第2節(jié)收斂數(shù)列的性質(zhì)

證limxna,又limxn證 0,N1,N2.使當(dāng)nN時(shí)恒有

當(dāng)nN時(shí)恒有

b 取NmaxN1,N 則當(dāng)nN時(shí)有ab(xnb)(xn xnbxna2上式僅當(dāng)ab時(shí)才能成立.故極限唯一

anM,n 數(shù)列

數(shù)列{2n證設(shè)lim 取則N使得當(dāng)nN時(shí)恒有xna1xna

M

x1,,

,a

a則對(duì)一切自然數(shù)n,

M

(1o設(shè)lim a,且a,則N,n nN時(shí),有an 設(shè)lim a,lim b,且ab,則n n 當(dāng)nN時(shí),anbn; 設(shè)lim a,lim b,若N,當(dāng)nN時(shí)n n anbn則有a (1)取a,N,當(dāng)nN,|

aa 同理，取a,N,當(dāng)nN,a 取NmaxN1N2當(dāng)nN,an(2)令ba,2N,當(dāng)nN時(shí),| aab N,當(dāng)nN時(shí),|bb|,即bbab 取NmaxN1N2當(dāng)nN,由上得anbn. (3)中即使有anbn,也可有a

limanalimbnb, lim[anbn]an 證(1)由絕對(duì)值的三角不等式可得|anbnab||anbnabnabnab|ana||bn||a||bnb|

,n

,|

a| 2(M1)N2

,n

,|

b| 2(a1)取NmaxN1N2當(dāng)nN,得|anbnab|lim1n 對(duì)于|b|0,Ns.t當(dāng)nN時(shí) |bb||b

且此時(shí)| ||b| 所以當(dāng)nN1時(shí)|11

|bnb

2| b|

b2 由于limbnb對(duì)0,N2s.tnN2時(shí)有|bnb|2|11|2|bb| 即證得,lim11 n

求

2n23n4n5n24n23 原式

5 lim2lim3lim n n lim5lim4lim n n例2設(shè)|q|1計(jì)算極限lim(1qq2qn1 lim(1qq2...qn1

n1

n1 limqn

定理2.5：若數(shù)列{an},{bn},{cn}anbncnn1,2,3,,且limanlimcn, limanlimbn 設(shè)limanlimcna, 0,N10,N20,使得

an

NmaxN1N2即aana,nN時(shí),恒有

acnaaanbncna 即ba lim n例3求n

n2

n2

n2n解 n2nn

n2

n2 n2

n2

1nn

n2

1

n2

n2

n2

)1limann

1對(duì)a1成立再設(shè)a(01這時(shí)a111lim 11.1 1n

lim

1 證設(shè)a證設(shè)a0求證liman先設(shè)a1,當(dāng)na時(shí)11a1由于limnn1n由1na 設(shè)0a1a2aklimnananan

nannananannkan limnananan

a,求證lima1a2

a1a2ann(a1a)(a2a)(ann12L12L

n

則 0，上式變0,NN*,nN 112Nn|12N|(nN) 112N

N*使nN

|12N|

1212 證明若li（a1a2an）（a12a2nan）證明：saa

a則lims

由例

i （a12a2nan）sn（a2(n1) sn(sns1)(sns2)(snsn1nnsn(s1s2sn) (s1L sn }n nk定理2.6如果數(shù)列{ana，那么它的任一子數(shù)列也收斂于a.n證n證k

}xn}limxn

存在著正整數(shù)N 當(dāng)nN時(shí)|xna| 取KN 當(dāng)kK時(shí) nKnNN于是｜xna| 證得lim 這個(gè)數(shù)列稱為無(wú)窮小列,簡(jiǎn)稱無(wú)窮小.定理

1o{a}為無(wú)窮小的充要