搜索
5p Œ˘61 SK) SSSKKK10.11.|^ ´æ ny†:(1)?Œ∞Pn=1cosnn(n+1)´æ.y.|n+pPk=n+1coskk(k+1)|≤n+pPk=n+11k(k+1)=n+pPk=n+1(1k−1k+1)=1n+1−1n+p+11n.Ø?¿ε0,-N=[1ε],K nN ,|n+pPk=n+1coskk(k+1)|1nε.d ´æ n,?Œ∞Pn=1cosnn(n+1)´æ.(2)?Œ∞Pn=11√nu .y.|n+pPk=n+11√k|≥p√n+p. Nıo , n=p=N+1 ,|n+pPk=n+11√k|≥qN+12≥1√2.d ´æ n,?Œ∞Pn=11√nu .(3) ?Œ∞Pn=1an ∞Pn=1bn ´æ, 3 ŒN, n≥N kan≤un≤bn,K?Œ∞Pn=1unǑ´æ.y.Ø?¿ ε0,ˇǑ?Œ∞Pn=1an ∞Pn=1bn´æ,⁄– 3N′N, nN′ ,|n+pPk=n+1ak|ε,|n+pPk=n+1bk|ε., ¡,n+pPk=n+1ak≤n+pPk=n+1uk≤n+pPk=n+1bk.⁄–|n+pPk=n+1uk|ε.⁄–?Œ∞Pn=1un´æ.2.fi ?Œ∞Pn=1an´æ,q ?Œ∞Pn=1bnu ,fl?Œ∞Pn=1(an±bn)·˜´æ? : ‰u .˜K∞Pn=1bn=±∞Pn=1an±∞Pn=1(an±bn)Ǒ´æ.3. e ?Œ·˜´æ.(1)∞Pn=1(√n+1−√n).ˇǑ ' nPk=1(√k+1−√k)=√n+1−1,3n→∞ vk4 ,⁄–?Œu .(2)∞Pn=11(2n−1)(2n+1).ˇǑnPk=11(2k−1)(2k+1)=nPk=112(12k−1−12k+1)=12(1−12n+1),3n→∞ k4 ,⁄–?Œ´æ.(4)∞Pn=1cos2πn.ˇǑlimn→∞cos2πn=1,⁄–?Œu .1(7)∞Pn=1n√0.0001.ˇǑlimn→∞n√0.0001=1,⁄–?Œu .4. ?Œ∞Pn=1un ' S Ǒ{Sn}.en→∞ {S2n} {S2n+1} ´æ ´æ ~ŒA.y†?Œ∞Pn=1un´æ.y.Ø?¿ ε0,ˇǑlimn→∞S2n=limn→∞S2n+1=A,⁄– 3N0, nN ,|S2n−A|ε,|S2n+1−A|ε.u· nN ,|Sn−A|ε.⁄–limn→∞Sn=A,?Œ∞Pn=1un´æ.5. ?Œ∞Pn=1un´æ, un≥un+1≥0(n=1,2,...),y†:limn→∞nun=0.y.Ø?¿ ε0,ˇǑ?Œ∞Pn=1un´æ,d ´æ n, 3N, nN ,n+pPk=n+1unε2.u· nN ,(2n)u2n≤22nPk=n+1ukε.(2n+1)u2n+1≤2nPk=n+1uk+2n+1Pk=n+1ukε.⁄–limn→∞nun=0.SSSKKK10.21.? e ?Œ æ 5.(1)∞Pn=12nsinπ4n.ˇǑlimn→∞2nsinπ4n/12n=π,?Œ∞Pn=112n´æ,⁄– ?Œ´æ.(2)∞Pn=11√2n3+1.ˇǑlimn→∞1√2n3+1/1√n3=1√2,?Œ∞Pn=11√n3´æ,⁄– ?Œ´æ.(3)∞Pn=11n√n.ˇǑlimn→∞1n√n=1,⁄–?Œu .(4)∞Pn=14nn2+4n−3.ˇǑlimn→∞4nn2+4n−3/1n=4,?Œ∞Pn=11nu ,⁄– ?Œu .(5)∞Pn=1nn(n2+3n+1)n+22.ˇǑlimn→∞nn(n2+3n+1)n+22/1n2=limn→∞(1+3n+1n2)−n+22=e−32,?Œ∞Pn=11n2´æ,⁄– ?Œ´æ.(6)∞Pn=2n(lnn)lnn.ˇǑlimn→∞ln[n(lnn)lnn/1n2]=limn→∞(3−lnlnn)lnn=−∞,⁄–limn→∞n(lnn)lnn/1n2=0.?Œ∞Pn=11n2´æ,⁄– ?Œ´æ.(7)∞Pn=1ntan13n.ˇǑlimn→∞nqntan13n=13,d O{,?Œ´æ.2.? e ?Œ æ 5.2(1)∞Pn=1n5n!.ˇǑlimn→∞(n+1)5(n+1)!/n5n!=0,d K O{,?Œ´æ.(2)∞Pn=1n!3n2.ˇǑlimn→∞n!3n2=+∞,?Œu .(3)∞Pn=13n·n!nn.ˇǑlimn→∞3n+1·(n+1)!(n+1)n+1/3n·n!nn=limn→∞3(1+1n)−n=3e−11,d K O{,?Œu .(4)∞Pn=11n1+1/n.ˇǑlimn→∞1n1+1/n/1n=1,?Œ∞Pn=11nu ,⁄– ?Œu .(5)∞Pn=1n2(3−1n)n.ˇǑlimn→∞nqn2(3−1n)n=13,d O{,?Œ´æ.(6)∞Pn=2n(lnn)n.ˇǑlimn→∞nqn(lnn)n=0,d O{,?Œ´æ.(7)∞Pn=11000nn!.ˇǑlimn→∞1000n+1(n+1)!/1000nn!=0,d K O{,?Œ´æ.(8)∞Pn=1(n!)2(2n)!.ˇǑlimn→∞((n+1)!)2(2n+2)!/(n!)2(2n)!=14,d K O{,?Œ´æ.(9)∞Pn=113n(n+1n)n2.ˇǑlimn→∞nq13n(n+1n)n2=e31,d O{,?Œ´æ.(10)∞Pn=21n(lnn)p(p0). p1 ,¨'R+∞21x(lnx)p=(lnx)1−p1−p|+∞2´æ,⁄–?Œ´æ. p=1 ,¨'R+∞21x(lnx)p=lnlnx|+∞2u ,⁄–?Œu . 0p1 ,¨'R+∞21x(lnx)p=(lnx)1−p1−p|+∞2u ,⁄–?Œu .(11)∞Pn=31n(lnlnn)q(q0).ˇǑlimn→∞1n(lnlnn)q/1n(lnn)1/2=limn→∞(lnn)1/2(lnlnn)q=limx→+∞x1/2(lnx)q=+∞.?Œ∞Pn=31n(lnn)1/2u ,⁄– ?Œu .3.y†:e ?Œ∞Pn=1un´æ,K?Œ∞Pn=1u2nǑ´æ. ‰⁄Æ,` ~‘†.y.ˇǑ?Œ∞Pn=1un´æ,⁄–limn→∞un=0,⁄–S {un}k.. unM.Ku2n≤Mun.d’ O{,?Œ∞Pn=1u2n´æ. ‰⁄Æ,~X?Œ∞Pn=1(1n)2´æ, ∞Pn=11nu .4.y†:e?Œ∞Pn=1a2n ∞Pn=1b2n ´æ,K?Œ∞Pn=1|anbn|,∞Pn=1(an+bn)2,∞Pn=1|an|nǑ´æ.y.dK ^ ,?Œ∞Pn=1(a2n+b2n)´æ.du|anbn|≤12(a2n+b2n),(an+bn)2≤32(a2n+b2n),⁄–?Œ∞Pn=1|anbn|,∞Pn=1(an+bn)2Ǒ´æ. bn=1n,= ?Œ∞Pn=1|an|n´æ.5.y†:e ?Œ∞Pn=1un ∞Pn=1vn ´æ,fle ?Œ·˜u ?(1)∞Pn=1(un+vn);(2)∞Pn=1(un−vn);(3)∞Pn=1unvn. .(1) ‰u ,ˇǑun+vn≥un, ’ O{,?Œ∞Pn=1(un+vn)u .(2) ‰,’Xun=vn=1n ,?Œ∞Pn=1(un−vn)´æ 0.(3) ‰,’Xun=vn=1n ,?Œ∞Pn=1unvn´æ.6. limn→∞nun=l, ¥0l+∞.y†?Œ∞Pn=1u2n´æ, ∞Pn=1unu .y.˜klimn→∞nun=l0¿ X n¿' un0.⁄– –b un0.ˇǑlimn→∞u2n/1n2=l2,?Œ∞Pn=11n2´æ,⁄–?Œ∞Pn=1u2n´æ.ˇǑlimn→∞un/1n=l,?Œ∞Pn=11nu ,⁄–?Œ∞Pn=1unu .SSSKKK10.31. e ?Œ·˜´æ?^ ´æ · Ø´æ?(1)∞Pn=1(−1)n−1(2n)2.ˇǑ∞Pn=1|(−1)n−1(2n)2|=∞Pn=11(2n)2´æ,⁄– ?Œ Ø´æ.(2)∞Pn=1(−1)n+1(2n−1)p(p0).S {1(2n−1)p} N“u0,⁄–T ?Œ´æ.?Œ∞Pn=11(2n−1)p = p1 ´æ,⁄– ?Œ p1 Ø´æ,˜K^ ´æ.(3)∞Pn=2(−1)nnlnn.S {1nlnn} N“u0,⁄–T ?Œ´æ. ·?Œ∞Pn=21nlnnu ,⁄– ?Œ^ ´æ.(4)∞Pn=1(−1)n√n−1n. “=∞Pn=1(−1)n1√n−∞Pn=1(−1)n1n,m ?Œ ´æ,⁄– ?Œ´æ. ·limn→∞√n−1n/1√n=1,⁄–?Œ∞Pn=1√n−1nu .⁄– ?Œ^ ´æ.(6)∞Pn=1(−1)nn!3n2.ˇǑlimn→∞(n+1)!3(n+1)2/n!3n2=limn→∞n+132n+1=0.d K O{,?Œ Ø´æ.4(7)∞Pn=2(−1)n1nsinπn.ˇǑlimn→∞1nsinπn/1n2=π,⁄–?Œ Ø´æ.(8)∞Pn=1(−1)n+1tanϕn(−π2ϕπ2). ϕ=0 ,?Œw, Ø´æ.˜KT?ŒǑ ?Œ, |tanϕn| N“u0,ˇd?Œ´æ. ·limn→∞|tanϕn|/1n=|ϕ|,?Œ Ø´æ.(10)∞Pn=1sin(π√n2+1).sin(π√n2+1)=(−1)nsinπ(√n2+1−n)=(−1)nsinπ√n2+1+n.ˇdT?ŒǑ´æ ?Œ. ·limn→∞sinπ√n2+1+n/1n=π2,?Œ Ø´æ.2.fi ?Œ∞Pn=1un´æ,y†?Œ∞Pn=1unnp(p0) ∞Pn=1nn+1un ´æ.y.Œ {1np}9{nn+1} Nk.,qˇǑ?Œ∞Pn=1un´æ,dC O{,?Œ∞Pn=1unnp ∞Pn=1nn+1un ´æ.3.y†?Œ∞Pn=1cosnϕnp(0ϕ2π) p1 Ø´æ, 0p≤1 ^ ´æ.y.(i) p1 ,?Œ∞Pn=11np´æ.du|cosnϕnp|≤1np, ?Œ∞Pn=1cosnϕnp Ø´æ.(ii) 0p≤1.duŒ {1np} N“u0, ' nPk=1coskϕk.,d)| X O{,?Œ∞Pn=1cosnϕnp´æ. n y ϕ6=π ,?Œ∞Pn=1cos2nϕ2np´æ.qdu?Œ∞Pn=112npu ,⁄–?Œ∞Pn=11+cos2nϕ2npu .ˇǑ|cosnϕnp|≥cos2nϕnp=1+cos2nϕ2np,?Œ∞Pn=1cosnϕnp Ø´æ.5./X∞Pn=1annx ?Œ¡ )| X?Œ.y†§ke 5 :e?Œ∞Pn=1annx0´æ(u ),o xx0(xx0) ,?Œ∞Pn=1annxǑ´æ(u ).y. ?Œ∞Pn=1annx0´æ. xx0 ,Œ {nx0−x} Nk..dC O{,?Œ∞Pn=1annx=∞Pn=1annx0nx0−x´æ.6. ?Œ∞Pn=1un Ø´æ,y†?Œ∞Pn=12n−1nunǑ Ø´æ.y.|2n−1nun|≤2|un|,⁄–?Œ∞Pn=1|un|´æ¿ X?Œ∞Pn=12n−1n|un|Ǒ´æ.SSSKKK10.451.e ?Œ ´æ .(1)∞Pn=1(lnx)n. = |lnx|1 ?Œ´æ,⁄–´æ Ǒ(e−1,e).(2)∞Pn=112n−1(1−x1+x)n. = 1−x1+x∈[−1,1) ?Œ´æ,⁄–´æ Ǒ(0,+∞).(3)∞Pn=11xnsinπ3n. |x|≤13 ,limn→∞1xnsinπ3n6=0,⁄–?Œu . |x|13 ,|1xnsinπ3n|≤1|x|nπ3n,d’ O{ ,?Œ Ø´æ.⁄–´æ Ǒ(−∞,−13)∪