为??1,3?。
n?1n?Un?1x2x2n?32n?1x2x2?1时，即19．解：∵，?n???，当?n2n?1??2Un222xx2?1，即x?2或x??2时发散，∴R?2。当?2?x?2时收敛，当2x??2时原级数为??22，发散，故收敛区间为?2,2。
n?1??? 6
2an?1?n?1?3n1?n?1?1?n?1??20．解：∵，?n???，∴R?3，当x??3??an33n23?n?2时，原级数???1?n2，发散。故收敛区间为??3,3?。
nn?1?21．解：设f?x???nxn?1，|x|?1，
n?1??x0??xx??n?1?n?1 f?x?dx????nx?dx???nxdx??xn?001?xn?1n?1?n?1?x?1?x?∴f?x???，|x|?1。 ??21?x??1?x??22．解：设f?x???1x2n?1，|x|?1，则
n?12n?1????1x2??1?2n?12n?1?2nx???x??x? f??x???? 22n?12n?11?xn?1?n?1?n?1???x0x2f??x?dx??dx，
01?x2xx?1?11??即f?x??f?0????????1?dx， 021?x1?x????11?x11?x?x?1?x?ln∴f?x??f?0??ln，|x|?1。
21?x21?xex?e?x1??1n?11?12kn?23．解： ???x????x????x，????x????。22?n?0n!n?0n!?2k?0?2k?!??11?1n24．解：cosx??1?cos2x???1????1??2x?2n?
?2n?!22?n?0?2n11?n2 ?????1?x2n，????x????。
?2n?!22n?025．解：?1?x?ln?1?x???1?x????1?n?1?n?1xn，|x|?1 nnn?1 ????1?n?1?n?1???1?n???1??n?1?xn?1。 xn?n?1????1?x??nn?1n?n?1?n?1 7
11126．解：??xx?3?33?11?n?x?3?n1n????1???????1?n?1?x?3? x?33n?03?3?n?01?3nx?3?1，即0?x?6 3x为偶函数，∴bn?0，?n?1,2,?? 21?x2?xa?coscosnxdx?coscosnxdx n????0?2?227．解：∵f?x??cos ?1???0??1??1??cos?nx?cos?n????x?dx ???2????2x??1?11?1??1??sin??n?x?sin??n?x? ??1??1?n?2?2???n?2?2?0 ?2??cosnxcosnx?11?n?12????????1???
??2n?12n?1???2n?12n?1?n?1 ???1?令n?0，得a0?41?4n2?1，n?0,1,2,?
x在???,??上连续 24?，且f?x??cosx24?n?1cosnx∴cos?????1?，????x???。
2??n?14n2?128．解：由于f?x???2x是奇函数，故an?0，n?0,1,2,?
2?1?1?? bn???2t?sinntdt?????xcosnx?sinx??
?????n?n????1?? ???1?∴f?x??4?n?1?n4 n??1?nsinnx。
n29．解：an?13n?10n?13n???fxcosxdx?2xcosxdx?xcosxdx 3??333??333?03036?n?xn?xn?x?3?n?xn?xn?x??sin?sin ?22?cos??22?cos? 333??3n??333?0n?? ?
3n?22?1???1??，n?2k时，an2k?0。
8
n?2k?1时，a2k?1? a0?6 22?2k?1??3131?0???3 ??fxdx?2xdx?xdx?0????3?3??33?213n?10n?13n?xxdx??2xsinx??xsindx bn??f?x?sin?3303333333n?3n??1?n?3n?? ???sinx??sinx? ?xcos??xcosx?n??xn?3??3n??3n?3?003 ?9??1?n?1，所以 n?36f?x????24??12k?1?9?n?n?11??cosx??1sinx ??23?n3n?0?2k?1?n?1?除x?3?2n?1?上均成立。?n?0,?1,?2,??
30．解：1）正弦级数，注意到f?0??0，作奇延拓F?x?，x???l,l?使在?0,l?上恒有F?x??f?x?。再将F?x?周期延拓得G?x?，x????,???，G?x?是一个以2l为周期的连续函数，G?x??F?x?，x???l,l?，计算付氏系数如下：
an?0，(n?0,1,2,?)
ll2?2n?xn?x?bn???xsindx??l?l?x?sindx?
0l?llx? ?4ln2?sin2n?,n?1,2,? 2?∴f?x??1n?n?x,?0?x?l?. sinsin2?22l?n?1n4l2)余弦函数
作偶延拓设F?x?，x???l,l?使在?0,l?上恒有F?x??f?x?。再将F?x?周期延拓得G?x?，x????,???，G?x?是一个以2l为周期的连续函数，G?x??F?x?，
x???l,l?，计算付氏系数如下：
ll?l2?2a0???xdx??l?l?x?dx??
l?02?x 9
ll2?2n?xn?x?an???xcosdx??l?l?x?cosdx?
l?0llx?2?2l2n?l2l2?n ??22cos?22???1?22?,n?1,2,?
l?n?2n?n??bn?0
l2l∴f?x???24??nn?1?1?n?n?xn?,?0?x?l?. ??2cos?1??1cos2??2l??(B)
n11?11?1?11?11．解：∵Sn????????????,
3n?4?3?43n?4?12k?1?3n?1??3n?4?k?13?3n?1n?n???，∴原级数收敛且和为
?1。 12?11?11?2．解：∵Sn???????
????k?2?k?1kk?1k?2k?1k?k?1?111?11??11?111????????????1???1???
kk?12kk?2k22k?1k?2?????k?1??1?311?，?n???，∴原级数收敛且和为。 444n?3．解：∵Sn??k?1??k?2?k?1???k?1?k????n?2?2????n?1?1
??n?2?n?1?1?2???1?1?2?1?2，
n?2?n?1??n???，∴原级数收敛且和为1?2。
Un?12n?1?n?1?!nn224．解：∵????1，?n???∴由比值判别n?1nnUne?n?1?2n!?1?1????n?法知原级数收敛。
?2n?5．解：∵nUn???3n?1??n2n?34?2n??2n?????????1，?n???∴由
9?3n?1??3n?1?2n3根值判别法知原级数收敛。
n2?Un?6．解：∵当n充分大时有
2n?1
10
nnnnn?ann?，而
?2n?1?n2n?1