[优质文档]第一讲__数列的极限典型例题
第一讲 数列的极限
一、内容提要
1.数列极限的定义
limx,a,,,,0,,N,,,,n,N,有. x,a,,nn,,n
注1 的双重性.一方面,正数具有绝对的任意性,这样才能有 ,,
无限趋近于 a,x,a,,(n,N),,xnn
另一方面,正数又具有相对的固定性,从而使不等式.还
表
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明数列无限趋近x,a,,,,x,nn于的渐近过程的不同程度,进而能估算趋近于的近似程度. ,,xaan
limxNN注2 若存在,则对于每一个正数,总存在一正整数与之对应,但这种不是,n,,n
NN唯一的,若满足定义中的要求,则取,作为定义中的新的一个也必须N,1,N,2,?
NN满足极限定义中的要求,故若存在一个则必存在无穷多个正整数可作为定义中的(
注3 (n,,)的几何意义是:对的预先给定的任意邻域,在中,,x,ax,,aU(a,,)nn至多除去有限项,其余的无穷多项将全部进入( U(a,,)
limx,a,,,,0,,N,,,,n,N注4 ,有x,a,,.00nn00,,n
2. 子列的定义
,,,,在数列x中,保持原来次序自左往右任意选取无穷多个项所得的数列称为x的子列,nn
k,,xx记为,其中表示在原数列中的项数,表示它在子列中的项数(nnnkkk
kn,k注1 对每一个,有( k
h,kh,kn,nn,n注2 对任意两个正整数,如果,则(反之,若,则(h,khkhk
limx,a,,,,0,,K,,,,k,Kx,a,,注3 ,有. nnkk,,n
limx,a,,,,,xxa注4 的任一子列收敛于. nnnk,,n
3.数列有界
,M,0,n,Nx,M,,,,xx对数列,若,使得对,有,则称数列为有界数列(nnn
4.无穷大量
,G,0对数列,如果,,有,则称为无穷大量,记x,G,,,,xx,N,,,,n,Nnnnlimx,,作( nn,,
limx注1 只是一个记号,不是确切的数(当为无穷大量时,数列是发散的,即,,,,xx,nnnn,,不存在(
limx,,注2 若,则无界,反之不真( ,,xnnn,,
注3 设与为同号无穷大量,则为无穷大量( ,,,,,,xyx,ynnnn
注4 设为无穷大量,有界,则为无穷大量( ,,,,,,xyx,ynnnn
,,,0,n,N注5 设为无穷大量,对数列,若,使得对,有,y,,,,,,xy,N,,,nnn
则为无穷大量(特别的,若,则为无穷大量(,,,,xyy,a,0xynnnnn
5.无穷小量
limx,0若,则称为无穷小量( ,,xnnn,,
limx,0limxy,0注1 若,有界,则( ,,ynnnnn,,n,,
1limx,,limx,0,n,N注2 若,则;若,且使得对,,x,0lim,0,N,,,nnnn,,n,,n,,xn
1lim则( ,,n,,xn
6.收敛数列的性质
,,,,xx(1)若收敛,则必有界,反之不真( nn
,,x(2)若收敛,则极限必唯一( n
limx,alimy,ba,b,N,,n,Nx,y(3)若,,且,则,使得当时,有(nnnn,,n,,n
注 这条性质称为“保号性”,在理论分析论证中应用极普遍(
limx,alimy,b,N,,a,bn,Nx,y(4)若,,且,使得当时,有,则(nnnn,,n,,n
注 这条性质在一些参考书中称为“保不等号(式)性”(
,,,,,,,,xyx,yx,y(5)若数列、皆收敛,则它们和、差、积、商所构成的数列,,nnnnnn
,,xnlimy,0,()也收敛,且有 ,,xy,,nnnn,,yn,,
,,limx,y,limylimx,,,,,,,,,,,,,,,,,,, , nnnnn,,n,,,,n
limx,y,limx,limy,,,,,,,,,,,,,,,,,, , nnnnn,,n,,n,,
limxnxn,,nlimy,0,,,,,,,,,,,,,,,,,,()( lim,nn,,n,,limyynnn,,
7. 迫敛性(夹逼定理)
,limz,alimx,a,N,,limyn,N若,使得当时,有,且,则(y,x,znnnnnnn,,,,,,nn
8. 单调有界定理
单调递增有上界数列必收敛,单调递减有下界数列必收敛( ,,,,xxnn
9. Cauchy收敛准则
数列收敛的充要条件是:,有x,x,,(,,x,,,0,,N,,,,n,m,Nnmn
注 Cauchy收敛准则是判断数列敛散性的重要理论依据(尽管没有提供计算极限的
方法
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,但它的长处也在于此――在论证极限问题时不需要事先知道极限值( 10.Bolzano Weierstrass定理
有界数列必有收敛子列(
n1,,11. lim1,,e,2.7182818284?,,,,nn,,
12.几个重要不等式
22a,b,2ab, sinx , 1. sinx , x .(1) (2) 算术,几何,调和平均不等式:
, 对 记 ,a,a,?,a,R,12n
na,a,?,a112n (算术平均值)M(a), , a,,iinn,1i
1nn,,nG(a),aa?a,a,,, (几何平均值),i12ni,,,,,i1
nn1Ha(),,,. (调和平均值)inn111111?,,,,,aaanaa,1,112iinii
有均值不等式: 等号当且仅当时成立.H(a) , G(a) , M(a),a,a,?,aiii12n
(3) Bernoulli 不等式: (在中学已用
数学
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归纳法证明过)
对 由二项展开式 ,,x0,
nnnnn(1)(1)(2),,,nn23 (1)1,,,,,,,,xnxxxx?2!3!
n ,(1,x),1,nx,(n,1)
(,)Cauchy,Schwarz 不等式: (),有 ,a,bk,1,2,?,nkk
22nnnn,,,,22,,,,,,,,, ab,ab,ab,,,,,,,,kkkkkkk,1k,1k,1k,1,,,,
111,n,N,ln(1,),(,), n,1nn
13. O. Stolz公式
二、典型例题
,,NG,N1(用“”“”证明数列的极限((必须掌握)
例, 用定义证明下列各式:
2nn3,5,1lim,1(,); 2n,,nn3,,6
limx,alimx,a(,)设x,0,,则;(97,北大,10分)nnn,,n,,n
nlnlim,0(,) (,,0),n,,n
,,,0证明:(,),欲使不等式
23n,5n,16n,56n6n6,1,,,,,,,,,,,,,,, 2222n3n,n,63n,n,63n,nn
66,,,0n,Nn,N,[],1成立,只须,于是,,取,当时,有,,
23n,5n,16,1,,,,,,,,,,,, 2n3n,n,6
2nn3,5,1即 ( lim,12n,,nn3,,6
limx,a(,)由,,知,有,则x,0x,a,a,,,,0,,N,,,,n,Nnnn,,n
x,ax,ann x,a,,,,nx,aan
x,an于是,,有, x,a,,,,0,,N,,,,n,N,,na
limx,a即,,,,,,,,,,,,,,,,,, ( n,,n
n,lnn(,)已知,因为
,,,,,,,,,,,,2222,,,,,22lnn,1n,122,,,,lnn,2[n],,,,2,,lnn44n,,,,,,,,,,,,,,,0,,,,,,,,,,,,nnnnnn2,n
2
,44lnnnln,,,,,,,0,所以,,欲使不等式成立,只须(n,,0,,,,,,n,,n,,2n,
2,,,4,,,,,1,,,0N,n,N,,,于是,,取,当时,有 ,,,,,,,,,,
4lnnnln,,,,,,,,,,,,, ,0,,,,nn2n,
nlnlim,0即,,,,,,,,,,,,( ,n,,n
x,ax,a,g(n),,评注, 本例中,我们均将做了适当的变形,使得,从而从解不nn
Nx,ag(n),,g(n)n,,等式中求出定义中的(将放大时要注意两点:?应满足当n
g(n),0g(n),,g(n)g(n),,时,(这是因为要使,必须能够任意小;?不等式容易
求解(
,,,0(n,,)x,aN(,)评注, 用定义证明,对,只要找到一个自然数,使得当n
时,有即可(关键证明的存在性( x,a,,n,N(,)N(,),,n
评注, 在第二小题中,用到了数列极限定义的等价命题,即:
(,),有(为任一正常数).Mx,a,M,,,,0,,N,,,,n,Nn
k(,),有. x,a,,,,,0,,N,,,,n,N(k,N)n
例, 用定义证明下列各式:
nlimn,1(,);(92,南开,10分) n,,
kn(,) lim,0(a,1,k,N)nn,,a
nnn,1,,0n,1n,1,,证明:(,)(方法一)由于(),可令(),则
n(1),(1)nnnn,n2n2n(1)1n,2,,,,,,,,,,,,?,,,,()nnn22
nn,2当时,n,1,,有 2
22(1)nn,nn222n,,,,(n,1),,,,,,,,, n,244
2nn即,,,,,, ( 0,,1,
n
42nn,,,0n,n,1,,欲使不等式成立,只须(n,,,,12,n
,,4,,,,,0n,N,max,1,2于是,,取N,当时,有 ,,2,,,,,,,
2nn,1,,,,
n
nlimn,1即,,,,,,,,,,,,( n,,
(方法二)因为
n,2个1,,,,,nnnn,,,,,,1?1222nnnnn,,,,,,,,,,,1(11?1)1,nnn
2n所以, n,1,
n
42nn,,,0,欲使不等式成立,只须(n,1,n,n,,,,12,n
4,,,,,0n,N于是,,取,当时,有 N,,12,,,,,
2n, n,1,,,
n
nlimn,1即,,,,,,,,,,,,( n,,
k,1a,1a,1,,,,0(,)当时,由于,可记(),则
(1),(1)nnnn,nn2n2(1)1n,2,,,,,,,,,?,,,,()an22
nn,2n,1,当时,,于是有 2
nn40,, ( ,n2nn,(1)an,2,
2
n44n,,,0,,,,n, ,欲使不等式 成立,只须(,0n22nan,,,a
,,4,,,,,0n,N,max,1,2 对,取,当时,有 N,,2,,,,,,,,
n4n,,,, ( ,0n2nan,a
k,,1knn,,kk,1a,1a,1,当时,(),而( 1n,,ank,,(a),,
n0则由以上证明知,,,0,,N,,,,n,N,有,,,,即1nk(a)
knk0,,, , na
kn故 ( lim,0nn,,a
,,,0N评注, 在本例中,,要从不等式中解得非常困难(根据的特征,x,a,,xnn
,,利用二项式定理展开较容易(要注意,在这两个小题中,一个是变量,一个是定值(
评注, 从第一小题的方法二可看出算术,几何平均不等式的妙处(
评注, 第二小题的证明用了从特殊到一般的证法(
naa,0例 用定义证明:()(山东大学) lim,0,,nn!
0,a,1证明:当时,结论显然成立(
,,naaaaaaaaaa,1当时,欲使0??成立,,,,,,,,,,,,,,,,,,,!121!naa,nan
,,a,1,,a,1,,aa,,,0N,n,N只须(于是,取,当时,有,1n,,,,,a,!,,a!,,,
n,,aaaa,,,,,0 ,,n!a!n
nalim,0即,,,,,,,,,,,,,,,,,,,,, ( ,,nn!
,,lim[(n,1),n],0,,1,,N例 设,用“”语言,证明:( n,,
,,0证明:当时,结论恒成立(
0,,,1,,,0当时,,欲使
111,,,,,nnn(,1),,0,[(1,),1],,,,,,n(11) 1,,nnn
,,11,,,1,,,0N,n,N只须n,(于是,取,当时,有11,,1,,1,,,,,,
1,,,,(n,1),n,0, 1,,n
,,lim[(n,1),n],0即,,,,,,,,,,,,,,,,,,,,, ( n,,
2.迫敛性(夹逼定理)
项和问题可用夹逼定理、定积分、级数来做,通项有递增或递减趋势时考虑夹逼定理(n
,,有界,但不能说明有极限(使用夹逼定理时,y,x,zy,bz,c,{x}xnnnnnnn要求趋于同一个数( y,znn
na例 求证:(为常数)( lim,0a,,nn!
naaaaaaa分析:,因为固定常数,必存在正整数,使,,,,?,,,?,amnmmn!123,1
aaaam,a,m,1,因此,自开始,,, ,且n,,,1,1?,,1m,1m,2m,1n
a时,( ,0n
n,m,1a,m,1证明:对于固定的,必存在正整数,使,当时,有am
nmaaaaaaaaa, 0,,,,,?,,,?,,,mn!n!123mm,1n
mnaaalim由于,由夹逼定理得, ,,0lim,0n,,,,nnm!n!
nalim,0即 ( ,,nn!
评注 当极限不易直接求出时,可将求极限的变量作适当的放大或缩小,使放大、缩小所得
的新变量易于求极限,且二者极限值相同,直接由夹逼定理得出结果(
aana,2,,?12n{a}例 若是正数数列,且,则 lim,0n,,nn
nlimn,a,?,a,0 ( n1n,,
aana,2,?,12nn,,,,,,1a,2a,?,na证明:由,知 ,12nn
aana,2,?,12nnn n!,a,a,?,a,12nn
a2ana1,,?,n12n即 (a,a,?,a,,12nnnn!
a2ana1,,?,n12n于是,,而由已知0,na,a,?,a,,12nnnn!
aana,2,,?112nlim及 ,0lim,0nn,,,,nn!n
aa?na,2,,1n12lim故 ,,0nn,,nn!
nlimn,a,?,a,0由夹逼定理得 ( n1n,,
评注1 极限四则运算性质普遍被应用,值得注意的是这些性质成立的条件,即参加运算各
变量的极限存在,且在商的运算中,分母极限不为0(
评注2 对一些基本结果能够熟练和灵活应用(例如:
1nlimq,0a,0lim,0q,1(1)() (2)()an,,n,,n
nnlima,1a,0limn,1(3)() (4) n,,n,,
na1a,0limlim,0(5)() (6) ,0nn,,,,nn!n!
limx,a例 证明:若(有限或),则 a,,n,,n
x,x,,x?12nlim,a(有限或)( a,,,,nn
,x,a,limx,a证明:(,)设a为有限,因为,则,有.,,,0,,N,,,,n,Nnn11,,n2
xaxaxax,x,,x,,,,,,?,,,,?,,12n12n,a,于是 nn
xaxax,a,x,a,,x,a?,,?,,12NN,1n11,, nn
n,NAA,1,,,,( ,2nnn
其中为非负数( A,x,a,x,a,?,x,a12N1
,AA,因为,故对上述的,有(lim,0,,0,,N,,,,n,N22n,,n2n
n,N取当时,有 N,max{N,N}12
x,x,,x?,,12n ,a,,,,n22
x,x,,x?12nlim,a即,,,,,,,,,,,,,,,,,,,,, ( ,,nn
limx,,,(,)设,因为,则,有,x,2Ga,,,,G,0,,N,,,,n,Nnn11n,,
(于是 且x,x,?,x,012N1
xxx,x,,x?,?,x,x,,x?12NN,1n12n11,, nnn
x,,x?Gn,NN2()2N,1n111,,,G,G,,,,,,,,,,,,,,, 2nnn
2N1n,NG,G取,当时,,于是 N,2N1n
x,x,,x?12n,2G,G,G( n
xxx,,,?12nlim,,,即,,,,,,,,,,,,,,,,,, ,,nn
(,)a,,,时证法与(,)类似( 评注, 这一结论也称Cauchy第一定理,是一个有用的结果,应用它可计算一些极限,例
如:
111,,,?1n2lim,0(,)lim,0(已知); n,,n,,nn
3nn1,2,3,,?nlimn,1lim,1(,)(已知)( n,,,,nn
评注, 此结论是充分的,而非必要的,但若条件加强为“为单调数列”,则由{x}n
x,x,,x?12nlimx,alim,a可推出( n,,n,,nn
评注, 证明一个变量能够任意小,将它放大后,分成有限项,然后证明它的每一项都能任
意小,这种“拆分方法”是证明某些极限问题的一个常用方法,例如:
lima,a0,,,1若,(为有限数),证明: an,,n
a2n,,,lim(,,,,),( ?aaaa,1,20nnn,,n1,,
2n分析:令,则 x,a,,a,,a,?,,a,1,20nnnn
21nn,((1,,)x,a,,(a,a),,(a,a),?,,(a,a),,a121010nnn,nn,n,
2n只须证,,,,,,()n,,,(a,a),,(a,a),?,,(a,a),0,1,2,101nnnn
lima,a由于,故,有a,a,,(于是 ,N,,,,n,Nnnn,1,,n
2n ,(a,a),,(a,a),?,,(a,a),1,2,101nnnn
2,1NNn,,a,a,,a,a,?,,a,a,,a,a,?,,a,a,1,2,1,,1,,,,101nnnnnNnNnNnN
n0,,,1lim,,0再利用()即得( n,,
例 求下列各式的极限:
n12,,,lim(?)(,) 222n,,n,n,n,n,n,n,n12
11n?lim1,,,(,) n,,2n
1,3,5,?,(2n,1)nlim(,) ,,n2,4,6,?,2n
12?121,2,?,n,,,nn?,,,,,解:(,)22222n,n,112n,n,nn,n,n,n,n,n,n
(1)nn,
,,,n112?2,limlim?, ,22n,,n,,n,n,n2n,n,n
(1)nn,
1,2,?,n12,limlim, ,22n,,n,,n,n,121n,n,
由夹逼定理,
12n1lim()? ,,?,,222n,,2nn1nn2nnn,,,,,,
11nnn1,1,,?,,1,1,?,1,n(,) 2n
nlimn,1?,由夹逼定理, n,,
11nlim1,,?,,1?( n,,2n
1352n,111,3,5,?,(2n,1)132n,1(,)?,,,?,,,,,,?,,1,2n242n,22n2,4,6,?,2n242n
?n11,3,5,,(2,1)n?( ,,1nn?n2,4,6,,2n2,
1lim,1?,由夹逼定理, nnn,,n2,
?n1,3,5,,(2,1)nlim,1?( n,,?n2,4,6,,2
2n,1评注 的极限是,,用此法体现了“,”的好处,可以放前,也可放后(若极限不2n
是,,则不能用此法,例如:
2,3,?,(n,1)limxx,,求( nn,,n3,5,?,(2n,1)
,,,,x,0xx解:?,单调递减,单调递减有下界,故其极限存在(nnn
n,2limx,ax,x,令,? nn,1n,,n2n,3
n,21limxlimx,lima,a?, , nn,1,,n,,nn,,2n,32a,0?,
limx,0即,,,,,,,,,,,,,,,,,, ( nn,,
11,,?,lim(1)(中科院) n,,,,,?,1212n
111评注 拆项:分母是两项的积, ,,n(n,1)nn,1
nn,1,11,,, 插项:分子、分母相差一个常数时总可以插项(,,1,n,1n,1n,1
,单调有界必有极限
x,1n常用方法:?;?;?归纳法;?导数法( x,xn,1nxn
,,,,,,,,,,,,,,,,单调递增 x,f(x)f(x),0f(x)n,1n
,,,,,, x,xx,xf(x),f(x)322121
,,,,,, x,xx,xf(x),f(x)322121
,,,,,,, 单调递减 f(x),0f(x)
,,,,,, x,xx,xf(x),f(x)322121
,,,,,, 不解决决问题( x,xx,xf(x),f(x)322121
命题:,若单调递增,且(),则,,单调递增(单调x,f(x)xx,xx,xf(x)n,1nn2121递减)(
例 求下列数列极限:
A1A,0x,x,(1)设,,;(98,华中科大,10分)()x,0n,1n1x2n
3,3xnx,(2)设,;(04,武大) x,0n,113,xn
x,xn,1n,2x,x,a(3)设,x,b,()((2000,浙大)n,2,3,?n012
1A1Ax,(x,),,2x,,A,,x解:(1)首先注意,所以为有下界数列(n,1nnn2x2xnn
另一方面,因为
1A1A (x,x,(x,),x,(,x),0n,1nnnn2x2xnn
A1,2x1A,,An,1(或) ,(1,),,1222xxnn
limx故为单调递减数列(因而存在,且记为( ,,xann,,n
A1 由极限的四则运算,在两端同时取极限,得x,x,()n,,n,1nx2n1Aa,(a,)(并注意到,解得( a,Ax,A,0n2a
xx3,33(1,)nnx(2)注意到,于是为有界数列(0,,,,3,,xn,1nxx3,3,nn
另一方面,由
2,,3,3xn,1,,3,,,223,x3,3x3,x2(3,x)n,1,,nnn,1x,x,,x,,, n,1nn3,3x3,x3,x(3,x)(4,2x)n,1nnn,1n,13,3,xn,1
23,xn,1, (3,x)(2,x)n,1n,1
2x3,n,1
xx,xx(3,)(2,)1n,1nn,1n,1,知( ,,02xx,x2,3,xnn,1n,1n,1
x3,n,1
limx,,x,xx,xx即与保持同号,因此为单调数列,所以存在(记为)(ann,1nnn,1n,,n
3,3x3,3ana,x,n,,,,,由极限的四则运算,在两端同时取极限,得(并注意到n,13,a3,xn
a,30,x,3,解得( n
,,,,,xxxxxxxxbann,nn,111021,,,,,,,,,?xxx(3)由于,n,nn1n,nn1,,,22(2)(2)(2)
1n1(),,n,1,1n12又,()()x,(x,x),x,x,b,a,a,b,a,a,,nm,1m0nm1(2),,0m,0m1(),,2
1n1(),,2(ba)2ba,,2limx所以 ((ba)limaa,,,,,,n,,n,,n1331(),,2
评注, 求递归数列的极限,主要利用单调有界必有极限的原理,用归纳法或已知的一些基本结果说明数列的单调、有界性(在说明递归数列单调性时,可用函数的单调性(下面给出一个重要的结论:设(),若在区间I上单调递增,x,f(x)x,In,1,2,?f(x)n,1nn
且(或),则数列单调递增(或单调递减)( ,,xx,xx,xn2121
评注2 第三小题的方法较为典型,根据所给的之间的关系,得到与x,x,xx,xn,1nn,1n,1n
的等式,再利用错位相减的思想,将数列通项写成级数的表达式(x,xxnn,1n
ab2n,1n,1a,例 设为任意正数,且,设,(),b,aba,ba,bn,2,3,?nnn,1n,11111a,bn,1n,1
则,,,,,收敛,且极限相同( abnn
ab22abn,1n,1n,1n,1a,,证明:由,ab,b,知 nn,1n,1nab,2abn,1n,1n,1n,1
,,,,,,,,,,,,,,,,,,b,ab,bb,b( nn,1n,1n,1n,1n,1
,,则0,b,bb,即为单调有界数列( n1n
0,a,b,b又,且 nn1
2abaababa(,)ab2,,2n,1n,1n,1n,1n,1n,1n,1n,1n,1n,1a,a,,a,,0,,nn,1n,1aba,bab,,n,1n,1n,1n,1n,1n,1
,,a所以亦为单调有界数列( n
limalimbba由单调有界必有极限定理,与存在,且分别记为与(在nn,,,,nn
ab22abn,1n,1a,与两端同时取极限,得与(b,aba,b,abn,,nnn,1n,1a,ba,bn,1n,1
考虑到为任意正数且( 0,a,a,b,ba,b1nn111
a,b,0即得(
1limx例 (1)设,,求; x2,,x,2nn,11,,nxn
limx(2)设,,且(),求(3x,x,2x,0x,0x,2n,2,3,?nn,1nn,112,,n
1limx解:(1)假设存在且等于,由极限的四则运算,在两端同时取极限x2,,ann,1,,nxn
1a,2,,得,即a,1,2( n,,a
又,故a,1,2( x,2n
趋于零()(由于 下面只须验证数列,,x,an,,n
nx,a,,1111,,,,n,,022,,x,a,,,,,,x,a,?,x,a,,,,n,1n1,,4xaxa4,,,,nn,,
n1,,limlimx,a,1,2而,由夹逼定理得( x,a,0,,n1n,,n,,4,,
(2)由3x,x,2x,0,知 n,1nn,1
,,,,,,3x,2x,3x,2x,3x,2x,?,,3x,2x,6n,1nnn,1n,1n,221
2x,,x,2则,,,,,,,,,,,,( n,1n3
6limxa,假设存在且等于a,由极限的四则运算,得( n,,n5
6,,n,,下面只须验证数列,趋于零()(由于 x,,n5,,
n,1n,1262626626,,,,,,,,,,x,,,,x,,?2(x,,,x,,,,,,,,,,,n,11nn,1353535535,,,,,,,,
n,1626,,lim显然,由夹逼定理得( lim,x,,0,,nn,,n,,535,,
评注, 两例题中均采用了“先求出结果后验证”的方法,当我们不能直接用单调有界必有
limx,a极限定理时,可以先假设,由递归方程求出,然后设法证明数列趋于,,x,aann,,n
零(
0,k,1评注, 对数列,若满足(),其中,则必有,,xx,a,kx,an,2,3,?nnn,1limx,a(这一结论在验证极限存在或求解递归数列的极限时非常有用(n,,n
评注, 本例的第二小题还可用Cauchy收敛原理验证它们极限的存在性(
a1n设>0,,,,证明,1(04,上海交大)limaaan,11nn,,a2nn
2aann 证 (1)要证lim,1 ,只要证, lim1,n,,n,,2n2n
22aa,22nn,1lim1,即只要证,即证 lim()2aa,,nn,1n,,n,,(22)2nn,,
1a11n,1aa,,,,,(2)因,,,故,1a0an,1nnn,12aaaannnn
aa,1122nn,1aaaaaa,,,,,,,,,,()()112 nnnnnn,,,11122aaannn
1,lim0因此只要证,即只要证lima,, n2n,,,,nan
1aa,,,0(3)由知,单调增加,假如有上界,则{}a{}a{}ann,1nnnan
111aa必有极限,由,,知,,,,因此,矛盾.aaa,0n,1naaan
lima,,这表明{}a单调增加、没有上界,因此. (证完)nn,,n
, 利用序列的Cauchy收敛准则
2xxxn,1limx0,x,1例 (1)设(),,求; x,,,xn1n,,n222
xn(2)设,,,求; x,x,2yy,x,yx,y,1limn,1nnn,1nn11n,,yn
11x120,x,1解:(1)由,(),得x,(假设x,则x,(有x,11kk2224
2xx112kx,x, x,,,kk1,2222
1,由归纳法可得,,,,,,,,,,,, x( n2
22x,,xxx,,np1,n1,,于是,,,,,, ,,,,, xx,npn,,2222,,
x,xx,x1n,p,1n,1n,p,1n,1,,,,,,,,,,,,,,,,,, ,,x,xn,p,1n,122
11,?,x,x,,0,,,,,,,,,,,,,,,,,,()( n,,p,11n,1n,122
2xxn,1limx,,由Cauchy收敛准则知:存在并记为,由极限的四则运算,在x两端ann,,n22
2a,2a,x,0同时取极限,得( n,,
1limx,a,,1,1,x,x注意到,故( nn,,n2
xna,a,1(2)设,显然. nnyn
xx,2y1n,1nna,,,1,由于,则 n,1yx,y1,an,1nnn
11a,a,, n,1n1,a1,ann,1
aa,11nn,1a,a,?,,aa . ,,nn,121n,144,,,,1a1a,,nn,1
a,a,a,a,?,a,aa,a,于是 n,pn,p,1n,p,1n,p,2n,1nn,pn
,a,a,a,a,?,a,an,pn,p,1n,p,1n,p,2n,1n
11,p111,,4 aaaa,,,,,?,,,,2121,1n,p,2n,1n1444,,1,4
11 (). ,,a,a,0n,,21n,134
limx由Cauchy收敛准则知:存在并记为. an,,n
12由极限的四则运算,在两端同时取极限,得(a,2a,1,n,,n,11,an
xnlima,2注意到,故( a,1lim,nnn,,n,,yn
评注1 Cauchy收敛准则之所以重要就在于它不需要借助数列以外的任何数,只须根据数列各项之间的相互关系就能判断该数列的敛散性. 本例两小题都运用了Cauchy收敛准则,但细节上稍有不同.其实第一小题可用第二小题的方法,只是在第一小题中数列有界,因此有,,xn
.保证了定义中的N仅与有关. x,x,x,x,1,p,11p,11
,,limx,x,0评注2 “对有”这种说法与Cauchy收敛准则并不一致(这里,p,Nn,pnn,,
x,x,,n,N要求对每个固定的p,可找到既与又与p的关的,,当,有(而,n,pnCauchy收敛准则要求所找到的,只能与任意的有关( ,
, 利用Stolz定理计算数列极限
例 求下列极限
333,,12,,,?nn,,(1) lim,3,,n,,4n,,
aana,,,2...a12n(2)假设(00,大连理工,10)(04,上海交lim,limaa,,证明:n2,,,,nnn2大)
证明:Stolz公式
aanaaananaaana,,,,,,,,,,,,2...(2...(1))(2...)1212112nnnn,limlim,222nn,,,,nnn(1),,
(1)na,an,1,,limn,,212n,
111,,,?2n(3) limn,,lnn
n3,1?2,,,n2(4) limn,,n
2na,1(5)() limn,,na
, 关于否定命题的证明 (书上一些典型例题需背) limx,a n,,n
,,x发散 n
111x例 证明:,1,,,?,发散( nn23
an,1lima,0例 设(),且,若存在极限,则((北大,l,1a,0n,1,2,?lim,lnnn,,n,,an
20)
, 杂例
111lim,,,(1) ?n,,1,22,3n(n,1)
(2) (04,武大)
12nlim(...),(1),,,,a2nn,,aaa
1n 1(),1naa,,,lim()n2n,,1aaa,,1(1)1,a
22nlim(1,x)(1,x)?(1,x)x,1(3) (); ,,n
2(4)设,(),求: a,3n,1,2,?a,a,a1n,1nn
,,111,,( llim,,,?,,,n,,1a1a1a,,,12n,,