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高等数学下(修订版)高纯一(复旦出版社)
习题十二答案详解
1.写出下列级数的一般项: (1)1?13?15?17??; (2)x2?x2?4?xxx22?4?6?2?4?6?8??; a3a5a7a9(3)3?5?7?9??;
解:(1)Un?12n?1; n2
(2)Uxn??2n?!!;
?1
(3)Un???1?n?1a2n2n?1; 2.求下列级数的和: ?(1)
?1?x?n?1??x?n??x?n?1?;
n?1(2)
???n?2?2n?1?n?;
n?1(3)
15?1152?53??; 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?n?12??x?x?1??1?x?1??x?2??1?x?1??x?2??1?x?2??x?3? ???1?x?n?1??x?n??1??x?n??x?n?1????1?12??x?x?1??1??x?n??x?n?1???因此limS1n??n?2x?x?1?,故级数的和为
12x?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Sn?15?1152???5n 1?1???1?n??5???5????1?15?1?4??1???1?n?5?????从而limS1n??n?4,即级数的和为14. 3.判定下列级数的敛散性: ?(1)
??n?1?n?;
n?1(2)
111?6?6?11?111?16???1?5n?4??5n?1???; (3) 23?2223n?12n33?33?????1?3n??; (4)
15?15?1135???n5??;
284
解:(1) Sn??2?1???3?2?????n?1?n?
?n?1?1从而limn??Sn???,故级数发散.
(2) S1?5??1?16?16?111?111?116???15n?4?1?n?5n?1??
?1?1?5??1?5n?1??从而limS1n?5,故原级数收敛,其和为15.
n??(3)此级数为q??23的等比级数,且|q|<1,故级数收敛.
(4)∵Un?1n5,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?n?1?(1) ?;
(2)
n?1n?cosnxn; n?12?(3)
???111n?1?3n?1?3n?2??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??1n?1???n?2?n?3???????n?p?2?1?1n?p?1???n?p?1n?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??ε>0,取N?11?, n?1n?1??1,则当n>N时,对任何自然数P恒有Un?1?Un?2???Un?p??成????????1?n?1立,由柯西审敛原理知,级数?收敛.
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???于是, ?ε>0(0<ε<1),?N=?log2??1?,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
Un?1?Un?2???Un?p111111????????????3?2n?13?2n?23?2n?3?3?n?1??13?n?1??23?n?1??3?11????3?n?1??13?2n?1n?6?n?1?1 ?12从而取?0?1,则对任意的n∈N,都存在P=n所得Un?1?Un?2???Un?p??0,由柯12西审敛原理知,原级数发散.
5.用比较审敛法判别下列级数的敛散性. (1)
111??????;
????4?65?7n?3n?5(2)1??1?21?31?n?????? 2221?21?31?n(4)
π(3)?sinn;
3n?11(5)?nn?11?a??n?1??12?n1n3;
?a?0?;
(6)
??2n?1?1?.
解:(1)∵ Un??11?2
?n?3??n?5?n?1而?2收敛,由比较审敛法知?Un收敛. n?1nn?1(2)∵Un??1?n1?n1?? 1?n2n?n2n而
1发散,由比较审敛法知,原级数发散. ?n?1nππsinnn33(3)∵lim?limπ??π
n??n??1π3n3nsin?ππ而?n收敛,故?sinn也收敛.
3n?13n?1? 287