我愛學(xué)習(xí)數(shù)學(xué)無窮級(jí)數(shù)是研究函數(shù)工具

￥￥

￥若原級(jí)數(shù)收斂,但取絕對(duì)值以后的級(jí)數(shù)發(fā)散,￥￥￥

(-1)n-

n￥(

均為絕對(duì)收斂

(n-1)!

￥若級(jí)數(shù) ￥設(shè)

令vn2(unun￥則0￡vn

un

由比較判別法知vn收斂un

un,2vn ￥

=

2 ￥unun ￥sin

n(1)

n

sin ￥證:

￡n4

,

￥\￥

sin n4因此sin

n4(2)

=un =lim

=

(nnfi￥

nfi

1n+12 =

=<1nfi￥e n

n\

收斂,因此(-

定

=|an| =|an|- 則an絕對(duì)收斂 un與vn均收斂． “”由于0￡un￡|an| 0￡vn￡|an|由比較判別法可知：|an|收斂時(shí)un與vn“”由于|an|=un+vn 而un與vn均收斂所以|an|收斂 即an絕對(duì)收斂． 對(duì)于絕對(duì)收斂級(jí)數(shù)a an=un-

*定理.絕對(duì)收斂級(jí)數(shù)不因改變項(xiàng)的位置而改變其和P412定理13 設(shè)級(jí)數(shù)un vn都絕對(duì)收斂,其和分別為S 則對(duì)所有乘積uivj按任意順序排列得到的級(jí)數(shù)也絕對(duì)收斂,Ss(P414定理14)

u1v1+u1v2+u2v2+u2v1+u1v3+u2v3+u3v3+u3v2+u3v1+

u1v1+u1v2+u2v1+u1v3+u2v2+u3v1+un0n=1,2

u1-u2+u3-+(-1)n-1un稱為定理10Leibnitz)若交錯(cuò)級(jí)數(shù)滿足條件un?

(n=1,2,);limun=0,￥nfi￥,,其余項(xiàng)滿 rn￡un+1

2) 證明思路：lim證明思路：limS2nS,limS2n-1SlimSnnfinfi1o先證部分和數(shù)列S2n單調(diào)增加且有上界 S2n=(u1-u2)+(u3-u4)++(u2n-1-u2n0￡un=S2n-2+(u2n-1-u2n)?0￡un+{S2n}\limS2n=S￡nfi2o再證limS2n-1nfilimS2n+1limS2nu2n+1limS2nnfi nfi nfi\limSn=Snfi故級(jí)數(shù)收斂于S,且S￡u1Sn的余項(xiàng)rn=S-Sn=–(un+1

-un+

+ =un+1-un+2+￡un+1

注 un+1￡￥

(n?N2{un不單調(diào)

(-1)n-1

(un0)發(fā)散( ( u=2+(-1)n> =

= < =3

= > =

uu=2+(-n n-12+(- nn=13{un}￥

=

)n-12

2n (-1)n-1

(un0)發(fā)散

(limun? nfi1-1+1-1++

n+1 1-1+1-1

=n=10n

= n+1 1-

+ -

+ 1)1

1 3) n

￥￥￥￥

￥￥概念:設(shè)un 若un收斂,稱un 絕對(duì)收 若un發(fā)散 稱un條件收

limS

nnfin

limun?nfi

判

收斂

￥

￥u判

un

un?un+1>limun=nfi

￥￥pp-級(jí)數(shù):n=1￥1p判別級(jí)數(shù)的斂散性 ln(n+1)

nn

解:(1)ln(n+1n,￥

>ln(n n(2)nfi￥

= =1n1nnn1nn1nn,設(shè)un?0(n=1,2, 且 n=1,則級(jí) nfi￥ )(

(D)收斂性根據(jù)條件不能確定分析:由 n=1