1??1??11?? ???1??????
8??2n?1??32n?3????121???n???, 12?2n?1??2n?3?121。 12故原级数收敛，且和为
2．证：
Un?1a?n?1?1，由比较判别法知原正项级数收敛。 Un1?an?1Un?13n?1?n?1?!nn133．解：∵???3???1，?n???，∴由比n?1nnUne3?n!?n?1??1??1???n?值判别法知，原级数发散。
4．解：考虑函数f?x??x?12?1??2?由f??x??0x??0,???，f?x???1?lnx?x，lnx，
?2?13?22得x?e，易知fe?时f?x?的最大值，所以当n?1地，n2lnn??1，∴
e32??2Un?n?12?lnn11?n，但?n为收敛的几何级数，∴原级数也收敛。 2n2n?121lnn5．解：an?nnx2?1?1?en2?1?1，∵n?2有0?lnnlnn?1；而当0?x?1时，2n?1有e?1?e?x，∴当n?2时，0?e是收敛的，∴原级数收敛。
6．解：因为已知级数???1?n?1?n?1n2?1?lnnlnn，而级九?2可判别其?1?e?2n?1n?1n?11 条件收敛的级数。设其部分和数Sn极
?2n?1?1n?111限为S，则有limSn?S，而级数?sin?1?0??0??0???，取其前
n??2357n?1n?2n项，其和与???1?n?1?n?11的部分和相等且为Sn，当n??时，2n??，故2n?1原级数收敛且和为S。
U7．解：n?1?Un
?n?2?n?12n?12n?1?n22，?n???，x?2x?2xnn?1?n2n?2?n?1???16
当2x2?1，即|x|?级数为?n?1?2222时，收敛；当|x|?时发散。故R?，当x??时，
2222?n?1?n??n?1???22?1?。 ?,发散，故原级数收敛域为???n?1?n?22?nnn11n11?n?8．解：an???1??1?????，由于1?1?|an|?1?????n，
n?2n?211?n?而当lim|an|?1，故R?1；当x??1时，原级数为????1??1?????，由
n??n??2n?1n?于通项不以零为极限，故发散。所以原级数的收敛域为??1,1?。
n?1n?1???1???1?2n9．解：当|x|?1时，级数?x收敛。设f?x???x2n，
n?1n?2n?1?n?1n?2n?1???1n?12n?1?|x|?1?，?|x|?1?，f???x??2???1?n?1x2n?1?22，则f??x??2???1?x，
1?xn?12n?1n?1??|x|?1?，两边积分得：
f??x??2?1dx?2a?tgx，(∵f??0??0)； 01?x2x再积分一次
f?x???2arctgxdx?2xarctgx?ln1?x2，(∵f?0??0)；
0x??∴?Un?limf?x??n?1x?1?0??2?ln2，即原级数的和S??2?ln2。
?x??????1de?1112??10．解：∵ex??xn，∴??1?x?x????? ????dx?x??x?2!n?0n!??????xx12n?1n?1n? ??1??????x???x???xn?1 ??2!3!?2!3!n!n?1?n?1?!??2?nxn?1n?n?1?|x|?0 因为当n??时，?n?1?n?1??n?1?n!d?ex?1?xex?ex?11?又当x?o时，??? 2??dx?x?2x故展开式对所有的x均成立，在展开式中令x?1，得
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nxex?ex?1 ???1。 2??n?1!xn?1x?1?Un?1?n?1?2x3?n?1??1n?1211．解：???2|x|3?2|x|3，(n??)，故nUnn23n?1n2x1n?12当2|x|?1，即当|x|?2时级数收敛，当x??2时级数发散，因此原级数收
1????166??2,2敛区间为???，且 ??3?16?16?n2x3n?1??2nx3n?1n?1?n2?n2??n?11x2??2n?13?n2??x??3nn?1???1?32?????2?x3?n?1????????n? ???n1??1?2x3?2x2?3?? ???2x???3??3?n?1?3?1?2x?1?2x3????21???6?，??|x|?2?。
??12．解：先求正弦级，将f?x?在???,0?作奇延拓，有an?0，
bn?2???0f?x?sinnxdx?2???20???x???sinnxdx
2?????????22????cosnx21??n??sinx??x????2cosnxdx????cos??2? ?0???n???2?n0n22nn0????? ?12n?2n??cos?2sin nn2n?2由狄里赫勒收敛定理知
?????fx,0?x?或?x???22?????bsinx?,x? ??n2n?1?2?0,x??,0??
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????∴f?x???bnsinnx，?0?x?,?x???
22??n?1?再求余弦级数，将f?x?在???,0?作偶延拓，有
?an?2???0f?x?cosnxdx?2??20???x???cosnxdx
2??2n?2 ?sin?2n2n?n????1?，n?0 ?cos2??an?2???20??3?x?dx??，bn?0 ??2?4?a0?∴f?x????ancosnx
2n?1?3n?2n?2????2?? ?????sin?2cos，cosnx0?x?,?x????? 282n?2n??22??n?1?n13．解：an?1???f?x?cosnxdx?0
?? bn?1???f?x?sinnxdx??????1?sinxdx??????101?0sinnxdx
1?cosnx?1?cosnx?1 ???1?cosn??cosn??1? ????????n?????n?0n??4,n?1,3,5,?2?n ? 1???1???n?n???0,n?2,4,6,?0???4?11?4?sin?2n?1?x所以f?x???sinnx?sin3x??? sin?2n?1?x???????32n?1?2n?1?n?1∵f1?x??a?ba?b?f?x? 22a?bb?a4?sin?2n?1?xa?b2?a?b??sin?2n?1?x∴f1?x?? ??????22?n?12n?1222n?1n?1f2?x???f?x?dx?0?x4???0??sin?2n?1?x???2n?1?dx ?n?1?x?4?1 ????cos?2n?1?x?
?n?1??2n?1?2?0
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?14?1 ?? ??cos2n?1x??23?n?1?2n?1??n?1?2n?1?4??4?1????x?时，f???代入上式有，??，即求得和式，且 222?n?1?2n?1??2?2?f2?x??
?2?1cos?2n?1?x，????x???。 ??n?1?2n?1?34? 20