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1、Problem IntroductionHow to approximate the area of a circle with the area of regular polygons inside a circle?R012nAaaaa,this sum is approaching the area of the circle:n Then the area of an inscribed regular polygonwithsides in the circle is012.naaaa3 2nSetto represent the area of an inscribedequila
2、teral triangle andto represent the areaincreased when the number of sides increases.0akaDefinition of the Infinite seriesInfinite series:Partial sum:-th partial sum:1231nnknkSaaaaa12341kkaaaaathe sum of a finite number of the terms at the beginning of the series.DefinitionDefinition of the Infinite
3、seriesRemark 1:For any point ,written in some forms:The infinite series converges and has sum S if the sequence of partial sums converges to S.1kkanSIf diverges,then the series diverges.A divergent series has no sum.nSDefinition of the Infinite series1231nnnaaaaa(constant term)infinite seriesGeneral
4、 termEg:333;1010010n111111(1);234nn 11 1 1 1(1).n These are all(constant term)infinite series.Example 1Show that a geometric series converges,and has sum if ,but diverges if .Let 21nnSaarararIf ,nSnadivergeIf ,212nnnnnSrSaararararararaar111nnnaaraaSrrrrIf ,lim0,nnrconvergelim.1nnaSSrIf or ,diverge.d
5、iverge(1)Sar|1r|1r 1r 1r|1r|1r 1r Example 2Discuss the seriesconverges or not.1111 33 5(21)(21)nn1(21)(21)nnnu 111()2 2121nn1111 33 5(21)(21)nnns111 11111(1)()()232 352 2121nn11=(1-)22n+1Example 2Discuss the seriesconverges or not.1111 33 5(21)(21)nn11limlim(1)221nnnsn12converge,Remainder term:nnrss
6、11112221nNamely，12s 1the sum is.2112 21nA General Text for DivergenceConsider the geometric seriesonce more.21naarararExample 1 shows that a geometric series converges if and only if lim0.nnaIts-th term is given by .1nnaarTheorem A:th-Term Test for DivergenceIf the series converges,then 1nnalim0.nna
7、Equivalently,if or if does not exist,then the series diverges.lim0nnalimnnaLet be the-th partial sum and Proof:.limnnSS1nnnaSSand-1lim=lim=nnnnSSSSo,-1lim=limlim0nnnnnnaSSSSnSExample 3Show that diverges.332lim32nnnnlimnna1lim32nn13By the th-Term Test,the series diverges.3321 32nnnnThe Harmonic Serie
8、s11111123nnn shows that this is false.1lim=lim0nnnanHowever,the series diverges,as we now show.Clearly,The harmonic series:Student invariably wants to turn Theorem A around and make it say that implies convergence of .0na naThe Harmonic Series11111123nnn divergeln(1)xx(0)x,namely,nn11ln1 nknkS11nn1l
9、n34ln23ln2ln nn134232ln)1ln(n nnSlimdiverge nkk111ln)1ln(lim nn Proof:11001,()yxx 1ln()yxxThe Harmonic Series0 x 00()()y xy10ln()yxx1ln().xx10ln()()xxxExample 4Discuss these series converge or not:(1)(2)3125(21)(21)(23)nnnnnn11ln 333nnnnThe necessary condition for series convergence:It can be used t
10、o determine the divergence of the series.lim0,nnuSolution:Example 4(1)3125(21)(21)(23)nnnnnn nnulim81 diverge0)32)(12)(12(52lim3 nnnnnnExample 4(2)13(1)nnnnnlimnnu13lim11nnn0diverge3eTheorem B:Linearity of Convergent Series11;kkkkcaca(i)111.kkkkkkkabab(ii)If and both converge,and if is a constant,th
11、en and also converge,and1kka1kkb1kkca1kkkabTheorem CIf diverges and ,1kka1kkcathen diverges.0c Summary of the Infinite SeriesDefinition of the Infinite seriesA General Test for Divergencenth-Term Test for DivergenceThe Harmonic SeriesLinearity of Convergent SeriesTheorem CQuestions and Answers13lnnnais a geometric series,lnraSo,1when,aee,1|ln|aconvergediverge1when 0ae,1|ln|a1when1,convergewhen1,divergennrarror,aeDiscuss the series converges or not.13ln(0)nna aInfiniteSeries