高等数学上（修订版）（复旦出版社）
习题六 无穷数级 答案详解
1．写出下列级数的一般项： (1)1?1?1?1357??；
(2)x2?x2?4?xx2?4?6?x22?4?6?8??；
(3)a33?a55?a77?a99??；
解：(1)U1n?2n?1； n (2)Ux2n??2n?!!；
(3)?1a2n?1Un???1?n2n?1； 2．求下列级数的和： (1)??1n?1?x?n?1??x?n??x?n?1?；
(2)
???n?2?2n?1?n?；
n?1(3)15?152?153??； 解：(1)u1n??x?n?1??x?n??x?n?1?
?1?112???x?n?1??x?n????x?n??x?n?1??? 283
从而S1?1111n?2??x?x?1???x?1??x?2???x?1??x?2???x?2??x?3?
???11??x?n?1??x?n???x?n??x?n?1????1?112??x?x?1????x?n??x?n?1???因此lim11n??Sn?2x?x?1?，故级数的和为
2x?x?1?
(2)因为Un??n?2?n?1???n?1?n? 从而Sn??3?2???2?1???4?3???3?2???5?4???4?3?????n?2?n?1???n?1?n??n?2?n?1?1?2?1n?2?n?1?1?2所以limn??Sn?1?2，即级数的和为1?2． (3)因为S111n?5?52???5n 1?1???1?n???5???5???1?15?1?4??1???1?n?5?????从而limn??S?1，即级数的和为1n44． 3．判定下列级数的敛散性： (1) ???n?1?n?；
n?1(2)
11?6?16?11?111?16???1?5n?4??5n?1???； (3) 22223n3?n?1233?33?????1?3n??； 284
(4)1?155?1135???n5??； 解：(1) Sn??2?1???3?2?????n?1?n?
?n?1?1从而limn??Sn???，故级数发散． (2) S11111111?n??5??1?6?6?11?11?16???5n?4?5n?1?? ?1?1?5??1?5n?1??从而limS11n??n?5，故原级数收敛，其和为5． (3)此级数为q??23的等比级数，且|q|<1，故级数收敛．(4)∵U1n?n5，而limn??Un?1?0，故级数发散． 4．利用柯西审敛原理判别下列级数的敛散性：
(1) ????1?n?1；
(2)
??cosnxn?1nn?12n； (3)
????11n?1?3n?1?3n?2?1?3n?3??． 解：(1)当P为偶数时，
Un?1?Un?2???Un?p??1?n?2??1?n?3??1?n?4??1?n?p?1?n?1?n?2?n?3???n?p?1n?1?1n?2?1n?3???1n?p?1?11??11?1n?1???n?2?n?3???????n?p?2?n?p?1???n?p?1
n?1当P为奇数时，
285
Un?1?Un?2???Un?p??1?n?2??1?n?3??1?n?4??1?n?p?1??????n?1n?2n?3n?p??1111?????n?1n?2n?3n?p11?1?1??1?????????n?p?1n?p?n?1?n?2n?3???1?n?1
因而，对于任何自然数P，都有
Un?1?Un?2???Un?p?11?， n?1n1??ε>0，取N???1，则当n>N时，对任何自然数P恒有?????Un?1?Un?2???Un?p??1?n?1收敛． ??成立，由柯西审敛原理知，级数?nn?1?(2)对于任意自然数P，都有
Un?1?Un?2???Un?p??cos?n?p?xcos?n?1?xcos?n?2?x????2n?12n?22n?p111????2n?12n?22n?p1?1?1???2n?1?2p?11? 21?1?1???2n?2p?12n????1?log于是， ?ε>0(0<ε<1)，?N=??2?，当n>N时，对任意的自然数P??都有Un?1?Un?2???Un?p??成立，由柯西审敛原理知，该级数收敛． (3)取P=n，则
286
