In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008

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1 I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces (8.). Elemetary Cocepts Simply speakig, a sequece is simply a list of umbers, writte i a de ite order: fa ; a 2 ; a 3 ; :::a ; a + ; :::g where the elemets a i represet umbers. I this sectio, we oly cocetrate o i ite sequeces. Here are some geeral facts about sequeces studied i this class: Every umber a i the sequece has a successor a + ; the sequece ever stops sice we study i ite 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, 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 de e a fuctio f such that a = f () where is a atural umber. I this class, we will cocetrate o i ite sequeces of real umbers. A more formal de itio of a sequece is as follows: De itio (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 the form f 0 ; 0 + ; :::g where 0 is a iteger such that The umbers a = f () are called the terms of the sequece.

2 The typical otatio for a sequece is (a ), or fa g or fa g = where a deotes the geeral term of the sequece. Remark Whe a sequece is give by a fuctio ff ()g = 0, the fuctio f must be de ed for every 0. A sequece ca be give di eret 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 2 The examples below illustrate sequeces give by listig all the elemets (it is really ot possible sice these are i ite sequeces),. f; 5; 0; 4; 98; 000; 0; 2; :::g. 2. ; 2 ; 3 ; :::. Though the elemets are listed, we ca also guess a formula to geerate them, what is it? 3. f ; ; ; ; ; ; :::g. Though the elemets are listed, we ca also guess a formula to geerate them, what is it? Example 3 The examples below illustrate sequeces give by a simple formula. Notice that does ot have to start at... The elemets of the sequece are: ; 2 ; 3 ; :::. The geeral = term of the sequece is a =. 2. p 3 =3. The elemets are 0; p ; p 2; p 3; :::. I this case, could ot start at. 3. f( ) g =2. The elemets of the sequece are f; ; ; ; :::g. Example 4 The examples below illustrate sequeces give by a recursive formula. a. =. We ca use this formula to geerate all the terms. a = 2a + 5 But the terms have to be geerated i order. For example, i order to get a 0, we eed to kow a 9, ad so o. Usig the formula, we get that a 2 = 2a + 5 = 7 2

3 Havig foud a 2, we ca ow geerate a 3 Ad so o. a 3 = 2a = 9 2. A special sequece geerated recursively is the Fiboacci 8 sequece, amed < f = after a Italia mathematicia. It is de ed as follows: f 2 =. : f = f + f 2 We ca use this formula 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 5 Other sequeces. Arithmetic sequece. A sequece is arithmetic if the di erece betwee two cosecutive terms is costat. Let r be this costat. The, we have a a 0 = r =) a = a 0 + r a 2 a = r =) a 2 = a + r = a 0 + 2r ::: a a = r =) a = a + r = a 0 + r Thus, there is a de ig formula for arithmetic sequeces. For example a arithmetic sequece startig at 2, such that the di erece betwee two cosecutive terms is 3 is de ed by: a = Geometric sequece. A sequece is geometric if the ratio of two cosecutive terms is costat. Let q deote this costat. The, we have a a 0 = q =) a = qa 0 a 2 a = q =) a 2 = qa = q 2 a 0 ::: a a = q =) a = qa = q a 0 Thus, there is a de ig formula for geometric sequeces. For example a geometric sequece startig at 2, such that the ratio betwee two cosecutive terms is 3 is de ed by: a = 2 3. Sequeces ca be plotted. However, the plot of a sequece will cosists of dots, sice they are oly de ed at the itegers. Figures, 2, 3, ad 4 show some sequeces beig plotted. 3

4 Figure : Plot of a =.2 Limit of a Sequece Give a sequece fa g, 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 d lim! a. De itio 6 (limit of a sequece) A sequece fa g coverges to a umber L as goes to i ity 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 ite umber, we say that fa g coverges. Otherwise, it diverges. Sometimes, we will make the distictio betwee diverges to i ity ad simply diverges. I the rst case, we still kow what the sequece is doig, it is gettig large without bouds. A sequece may diverge for several reasos. Its geeral term could get arbitrarily large (go to i ity), as show o gure 3. Its geeral term could also oscillate betwee di eret values without ever gettig close to aythig. This is the case of f( ) g, or f( ) l g as show o gure 4. 4

5 Figure 2: Plot of a = si Aother way to uderstad this is that if lim a = L the ja Lj goes to 0! whe!. It should also be oted that if lim a = L, the lim a + = 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 gure 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 gure 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 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 > 0. For the smaller regio, it happes whe > 22 (approximately). 5

6 Figure 3: Plot of l Let us rst state, but ot prove, a importat theorem. Theorem 7 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 di eret 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 dig the limit of a fuctio, to d the limit of a sequece. This is what the ext theorem tells us. Theorem 8 If lim x! f(x) = L ad a = f() the lim! a = L 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 sequece coverges to the same limit. We kow how to do the latter from Calculus I. 6

7 Figure 4: Plot of ( ) l Example 9 Fid lim +. The fuctio geeratig this sequece is f (x) = (l Hôpital s rule), lim + =. x. Sice lim x + x! Be careful, this theorem oly gives a de ite aswer if x x + = lim f(x) = L. If x! the fuctio diverges, we caot coclude. For example, cosider the sequece a = cos 2. The fuctio f such that a = f () is f (x) = cos 2x. This fuctio diverges as x!. Yet, a = cos 2 = for ay. So, lim a =. Theorem 0 If a b c for 0 ad lim b = L! lim a = lim c = L the!! This is the equivalet of the squeeze theorem. 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 Fid lim! 7

8 Figure 5: Limit of a sequece Sice! is oly de ed for itegers, we caot d a fuctio f such that a = f (). Thus, we caot use the previous theorem. We use the squeeze theorem istead. We otice that: 0! = 2 3 ::: ::: By the squeeze theorem, it follows that lim! = 0. Theorem 2 If lim! ja j = 0 the lim! a = 0 This is a applicatio of the squeeze theorem usig the fact that ja j a ja j. This theorem is ofte useful whe the geeral term of a sequece cotais ( ) as suggests the ext example. 8

9 Example 3 Fid lim a for a = ( ). First, we otice that ja j = ( ) = which coverges to 0 by the previous example. Hece, by theorem 2, 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 di cult. Theorem 4 The sequece fx g coverges to 0 if jxj <. It coverges to if x =. It diverges otherwise. Proof. We cosider several cases. case : x =. The, x = =. Thus, the sequece coverges to. case 2: x =. The, x = ( ) which diverges. case 3: x = 0. The, x = 0 = 0. Thus, the sequece coverges to 0. case 4: jxj < ad x 6= 0. The, l jxj < 0. Thus, l jxj = l jxj!. Thus, jxj! 0 as!. Hece, by the previous theorem, x! 0 as!. case 5: x >. The, x!. case 6: x <. x oscillate betwee positive ad egative values, which are gettig larger i absolute value. Thus it also diverges. Fially, we state a theorem which is the equivalet for sequeces of the limit rules for fuctios. Theorem 5 Suppose that fa g ad fb g coverge, ad that c is a costat. The:. lim! (a + b ) = lim! a + lim! b 2. lim (a b ) = lim a lim!! 3. lim (a b ) = lim a!! 4. lim a! b = lim! a lim! 5. lim! ca = c lim! a 6. lim! c = c 7. lim! ap = h! b lim! b providig lim b b 6= 0! i p lim a if p > 0 ad a > 0.! 9

10 Sice we oly cosider limits as!, we will omit it ad simply write lim a. The way we d lim a depeds greatly o how the sequece is give. Example 6 Fid lim a for a =. I this case, the fuctio which geerates the terms of the sequece is f (x) = x. Sice lim x! x = 0, it follows that lim a = 0 by theorem 8. Example 7 Fid lim l. From a previous example, lim + + =, therefore lim l = l = 0. + Example 8 Fid lim ( ). Sice the terms of this sequece are f ; ; ; ; :::g, they oscillate but ever get close to aythig. The sequece diverges. I cotrast, the sequece of the rst example also oscillated. But it also got closer ad closer to 0. Example 9 Fid lim si. Though we ca d a fuctio to express the geeral term of this sequece, sice si x diverges, we caot use theorem 8 to try to compute the limit of this sequece. We ote that si Therefore Sice lim 0. si = 0, by the squeeze theorem for sequeces, it follows that lim si = Example 20 Fid lim l The fuctio de ig the geeral term is f (x) = l x x. Sice l x lim x! x It follows that lim l = 0. = lim x! = lim x! = 0 x x by l Hôpital s rule 0

11 Example 2 Assumig that the sequece give recursively by ( a = 2 a + = 2 (a + 6) coverges, d its limit. Let L = lim a. If we take the limit o both sides of the relatio de ig the sequece, we have lim a + = lim 2 (a + 6) So, lim a = 6. L = (L + 6) 2 2L = L + 6 L = 6.3 Icreasig, Decreasig ad Bouded Sequeces De itio 22 (icreasig, decreasig sequeces) A sequece fa g is said to be. icreasig if ad oly if a < a + for each oegative iteger. 2. o-decreasig if ad oly if a a + for each oegative iteger. 3. decreasig if ad oly if a > a + for each oegative iteger. 4. o-icreasig if ad oly if a a + 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:. Show that a < a + for all. 2. Show that a a + < 0 for all. 3. If a > 0 for all, the show that 4. If f () = a, show that f 0 (x) > 0 5. By iductio. Example 23 Let a = a a + < for all. +. Show fa g is icreasig.

12 Method : look at a a + Method 2: Let f (x) = a a + = = ( + 2) ( + ) ( + ) = < x x +. The f 0 (x) = (x + ) 2 > 0 De itio 24 (bouded sequeces) A sequece fa g 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 fa g 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. Example 25 Cosider the sequece a = cos. Sice cos, a is bouded from above by ad bouded from below by. So, a is bouded. Example 26 Cosider the sequece. We have 0 <. Thus the = sequece is bouded below by 0 ad above by. Remark 27 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 i itely may upper bouds. Similarly, if a sequece is bouded below by m, the ay umber less tha m is also a lower boud. Theorem 28 coverge. Theorem 29 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. Before we do this, let us review iductio. This cocept is ofte eeded to show a sequece is icreasig or bouded. Theorem 30 (Iductio) Let P () deote a statemet about atural umbers with the followig properties: 2

13 . The statemet is true whe = i.e. P () is true. This is call the base case. 2. P (k + ) is true wheever P (k) is true for ay iteger k. The, P () is true for all 2 N. Example 3 Prove that the sequece de ed by a = 2 ad a + = 2 (a + 6) is icreasig ad bouded. Fid its limit. Icreasig: We show by iductio that a + > a. Base case. a 2 = 2 (a + 6) = 2 (2 + 6) = 4 > a. Assume the result is true for ay iteger k, that is assume a k+ > a k. Show the result is true for k + that is a k+ > a k+. a k+2 = 2 (a k+ + 6) > 2 (a k + 6) sice a k+ > a k = a k+ The result follows by iductio. Bouded: We show by iductio that a 6. Base case: a = 2 6. Assume the result is true for ay iteger k, that is a k 6, show the result is also true for k + that is a k+ 6. The result follows by iductio. a k 6 =) a k =) 2 (a k + 6) 6 =) a k+ 6 Limit: We have already computed the limit of such a sequece. Now that we kow the limit exists sice fa g i icreasig ad bouded above, the idea is to give it a ame, say lim a = L ad d what L is usig the limit 3

14 rules as follows: a + = 2 (a + 6) () lim (a + ) = lim 2 (a + 6) () L = 2 lim (a + 6) () L = 2 (lim a + lim 6) () L = (L + 6) 2 () 2L = L + 6 () L = 6 So lim a = 6 We ish with a importat remark. Remark 32 Cosider a sequece fa g. If fa g 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..4 Thigs to Kow ad Problems Assiged 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 d its limit. Be able to tell if a sequece is icreasig or decreasig. Related problems assiged:. o pages 565, 566 #, 3, 5, 7, 9,, 3, 5, 7, 9, 2, 23, 25, 35, 39, 4, Let fa g be a sequece such that a = f () for some fuctio f. Suppose that lim f (x) does ot exist. Does it mea that fa g x! diverge? Explai or give a couter example. 3. Explai why if a sequece has a limit the it must be bouded. 4. Give examples of bouded sequeces which do ot have limits. 4

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