《微积分》课程教学课件（Calculus）10. Infinite Sequences and Series

CALCULUSTHOMAS'CALCULUSARLYTRANSCENChapter 10JAMESSTEWARTInfinite Sequences and SeriesERMSNA10.11.Representing SequencesSequences广EMethod for representing sequcnces:AseguenceisalistofnumberMethod 1, Writing out the rulesO.o2.....The sequence2,4,6,8...can be described by the fomula a, 2n,in a given order. Each of aj, a2, ay and so on represents a tiumber. These are the terms ofwhere the index starts at I and increases.the sequence, For example, the sequenccThe sequence12,14,16,18...can be described by the formula福1a, isthe h term ofe sequnce,te ineger nis caled the idex fa,=10+2n, where the index starts at 1 and increases. It can bea,and ndicates wherea, occues n the lstdescribed by the formula b,=2n, where the index starts at 62. Onder is mportart.The sequenee 2,4,6,8.., B mot the sane as theand increases.sequence 4,2,6,8.3. An indinie sequence of nmbers sa finction wiose domain tlesetMethod 2. Listing termsof positive imtegers(a.)(2,4,6,8..,2n.,-.)白目

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 10 Infinite Sequences and Series Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 10.1 Sequences Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 1. Representing Sequences 1. an is the nth term of the sequence, the integer n is called the index of an , and indicates where an occurs in the list. 2. Order is important. The sequence 2,4,6,8. is not the same as the sequence 4,2,6,8. 3. An infinite sequence of numbers is a function whose domain is the set of positive integers. The sequence 2,4,6,8. can be described by the formula an=2n, where the index starts at 1 and increases. The sequence 12,14,16,18. can be described by the formula an=10+2n, where the index starts at 1 and increases. It can be described by the formula bn=2n, where the index starts at 6 and increases. Method for representing sequences: Method 1. Writing out the rules Method 2. Listing terms {an}={2,4,6,8,.,2n,.}

Method 3. Showing terms graphically2.Convergence and DivergenceI-AFIGuRE 10.1Sequencescanberepreseorad as pounts on the teal bpoinas in the planddaisuisharaftMHetemandtheveticalasisaisitsalae(s) (.i. ... ......yh→aa→wdiergeslima, =.0(5) {-+-(1 (1 —0 a n→converges3.CalculatingLimits oflimb,=0Sequences二→(e)-2多.comvergeslimc,=1(d.) (,,, .1, .-.(-*-3 dinerges((lyly boumnceback arad fonth benweenland-las n-→a)limd, doesn't exist.广A3Since sequences are fiunctions with domain restricted to the positive(0) ,m (-)=-I-mI =-→-0 =0imtegers, it is not surprising that the theorems on limits of functionshave versions for sequences,(b)m()m(1)-m1m-1-0-1THEOREM1Let (a-) and (b,) be sequences of real sumbers, and let 4and B be real mumbers, The following rules hold if limoa, 4 an55:0-0-0limcb,=B1. Sum Rafe:lim-a(au +b) =A + Boa-m--2. Diference Rule:limec(a, h) =A B3. Constanr Mulriple Rale:lime-aa(Ab,) =k+B(any number A)4. Product Rule:lim-a(a,-b,), A-B5. Quotient Rule:lim%=ifB*0A百

2016/11/15 2 Method 3. Showing terms graphically. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Convergence and Divergence n as n     1 1 ( 1) 0 n as n n      1 1 n as n n     1 {( 1) } 1 1 n bounce back and forth between and as n      converges converges diverges diverges lim n n a   lim 0 n n b   lim 1 n n c   lim ' . n n d doesn t exist  Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 3. Calculating Limits of Sequences Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Since sequences are functions with domain restricted to the positive integers, it is not surprising that the theorems on limits of functions have versions for sequences

THEOREM 3-The Continuous Function Theorem for SequencesTHEOREM 2The Sandwich Theorem for Sequences,.Let (a), (b,),and (c.)Let fa,) beasequence ofreal mumbers. Ifa, → L and if y is a fiunction that is continuous atbe sequences ofreal numbers.Ifa,S b,S c,holdsfor all n beyond some indexand defined at all as, then f(oa) → f(L).N, and iflim, a, = lim, c, = L, then limb, = Lalso.EXAMPLE5Show that V(n + T)/n ~→1Solution_ We know that (w + 1)/n →1. Taking f(x) = Vx and L =: 1 in Theorem 3gives Vin + 1/n-→ Vi =1.EXAMPLE6The sequence (1/n]) comerges to 0, By taking % = 1/n, J(x) = 2', and0L 0 in Theorem 3, we see thar 2/h = J(1/n) → f(L) = 2. 1. The sequence [2/g]FIGURE10.4The tems ofcomverges to 1 (Figure 10.5),sequence (b,] are sandhwichedbetween those of (a,) and (cal,forcing them to the sameconmon limit LAAETHEOREM4Suppose that f(x) is a function defined for all x ≥ sg and that[a,) is a sequcence of real numbers such that a, = f(n) for n ≥ np- Thena-L[im, () = La4.Using L'hopital's RuleEXAMPLE7Show thatn-0EXAMPLE8%- (3)coeve?o, find limaThe limitleadsthe indeteminateform Wecapply HpitalRuleiSolutionwe first change the form to oo-0 by taking the natural logarithm ofas:na =() --(e-)()5.CommonlyOccurringLimitsTha lng (e)linm1/n2n2/(2 .1)LHipealv Ride diinetiaielim2-lim1/n2Since In a, → 2 and f(x) e' is continuous, Theorem 4 tells us thatd, ghaelThe sequence (a,) converges toe白Slide 4-18

2016/11/15 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 4. Using L’hÔpital’s Rule Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 18 5. Commonly Occurring Limits

THEOREMSThe folowingsixseqaencescomerge to the timits listod below14-02 ln Vi=13. lm ** = 1(> 0)4.Jm*=0(/1)5(1+)-*m5-0 (amys)(any x)In Fomulas (3)toug (6),rmain fed06.RecursiveDefinitionsEXAMPLE9These arectamplesote limits iThocem5,a)a22号4→2:0=0Nemiat(b) VR =-= ()(P -1Fiemils [(c) Vm = 3)(g)→1-1 = 1Fomh 3 with r=3md Fomla 2(a (-)-0Femh4mith(()-(+%)-2fimma 5 wih--5Fomlb o wit= 0oSlde 4- 19slide 4- 20EXAMPLE10() Thesatemts a) = I da a-+ I for >I defie the seqacnce 1,3.of positive integers. With a = 1, we have og = aj + 1 = 2, ay = + 1 = 3,and so on,(b) The statements ay=1 md a, = na-- for >1 define the sequence1,2,6,24,..-,l..of factorials.With a 1,we hane 2-a 2.dy 3.a = 6, a 4.ay 24,and so on.(c) The statements a = 1, a= 1,and a1= a, + au-1 foe > 2 define the se-7.Bounded Monotonicquence1.2.3,5,..ofFibaccl mbs.Wihaand1,weh)=1+1=2.m=2+1=3,0=3+2=5.amdsoom.Sequences(d) As we can see by applying Newton method (see Exercise 133),the statementsxo=1and xe+1=,[(sinxx)/(cosx2xa)] forn>0defineasequencethat, wben it comverges, gives.a solution to the equation sin x x= 0..slide 4- 21side 4- 22aEXAMPLE11Two concepts that play a key role in determining the comvergence ofasequence are thoseof a bounded sequence and a monotonic sequence.(a)The sequence 1,,3hasno upperbound since iteventuly surpassesevery number M. However, it is bounded below by every real mumber less than orequal to 1, The number m 1 is the greatest lower bound of the sequence.DEFINITIONSAsequence fa,]isboundedfrom above if thereexists anumbrMsuchthatMfoallThenumbrMianuppundhe(a.).I Mis an upper bound for (a.) but po mumber less than Mis an upperthan or equal to I. The upper bound M = 1 is the least upper bound (Exerecise 125).boundfor (a.)thenMisthe leastupperbound for (a.]Asequence (a,)isboundedfrombelow ifthereexistsanumberm such thatThe sequence is also bounded below by every mumber lessthanorequal to whichisa, ≥ m for all n. The ntmber m is a lower bound for (da]. If m is a lower boundits greatest lower bound..for (a)butnonumbergreater thanm isalowerbound for (a.l,thenm isthegreatest lower bound for (a_).If (a.) is bounded from above and below, then (a.) is bounded. If(au) is notbounded, then we say that (ae) is an unbounded sequenceSlide 4- 23Slide 4- 24

2016/11/15 4 Slide 4 - 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 20 6. Recursive Definitions Slide 4 - 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 22 7. Bounded Monotonic Sequences Slide 4 - 23 Slide 4 - 24

EXAMPLE12Finding a Sequesce's FormulExercises(a) The sequence 1, 2, 3,..., n,..is nondecreasingInEserciscs1326,findaformulaforthe nthtermofthesoqucocc13, The sequence 1, 1, I, 1, 1,...wialien)hese.odain14, The seqoenee 1, 1, 1, 1, 1,.Dwoalenmt(hen...onnsing15, The sequence 1, 4, 9, 16, 25,har(d)hca3bhoningdinie16nesegpence16(e) The sequence 1, 1, 1, 1, 1, 1,... is not monotonic1品品#蛋ofConvergence and OivergenceTHEOREM 6—The Monotonic Sequence TheoremIf a sequence (ou) is bothWhich of the sequences (a-] in Exerises 2790 comverg,and whichbounded and monotonic, then the sequence comverges.diverge? Find the limit of each comvergemt sequence#+ (-1)Ineorem 6 doesn't say that convergent sequences are monotonic.27. 4, = 2 + (0.1)28. 4,Forexample, the sequence (-1p*1/n) comverges and is bounded, but i isn2v29.4-1-230, a,1-3Vnmonotonic since it altemates between positive and negative values as i tendstowards zero.1-5g3L. a, =32. a, 7+5+6n+8lide 4- 25alide 4- 2810.21.n-thPartial SumInfinite SeriesSlide 4- 27Slide 4- 28An infinite serigs is the sum of an infinite sequence of numbersDEFINITIONSGiven a soquence of numbers (a,], an expressioe of the forina++a+ay+.+a,+..=++0nthpartial sumis an infinite series. The number o, is the ath term of thcThe:seqaeno(s,) defined b1+++++++#+-2431 0)=a+gSuggestive expresdion282lm..=lim(2-Valuefor partial sum,:The sum of the ifinjte series 1+-71！5-14号Second14is the sequence of partial sums of the series,the mumber s, being the ath partialsum.Iftheseqenceofpartial sums comernges to alimitL,wesay that tbe series1/2comverges and thut its sum is L, ln this case, we also writeFGURE10.8Ahoghs addedyne2?L+dapprooches 2,The partial sums formquenee whose.wthterm isIf the se say that thofpalseries diverges$-2-寸Slide 4- 29Slide 4-30

2016/11/15 5 Slide 4 - 25 Theorem 6 doesn’t say that convergent sequences are monotonic. For example, the sequence {(-1)n+1/n} converges and is bounded, but it isn’t monotonic since it alternates between positive and negative values as it tends towards zero. Slide 4 - 26 Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 27 10.2 Infinite Series Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 28 1. n-th Partial Sum Slide 4 - 29 nth partial sum 1 1 lim lim(2 ) 2 2 n n n n S       1 1 1 1 The sum of the infinite series1 is 2. 2 4 2n      1 n n a     Slide 4 - 30

Geometric series are series of the formin which a and r are fixed real nuimbers and a. *.0Theratiercambepoutive,usi1+#+++1/2,1=2.Geometric Seriesornegatiwg as in1-++--+(-1/3,a-I teh pantial sum ofthe gometricseriesiSa+ a(1) + a(1) +.. +o1-1 m,and the series divenges because lime s, - ±oo, depending on the sign ofa. Ifr = 1,the series diverges because the wth partial sums alternatebetween a and 0Slide 4-31Slide 4-32+aIfir/≤1, thensa+ar+ar+EXAMPLE1The geometric series with a = 1/9 and r 1/3 is+ar ++.+ arr.Muhtiphy J, by1/9+++++*()1-0/) -Iudrsfmm.Motoare5(1 r) = a(1 -uEXAMPLE2The seriesa(l -r)(r→1)2-5--+-+Ii 1,04thenjr→ oo and the series diverngesisageometricseries wih= Sandr=1/4.Itcovergesto1-,"1+ (/4) = 4,1f]1] < 1, the geometric' series a--comvergestoa/(1 r):aI/<1.-rIf|r] ≥ 1, the series divengesSlide 4- 33Slide 4- 34EXAMPLE3You drop a ballfirom a mcters abovea flat surface. Each time the ballhitsExercisesthesurfaee ifterfllingadistanceh,itreboundsadistance rh, whererispositivebutlessFinding nth Partial Sumsthan 1. Find the total distance the balltravels up and down (Figure 10.9),In Exercises I-6, find a formula for the nth partial sum of each seriesand use it to find the series'sum if the series converges.1.2+1+#+#+++..2++100+100100210033.1-+++-+++(1)-1 +..2814. 1 2 + 4 8 + -+ (1)1 2*1 + ..The total distance isIn Exercises 1518, determine if the geometric series converges or di-inghaBifdel=a+20r+2++2verges.Ifaseries comverges,find its sum15. 1+()+() + () + (图) +If a = 6 m and r = 2/3, foe instance, the distance i+ (2/3)(5/3)16. 1 + (3) + (3) + (3) + (3)* + 4.= 30 -(2/3)-()slide 4. 35Slide 4-38

2016/11/15 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 31 2. Geometric Series Slide 4 - 32 Slide 4 - 33 Slide 4 - 34 Slide 4 - 35 Slide 4 - 36 Exercises

PxerEisesFind thesum ofthle 'elescoping"seriesEXAMPLE5+1)Find the sum of each series in Exercises 4148SolutionWe look for a pattern in the sequence of partial sums that might lead to a for-8W546mula for Sg, The key observation is the purtial fraction decomposition41.42.台 (4n 3)(4n + 1)台 (2n 1)(2n + 1)+---405_2n +1so43.44. (2n 1)(2# + 1)2台n（n+1)2高品烹(-+)and#-（1)+（送-）+（-）++（Removingparenlsign collapses the sum tcndcingadacfop$-1-本We now seethat S1as k→ oc,The series converges,and its sum is 1:S2+=."Slide 4-37Slide 4-38EXAMPLE6The series+-++++diverges because the partial sums eventually outgrow every preassigned number. Eachtemgreaterhanoesumoftesigeaterth3.Thenth-TermTestforaTHEOREM?Sges, then a,Divergent SeriesThe nth-Term Test for DihergenceiieaSlide 4- 39Slide 4- 40ExercisesEXAMPLE7The following are all examples of divergenit'series.Using the nth-Term TestNr diverges eausen'- 0(a)In Exercises 27-34, use the nth-Term Test for divergence to show thatthe series is divergent, or state that the test is inconclusive.FA+IIdivegsbcause "t!1(b)in on(n+1)5127. 28.高+102(m+2)(n+3)(c)S30.2129.台28+4+3An(d)22 diegs bease im 21 - 0.Slide 4- 41Slide 4- 42

2016/11/15 7 Slide 4 - 37 Slide 4 - 38 Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 39 3. The nth-Term Test for a Divergent Series Slide 4 - 40 Slide 4 - 41 Slide 4 - 42 Exercises

THEOREM8I Ea, =Aand b, =Bare comvergeut series, thenLSum Rale:Ea, +b)-Ea, +h,=A+82. Diference Rule:E(a,h)-Ea,-Eh, - A -B3,Constamt Mulriple RaleSo,-a-4(any mumber ),CorollariesEvery nonzero constant multiple ofadivengent seriesdivenges4.Combing Series2. If Za, comerges and Eb, dinerges,them (a, + h) and (a, h) bothdivergeRemember thar E(a+ b) can convege when Za, and b, both divergeCantionForexample,Za,=1+1+1++ and Zb, = (1) + (1) + (1) +--divernge.whereas (a, + b,)=0 + 0 + 0+--coverges to 0.lide 4- 43Slide 4- 44EXAMPLE9Findthe sums ofthefollowing series-()6--Differesce.Ral10.3a1(/2) ~1 (0/6)-2-号-号024-42The Integral TestMunipirRt-(r-(s)cotdc mries wie a.= 1,r = 1/2-8slide 4- 45Slide 4- 48Suppose that Zht a, is an infinite series with a, ≥ 0 for all n. Then each partial sum isgreater than or equal to its predecessor because S+1 S, + ae:SSS≤s≤Since the partial sums form a nondecreasing sequence, the Monotonic Sequence Theorem(Theorem 6, Section 10.1) gives the following result.1.NondecreasingPartial SumsAseriesfoetemmesCarollary af Theorem 6and only if its partial sums are bounded from aboveSlide 4- 47Slide 4.48

2016/11/15 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 43 4. Combing Series Slide 4 - 44 Corollaries Slide 4 - 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 46 10.3 The Integral Test Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 47 1. Nondecreasing Partial Sums Slide 4 - 48

EXAMPLE1The series-++is called the harmonic series. The harmonic series is divergent, but this doesn't followfhmhssreason it diverges is because there is no upper bound for its partial sums. To see whys2. The Integral Test+++($+)+(#+++)+(++++)+兰lide 4- 49Slide 4- 50EXAMPLE2Does the followerge营-+++忘+THEOREM 9-TheIntegral TestLet (a) be a sequence of positive termsSuppose that a,=f(n),where fis acontinuous,positive, decreasing function ofWe detemine the comergence of ;(1/n') by comparing it withSolutionXfor all ≥ N (Napositive integen),Then theseries mxdand the integnal(/yttaJy f(x) dr both comverge or both diverge.the function f(x) = 1/r* and interpret these values as the areas of rectangles under thecurve y = 1/rAs Figure 10.10 shows,.AiThe series and integral need not have the same value in the.++++++++convergent case.CunaRo"= y(1) + (2) + J(3) +++ Nn)+hen1/oiidcingfinetiSin10.4广-[]()古（-1吉b1Comparison TeststhestiesCepy thelmeal TeWeempasizsthaesmoeses i1/1)TheseriescoexbtwedonknowthevaluooivgoIfp 0and【01---0The series diverges by the lntegral TestIf p = 1, we have the (divergest) harmonic serics1++++++++.We haveergence for p > 1 but idivergence forallother values of p.Slide 4- 53Slide 4-54

2016/11/15 9 Slide 4 - 49 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 50 2. The Integral Test Slide 4 - 51 Slide 4 - 52 The series and integral need not have the same value in the convergent case. Slide 4 - 53 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 54 10.4 Comparison Tests

THEOREM 10—The Conparison TestLetSaScandSd,be serieswithnonnegative lerms, Suppose that for some inmtegerNfor all">Nd, SoS c,(a) Ir c comerges,then aalsocomeges.(b) Ir Ed, diverges,then Za, also divenges1.TheComparisonTestEXAMPLE1We apply Theorem 10 to several series.(a) The seriesMdiverges because its nth term5#-1is greater than the wth tem of the divergent harmonicseriesslide 4-55Slide 4- 58(b) The series含克-1++++converges because its tems are all positive and less than or equal to the correspon-ding terms of1+#-1+1+++++42.TheLimit Comparison Testft comverges and we haveThe geometricsonVI++-1+1-(/2) -3.0The fact thut 3 ibs an upper bound fiur the partial sums of Zr (1/et) does not meanthat the series comverges to 3. As we wll see in Section 10.9,the series converges to eSlide 4- s7Slide 4- s8EXAMPLE2Which ofthe following series comenge,and whichdn+++县+-高+3+THEOREN 11-LImit Comparisen TestSuppose that a, > 0 and b, > 0 forall n ≥ N (Nan integer).(6)+++++++-thnEdbohcoeobhdieI+an#( ++gm2,++3+1+h+++2SolutionWeapplytheLimitComparisoTeschserie(a) Let g = (2n + 1/(P + 2n + 1),For lange m.we expect a, to behune liko2n/e = 2/n since the leading terms dominate for large n, so we let b_= 1/n. SinceSn-ZlahmeandwF+2h+1*2.Za diverges by Pat I of the Limit Comparison Test We could just as well havetaken h, = 2/n, but 1/w is simplerslide 4- 60Slide 4- 59

2016/11/15 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 55 1. The Comparison Test Slide 4 - 56 Slide 4 - 57 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 58 2. The Limit Comparison Test Slide 4 - 59 Slide 4 - 60