Sequences II. Chapter Convergent Sequences


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1 Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0., ǫ = 0.0 ad ǫ = 0.00 fid a N such that a < ǫ wheever > N. Defiitio Let a R. A sequece a ) teds to a if, for each ǫ > 0, there exists a atural umber N such that a a < ǫ for all > N. See figure 3. for a illustratio of this defiitio. We use the otatio a ) a, a a, as ad lim a = a ad say that a ) coverges to a, or the limit of the sequece a ) as teds to ifiity is a. Example Prove a ) = ) +. Let ǫ > 0. We have to fid a atural umber N so that a = + < ǫ whe > N. We have + = + = + <. Good Nough Ay N that works is good eough  it does t have to be the smallest possible N. Recycle Have a closer look at figure 3., what has bee chaged from figure2.6ofchapter2? Itturs out that this defiitio is very similar to the defiitio of a ull sequece. Elephats Revisited A ull sequece is a special case of a coverget sequece. So memorise this defiitio ad get the other oe for free. Hece it suffices to fid N so that < ǫ wheever > N. But < ǫ if ad oly if ǫ < so it is eough to choose N to be a atural umber with N > ǫ. The, if > N we have a = + = + = + < < N < ǫ. Lemma a ) a if ad oly if a a) 0. 27
2 28 CHAPTER 3. SEQUENCES II e e a 0 N Figure 3.: Coverget sequeces; first choose ε, the fid N. Proof. We kow that a a) 0 meas that for each ǫ > 0, there exists a atural umber N such that a a < ǫ whe > N. But this is exactly the defiitio of a ) a. We have spoke of the limit of a sequece but ca a sequece have more tha oe limit? The aswer had better be No or our defiitio is suspect. Theorem Uiqueess of Limits A sequece caot coverge to more tha oe limit. Exercise Prove the theorem by assumig a ) a, a ) b with a < b ad obtaiig a cotradictio. [Hit: try drawig a graph of the sequeces with a ad b marked o] Theorem Every coverget sequece is bouded. Exercise 2 Prove the theorem above. Coectio It wo t have escaped your otice that the Sum Rule for ull sequeces is just a special case of the Sum Rule for sequeces. The same goes for the Product Rule. Why do t we have a Quotiet Rule for ull sequeces? 3.2 Algebra of Limits Theorem Sum Rule, Product Rule ad Quotiet Rule Let a,b R. Suppose a ) a ad b ) b. The ca +db ) ca+db a b Sum Rule for Sequeces a b ) ab Product Rule for Sequeces ) a b, ifb 0 Quotiet Rule for Sequeces
3 3.2. ALGEBRA OF LIMITS 29 There is aother useful way we ca express all these rules: If a ) ad b ) are coverget the lim lim ca +db ) = c lim a +d lim b lim a b ) = lim a lim b a b ) = lim a ), if lim lim b ) b ) 0 Sum Rule Product Rule Quotiet Rule Example I full detail 2 +)6 ) lim = lim ) ) usig the Quotiet Rule [ + 2 ) 6 = lim lim 2+ 5 = 3 ) )] usig the Product ad Sum Rules )) +lim 2 6 lim )) = +0)6 0) 2+0 = 3 2+5lim 3 ) Uless you are asked to show where you use each of the rules you ca keep your solutios simpler. Either of the followig is fie: ) ) 2 +)6 ) + 6 lim 2 3 = lim = +0)6 0) = or 2 +)6 ) = ) ) )6 0) = Bigger ad Better By iductio, the Sum ad Product Rules ca be exteded to cope with ay fiite umber of coverget sequeces. For example, for three sequeces: lim abc) = lim a lim b lim c Exercise 3 Show that a a)b b)+ab b)+ba a) = a b ab Exercise 4 Use the idetity i Exercise 3 ad the rules for ull sequeces to prove the Product Rule for sequeces. Exercise 5 Write a proof of the Quotiet Rule. You might like to structure your proof as follows.. Note that bb ) b 2 ad show that bb > b2 2 for sufficietly large. Do t Worry You caot use the Quotiet Rule if some of the b s are zero. Fortuately, the fact that b 0 esures that there ca oly be a fiite umber of these. Ca you see why? So you ca apply the Rule to the shifted sequece a N+/b N+) for some wisely chose N.
4 30 CHAPTER 3. SEQUENCES II 2. The show that evetually 0 b. 3. Now tackle a b = a b. b b ) 2 b b b 2 adtherefore b Cuig Required Do you kow a cuig way to rewrite ? Exercise 6 Fid the limit of each of the sequeces defied below ) 2) Coectio The Sadwich Rule for ull sequeces represets the case whe l = Further Useful Results Theorem Sadwich Theorem for Sequeces Suppose a ) l ad b ) l. If a c b the c ) l. This improved Sadwich theorem ca be tackled by rewritig the hypothesis as 0 c a b a ad applyig the earlier Sadwich theorem. Exercise 7 Prove the Sadwich Theorem for sequeces. Max ad Mi I your proof you may well use the fact that each fiite set has a maximum ad a miimum. Is this true of ifiite sets? There are goig to be may occasios whe we are iterested i the behaviour of a sequece after a certai poit, regardless of what wet o before that. This ca be doe by choppig off the first N terms of a sequece a ) to get a shifted sequece b ) give by b = a N+. We ofte write this as a N+ ), so that a N+ ) = a N+, a N+2, a N+3, a N+4,... which starts at the term a N+. We use it i the defiitio below. Defiitio A sequece a ) satisfies a certai property evetually if there is a atural umber N such that the sequece a N+ ) satisfies that property. For istace, a sequece a ) is evetually bouded if there exists N such that the sequece a N+ ) is bouded. Lemma If a sequece is evetually bouded the it is bouded. Exercise 8 Prove this lemma. The ext result, called the Shift Rule, tells you that a sequece coverges if ad oly if it coverges evetually. So you ca chop off or add o ay fiite
5 3.3. FURTHER USEFUL RESULTS 3 umber of terms at the begiig of a sequece without affectig the coverget behaviour of its ifiite tail. Theorem Shift Rule Let N be a atural umber. Let a ) be a sequece. The a a if ad oly if the shifted sequece a N+ a. Proof. Fix ǫ > 0. If a ) a we kow there exists N such that a a < ǫ wheever > N. Whe > N, weseethatn+ > N, therefore a N+ a < ǫ. Hece a N+ ) a. Coversely, suppose that a N+ ) a. The there exists N 2 such that a N+ a < ǫ wheever > N 2. Whe > N + N 2 the N > N 2 so a a = a N+ N) a < ǫ. Hece a ) a. Corollary Sadwich Theorem with Shift Rule Suppose a ) l ad b ) l. If evetually a c b the c ) l. Example We kow / 0 therefore /+5) 0. Exercise 9 Show that the Shift Rule also works for sequeces which ted to ifiity: a ) if ad oly if a N+ ). If all the terms of a coverget sequece sit withi a certai iterval, does its limit lie i that iterval, or ca it escape? For istace, if the terms of a coverget sequece are all positive, is its limit positive too? Lemma Suppose a ) a. If a 0 for all the a 0. Exercise 0 Prove this result. [Hit: Assume that a < 0 ad let ǫ = a > 0. The use the defiitio of covergece to arrive at a cotradictio.] Exercise Prove or disprove the followig statemet: Suppose a ) a. If a > 0 for all the a > 0. Theorem Iequality Rule Suppose a ) a ad b ) b. If evetually) a b the a b. Exercise 2 Prove this result usig the previous Lemma. [Hit: Cosider b a ).]
6 32 CHAPTER 3. SEQUENCES II Limits o Limits Limits caot escape from closed itervals. They ca escape from ope itervals  but oly as far as the ed poits. Cautio Note that the subsequece a i ) is idexed by i ot. I all cases i i. Why is this?) Remember these facts whe subsequeces appear! Prove the obvious It may seem obvious that every subsequece of a coverget sequece coverges, but you should still check that you kow how to prove it! Corollary Closed Iterval Rule Suppose a ) a. If evetually) A a B the A a B. If A < a < B it is ot the case that A < a < B. For example 0 < + < but Subsequeces A subsequece of a ) is a sequece cosistig of some or all) of its terms i their origial order. For istace, we ca pick out the terms with eve idex to get the subsequece a 2,a 4,a 6,a 8,a 0,... or we ca choose all those whose idex is a perfect square a,a 4,a 9,a 6,a 25,... I the first case we chose the terms i positios 2,4,6,8,...ad i the secod those i positios,4,9,6,25,... I geeral, if we take ay strictly icreasig sequece of atural umbers i ) =, 2, 3, 4,... we ca defie a subsequece of a ) by a i ) = a,a 2,a 3,a 4,... Defiitio A subsequece of a ) is a sequece of the form a i ), where i ) is a strictly icreasig sequece of atural umbers. Effectively, the sequece i ) picks out which terms of a ) get to belog to the subsequece. Thik back to the defiitio of covergece of a sequece. Why is it immediate from the defiitio that if a sequece a ) coverges to a the all subsequece a i ) coverge to a? This is a fact which we will be usig costatly i the rest of the course. Notice that the shifted sequece a N+ ) is a subsequece of a ). Exercise 3 Let a ) = 2 ). Write dow the first four terms of the three subsequeces a +4 ), a 3 ) ad a 2 ). Here is aother result which we will eed i later chapters. Exercise 4 Suppose we have a sequece a ) ad are tryig to prove that it coverges. Assume that we have show that the subsequeces a 2 ) ad a 2+ ) both coverge to the same limit a. Prove that a ) a coverges. Exercise 5 Aswer Yes or No to the followig questios, but be sure that you kow why ad that you are t just guessig.
7 3.5. * APPLICATION  SPEED OF CONVERGENCE * 33. A sequece a ) is kow to be icreasig, but ot strictly icreasig. a) Might there be a strictly icreasig subsequece of a )? b) Must there be a strictly icreasig subsequece of a )? 2. If a sequece is bouded, must every subsequece be bouded? 3. If the subsequece a 2, a 3,..., a +,... is bouded, does it follow that the sequece a ) is bouded? 4. If the subsequece a 3, a 4,..., a +2,... is bouded does it follow that the sequece a ) is bouded? 5. If the subsequece a N+, a N+2,..., a N+,... is bouded does it follow that the sequece a ) is bouded? Lemma Every subsequece of a bouded sequece is bouded. Proof. Let a ) be a bouded sequece. The there exist L ad U such that L a U for all. It follows that if a i ) is a subsequece of a ) the L a i U for all i. Hece a i ) is bouded. You might be surprised to lear that every sequece, o matter how boucy ad illbehaved, cotais a icreasig or decreasig subsequece. Theorem Every sequece has a mootoic subsequece. 3.5 * Applicatio  Speed of Covergece * Ofte sequeces are defied recursively, that is, later terms are defied i terms of earlier oes. Cosider a sequece a ) where a 0 = ad a + = a +2, so the sequece begis a 0 =,a = 3,a 2 = 3+2. Exercise 6 Use iductio to show that a 2 for all. Sie Time Agai The fact that a sequece has a guarateed mootoic subsequece does t mea that the subsequece is easy to fid. Try idetifyig a icreasig or decreasig subsequece of si ad you ll see what I mea. Now assume that a ) coverges to a limit, say, a. The: a = lim a = lim a+ ) 2 2 ) = ) 2 lim a + 2 = a 2 2 So to fid a we have to solve the quadratic equatio a 2 a 2 = 0. We ca rewrite this as a+)a 2) = 0, so either a = or a = 2. But which oe is it? The Iequality Theorem comes to our rescue here. Sice a for all it follows that a, therefore a = 2. We will ow ivestigate the speed that a approaches 2.
8 34 CHAPTER 3. SEQUENCES II Exercise 7 Show that 2 a + = 2 a 2+. Use this idetity ad iductio 2+a to show that 2 a ) 2+ for all. How may iteratios are eeded so 3) that a is withi 0 00 is its limit 2? A excellet method for calculatig square roots is the NewtoRaphso method which you may have met at Alevel. Whe applied to the problem of calculatig 2 this leads to the sequece give by: a 0 = 2 ad a + = a + a 2. Exercise 8 Use a calculator to calculate a,a 2,a 3,a 4. Compare them with 2. Exercise 9 Use iductio to show that a 2 for all. Assumig that a ) coverges, show that the limit must be 2. We will ow show that the sequece coverges to 2 like a bat out of hell. Exercise 20 Show that a + 2) = a 2) 2 2a. Usig this idetity show by iductio that a How may iteratios do you eed before you ca guaretee to calculate 2 to withi a error of 0 00 approximately 00 decimal places)? SequecesasiExercise7aresaidtocovergeexpoetially adthoseasi Exercise 20 are said to coverge quadratically sice the error is squared at each iteratio. The stadard methods for calculatig π were expoetial just as is the Archimedes method) util the mid 970s whe a quadratically coverget approximatio was discovered. Check Your Progress By the ed of this chapter you should be able to: Defie what it meas for a sequece to coverge to a limit. Prove that every coverget sequece is bouded. State, prove ad use the followig results about coverget sequeces: If a ) a ad b ) b the: Sum Rule: ca +db ) ca+db Product Rule: a b ) ab Quotiet Rule: a /b ) a/b if b 0 Sadwich Theorem: if a = b ad a c b the c ) a Closed Iterval Rule: if A a B the A a B Explai the term subsequece ad give a rage of examples.
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