高等数学第十章习题详细解答答案

更新时间:2024-03-02 14:04:01 阅读量: 综合文库 文档下载

说明:文章内容仅供预览,部分内容可能不全。下载后的文档,内容与下面显示的完全一致。下载之前请确认下面内容是否您想要的,是否完整无缺。

习题10.1

1. 写出下列级数的前五项:

?1?3??(2n?1)n(1)?; (2)?; 22?4??(2n)(2?n)n?1n?1?n!(?1)n?1(3)?; (4)?. nn?1(n?1)n?110n12345解 (1)2?2?2?2?2??

3456711?31?3?51?3?5?71?3?5?7?9????? (2) ?22?42?4?62?4?6?82?4?6?8?1011111?????? (3)

10203040501!2!3!4!5!(4)1?2?3?4?5??.

23456??2. 写出下列级数的一般项: (1)

111????; 2461aa2a3?????; (2)

1?53?75?97?1135791113????; (3) ????149162536xxxxx2????? (x?0). (4) 22?42?4?62?4?6?8解(1)因为

1111111??,因此一般项un?,?,? 21?242?263?22n1a0aa1 (2) 因为 , ??1?5(2?1?1)?(2?1?3)3?7(2?2?1)?(2?2?3)a2a2an?1 ??因此一般项un?5?9(2?3?1)?(2?3?3)(2n?1)(2n?3)(3) 因为

3(2?1?1)572(2?2?1)3(2?3?1)?(?1)1??(?1)? ,?(?1),11?1492232n(2n?1)因此一般项un?(?1)

n2?(4)因为

xxxxxxx???, ?,,

21?22?42?42?4?62?4?6122232因此一般项un?xxx?n?n.

2?4?6??(2n)2(1?2?3??n)2n!1;

n?1(2n?1)(2n?1)?n2n2n23. 判定下列级数的敛散性:

(1)?(n?1?n); (2)?n?1?111??????; 1?22?3n(n?1)π2πnπ???sin??; (4)sin?sin666(3)

(5)?(n?2?2n?1?n); (6)

n?1?1111??3?4??; 3333(7)(?)?(111111?)????(?)??;

3232223n2n13572n?1??; (8)??????35792n?1(9)?(2n?1a?2n?1a) (a?0);

n?1?11解(1)因为

1?(10)

1?111??????. 12131n(1?)(1?)(1?)23nSn?(2?1)?(3?2)?(4?3)???(n?1?n)?n?1?1 当n??时,Sn??,故级数发散. (2)因为

1111?(?)

(2n?1)(2n?1)22n?12n?1Sn?1111????? 1?33?55?7(2n?1)(2n?1)111111?[(1?)?(?)??(?)] 23352n?12n?111?[1?], 22n?11当n??时,Sn?,故级数收敛.

2

本文来源:https://www.bwwdw.com/article/01fa.html

Top