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1、§3数列极限存在的条件1n1、利用lim(1+)=e求下列极限:n→∞n1n1n+11n(1)lim(1−);(2)lim(1+);(3)lim(1+);n→∞nn→∞nn→∞n+11n1n(4)lim(1+)(5)lim(1+)n→∞2nn→∞n21nn−1n1解(1)lim(1−)=lim()=limn−1n→∞nn→∞nn→∞111+⋅1+n−1n−111==n−1e11lim1+⋅1+n→∞n−1n−11n+11n1(2)lim(1+)=
2、lim(1+)lim(1+)=e⋅1=en→∞nn→∞nn→∞n1n1n+11−1(3)lim(1+)=lim(1+)(1+)=en→∞n+1n→∞n+1n+111n12n⋅212n(4)lim(1+)=lim(1+)=lim(1+)=en→∞2nn→∞2nn→∞2n1(5)设a=1+,则a>0,lima=1>0,由第二节习题知n2nnnn→∞211n1n⋅1n2lim(1+)=lim(1+)n=limn(1+)=1n→∞n2n→∞n2n→∞n2注:以上的(4)(5)用到事实;若lima=a>0则limna=1.这一点请读者自行证明.n
3、nn→∞n→∞2、试问下列解题方法是否正确?n求:lim2.n→∞nn解设a=2及lim2=a,由于a=2a,两边取极限(n→∞),则有a=2a,所以nnn−1n→∞a=0.n答以上解题方法不正确,因为只有证明了{a}的极限存在以后才可设lima=a.而{2}nnn→∞是递增且无上界的数列,它不存在极限,所以以上解题方法不正确.3、证明下列极限存在并求其值:(1)a=2,a=2a;1n+1n(2)设a=c>0,a=a+c;1n+1nnc(3)a=(c>0).nn!证(1)a=2<2,设a<2,则a=2a<2;所以{a}有上界2;1nn+
4、1nna(2−a)nn而a−a=2a−a=>0;因之{a}是递增且有上界的数列.由单调有界n+1nnnn2a+ann2定理知{a}极限存在,设其为a,对等式a=2a令n→∞取极限,有α=2α,解之得nn+1nα=0(舍去),α=2,故lima=2.12nn→∞(2)由a=c>0,知a=c0时,a=c<1+c,设nn+1n12a
5、<1+c,则a=a+c=1+c+c<1+2c+c=1+c,{a}有上界1+c,故有单nn+1nn2调有界定理知:{a}极限存在,设其为a,对a=a+c,两端令n→∞取极限,得nn+1n21±1+4cα=α+c,解得α=,由于a>0,所以α>0,故n21+1+4clima=nn→∞2(3)易见n+1nncccca−a=−=(−1).n+1n(n+1)!n!n!n+1ncc取自然数N,使cN时,有a−a=(−1)<0n+1nn!n+1∞故{a}(不计前N项)为递减数列.又a>0,可见{a}有下界.由单调有界定理知{a}nN+
6、1nn=N+1nc极限存在,设其为α,对a=a⋅两端令n→∞去极限,得α=α⋅0,n+1nn+1故lima=α=0nn→∞114、利用1+为递增数列的结论,证明:1+为递增数列.nn+1nn+1111证因1+是递增数列,所以对任意的自然数n都有1+<1+即nnn+1n−1n11111+1+<1+1+nnn+1n+11又因1+>0,从而有n1n−1n1+11n+11+<1+n
7、n+111+nn2n1n+2n1=1+<1+2n+1n+2n+1n+11故1+是递增数列.n5、利用柯西收敛准则,证明以下数列{a}收敛:nsin1sin2sinn(1)a=+++n2n222111(2)a=1++++n22223n11证(1)因为lim=0,于是对任给ε>0,必存在N,当n>N时,<ε,所以当n>Nn→∞2n2n时,对任意的自然数p,有sin(n+1)sin(n+2)sin(n+p)111a−a=+++<+++n+pnn+1n+2n+pn+1n+2n+p2222
8、22111=(1−)<<ε,npn222故由柯西收敛准则知{a}收敛.n1(2)对任给ε>0,取N=+1,当m>n>N时ε111a−a=+++mn222(n+1)(n+2)m111111<++