MA131 - Analysis 1. Workbook 3 Sequences II

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1 MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces Algebra of Limits Further Useful Results Subsequeces * Applicatio - Speed of Covergece *

2 e a -e 0 N Figure : Coverget sequeces; first choose ε, the fid N. 2.8 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 ɛ = 0., ɛ = 0.0 ad ɛ = 0.00 fid a N such that a < ɛ wheever > N. Defiitio Let a R. A sequece (a ) teds to a if, for each ɛ > 0, there exists a atural umber N such that a a < ɛ for all > N. See figure?? 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 ɛ > 0. We have to fid a atural umber N so that a = + < ɛ whe > N. We have + = + = + <. Good N-ough Ay N that works is good eough - it does t have to be the smallest possible N. Recycle Have a closer look at figure??, what has bee chaged from figure 6 of workbook 2? It turs 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 < ɛ wheever > N. But < ɛ if ad oly if ɛ < so it is eough to choose N to be a atural umber with N > ɛ. The, if > N we have a = + = + = + < < N < ɛ. Lemma (a ) a if ad oly if (a a) 0.

3 Proof. We kow that (a a) 0 meas that for each ɛ > 0, there exists a atural umber N such that a a < ɛ 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. Assigmet 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. Assigmet 2 Prove the theorem above. 2.9 Algebra of Limits Theorem 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, if b 0 Quotiet Rule for Sequeces 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 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? Polly Wat a Cracker? If you have a parrot, teach it to say: The limit of the sum is the sum of the limits. The limit of the product is the product of the limits. The limit of the quotiet is the quotiet of the limits. 2

4 Example I full detail ( 2 + )(6 ) lim = lim ( ) ( ) usig the Quotiet Rule [( + 2 ) ( 6 = lim ( lim = 3 ) )] usig the Product ad Sum Rules ( ( + )) ( ( lim 6 2 lim )) ( + 0)(6 0) = = lim ( 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 2 ( + 0)(6 0) = = or ( 2 + )(6 ) = ( ) ( ) ( + 0)(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 Assigmet 3 Use the Sum Rule for ull sequeces to prove the Sum Rule for sequeces. Exercise Show that (a a)(b b) + a(b b) + b(a a) = a b ab Assigmet 4 Use the idetity i Exercise ad the rules for ull sequeces to prove the Product Rule for sequeces. Assigmet 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. ( ) 2. The show that evetually 0 b 2 b b b 2 ad therefore b b. 3. Now tackle a b = a b. b Do t Worry You ca still use the Quotiet Rule if some of the b s are zero. The fact that b 0 esures that there ca oly be a fiite umber of these. Ca you see why? Ad evetually, the sequece leaves them behid. 3

5 Assigmet 6 Fid the limit of each of the sequeces defied below ( +3)( 2) Cuig Required Do you kow a cuig way to rewrite ? 2.0 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. Coectio The Sadwich Rule for ull sequeces represets the case whe l = 0. Assigmet 7 Prove the Sadwich Theorem for sequeces. 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. Assigmet 8 Prove this lemma. 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? 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 umber of terms at the begiig of a sequece without affectig the coverget behaviour of its ifiite tail. 4

6 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 ɛ > 0. If (a ) a we kow there exists N such that a a < ɛ wheever > N. Whe > N, we see that N+ > N, therefore a N+ a < ɛ. Hece (a N+ ) a. Coversely, suppose that (a N+ ) a. The there exists N 2 such that a N+ a < ɛ wheever > N 2. Whe > N + N 2 the N > N 2 so a a = a N+( N) a < ɛ. 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 2 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. Assigmet 9 Prove this result. [Hit: Assume that a < 0 ad let ɛ = a > 0. The use the defiitio of covergece to arrive at a cotradictio.] Assigmet 0 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. Assigmet Prove this result usig the previous Lemma. [Hit: Cosider (b a ).] 5

7 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,... 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! 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 ). Assigmet 2 Let (a ) = ( 2 ). Write dow the first four terms of the three subsequeces (a +4 ), (a 3 ) ad (a 2 ). 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! Here is aother result which we will eed i later workbooks. Assigmet 3 Suppose we have a sequece (a ) ad are tryig to prove that it coverges. Assume that we have have show that the subsequeces (a 2 ) ad (a 2+ ) both coverge to the same limit a. Prove that (a ) a coverges. Exercise 3 Aswer Yes or No to the followig questios, but be sure that you kow why ad that you are t just guessig. 6

8 0 Figure 2: Floor terms are lower bouds for the rest of the sequece.. 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 ill-behaved, cotais a icreasig or decreasig subsequece. Theorem Every sequece has a mootoic subsequece. We say a f is a floor term of the (a ) if a a f for all f. So each floor term is evetually a lower boud. Exercise 4 Write dow the floor terms of the sequeces: 7

9 . (( ) ) 2. 0,, 3, 2, 5, 4, 7, 6, ( Idetify a mootoic subsequece of each. ) 8

10 Exercise 5. If there is a ifiite umber of floor terms, show that they form a mootoic icreasig subsequece. 2. If there is a fiite umber of floor terms ad the last oe is a F, costruct a mootoic decreasig subsequece with a F + as its first term. 3. If there are o floor terms, costruct a mootoic decreasig subsequece with a as its first term. Assigmet 4 Tur your aswers to Exercise 5 ito a proof of the previous theorem. 2.2 * 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 = 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. Assigmet 5 Show that 2 a + = 2 a 2+. Use this idetity ad iductio to show that 2+a (2 a ) (2+ 3) for all. How may iteratios are eeded so that a is withi 0 00 is its limit 2? A excellet method for calculatig square roots is the Newto-Raphso method which you may have met at A-level. Whe applied to the problem of calculatig 2 this leads to the sequece give by: a 0 = 2 ad a + = a + a 2. Exercise 7 Use a calculator to calculate a, a 2, a 3, a 4. Compare them with 2. Assigmet 6 Use iductio to show that a 2 for all. Assumig that (a ) coverges, show that the limit must be 2. 9

11 We will ow show that the sequece coverges to 2 like a bat out of hell. Assigmet 7 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)? Sequeces as i Assigmet 5 are said to coverge expoetially ad those as i Assigmet 7 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 Workbook 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. 0