第十一章 无穷级数（答案）

1lnn5．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当2x2?1，即|x|?级数为?n?1??n?2?n?12n?12n?1?n22，?n???，x?2x?2xnn?1?n2n?2?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，??1?x，则f??x??2?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

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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?x?2?ln2。

????1d?e?1?1?12???10．解：∵ex??xn，∴??1?x?x???? ?x??x?n!dx2!n?0???????2???xx12n?1n?1nn?1? ?? 1??????x???x???x??2!3!?2!3!n!n?1?n?1?!???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

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nUnnn22x3n?11n?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，

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bn?2???0f?x?sinnxdx?2???20????x??sinnxdx

2?????????22????cosnx21??n??sinx??x????2cosnxdx????cos??2? ?0n???n???2?n022nn0???? ?1n?2ncosn?2?2n?n2?sin2 由狄里赫勒收敛定理知

??f?x?,0?x???2或2?x?????bnsinx????,x?? n?1?22??0,x??,0??∴f?x???b??0?x??,??x??nsinnx，n?1?22???

再求余弦级数，将f?x?在???,0?作偶延拓，有

2?a?n????0f?x?cosnxdx?2??2???0?x?2??cosnxdx

?2sinn?2???cosn?n2?n2?2?1???，n?0 an?2?2???x???3??02??dx?4?，bn?0

f?x??a?∴02??ancosnx

n?1 ?38??????2sinn??2n?2?2cos，n?1?n2n?2n2???cosnx13．解：a?n?1????f?x?cosnxdx?0

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????0?x??2,??2?x???? bn?1???f?x?sinnxdx??????1?sinxdx??????101?0sinnxdx

1?cosnx?1?cosnx?1 ???1?cosn??cosn??1? ???????n??nn??????0?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?12n?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?0x4???0??sin?2n?1?x???2n?1?dx ?n?1?x??1?? ????cos2n?1x? 2?n?1??2n?1??04??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 22222???2n?1??n?1?f2?x??

?2?1cos?2n?1?x，????x???。 ?3?n?1?2n?1?4? 16