Infinite Sequences and Series

Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet umbers. I this sectio, we oly cocetrate o ifiite sequeces. Here are some geeral facts about sequeces studied i this class: Every umber a i the sequece has a successor a +1 ; the sequece ever stops sice we study ifiite sequeces. Thik of the idex of a particular elemet as idicatig the positio of the elemet i the list. The idex ca also be associated with a formula whe the elemets i the list are geerated by oe (see below). As such, the idex of the umbers does ot have to start at 1, though it does most of the time. If we call 0 the value of the startig idex, the there is a umber a for every 0. Thus, we ca defie a fuctio f such that a = f () where is a atural umber. I this class, we will cocetrate o ifiite sequeces of real umbers. A more formal defiitio of a sequece is as follows: Defiitio 6.1.1 (sequece) A sequece of real umbers is a real-valued fuctio f whose domai is a subset of the o-egative itegers, that is a set of 275

276 CHAPTER 6. INFINITE SEQUENCES AND SERIES the form { 0, 0 + 1,...} where 0 is a iteger such that 0 0.. The umbers a = f () are called the terms of the sequece. The typical otatio for a sequece is (a ), or {a } or {a } =1 where a deotes the geeral term of the sequece. Remark Whe a sequece is give by a fuctio {f ()} = 0, the fuctio f must be defied for every 0. A sequece ca be give differet ways The elemets of the sequece are give. A formula to geerate the terms of the sequece is give. A recursive formula to geerate the terms of the sequece is give. Example 6.1.2 The examples below illustrate sequeces give by listig all the elemets (it is really ot possible sice these are ifiite sequeces), 1. {1, 5, 10, 4, 98, 1000, 0, 2,...}. 2. {1, 12, 13 },.... Though the elemets are listed, we ca also guess a formula to geerate them, what is it? 3. { 1, 1, 1, 1, 1, 1,...}. Though the elemets are listed, we ca also guess a formula to geerate them, what is it? Example 6.1.3 The examples below illustrate sequeces give by a simple formula. Notice that does ot have to start at 1. { } 1 1.. The elemets of the sequece are: {1, 12, 13 },.... The geeral =1 term of the sequece is a = 1. 2. { 3 } =3. The elemets are { 0, 1, 2, 3,... }. I this case, could ot start at 1. 3. {( 1) } =2. The elemets of the sequece are {1, 1, 1, 1,...}. Example 6.1.4 The examples below illustrate sequeces give by a recursive formula. { a 1. 1 = 1. We ca use this formula to geerate all the terms. a = 2a 1 + 5 But the terms have to be geerated i order. For example, i order to get a 10, we eed to kow a 9, ad so o. Usig the formula, we get that a 2 = 2a 1 + 5 = 7

6.1. INFINITE SEQUENCES 277 Havig foud a 2, we ca ow geerate a 3 Ad so o. a 3 = 2a 2 + 5 = 19 2. A special sequece geerated recursively is the Fiboacci sequece, amed after a Italia mathematicia (Leoardo Fiboacci, 1170-1250). It is defied as follows: f 2 = 1. We ca use this formula f 1 = 1 f = f 1 + f 2 for > 2 to geerate the terms of the sequece. This sequece was devised to model rabbit populatio. f represets the umber of pairs of rabbits after moths, assumig that each moth, a pair of rabbit produces a ew pair which becomes productive at age 2 moths. Example 6.1.5 Other sequeces 1. Arithmetic sequece. A sequece {a } =0 is arithmetic if the differece betwee two cosecutive terms is costat. Let r be this costat. The, we have a 1 a 0 = r = a 1 = a 0 + r a 2 a 1 = r = a 2 = a 1 + r = a 0 + 2r... a a 1 = r = a = a 1 + r = a 0 + r Thus, there is a defiig formula for arithmetic sequeces. For example a arithmetic sequece startig at 2, such that the diff erece betwee two cosecutive terms is 3 is defied by: a = 3 + 2 2. Geometric sequece. A sequece {a } =0 is geometric if the ratio of two cosecutive terms is costat. Let q deote this costat. The, we have a 1 = q = a 1 = qa 0 a 0 a 2 = q = a 2 = qa 1 = q 2 a 0 a 1... a = q = a = qa 1 = q a 0 a 1 Thus, there is a defiig formula for geometric sequeces. For example a geometric sequece startig at 2, such that the ratio betwee two cosecutive terms is 3 is defied by: a = 2 3. Sequeces ca be plotted. However, the plot of a sequece will cosists of dots, sice they are oly defied at the itegers. Figures 6.1, 6.2, 6.3, ad 6.4 show some sequeces beig plotted.

278 CHAPTER 6. INFINITE SEQUENCES AND SERIES Figure 6.1: Plot of a = 1 6.1.2 Limit of a Sequece Give a sequece {a }, oe of the questios we try to aswer is: what is the behavior of a as? Is a gettig closer ad closer to a umber? I other words, we wat to fid lim a. Defiitio 6.1.6 (limit of a sequece) A sequece {a } coverges to a umber L as goes to ifiity if a ca be made as close as oe wats to L, simply by takig large eough. I this case, we write lim a = L. If lim a = L ad L is a fiite umber, we say that {a } coverges. Otherwise, it diverges. Sometimes, we will make the distictio betwee diverges to ifiity ad simply diverges. I the first case, we still kow what the sequece is doig, it is gettig large without bouds. Remark 6.1.7 Sice the oly limit we cosider for sequeces is as, istead of writig lim a, we ofte write lim a. A sequece may diverge for several reasos. Its geeral term could get arbitrarily large (go to ifiity), as show o figure 6.3. Its geeral term could

6.1. INFINITE SEQUENCES 279 Figure 6.2: Plot of a = si also oscillate betwee differet values without ever gettig close to aythig. This is the case of {( 1) }, or {( 1) l } as show o figure 6.4. Remark 6.1.8 Aother way to uderstad this is that if a L goes to 0 whe. lim a = L the Remark 6.1.9 It should also be oted that if lim a = L, the lim a +1 = L. I fact, lim a +p = L for ay positive iteger p. Graphically, the meaig of lim a = L is as follows. Cosider the sequece show o figure 6.5 whose plot is represeted by the dots. The sequece appears to have 3 as its limit, this is idicated by the horizotal solid lie through 3. If we draw a regio havig the lie y = 3 at its ceter, the sayig that lim a = 3 meas that there exists a certai value of (deote it 0 ) such that if > 0, the all the dots correspodig to the plot of the sequece will fall i the regio. O figure 6.5, we drew two regios, oe with dotted lies, the other oe with dashdot lies. We ca see that i both cases, after a while, the sequece always falls i the regio. Of course, the more arrow the regio is, the larger 0 will

280 CHAPTER 6. INFINITE SEQUENCES AND SERIES Figure 6.3: Plot of l be. I other words, if we wat to guaratee that a is closer to 3, we have to look at a for larger values of. It appears that for the larger regio, the sequece falls i the regio if > 10. For the smaller regio, it happes whe > 22 (approximately). Let us first state, but ot prove, a importat theorem. Theorem 6.1.10 If a sequece coverges, its limit is uique. We ow look at various techiques used whe computig the limit of a sequece. As oticed above, a sequece ca be give differet ways. How its limit is computed depeds o the way a sequece is give. We begi with the easiest case, oe we are already familiar with from Calculus I. If the sequece is give by a fuctio, we ca, i may istaces, use our kowledge of fidig the limit of a fuctio, to fid the limit of a sequece. This is what the ext theorem tells us. Theorem 6.1.11 If a = f () ad lim f (x) = L the lim a = L. x This theorem simply says that if we kow the fuctio which geerates the geeral term of the sequece, ad that fuctio coverges as x the the

6.1. INFINITE SEQUENCES 281 Figure 6.4: Plot of ( 1) l sequece coverges to the same limit. Calculus I. We kow how to do the latter from Example 6.1.12 Fid lim 1. The fuctio geeratig this sequece is f (x) = 1 x hece lim 1 = 0. Example 6.1.13 Fid lim + 1. The fuctio geeratig this sequece is f (x) = (l Hôpital s rule), lim + 1 = 1. 1 ad we kow that lim x x = 0 x. Sice lim x + 1 x x x + 1 = 1 Remark 6.1.14 Be careful, this theorem oly gives a defiite aswer if lim x L. If the fuctio diverges, we caot coclude. For example, cosider the sequece a = cos 2π. The fuctio f such that a = f () is f (x) = cos 2πx. This fuctio diverges as x. Yet, a = cos 2π = 1 for ay. So, lim a = 1.

282 CHAPTER 6. INFINITE SEQUENCES AND SERIES Figure 6.5: Limit of a sequece Theorem 6.1.15 (Squeeze Theorem) If a b c for 0 ad lim = lim = L the lim = L This is the equivalet of the squeeze theorem studied for fuctios. You will otice i the statemet of the theorem that the coditio a b c does ot have to be true for every. It simply has to be true from some poit o. Example 6.1.16 Fid lim! Sice! is oly defied for itegers, we caot fid a fuctio f such that a = f (). Thus, we caot use the previous theorem. We use the squeeze theorem istead. We otice that: 0! = 1 2 3...... 1 By the squeeze theorem, it follows that lim! = 0.

6.1. INFINITE SEQUENCES 283 Theorem 6.1.17 If lim a = 0 the lim a = 0 This is a applicatio of the squeeze theorem usig the fact that a a a. This theorem is ofte useful whe the geeral term of a sequece cotais ( 1) as suggests the ext example. Example 6.1.18 Fid lim si. Though we ca fid a fuctio to express the geeral term of this sequece, sice si x diverges, we caot use theorem 6.1.11 to try to compute the limit of this sequece. We ote that 1 si 1 Therefore Sice lim 1 0. 1 si 1 = 0, by the squeeze theorem for sequeces, it follows that lim si = Example 6.1.19 Fid lim a for a = ( 1). First, we otice that a = ( 1) = 1 which coverges to 0 by the previous example. Hece, by theorem 6.1.17, a 0 also. We ow look at a theorem which is very importat. Ulike the other theorems, we ca prove this oe as it is ot very diffi cult. Theorem 6.1.20 Let x be a real umber. The sequece {x } coverges to 0 if x < 1. It coverges to 1 if x = 1. It diverges otherwise. Proof. We cosider several cases. case 1: x = 1. The, x = 1 = 1. Thus, the sequece coverges to 1. case 2: x = 1. The, x = ( 1) which diverges. case 3: x = 0. The, x = 0 = 0. Thus, the sequece coverges to 0. case 4: x < 1 ad x 0. The, l x < 0. Thus, l x = l x. Thus, x 0 as. Hece, by the previous theorem, x 0 as. case 5: x > 1. The, x. case 6: x < 1. x oscillate betwee positive ad egative values, which are gettig larger i absolute value. Thus it also diverges.

284 CHAPTER 6. INFINITE SEQUENCES AND SERIES The ext theorem is a theorem which is the equivalet for sequeces of the limit rules for fuctios. Theorem 6.1.21 Suppose that {a } ad {b } coverge, ad that c is a costat. The: 1. lim (a + b ) = lim a + lim b 2. lim b ) = lim lim ( ) ( 3. lim b ) = lim ( ) a b 4. lim = lim a lim b 5. lim ca = c lim a 6. lim c = c 7. lim ap = [ b lim b ) providig lim b 0 ] p lim a if p > 0 ad a > 0. Sice we oly cosider limits as, we will omit it ad simply write lim a. The way we fid lim a depeds greatly o how the sequece is give. Example 6.1.22 Fid lim 1 lim 1 = lim 1 = ( 1) (0) = 0 Example 6.1.23 Fid lim a for a = 1. We ote that 1 lim 1 = 1 1 hece ( = lim 1 1 ) = lim (1) lim = 1 0 = 1 ( ) 1 from the above theorem, part 2 Example 6.1.24 Fid lim 5 2 lim 5 ( 2 = (5) lim 1 ) ( lim 1 ) = (5) (0) (0) = 0

6.1. INFINITE SEQUENCES 285 Example 6.1.25 Fid lim 2 + 2 2 + 1 We could fid this limit by usig l Hôpital s rule o the fuctio f (x) = x 2 + x 2x 2. Here, we show how to compute this limit usig the theorem we just + 1 studied. We begi by factorig from both the umerator ad deomiator the term of highest degree. ( ) 2 lim 2 + 2 2 + 1 1 + 1 = lim ( 2 2 + 1 ) 1 + 1 = lim 2 + 1 = 1 + 0 2 + 0 = 1 2 The ext theorem broades the theorems we have already studied. It tells us that applyig a cotiuous fuctio to a coverget sequece produces a coverget sequece. Theorem 6.1.26 Let {a } be a sequece such that lim a = L. If f is a cotiuous fuctio the lim f (a ) = f (L). ( ) Example 6.1.27 Fid lim l. From a previous example, lim + 1 + 1 = ( ) 1, therefore lim l = l 1 = 0. + 1 Example 6.1.28 Fid lim ( 1). Sice the terms of this sequece are { 1, 1, 1, 1,...}, they oscillate but ever get close to aythig. The sequece diverges. I cotrast, the sequece of the first example also oscillated. But it also got closer ad closer to 0. Example 6.1.29 Fid lim l The fuctio defiig the geeral term is f (x) = l x x. Sice l x lim x x = lim x = lim x = 0 1 x 1 1 x by l Hôpital s rule

286 CHAPTER 6. INFINITE SEQUENCES AND SERIES It follows that lim l = 0. Example 6.1.30 Assumig that the sequece give recursively by { a 1 = 2 a +1 = 1 2 (a + 6) coverges, fid its limit. Let L = lim a. If we take the limit o both sides of the relatio defiig the sequece, we have ( ) 1 lim a +1 = lim 2 (a + 6) So, lim a = 6. L = 1 (L + 6) 2 2L = L + 6 L = 6 We fiish by listig five limits which are importat to remember. Theorem 6.1.31 Let x be a fixed real umber. 1. lim x = 1 if x > 0. 2. lim = 1. 3. lim x = 0 if x < 1. ( 4. lim 1 + x = e x ) x 5. lim! = 0 Example 6.1.32 Example 6.1.33 Example 6.1.34 Example 6.1.35 by formula 3. lim ( ) 1 = 0 by formula 3. 2 lim 3 = lim 3 = (1) (1) = 1 by formulas 1 ad 2. lim lim ( 1 3 = lim 5+2 ) ( = lim 1 + 1 ) = e 1 by formula 4. 3 5 2 5 = 1 25 lim 3 5 = 1 25 lim ( ) 3 = 0 5

6.1. INFINITE SEQUENCES 287 6.1.3 Icreasig, Decreasig ad Bouded Sequeces Defiitio 6.1.36 (icreasig, decreasig sequeces) A sequece {a } is said to be 1. icreasig if ad oly if a < a +1 for each oegative iteger. 2. o-decreasig if ad oly if a a +1 for each oegative iteger. 3. decreasig if ad oly if a > a +1 for each oegative iteger. 4. o-icreasig if ad oly if a a +1 for each oegative iteger. 5. mootoic if ay of these four properties holds. To show that a sequece is icreasig, we ca try oe of the followig: 1. Show that a < a +1 for all. 2. Show that a a +1 < 0 for all. 3. If a > 0 for all, the show that 4. If f () = a, show that f (x) > 0 5. By iductio. Example 6.1.37 Let a = Method 1: look at a a +1 Method 2: Let f (x) = a a +1 < 1 for all. + 1. Show {a } is icreasig. a a +1 = = + 1 + 1 + 2 ( + 2) ( + 1) ( + 1) 2 + 2 = 2 + 2 + 1 < 1 x x + 1. The f 1 (x) = (x + 1) 2 > 0 Defiitio 6.1.38 (bouded sequeces) A sequece {a } is said to be bouded from above if there exists a umber M such that a M for all. M is called a upper boud of the sequece. A sequece {a } is said to be bouded from below if there exists a umber m such that a m for all. m is called a lower boud of the sequece. A sequece is bouded if it is bouded from above ad below.

288 CHAPTER 6. INFINITE SEQUENCES AND SERIES Example 6.1.39 Cosider the sequece a = cos. Sice 1 cos 1, a is bouded from above by 1 ad bouded from below by 1. So, a is bouded. { } 1 Example 6.1.40 Cosider the sequece. We have 0 < 1 1. Thus =1 the sequece is bouded below by 0 ad above by 1. Remark 6.1.41 Clearly, if M is a upper boud of a sequece, the ay umber larger tha M is also a upper boud. So, if a sequece is bouded from above, it has ifiitely may upper bouds. Similarly, if a sequece is bouded below by m, the ay umber less tha m is also a lower boud. Theorem 6.1.42 coverge. Theorem 6.1.43 coverge. A sequece which is icreasig ad bouded from above must A sequece which is decreasig ad bouded from below must We use this theorem to prove that certai sequeces have a limit. ( ) Example 6.1.44 The sequece is icreasig (see above) ad bouded + 1 above by 1 hece it must coverge. We fiish with a importat remark. Remark 6.1.45 Cosider a sequece {a }. If {a } has a limit, it must be bouded. Try to explai why. Not every bouded sequece has a limit. Give a example of a bouded sequece with o limit. 6.1.4 Thigs to Kow Be able to write the terms of a sequece o matter which way the sequece is preseted. Be able to tell if a sequece coverges or diverges. If it coverges, be ale to fid its limit. Kow ad be able to use all the theorems about the limit of a sequece. Be able to tell if a sequece is icreasig or decreasig. Be able to tell if a sequece is bouded.

6.1. INFINITE SEQUENCES 289 6.1.5 Problems 1. List the first six terms of the sequece (a ) give by a = 2 + 1. Does the sequece appear to have a limit? If yes, fid it. 2. List the first six terms of the sequece give by a 1 = 1 ad a +1 = 1 1 + a for 1. 3. Determie if the sequeces of geeral term a below coverge or diverge. If they coverge, fid their limit. (a) a = 1 + 22 + 2 (b) a = 2 3 +1 (c) a = 1 2 (d) a = ( + 2)!! (e) a = ( 1) 2 + 1 (f) a = e + e e 2 + 1 (g) a = 2 e (h) a = cos2 (i) a = 2 ( 1 + 3 ) (j) a = {1, 1.1, 1, 1.1, 1, 1.1,...} 4. Determie if the sequeces of geeral term a below are icreasig, decreasig or ot mootoic. Are they bouded? (a) a = 1 2 + 5 (b) a = si π 2 (c) a = 2 3 3 + 4 (d) a = + 1 5. Let {a } be a sequece such that a = f () for some fuctio f. Suppose that lim x f (x) does ot exist. Does it mea that {a } diverge? Explai or give a couter example.

290 CHAPTER 6. INFINITE SEQUENCES AND SERIES 6. Let (a ) be a sequece. If we are told that (a ) is decreasig ad all its terms are betwee 1 ad 5, what ca we say about the covergece of this sequece ad why? 7. Explai why if a sequece has a limit the it must be bouded. 8. Give examples of bouded sequeces which do ot have limits. 6.1.6 Aswers 1. List the first six terms of the sequece (a ) give by a = 2 + 1. Does the sequece appear to have a limit? If yes, fid it. The first six terms are 1 3, 2 5,3 7, 4 9, 5 11, 6 13 ad the limit is 1 2. 2. List the first five terms of the sequece give by a 1 = 1 ad a +1 = 1 1 + a for 1. The first five terms are 1 2, 2 3, 3 5, 5 8, 8 13 3. Determie if the sequeces of geeral term a below coverge or diverge. If they coverge, fid their limit. (a) a = 1 + 22 + 2 Coverges to 2. (b) a = 2 3 +1 Coverges to 0 (c) a = 1 2 Coverges to 0 ( + 2)! (d) a =! Diverges (e) a = ( 1) 2 + 1 Coverges to 0 (f) a = e + e e 2 + 1 Coverges to 0 (g) a = 2 e Coverges to 0 (h) a = cos2 2 Coverges to 0

6.1. INFINITE SEQUENCES 291 ( (i) a = 1 + 3 ) Coverges to e 3 (j) a = {1, 1.1, 1, 1.1, 1, 1.1,...} Diverges 4. Determie if the sequeces of geeral term a below are icreasig, decreasig or ot mootoic. Are they bouded? 1 (a) a = 2 + 5 decreasig, bouded (b) a = si π 2 ot mootoic, bouded (c) a = 2 3 3 + 4 Icreasig, bouded (d) a = + 1 Icreasig, bouded below but ot above. 5. Let {a } be a sequece such that a = f () for some fuctio f. Suppose that lim f (x) does ot exist. Does it mea that {a } diverge? Explai x or give a couter example. 6. Let (a ) be a sequece. If we are told that (a ) is decreasig ad all its terms are betwee 1 ad 5, what ca we say about the covergece of this sequece ad why? 7. Explai why if a sequece has a limit the it must be bouded. 8. Give examples of bouded sequeces which do ot have limits.