微积分(下)英文教案资料

1、ChapterChapter1InfiniteSerieskk=0kk=0Generally,forthegivensequencea,aaa,the2,3nTOC o 1-5 h zexpressionformedbythesequencea,aaa,12,3na+a+a+a+，123niscalledtheinfiniteseriesoftheconstantsterm,denotedby艺a/thatisnn=1TOC o 1-5 h za=a+a+a+a+,n123nn=1Wherethenthtermissaidtobethegeneraltermoftheseries,moreov

2、er,thenthpartialsumoftheseriesisgivenbyS=a+a+a+.+a.n123nDeterminewhethertheinfiniteseriesconvergesordiverges.Whileitspossibletoaddtwonumbers,threenumbers,ahundrednumbers,orevenamillionnumbers,itsimpossibletoaddaninfinitenumberofnumbers.Toformaninfiniteserieswebeginwithaninfinitesequenceofrealnumbers

3、:a,a,a,a,wecannotformthe0123sumofallthea(thereisaninfinitenumberoftheterm),butkwecanformthepartialsumsS=a=为a00kkk=0k=0kk=0k=0S=a+a=工aTOC o 1-5 h z101kk=0S=a+a+a=为a012kk=0S=a+a+a+a=工a0123k二工ak二工akk=0S=a+a+a+a+an0123Definition1.1.1IfthesequenceSofpartialsumshasafinitelimitL,nWewriteL=Xakk=0andsaythatt

4、heseries艺aconvergestoL.wecallLthekk=0sumoftheseries.IfthelimitofthesequenceSofpartialsumsdontnexists,wesaythattheseries艺adiverges.kk=0Remarkitisimportanttonotethatthesumofaseriesisnotasumintheorderingsense.Itisalimit.EX1.1.1provethefollowingproposition:Proposition1.1.1:If|x1,thenthe艺xdiverges.kk=0Pr

5、oof:thenthpartialsumofthegeometricseries艺akk=0takestheforms=1+x1+x2+x3+.+xn-1MultiplicationbynxgivesxS=x(1+x1+x2+x3+xn-1)=x1+x2+X3+Xn-1+XnSubtractingthesecondequationfromthefirst,wefindthat(I-x)s二1-xn.Forx土1,thisgivesnIfIX1,IfIX丄=*nisunbounded.Sonnlims=limn=g/thereforetheseriesdiverges.nnsnsButlima=

6、lim丄=0ks*ksYkEX.1.1.3determinewhetherornottheseries:艺丄=o+1+2+3+4+Converges.k+12345k=0thisseriesSolutionsince込a=lim-=lim=10thisseriesks*ksk+1ksdiverges.EX.1.1.4Determinewhetherornottheseries艺丄2kk=0无x无xk二艺k=0k=0k，andx=21，bypropositionmweknowthatseriesconverges.Solution211S=1+n2411112n22seriesconverges

7、.Solution211S=1+n2411112n22223-(1-1)S-1-丄，S-2(1-12).2n2nn/2nlimSlim2(1-丄)2ns料ns2nBydefinitionofconvergesofseries,thisseriesconverges.+丄，2n-111.+2n-12nEX.1.1.5proofsthefollowingtheorem:Theorem1.1.2Iftheseries艺aand艺bconverges,thenkkk0k0艺(a+b)kkk=0series.alsoconverges,andisequalthesumofthetwo(2)IfCisar

8、ealnumber,then艺Caalsoconverges.kk=0Moreoverif艺akk=0=ithen艺Ca=Cl.kk=0Prooflets(1)工a,S2工bS工(a+b),S4工CaTOC o 1-5 h znknknkknkk0k0k0k0Notethats=S+Sands=CS(i)nnnnnSinceHmSG)=l,limS=m,nnnsnsThenlimS(3)=lim(S(1)+S(2)=limS(1)+limS(2)=l+mnnnnnnsnsnsnslimS(4)=limCS(1)=ClimS(1)=Cl.nnnnsnsnsTheorem1.1.4(squeeze

9、theorem)nabk,(knnnconvergestol.nnabk,(knnnconvergestol.nisafixedinteger),thenbalsonEx.1.1.6showthatlimnT8sin3n=0.nSolutionForn1,一丄(沁)丄,nnnsincelim(一)=0,andlim(-)=0,thengnn“ntheorem.resultfollowsbythesqueezensconvergetolandthatSupposethataandcconvergetolandthathavetheForsequenceofvariablesign,itishel

10、pfultofollowingresult.havetheEX1.1.7provethatthefollowingtheoremholds.Theorem1.15If呵a|=0,thenlima=0/TOC o 1-5 h znnnsnsProofsince一”|aa|,fromthetheorem1.1.4nnnNamelythesqueezetheorem,weknowtheresultistrue.Exercise11Anexpressionoftheforma+a+a+iscalled123Aseries+a+a+issaidtoconvergeifthesequence123scon

11、verges,whereS=nn1-Thegeometricseriesa+ar+ar2+convergesif;inthiscasethesumoftheseriesis2.Iflima丰0,wecanbesurethattheseriesa=n=n=1Evaluate艺厂(1_r)k,0r2k=0Evaluate艺(_1)kxk,_1x1.k=0Showthat艺diverges.k+1k=1Findthesumsoftheseries6-11艺17.艺18.艺19.无ATOC o 1-5 h z(k+1)(k+2)2k(k+1)k(k+3)10kk=3k=1k=1k=010.艺3k+4k

12、11.艺2k+35k3kk=0k=012.Derivethefollowingresultsfromthegeometricseries艺(_x2k=,ixi0zthenthepartialsumS=a+a+aareincreasing,n12niesssss123nn+1Iftheyaretoapproachalimitatall,theycannotbecomearbitrarilylarge.ThusinthatcasethereisanumberBsuchthatsS.Wetakeforgrantedthataleastupperboundexists.Thecollectionofn

13、umbersShasthereforealeastupperbound,ni.e.,thereisasmallestnumberssuchthatSSfora|n.innthatcase,thepartialsumsSapproachSasalimit.inothernwords,givenanypositivenumbers0,wehaveS_eS0forallnandb0fornnnnn=1n=1alln.Assumethatthereisanumbersc0,suchthatacbfornnalln,andthat艺bconverges,then艺aconverges,andnnn=1n

14、=1nnnn=1nn=1Proof:Wehavea+a+acb+cb+cb=c(b+b+b)c艺b.12n12n12nnn=1Thismeansthat正bisaboundforthepartialnn=1sums.nuiiia+a+a.12nTheleastupperboundofthesesumsistherefore0foralln.Assumethatnnnthereisanumberc0suchthatcbforallnsufficientlynnlarge,and艺bdoesnotconverge,then艺adiverges.nnn=1n=1Proof.Assumeacbforn

15、nnn,0since无bdiverges,wenn=Proof.Assumeacbfornnnn,0since无bdiverges,wenn=1canmakethepartialsumarbitrarilylarge另bn=n0as=bn0+b+bnQ+1Nbecomesarbitrarilylarge.But迓ain=n0迓cb=c迓bnnn=nn=n00Hencethepartialsum迓a=a+a+aarearbitrarilylargeasNbecomesarbitrarilyn12Nn=1large,arehence艺adiverges,aswastobeshown.Remarkn

16、n=1onnotationyouhaveeasilyseenthatforeachj0,艺aconvergesiff乞aconverges.Thistellsusthat,inkkk=0k=j+1determiningwhetherornotaseriesconverges,itdoesnotmatterwherewebeginthesummation,wheredetailedindexingwouldcontributenothing,wewillomititandwriteywithoutspecifyingwherethesummationbegins.Forinstance,itma

17、kessensetoyouthat工丄convergesand工1k2kdivergeswithoutspecifyingwherewebeginthesummation.Butintheconvergentcaseitdoes,however,affectthesum.Thusforexample艺丄二2，艺丄二1/艺丄二1/andsoforth.2k2k2k2k=0k=1k=2艺丄converges.n艺丄converges.n2n=1SolutionLetuslookattheseries:11111111111+12223242527282152162Welookatthegroups

18、oftermsasindicated.Ineachgroupofterms,ifwedecreasethedenominatorineachterm,thenweincreasethefraction.Wereplace3by2,then4,5,6,7by4,thenwereplacethenumbersfrom8to15by8,andsoforth.Ourpartialsumsthereforelessthanorequalto丄+丄+andwenotethat2TOC o 1-5 h z1222224242428282occurstwice,4occursfourtimes,8occurs

19、eighttimes,andsoforth.Ourpartialsumarethereforelessthanorequalto11111111+1222224242428282andwenotethat2occurstwice,4occursfourtimes,8occurseighttimes,andsoforth.Hencethepartialsumsarelessthanorequalto丄+2+上+A+.J+!+?12224282248Thusourpartialsumsarelessthanorequaltothoseofthegeometricseriesandarebounde

20、d.Henceourseriesconverges.Generallywehavethefollowingresult:Theseries艺丄二i+丄+丄+丄+丄+，wherepisanp2p3p4pnpn=1constant,iscalledap-series.Proposition】21Ifpi,thep-seriesconverges;andifp丄=11.Since艺1doesnotconverge,itfollowsn3+12n2nnn=1thattheseries艺n2doesnotconvergeeither.Namelythisseriesdiverges.n3+1series

21、diverges.Ex123Provetheseries万、n2+7converges.2n4一n+3n=1n=1Proof:Indeedwecanwriten2+n2+72n4一n+3iscertainlyiscertainlyFornsufficientlylarge,thefactorbounded,andinfactisnear1/2.Hencewecancompareourserieswith工丄toseeconverges,becauseE1convergesn2n2andthefactorisbounded.Ex.125Showthat工diverges.ln(k+b)Solut

22、ion1Weknowthatask*,Ink-0.Itfollowsthatkin(k+b)o,andthusthatin(k+b)=in(k+b)k+b0.Thusforkk+bkk+bksufficientlylarge,ln(k+b)kand1.Since工1kln(k+b)kdiverges,wecanconcludethat工diverges.ln(k+b)Solution2Anotherwaytoshowthatln(k+b)0SincefSincef(x)=丄0forallxo,x+bf(x)0forallx3.Itfollowsthatln(x+b)3-Wecomenowtoa

23、somewhatmorecomparisontheorem.Ourproofreliesonthebasiccomparisontheorem.Theorem125(TheLimitComparisonTest)Let工aandk工bbeserieswithpositiveterms.ksomepositivenumber,then工bbeserieswithpositiveterms.ksomepositivenumber,then工akIflim(ak/k)=i,whereand工bconvergekisordivergetogether.ProofChooseebetween0andPr

24、oofChooseebetween0andknowforallksufficientlylarge（foralll,wekgreaterthansomeak）Ik1|e.0bkForsuchkwehave18空i+8,andthusbk（1_E）ba（1+8）bthislastinequalityiswhatweneeded.kkkIf工aconverges,then工（18）bconverges,andthuskk工bconverges.k（2）If工bconverges,then工（1+8）bconverges,andthuskk工aconverges.kToapplythelimitco

25、mparisontheoremtoaseries工a,wekmustfirstfindaseries工bofknownbehaviorforwhichkakconvergestoapositivenumber.bkEx126DeterminewhethertheseriesEsin,convergeskordiverges.SolutionRecallthatas兀-沁-1.Ask-oandthussing；）diverges.thussing；）diverges.sik-i.Since工Ldiverges,soE兀kkEx127Determinewhethertheseries工5号100一

26、2k2k+9skconvergesordiverges.SolutionForlargevalueofk,5jkdominatesthenumeratorand2k2魚dominatesthedenominator,thus,forsuchk,5麻+1differslittlefrom5攸二.Since2k2長+9長2k27!2kJ5麻+100.5_10k2麻+200k2_1麻、12kJk+9jk示-100kk+4k-1+9And工_L=5工丄converges,thisseriesconverges.2k22k2Theorem1.2.6Let工aand工bbeserieswithpositi

27、vekktermsandsupposethus乞_0,thenbkIf工bconverges,then工aconverges.kkIf工adiverges,then工bdiverges.kkIf工aconverges,then工bmayconvergeordiverge.kkIf工bdiverges,then工amayconvergeordiverge.kkParts(3)and(4)explainwhywestipulatedl0intheorem1.2.5TheroottestandtheratiotestTheorem127(theroottest,Cauchytest)let工abea

28、kserieswithnonnegativetermsandsupposethatlim0=limat=P,ifp1,kT8kkT8kk工adiverges,ifp=1,thetestisinconclusive.kProofwesupposefirstp1andchoose卩sothat(k)3(k)3pu1.Since)；p,wehavea丄R,foraksufficientlykklargethus卩kforallksufficientlylargesince工卩kkconverges(ageometricserieswith0卩Rforallksufficientlylarge.kkT

29、husapkforallksufficientlylarge.k卩1)theSince工卩kdiverges(ageometricserieswiththeorem1.2.6tellusthat卩1)thekwhenp=i,Toseetheinconclusivenessoftheroottestnotethat(whenp=i,k二k二ikk工and工：(a)k二()k二()2t12二1，(a)k二(_)k2kkk2丄kkkkThefirstseriesconverges,buttheseconddiverges.EX.1.2.7Determinewhethertheseries工1conv

30、erges(lnk)kordiverges.SolutionFortheseries工1,applyingtheroottest(Ink)kwehavelim(a)1k二lim丄二0,theseriesconverges-kskkTglnkEX.1.2.8DeterminewhetherseriesE丄convergesordiverges.SolutionFortheseries工2k,applyingtheroottest,wekhave(a):二2(丄)3二2丄3T2X13二21.Sotheseriesdiverges.kkkkEX1.2.9Determineswhethertheser

31、ies工(1_1)kconvergesordiverges.Solutioninthecaseof工(1_1)k,wehave(akkkapplyingtheroottest,itisinconclusive.Butsincea=(1_1)kconvergesto1andnotto0,theseriesdiverges.kkeWecontinuetoconsideronlyserieswithterms0.Tocomparesuchaserieswithageometricseries,thesimplesttestisgivenbytheratiotesttheoremTheorem1.2.

32、8仃heratiotest,DAlemberttest)let工abekaserieswithpositivetermsandsupposethatlimn二九/ksakIf九1,工aconverges,if九1,工adiverges.kkIfthe九=,thetestisinconclusive.Proofwesupposefirstthatxhsincelim人=九NTOC o 1-5 h zn+iCThenaCaaCaC2aaN+1N,N+2N+1NninductionaCkaThusN+kN,andingeneralbyN+kandingeneralby工aa+ca+c2a+ckanN

33、NNNn=Nck)aa(1+c+c2ck)aNThusineffect,wehavecomparedourserieswithageometricseries,andweknowthatthepartialsumsarebounded.Thisimpliesthatourseriesconverges.Theratiotestisusuallyusedinthecaseofaserieswithpositivetermssuchthat恤4=xoandbo.ThelimitcomparisonTestsayskkthatifi-1foralln.showthattheseriesEadiver

34、ges.annn1.3Alternatingseries,AbsoluteconvergenceandconditionalconvergenceInthissectionweconsiderseriesthathavebothpositiveandnegativeterms.1.3.1AlternatingseriesandthetestsforconvergenceTheseriesoftheform%-u+u-u+iscalledthe1234alternatingseries,where%oforalln,heretwoexample:n11111岩(-1)n-i1+.,23456nn

35、112一一+一212一一+一2334+456+-助n1Weseefromtheseexamplesthatthenthtermofanalternatingseriesistheforma(1)n1uora(1)nu,whereunnnnnisapositivenumber(infactua.)nnThefollowingtestsaysthatifthetermsofanalternatingseriesdecreasetoward0inabsolutevalue,thentheseriesconverges.Theorem1.3.1(LeibnizTheorem)Ifthealternat

36、ingseries(_i)nusatisfy:nn=1(1)uu(n=1,2);(2)limu=0,nn+1nnsthentheseriesconverges.Moreover,itissumsu,andtheerrorrmakebyusingsofthefirstntermstoapproximatennthesumsoftheseriesisnotmorethanu,thatis,卩|un+1nn+1namely卩|=卜-s|u.nnn+1Beforegivingtheproofletuslookatfigure1.3.1whichgivesapictureoftheideabehindt

37、heproof.Wefirstplots=uona11numberline.Tofindswesubtractu,sosistheleftofs.Thentofind2221sweaddu,sosistotherightofs.But,sinceu，Sinceus,sinceussinceuu2n2n-22n-12n2n-22n2n-1Thus0sss.s2462nButwecanalsowrites=u(uu)(uu).(uu)u2n123452n-22n-12nEveryterminbracketsispositive,sosuforalln.therefore,2n1thesequenc

38、esofevenpartialsumsisincreasingand2nboundedabove.Itisthereforeconvergentbythemonotonicsequencetheorem.Letscallitislimits,thatis,lims2nnsNowwecomputethelimitoftheoddpartialsums:lims二lim(s+u)2n+12n2n+1nsns-lims+limu二s+0(bycondition(2)二s2n2n+1nsnsSinceboththeevenandoddpartialsumsconvergetos,wehaveiims=

39、s,andsotheseriesisconvergent.nnsEX.1.3.1showsthatthefollowingalternatingharmonicseriesisconvergent:1_1+1-1+(-1)n-1.234nn=1Solutionthealternatingharmonicseriessatisfies(1)11;u(1)11;u=u=n+1n+1nn(2)limu=lim=0nnnsns1QSotheseriesisconvergentbyalternatingseriesTest.Ex.132Testtheseries艺(j)”3nforconvergence

40、and4n一1n=1divergence.3n3n4n一1limu=limnnsnsSocondition(2)isnotsatisfied.Instead,welookatthelimitofthenthtermoftheseries:divergesbythetestfordivergence.limanns=limanns=limns(T)3n4n一1Thislimitdoesnotexist,sotheseriesEX.133Testtheseries艺(-1)nforconvergenceorn2+1n=1divergence.Solutionthegivenseriesisalte

41、rnatingsowetrytoverifyconditions(1)and(2)ofthealternatingseriestest.Unlikethesituationinexample1.3.1,itisnotobviousthesequencegivenbyu=nisdecreasing.Ifweconsiderthenn2+1relatedfunction，weeasilyfindthatx2+1x2+1一2x21一x2=1.(x2+1)2(x2+1)2Thusfisdecreasingon1,)andsof(n)f(n+1).Therefore,uisdecreasingnWema

42、yalsoshowdirectlythatuu,thatisn+1nkk=1k=1kk=1k=1(n+1)2+1n2+1Thisinequalityitequivalenttotheonewegetbycrossn+n+1(n+1)2+1o(n+1)(n2+1)n(n+1)2+1n2+1on3+n2+n+1n3+2n2+2no11,weknowthattheinequalityn2+n1istrue.Therefore,uuanduisdecreasing.n+1nnCondition(2)isreadilyverified:convergentbytheAlternatingseriesTe

43、st.limunnslimunns=limnsthusthegivenseriesisAbsoluteandconditionalconvergenceInthissectionweconsiderseriesthathavebothpositiveandnegativeterms.Absoluteandconditionalconvergence.Definition131supposethattheseries艺aisnotserieswithkk=1positiveterms,iftheseriesQaformedwiththeabsolutekk=1valueofthetermsaco

44、nverges,theseries艺aiscallednkk=1absolutelyconvergent.Theseries艺iscalledconditionallykk=1convergent,iftheseries艺aconvergesbut艺|adiverges.kkTheorem九左2if工converges,the工aconverges.kkProofforeachk,|aa|,andtherefore0a+|a|2|a|.ifkkkkkk工”|converges,then工2|a|=2工|aconverges,andtherefore,bytheorem1.2.3(theordi

45、narycomparisontheorem),工(a+|a|)converges.Sincea=(a+|a|)-|a|bythetheoremkkkkkk(1),wecanconcludethat工aisconvergence.kTheabovetheoremwejustprovedsaysthatAbsolutelyconvergentseriesareconvergent.Aswellshowpresently,theconverseisfalse.Thereareconvergentseriesthatarenotabsolutelyconvergent;suchseriesarecal

46、ledconditionallyconvergent.EX.1.3.4Provethefollowingseriesisabsolutelyconvergent11111+22324252ProofIfwereplacetermbyitsabsolutevalue,weobtaintheseries.,1,1,1,十十十十223242ThisisaPserieswithP=2.Itisthereforeconvergent.Thismeansthattheinitialseriesisabsolutelyconvergent.EX.1.3.5provesthatthefollowingseri

47、esisabsolutelyconvergent:1111111111+222232425262728Proofifwereplaceeachtermbyitsabsolutevalue,weobtaintheseries:111111111+=+222232425262728Thisisaconvergentgeometricseries.Theinitialseriesisthereforeabsolutelyconvergent.Ex.1.3.6provesthatthefollowingseriesisonlyconditionallyconvergent:111111弋(一l)n23

48、456nn=1Proofthegivenseriesisconvergent.Sinceu=-1andsothepowerseries(1.4.3)alwaysconvergeswhenx=aintoaxkkk=0Sinceasimpletranslationconvertsaintoaxkkk=0kk=0wecanfocusourattentiononpowerseriesoftheformaxkkk=0Whendetailedindexingisunnecessary,wewillomititandwriteaxk.Webegindiscussionwithadefinition.kDef

49、inition1.4.1ApowerseriesaxkissaidtoconvergekAtx,iffaxkconverges1k1Onthesetsiffaxkconvergesforeachxins.Thekfollowingresultisfundamentalkk=0nn=0Theorem1.4.1AbelTheorem(1)Ifthepowerseries工axkkconvergesatx丰othenitconvergesabsolutely.Foranyx1that|x|x|(2)Ifthepowerseries工that|x|xJ.卜|xJ.ProofIf工axkconverge

50、s,thenxktoInparticular,forkAndthusaxk=axkxkxkkk1x1x1k1followsbycomparisonwiththek1followsbycomparisonwiththesufficientlylarge,axkik1For|xx/wehave兰|x|suchthat工axkconvergence.TheexistenceofsuchanxwouldimplytheabsoluteconvergenceofEaxk.Thisprovesthesecondstatement.k1Iftheconvergencesetforapowerseries艺a

51、xnisnot=0nn=0northeline,thenthereexistsapositivenumberR,suchthatifxfR,艺axnisabsolutelyconvergent.nn=0ifx|R,艺axnisdivergent.nn=0ifx=Randx=-R,艺axnmayconvergenceordivergent.nThepositivenumberRsatisfiedaboveconditions.(1)and(2)iscalledtheradiusofconvergenceofthepowerseries艺axn.nn=0Ingeneral,foragivenser

52、ies艺a(x_a)n,ifthereexistsann=0positivenumberRsuchthattheseriesconvergesif|x_R.ThepositivenumberRiscalledtheradiusofconvergenceofthepowerseries艺a(x_a)n.nn=0Fromthetheoremwejustprovedwecanseethatthereareexactlythreepossibilitiesforapowerseries.Case1.Theseriesconvergesonlyat0.Thisiswhathappenswith工kkxk

53、.Asyoucantellfromtheroottest,thekthterm,k0kxxk,tendsto0onlyatx=0.Case2.Theseriesconvergeseverywhereabsolutely.ThisiswhathappenswithexponentialseriesE兰.k!Case3.Thereexistsapositivenumberrsuchthattheseriesconvergesabsolutelyforxr.Thisiswhathappenswiththegeometricseries工hisinstance,thereisabsoluteconve

54、rgencefor艸1.Associatedwitheachcaseinaradiusofconvergence:Incase1,wesaythattheradiusofconvergenceis0.Incase2,wesaythattheradiusofconvergenceis.Incase3,wesaythattheradiusofconvergenceisr.Ingeneralthebehaviorofapowerseriesat-randrisnotpredictable.Forexample,theseriesExkpredictable.Forexample,theseriesE

55、xkyXkkXky_xkallhaveradiusofconvergence1,butwhilethefirstk2seriesconvergesonlyon(-1,1),thesecondseriesconvergeson-1,1),thethirdon(-1,1,andtheforthon-1,1.Themaximalintervalonwhichapowerseriesconvergesiscalledtheintervalofconvergence.Foraserieswithradiusofconvergencer,theintervalsofconvergencehavefourp

56、ossibilities.Theycanbe-r,r,-r,r),(-r,ror(-r,r).Foraserieswithradiusofconvergence0.Theintervalofconvergencereducestoapoint,0.Thereis,however,anotherwaytodeterminewhetherthepowerseriesconvergenceatendpointsofinterval.Theorem1.4.2(Ablestheorem)supposethatyaxkkk=0convergeson(-c,c)andthererepresentsf(X).

57、Iff(X)iscontinuousatoneoftheendpoints(cor-c)andtheseriesconvergesatthatendpoint,thentheseriesrepresentsthefunctionatthatendpoint,namelyf(c)=yackorf(-c)=k艺a（_c）k.Inthefollowingwegivethesemethodsoffindingkk=0theradiusofconvergence.Theorem1.4.3Iflimh=Iflimh=p,wherensanan+1arethecoefficientsofpowerR=+0Theorem1.4.4Iflim|aR=+0Theorem1.4.4Iflim|a1=p,nnspowerseries艺axn,thenthewhereaisthecoefficientofnradiusofconvergenceisgivenseries艺axn.Theradiusofconvergenceisgivenbynn=0PH0p=0P=EX.1.4.1Verifythattheseries工（_i）kxkkhasintervalofconvergence（