The de57356nition of convergence The sequence converges to when this holds for any 57359 0 there exists such that for all Informally this says that as gets larger and larger the numbers get closer and closer to Butthe de57356nition is ID: 33175
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Cauchy'scriterionforconvergenceThedenitionofconvergenceThesequenceconvergestowhenthisholds:forany0thereexistssuchthatforallInformally,thissaysthatasgetslargerandlargerthenumbersgetcloserandcloserto.Butthedenitionissomethingyoucanworkwithprecisely.Ine ect,thedenitionsaysthatinordertoshowthataconvergestoyouhavetoexplainhowtogetfrom.Itisimportanttorealizethatyoudonothavendthebestpossiblevaluefor-justanythingthatworkswillenableyoutoverifyconvergence.ConvergentsubsequencesAnyboundedsequenceofrealnumbershasaconvergingsubsequence.Letthesequencebe.Wecanassumeallofonesign,becausewecanchooseaninnitesubseueqnceofonesign.Saynon-negative.Becauseofboundedness,wecanassumeallforsome.Aninnitenumberofthemusthavearstdigitincommon;throwawaytherest.Letbetherstinwhatisleft.Amongthoseleft,aninnitenumbermusthavethesameseconddigit.Letbetherstamongthese.Throwawaytherest.Etc.Thenaftertherstallthewillhavethesamerstdigits.Wecanconstructaparticularrealnumbertobethatnumberwhoserstdigitsagreewiththoseof.Thedi erencebetweenandwillbeatmost10.ThismakesiteasytoverifytheconvegencetoWhatthismeansforseriesAseriesisaformalexpressionanditcorrespondstothesequenceofpartialsumsTheseriesissaidtoconvergeifthesequenceofpartialsumsconverges.Ifaseriesconverges,thenitsindividualtermsmusthavelimit,butthisisnotasucientconditionforconvergence.Applyingthedenitionliterally,weseethattheseriesconvergestothenumberifforanythereexistssuchthatwhenever�nK.We'llseehowthisworksinpracticeinthenextsection.Thegeometricseries1,andconsidertheseriesAlgebratellsusthatthe-thpartialsumis=1+ Intuitively,thisshouldconvergeto)asgetslarger.Let'strytoverifythedenitioninthis Mathematics220-Cauchy'scriterion2 Wehaveexplicitly 1�x�1�xn 1�x=xn Sonowwehavetoverifythatforany0thereexistssuchthat .Butwecanpracticallytakeasgiveninthiscoursethatthisisso,orinotherwordsthatifthenthesequenceconvergesto0.Explicitly,wecansolve=(1;K ln(,andhenceSointhiscasewecanusethedenitiontoprovedirectlythatthegeometricserieswith1convergesto).Itisraretoknowexactlywhjataseriesconvergesto.Thegeometricseriesplaysacrucialroleinthesubjectforthisandotherreasons.Cauchy'scriterionThedenitionofconvergencereferstothenumbertowhichthesequenceconverges.Butitisraretoknowexplicitlywhataseriesconvergesto.Infact,thewholepointofseriesisoftenthattheyconvergetosomethinginterestingwhichyoumightnotknowhowtodescribeotherwise.Forexample,itisessentiallythedenitionofthatitisthenumbertowhichtheseries2+13!+converges.Thereforewhatisneededisacriterionforconvergencewhichisinternaltothesequence(asopposedtoexternal).Cauchy'scriterion.Thesequenceconvergestosomethingifandonlyifthisholds:foreverytheresuchthatwheneverThisisnecessaryandsucient.Toproveoneimplication:Supposethesequenceconverges,sayto.Thenbydenition,foreverywecanndsuchthatwhenever.Butthenifwearegiven0wecanndsuchthat=2for,andthen=forToprovetheother:Supposethecriterionholds.Weknowthatwehaveasubsequencewhichconvergestosome.Iclaimthatinfactthewholesequenceconvergestothissame.Weknowthatforanywecanndsuchthatfor.Wealsoknowthatifwearegiven0wecanndsuchthatforNowwewanttoprovethatforany0wecanndsuchthatforFirstchoosesuchthat=2for.Second,choosesuchthat= Mathematics220-Cauchy'scriterion3 for.Suppose.Choosesomewithbothand.Thenjj=ConvergencebycomparisonTheorem.Iftheseriesofnon-negativetermsconvergesandforeach,thentheseriesconvergesalso.Supposewearegiven0.ByCauchy'scriterion,weknowthatwecanndsuchthatfor.ButthenforthesameBecauseofthisLemma.(Cauchy'sinequality)WehavejjProvethisbyinduction,startingwith2.Wehavealreadyusedthisinequalitywithtwotermsinaprevioussection.Corollary.Iftheseriesconverges,thensodoesTheseriesissaidtoconvergeabsolutelyiftheseriesconverges.Corollary.forall,where,thentheseriesconverges.AnewexampleLet'snowlookattheseries Mathematics220-Cauchy'scriterion4 IclaimthatitconvergesforallNoneofthetechniquesmentionedsofarapplydirectlytoit.Itlookssomethinglikethegeometricseries,butthecoecientofgrowswithinsteadofremainingbounded.Weneedanewideatodealwithit.Itistruethatthecoecientsgrowwith,buttheydon'tgrowveryfast.Theyformanarithmeticprogres-sion,andinparticularTheorem.ThesequencegrowslessslowlythananygeometricsequenceThisisnotimmediatelyapparent.Ifiscloseto1thenthegeometricsequencestartsoutslowly,perhapsveryslowly.Nonethelesssoonerorlateritsurpassesthearithmeticprogression.Thisisnotquiteaseasytoseeasonemightlike,sinceitisnoteasytospecifythesmallestsuchthatLemma.thenforsomewehavewhenever=1+,andchooselargeenoughthatMx�N1.Thenbythebinomialtheorem,for=(1+=1++positivetermsMx�1)=Wecannowusethistoseethattheseriesconvergesfor1.Since1,wecannd1suchthat1also(say ).Wecanthenndsuchthatforall.ThisimpliesthatforwehaveButthenafterthersttermstheseriesisdominatedbythegeometricseriesfor,henceconvergesTheratiotestThesameargumentusedforoneseriesintheprevioussectioncanbeappliedtoprovethisaswell:Theorem.(Theratiotest)Supposeisaseriessuchthatthelimitofislessthan.Thentheseriesconverges.Thiswillshow,forexample,thattheseriesconvergesforTotestyourself:HowmanytermsofthisseriesarerequiredtocomputeitslimittowithinPowerseriesAnotherrelatedresultisthis:Theorem.Supposetheseriestoconverge.Thensodoallseries