数学分析课后习题答案2.2

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1、§2收敛数列的性质1、求下列极限:32nnn+3n+11+2n(−2)+3(1)lim;(2)lim;(3)lim;n→∞4n3+2n+3n→∞n2n→∞(−2)n+1+3n+1111+++2n2nnn222(4)lim(n+n−n);(5)lim(1+2++10);(6)lim.n→∞n→∞n→∞111+++2n33331321++n+3n+1nn31证明:(1)lim=lim=.n→∞4n3+2n+3n→∞2344++23nn1+2n12(2)lim=lim(+)=0.n→∞n2n→∞n2n2nnn(−)+1(−2)+331(3)lim=lim=.n→∞(−2)n+1+3n+

2、1n→∞23n(−2)⋅(−)+332n11(4)lim(n+n−n)=lim()=lim()=.n→∞n→∞n2+n+nn→∞121++1nnnn(5)lim(1+2++10)=1+1++1=10.n→∞1n1−()12⋅11121+++1−2222n21(6)limlim==2.n→∞111n→∞1n1+++1−()23323n13⋅311−32、设lima=a,limb=b,且a>b,nnn→∞n→∞证明:存在自然数N,当n>N时,a0,根据两个已知极限知分别存在N,N,01221当n>N时,a−a<ε,从而a

3、n021当n>N时,b−b<ε,从而b>b−ε=(a+b).2n0n021取N=max{N,N},当n>N时,必有a<(a+b)N时,有a0,使b≤M,n=1,2,;nnε由{a}为无穷小数列知,对任给的ε>0,必存在N,当n>N时,a<.nnMε因之当n>N时,ab−0=a⋅b<⋅M=εnnnnM所以limab=0,即{ab}为无穷小数列.nnnnn→∞4、求下列极限:111(1)lim(+++);n→∞1⋅22⋅3n⋅(n+1)482

4、n(2)lim(2⋅2⋅22);n→∞132n−1(3)lim(+++);n→∞2222n1(4)limn1−;n→∞n111(5)lim(+++);n→∞n2(n+1)2(2n)2111(6)lim(+++).n→∞222n+1n+2n+n11111111解:(1)lim(+++)=lim[(1−)+(−)++(−)]n→∞1⋅22⋅3n⋅(n+1)n→∞223nn+11=lim(1−)=1n→∞n+11111482n+++n1−n2(2)因为2⋅2⋅22=2242=22=122112nn而1<2<2=2→1(n→∞),482n2所以lim(2⋅2⋅22)=lim=

5、21n→∞n→∞n22132n−1(3)令a=+++,n2n2225572n+12n+32n+3则a=(3−)+(−)++(−)=3−n2n−1nn2222222n+3所以lima=lim(3−)=3.nnn→∞n→∞2111n(4)当n>2时,<1−<1,limn=lim1=12nn→∞2n→∞1由迫敛性定理知limn1−=1.n→∞n111111(5)由于0<+++<+++222222n(n+1)(2n)nnnn+111==+→0(n→∞)22nnn111由迫敛性定理知lim(+++)=0.n→∞n2(n+1)2(2n)2n111n(6)由于<+++<,22222n+n

6、n+1n+2n+nn+1nn而lim=lim=1,n→∞2n→∞2n+nn+1由迫敛性定理知:111lim(+++)=1n→∞222n+1n+2n+n5、证明:若{a},{b}中有一个是收敛数列,另一个是发散数列,则{a±b}是发散数列;又nnnnan问{ab}和{}(b≠0)是否也是发散数列?nnnbn证:不妨设lima=a,{b}发散.假设{a+b}收敛于b,则由极限性知nnnnn→∞b=(a+b)−a收敛于b−a,即limb=b−a,这与题中条件{b}发散矛盾,故由nnnnnnn→∞{a+b}发散.同理可得{a−b}发散.nnnna1n然而{ab}和{}不一定是发散数列.例如,

7、若取问a={},b={n},则a收敛,nnnnnbnnan{b}发散,但{ab}和{}都收敛.nnnbn6、证明下列数列不收敛nn(−1)nnπ(1){(−1)}(2){n}(3){cos}.n+14nnnn证(1)令a=(−1),则lima=1,lima=−1,由定理2.8知{a}={(−1)}n2n2n−1nn+1n→∞n→∞n+1不收敛.nn(−1)(−1)(2)因为{n}为无界数列,由定理2.3知{n}发散.nπ(3)令a=cos,则l