使用莱布尼兹审敛法证明交错级数敛散性的几种注记

JournalofAnshanNormalUniversity2005204,7(2):10-11(,114044):,,,.:;;;:O13:A:100822441(2005)0220010202SomeAnnotationsonCourseofProvingDisappearanceandDivergenceofanAlternateProgressionbyLeibnizePrincipleAIYi2min(AnshanUniversityofScienceandTechnology,AnshanLiaoning114044,China)Abstract:ThepaperpointsoutsomekeyswhendeduingbyLeibrizePrinciple,concludinghowtoproveitsdisappearanceanddivergencewhicharethetypicalfeaturesofanalternateprogression.ItalsopresentssomejudgementsofsuchfeatureswhenLeibnizePrinciplefailstobeused.Keywords:Alternateprogression;Disappearance;Divergence;LeibnizePrinciple,,.,.1.[1]6n=1(-1)n-1un6n=1(-1)nun,un0,.2.[1]()6n=1(-1)n-1un(un0).:i)limnun=0;ii)unun+1(n=1,2,).6n=1(-1)n-1un,sFu1,()rn|rn|Fun+1.16n=1(-1)n-1lnnn.(1)limnun=0,f(x)=lnxxlimx+f(x)=limx+lnxx=limx+1x=0,:2004-05-11:(1965-),,,.unn0.(2)unEun+1,(un)=lnnn=2-lnn2nn,ne2,(un)0,ne2,un.(1),(2),nEe2,n1FnFe2,,:,6n=1(-1)n-1lnnn.1,,(2),[2]:(1)un-un+10;(2)unun+11;(3)un=f(n)f(x),f(x).1,,.,.,..3.(1)limnun0i),,.(2)limnun=0,i),ii),.26n=2(-1)nn+(-1)n.i),limn1n+(-1)n=0,ii)u20.5774,u30.7072,,ii).:S2n=(1/3-1/2)+(1/5-1/4)++(1/2n+1-1/2n),S2n.S2n(1/4-1/2)+(1/6-1/4)++(1/2n+2-1/2n)=-1/2+1/2n+2-1/2,0S2n-1/2,|S2n|1/2,S2n.limnS2nS,limnS2n=S.limnS2n+1=limn(S2n+u2n+1)=limnS2n+limnu2n+1=S+limn(-1)2n+1(2n+1)+(-1)2n+1=S,limnSn=S,.:(1),limnun0,,.(2)limnun=0,unun+1,,,,.:[1],,.[M].:,1999.[2],.[M].:,1999.(:)112: