《高数双语》课件section 7-2.pptx

1、Convergence Tests for Series with Constant Terms1Convergence Tests for Series of Positive Terms2Definition (series of positive terms)A seriesis called a series of positive terms 正项级数正项级数(or series of non-negative terms)or simply positive series if the general term an 0,n=1,2,.1nnaA series of positiv

2、e terms is convergent iff the sequence of its partial sums is bounded above.The partial sums are non-decreasing because11nnnssaand0.na 3Integral TestShow that the series2211111114916nnn LLconverges.SolutionWe determine the convergence of this series by comparingit with211.dxx Think of the terms of t

3、he series as values of the function21()f xx and interpret these value as area of rectangles under the curve 21.yx 211.dxx 4Integral TestSolution(continued)22221111123nsnL(1)(2)(3)()ffff nL211(1)nfdxx 2111dxx 112.Thus the partial sum of the series are bounded from above(by 2)and the series converges.

4、The sum of the series is known to be 21.64493.6 It is easy to see that5Integral TestTheorem (Integral Test 积分判别法积分判别法)Let an be a sequence of positive terms.Suppose that an=f(n),where f is a continuous,positive,decreasing function of x for all x N(N is a positive integer).Then the series and the int

5、egral both converge or both diverge.nn Na ()Nf x dx 6Applying the Integral TestDoes converge?11 nn nSolutionThe integral test applies because1()f xxx positive,decreasing function of x for x 1.We have3/2111limkkdxxdxxx 1/21lim2kkx 2lim22.kkThe integral converges,so must the series.is a continuous,7p-

6、Series and Harmonic SeriesThe Integral Test can be used to settle the question of convergence for any series of the form ,p a real constant.Such series are called p-series.11pnn The p-series(p a real constant)111111123 pppppnnnLLconverges if p 1 and diverges if p 1.The p-series with p=1 is the harmo

7、nic series,and it isprobably the most famous divergent series in mathematics.8Direct Comparison TestSuppose and are positive series and N N+,such that for all n N.(b)diverges if diverges.nb(a)converges if converges.na nnab na na nb nb 9Direct Comparison TestIn part(a),the partial sums of are bounded

8、 above byna 121.nnn NMaaab LThey therefore form a non-decreasing sequence with a limit.LM So,if is a convergent series,thennb converges.na ProofProofSuppose and are positive series and N N+,such that for all n N.(b)diverges if diverges.(a)converges if converges.nnab nb nb na nb na na 10Direct Compar

9、ison TestProof In part(b),the partial sums of are not bounded from above.nb If they were,the partial sums for would be bounded byna*121.Nnn NMaaab Land would have to converge instead of diverge.na ProofSuppose and are positive series and N N+,such that for all n N.(b)diverges if diverges.(a)converge

10、s if converges.nnab nb nb na nb na na 11Direct Comparison TestTo apply the Direct Comparison Test to a series,we need not include the early terms of the series.We can start the test with any index N provided that we include all the terms of the series being tested from there on.ProofSuppose and are

11、positive series and N N+,such that for all n N.(b)diverges if diverges.(a)converges if converges.nnab nb nb na nb na na 12Applying the Direct Comparison TestDoes the following series converge?211111513723!4!k LLSolution We ignore the first four terms and compare the remainingterms with those of the

12、convergent geometric series .112nn We see that11111123!4!248LLTherefore,the original series converges by the Direct Comparison Test.To apply the Direct Comparison Test,we need to have on hand a list of series whose convergence or divergence we know.The next table shows a list of what we know so far:

13、13Some Important Convergent and Divergent SeriesConvergent SeriesDivergent SeriesGeometric Series with|1.q The harmonic series11.nn Any p-series with 11pnn 1.p Geometric Series with|1.q Telescoping Series like11.(1)nn n The Series01!nn Any p-series with 11pnn 1.p Any series for which thena limnna do

14、es not exist or lim0nna 14Applying the Direct Comparison Test Discuss the convergence of the following series:1111)sin;2).5(1)nnnn n (1)Since n N+,and the geometric series converges,by the Direct Comparison Test,the series converges as well.1sin5nn sin,55nn 15nn Solution 15Applying the Direct Compar

15、ison TestSolution(2)Since n N+,and the harmonic series diverges,by the Direct Comparison Test,diverges as well.11(1)nn n 111,12(1)nnn n 112nn Discuss the convergence of the following series:1111)sin;2).5(1)nnnn n 16Limit Comparison TestSuppose and are positive series,and bn0 for n N+,na lim.nnnab nb

16、(1)If ,then the two series converge or diverge simultaneously;(2)If and converges,then converges;(3)If =+and diverges,then diverges.0 0 na nb na nb 17Limit Comparison TestThus,for nN 22nnab 322nnab3,22nnnbabBy the Direct Comparison Test,we have the conclusion(1).Proof of Part 1Since ,there exists an

17、 integer N such that for02 .2nnanNb (1)If ,then the two series converge or diverge simultaneously;0 limnnnab 18Using the Limit Comparison Test(d)2212ln213ln314ln41ln914215nnnn L(a)221135792121491625(1)21nnnnnnnL(b)1111111371521nn LExample Discuss the convergence of the following series:(c)1111111ln2

18、ln(1)ln(1)ln(1)2233nnn L19Using the Limit Comparison Test(a)221135792121491625(1)21nnnnnnnLSolution like ,so we let 222nnn 1.nbn Let221.21nnann For n large,we expect an to behaveSince 111nnnbn diverges and 222limlim2,21nnnnannbnn na diverges by part 1 of the Limit Comparison Test.20Using the Limit C

19、omparison TestSolution Let1.21nna like ,so we let 12n1.2nnb For n large,we expect an to behaveSince 1112nnnnb converges and 21limlimlim1,211(1/2)nnnnnnnnabna converges by part 1 of the Limit Comparison Test.(b)1111111371521nn L21Using the Limit Comparison Test(c)1111111ln2ln(1)ln(1)ln(1)2233nnn LSol

20、ution Let11ln(1).nannSince11ln(1),nn to behave like ,we let 21n21.nbn we expect anSince 2111nnnbn converges and 21ln(1)1limlimlim ln(1)1,1nnnnnnannbnn na converges by part 1 of the Limit Comparison Test.22Using the Limit Comparison TestSolution Let21ln.5nnnan For n large,we expect an to behaveSince

21、111nnnbn diverges and 22lnlimlim,5nnnnannnbn na diverges by part 3 of the Limit Comparison Test.(d)2212ln213ln314ln41ln914215nnnn Llike ,which is greater than for ,so we take 2lnlnnnnnn 1.nbn 1n3n 23Ratio TestThe Ratio Test measures the rate of growth(or decline)of a series by examining the ratio .F

22、or a geometric series ,the rate is a constant(),and the series converges if and only if Its ratio is less that 1 in absolute value.The Ratio Test is a powerful rule extending that result.1nnaa naq 1()()nnaqaqq Let be a series of positive terms and suppose that .Then (1)the series converges if .(2)th

23、e series diverges if .(3)the test is inconclusive if .na 1limnnnaa 01 1 1 24Ratio TestProof of Ratio TestPart(1)01 Let q be a number between and 1.Since 1,nnaa That is,1NNaqa 221NNNaqaq a1mNmNmNarar aThese inequalities show that the terms of our series,after Nth term,approach zero more rapidly than

24、the terms in a geometric series with ratio1.q Then the number is qpositive.25Ratio TestProof of Ratio Testnc More precisely,consider the series ,where for and for all m,andnnca 1,2,.,nN 212,.,.mNNNNNmNcqacq acq a21211nNNNNncaaaaqaq a LL2121(1).NNaaaaqq LLThe geometric series converges because ,then2

25、1qqL|1q nc Since ,also converges.nnac na(continued)Part(1)01 converges.26Ratio TestProof of Ratio TestPart(3)1 From some index N on N+,11nnaa and12.NNNaaa LThe terms of the series do not approach zero as n becomes infinite,and the series diverges by the nth-Term Test.Part(2)1 The two series and show

26、 that some other11nn 211nn test for convergence must be used when .1 For :211nn 22121(1)1.11nnannann 11nn For :11(1)1.11nnannann 27Using the Ratio Test(a)1!5nnn (b)11nnn Discuss the convergence of the following series:Solution(1)By the Ratio Test,sincethe series diverges.11(1)!()555nnnnnn 11(1)!5nnn

27、 (2)By the Ratio Test,the series converges.11nnn (1)11110()1(1)(1)nnnnnnnn 28Root TestThe nth-Root Test is another useful tool for answering the question of convergence for series with nonnegative terms.We state the result here without proof.Let be a series with for ,and suppose that Then (1)the ser

28、ies converges if .(2)the series diverges if .(3)the test is inconclusive if .na lim.nnna 1 1 1 0na nN Exercise Which of the following series converge,and which diverge?(a)(b)212nnn 212nnn 29Alternating Series11111(1)12345nn LL1123456(1)nn LLA series in which the terms are alternately positive and ne

29、gative is an alternating series交错级数交错级数.111(1)or(1)(0)nnnnnnnaaa For example,30Alternating Series TestThe series of Leibniz form converges,and its sum1.Sa The alternating series is a series of Leibniz Form,if112341(1)(0)nnnnaaaaaa L1nnaa (1)for all ,for some integer N.nN 0().nan (2)Alternating Serie

30、s Test31Proof If n is an even integer,say ,then the sum of the first 2nm nterm is21234212()()()mmmsaaaaaa L1234522212()()()mmmaaaaaaaaL0 0 0 222mmss 21msa Since is nondecreasing and bounded from above,it has a limit,2ms2limmmsL If n is an odd integer,say ,then the sum of the first n termsis .Since ,

31、and as ,21nm21221mmmssa0na m 212210.mmmssuLL(2)(1)1lim,.nnsL LaCombining the results of equation(1)and(2)gives32Alternating Series Test11(1)nnn Discuss the convergence of the series(1)(2)11(1)21nnn Solution1)Let .The sequence is monotone decreasing and 1nan nalim0,nna so the given series is of Leibn

32、iz form.It is convergent.2)Let .The sequence is monotone decreasing and 121nan nalim0,nna so the given series is of Leibniz form.It is convergent.33General Series and Tests for ConvergenceA series converges conditionally if it converges but does not converge absolutely.The series converges absolutel

33、y (is absolutely convergent)if the corresponding series of absolute values,converges.na|na Absolute convergence is important for two reasons.First,we have good tests for convergence of series of positive terms.Second,if a series converges absolutely,then it converges.34Absolute Convergence Test If c

34、onverges,then converges.1|nna 1nna Proof For each n,|nnnaaaso0|2|nnnaaaIf converges,then converges and,by the Direct Comparison Test,1|nna 12|nna 1(|)nnnaa lets us express the series as the difference of|nnaa the nonnegative seriesconverges.The equality1111(|)(|)|.nnnnnnnnnnnaaaaaaaTherefore,converg

35、es.1nna two convergent series:35Absolute Convergence Test The alternating series is converges because it converges absolutely.1211(1)nnn If converges,then converges.1|nna 1nna Absolute Convergence Test36 Discuss the convergence of the following series;if it converges,is it absolutely or conditionall

36、y convergent?111(1)();(2)(1)(21).!nnnnnxxRn Solution(1)If x=0,obviously,the series converges.If x 0,set .Since!nnxan 110,as (1)!1nnnnxxxannnna the series converges.So,the given series converges absolutely.1!nnxn Absolute Convergence Test37Solution Set and .1(1)(21)21nnnna 1nbn(2)It is easy to see th

37、e given series is of Leibniz form.11(1)(21)nnn converges.1212111nnnnabnn Discuss the convergence of the following series;if it converges,is it absolutely or conditionally convergent?111(1)();(2)(1)(21).!nnnnnxxRn Absolute Convergence Test38Solutionso 00212 ln2limlimln2,1xxxxx 121limln2.1nnn(2)Let 1,

38、0 as.xxnn The series diverges,so the series is conditionally convergent.1nna 11(1)(21)nnn Discuss the convergence of the following series;if it converges,is it absolutely or conditionally convergent?111(1)();(2)(1)(21).!nnnnnxxRn 1212111nnnnabnn39Rearranging SeriesIf converges absolutely and is any

39、arrangementof the sequence ,then converges absolutely and1nna nbna1nnb 11.nnnnba CAUTION If we rearrange infinitely many terms of a conditionally convergent series,we can get results that are far different from the sum of the original series.40123111 112131112212223212331323331311 112131111231nnnnnn

40、nnnnnnnnnnnnnnnnnnbbbbbaa ba ba ba ba baa ba ba ba ba baa ba ba ba ba baabababababaa ba ba ba ba b LLLLLLLLMMMMMMLLLLMMMMMMCauchys Product The product of two series and is a series ,wherena nb nc 121121.nnnnnca ba baba bLThis type of product is called a Cauchys Product.41Multiplication of Absolutely

41、 Convergent Series Theorem(Convergence of the product)If the series and are both converge absolutely and their sums are A and B,respectively.Then the series of products is also converges absolutely and its sum is AB.na nb 11nnnnab This conclusion can be thought as a special result of the series of f

42、unctions,which we will learn next lecture.Multiplication of Absolutely Convergent Series 42 Discuss the convergence of the product00nnnnxx and find its sum.SolutionThe geometric series is absolutely convergent for .0nnx 1x The given product converges for ,and the Cauchy product is 1x 211000123nnnnnnnxxxxnxnx When ,.Hence,1x 011nnxx 2001if1.(1)nnnnxxxx