数列极限收敛准则(单调有界收敛准则)

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1、§2数列极限§2.1数列极限的定义§2.2收敛数列的性质§2.3极限存在准则§2.3数列极限存在的准则夹逼准则单调有界准则23数列及子数列1n1lim1e收敛准则nn定理5（单调有界准则）单调有界数列必有极限limaa(M)单调递增有上界Mn{a}nn单调递减有下界Nlimana(N)na1a2anan1MlimanaMn(单调递增有上界)aaaaaM12nn1limabNnn(单调递减有下界)b定理5（单调有界准则）单调有界数

2、列必有极限limaa(M)单调递增有上界Mn{a}nn单调递减有下界Nlimaa(N)nn证:设数列{a}递增有上界M,aaaM,n12n由确界原理知必存在上确界设为a,故aaM.n对0,N0,使aNa,当n>N时,有aaNanaa,即ana.则limaaMnna1aaanaMN二项式公式（其中n为正整数）特别地均值不等式几何平均算术平均例8.求证:数列的极限存在.证:利用单调有界定理证.先证数列递增:由均值不等式,

3、n11n111(1)n1(1)(1)(1)1nnnnn(11)1n11n1n1a(11)n(11)n1nnn1an1即aa(n1,2,),则数列递增.nn1再证数列有上界:利用二项式公式,有a(11)n1n1n(n1)1n(n1)(n2)1nn1!n2!23!3nnn(n1)(nn1)1!nnn11111(n2)2!3!n!1111112n111222232n11121

4、123即a3则数列有上界.2n1n数列单调递增有上界,limana3.n1nlim(11)1(常数)lim(1)ennnn其中e为无理数,其值为e2.7182818284590451n3例9.lim(1)nn1n13lim[(1)(1)]e1ennnlim(11)nlim11nnn1ne(1)nlim(11)n1n313lim[(1)(1)]enn3nn3n3k例10.利用单调有界

5、准则求证:limn0(a1,k常数)nnak(n1)knkk证:设xn,x(1)n1(11)kx1nann1an1nkananna11kx(1)x(*)递推公式n1nanxn111k1limlim(1)1,由不等式性质,nxnananxN0,当nN时,n11,即xn1xn,又x0，00nxn故xn递减有下界,则数列收敛.设limxncnk例10.利用单调有界准则求证:limn0(a1,k常数)nna证

6、:11kxn1a(1n)xn(*)设limxncn(*)式两边取极限,有1cc即(a1)c0c0akn则lim0(a1,k常数)nna常用极限11nlim0k0lim(1)ennknnn1limq0q1lim(1)1(常数)nnnklimna1a0limn0(a1,k常数)nnnanalimnn1lim0a0nn!n例11已知a11,an112an(n1,2,)求li

7、man时,下述作法是否正确?不正确!n设limaa,由递推式两边取极限得nna12aa1此处a:1,3,7,15,31,发散！n能假设极限式成立的前提：已证明极限存在!利用单调有界准则求极限的方法证明极限存在{a}递增有上界M(nN){a}收敛n递减有下界m0nan11(1)或aa0(0)数学归纳法an1nn求极限limaann建立数列的递推关系:an1f(an),两边取极限：liman1f(liman)af(a)解得a.nn32x

8、1n1例12x02,xn2(n1,2,),证明数列3xn1收敛,并求极限(P28例20)证:x1,n且0,故x递减且有下界，n则数列收敛.设limxna1,递推式两边取极限,有n32a13a即a1,a1,于是23a设x1(kxa)(n0,1,2,),a0,k正整数,x00,n1nkk1xn求limxn.(P339(2))(a1,k2，x02)n例13c0,xnccc证明数列收敛,