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