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


 Sabrina Green
 1 years ago
 Views:
Transcription
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 realvalued fuctio f whose domai is a subset of the oegative 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. odecreasig if ad oly if a a + for each oegative iteger. 3. decreasig if ad oly if a > a + for each oegative iteger. 4. oicreasig 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
MATH Testing a Series for Convergence
MATH 0  Testig a Series for Covergece Dr. Philippe B. Laval Keesaw State Uiversity October 4, 008 Abstract This hadout is a summary of the most commoly used tests which are used to determie if a series
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationLecture 23 : Sequences
A Sequece is a list of umbers writte i order. Lecture 23 : Sequeces a a 2 a 3... } The sequece may be ifiite. The th term of the sequece is the th umber o the list. O the list above a = st term a 2 = 2
More information2.7 Sequences, Sequences of Sets
2.7. SEQUENCES, SEQUENCES OF SETS 67 2.7 Sequeces, Sequeces of Sets 2.7.1 Sequeces Defiitio 190 (sequece Let S be some set. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For
More information4.4 Monotone Sequences and Cauchy Sequences
4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 3 4.4 Mootoe Sequeces ad Cauchy Sequeces 4.4. Mootoe Sequeces The techiques we have studied so far require we kow the limit of a sequece i order to prove the
More informationSection 1.3 Convergence Tests
Sectio.3 Covergece Tests We leared i the previous sectio that for ay geometric series, we kow exactly whether it coverges ad the value it coverges to. For most series, however, there is o exact formula
More informationnatural number contains 1 and also contains the successor of any number that it contains then the set is all natural numbers.
2. Proof by Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both the weak ad the strog versio, ad show how certai types of theorems ca easily be prove usig this techique. Itroductio
More informationExample: Consider the sequence {a i = i} i=1. Starting with i = 1, since a i = i,
Sequeces: Defiitio: A sequece is a fuctio whose domai is the set of atural umbers or a subset of the atural umbers. We usually use the symbol a to represet a sequece, where is a atural umber ad a is the
More informationIntegration by Parts
Itegratio by Parts Dr. Philippe B. Laval Keesaw State Uiversity August 2, 2008 Abstract This hadout describes a itegratio method called Itegratio by Parts. Itegratio by parts is to itegrals what the product
More informationA SHORT SUMMARY OF SEQUENCES AND SERIES. by John Alexopoulos
A SHORT SUMMARY OF SEQUENCES AND SERIES by Joh Alexopoulos Last updated o October 2 st, 2005 Cotets Sequeces 3. Defiitios, otatio ad examples...................................... 3.2 More defiitios ad
More informationMgr. ubomíra Tomková. Limit
Limit I mathematics, the cocept of a "limit" is used to describe the behaviour of a fuctio as its argumet either gets "close" to some poit, or as it becomes arbitrarily large; or the behaviour of a sequece's
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
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.,
More informationLesson 12. Sequences and Series
Retur to List of Lessos Lesso. Sequeces ad Series A ifiite sequece { a, a, a,... a,...} ca be thought of as a list of umbers writte i defiite order ad certai patter. It is usually deoted by { a } =, or
More information2.5. THE COMPARISON TEST
SECTION. THE COMPARISON TEST 69.. THE COMPARISON TEST We bega our systematic study of series with geometric series, provig the Geometric Series Test: ar coverges if ad oly if r. The i the last sectio we
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More informationTo introduce the concept of uniformity, and in particular uniform convergence and continuity, and show why compactness is critical for uniformity.
Sectio 5.5 1 Uiform Cotiuity Sectio 5.5 Uiformity ity ad Compactess Purpose of Sectio To itroduce the cocept of uiformity, ad i particular uiform covergece ad cotiuity, ad show why compactess is critical
More informationSAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More informationCombinatorics Dan Hefetz, Richard Mycroft (lecturer)
Combiatorics Da Hefetz, Richard Mycroft (lecturer) These are otes of the lectures give by Richard Mycroft for the course Commuicatio Theory ad Combiatorics (3P16/4P16) at the Uiversity of Birmigham, o
More informationn a n = L, then lim cos(2/n) = cos(0), lim arctan 2n = π/2, lim = limit does not exist. lim
Usig L Hospitals Rule fid the followig its e 0, 0, si(π/ ) π. (l ) If f(x) is a cotiuous fuctio, ad a L, the f(a ) f(l). Usig this fid the followig its 7 +9 +,. e/ e 0, To Determie the it of fractios like
More informationNOTES ON INFINITE SEQUENCES AND SERIES
NOTES ON INFINITE SEQUENCES AND SERIES MIGUEL A. LERMA. Sequeces.. Sequeces. A ifiite sequece of real umbers is a ordered uedig list of real umbers. E.g.:, 2, 3, 4,... We represet a geeric sequece as a,a
More informationSection 11.3: The Integral Test
Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult
More informationn a n = L, then lim, lim = limit does not exist. lim
Usig L Hospitals Rule fid the followig its e If f(x) is a cotiuous fuctio ad a = L the f(a ) = f(l). Usig this fid the followig its e/ cos(/) arcta +. To Determie the it of fractios like We take the 4
More informationThe Field of Complex Numbers
The Field of Complex Numbers S. F. Ellermeyer The costructio of the system of complex umbers begis by appedig to the system of real umbers a umber which we call i with the property that i = 1. (Note that
More informationCONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART I: THE COMPARISON AND INTEGRAL TESTS. Math 112
CONVERGENCE TESTS FOR SERIES: COMMENTS AND PROOFS PART I: THE COMPARISON AND INTEGRAL TESTS Math 112 The covergece tests for series have ice ituitive reasos why they wor, ad these are fairly easy to tur
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More informationMATH 1D, WEEK 3 THE RATIO TEST, INTEGRAL TEST, AND ABSOLUTE CONVERGENCE
MATH D, WEEK 3 THE RATIO TEST, INTEGRAL TEST, AND ABSOLUTE CONVERGENCE INSTRUCTOR: PADRAIC BARTLETT Abstract. These are the lecture otes from week 3 of Mad, the Caltech mathematics course o sequeces ad
More information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationLecture 25/26 : Integral Test for pseries and The Comparison test
Lecture 5/6 : Itegral Test for pseries ad The Compariso test I this sectio, we show how to use the itegral test to decide whether a series of the form (where p =a a ) coverges or diverges by comparig
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationwhen n = 1, 2, 3, 4, 5, 6, This list represents the amount of dollars you have after n days. Note: The use of is read as and so on.
Geometric eries Before we defie what is meat by a series, we eed to itroduce a related topic, that of sequeces. Formally, a sequece is a fuctio that computes a ordered list. uppose that o day 1, you have
More informationARITHMETIC AND GEOMETRIC PROGRESSIONS
Arithmetic Ad Geometric Progressios Sequeces Ad ARITHMETIC AND GEOMETRIC PROGRESSIONS Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives
More informationThe interval and radius of convergence of a power series
The iterval ad radius of covergece of a power series Bro. David E. Brow, BYU Idaho Dept. of Mathematics. All rights reserved. Versio 1.1, of April, 014 Cotets 1 Itroductio 1 The ratio ad root tests 3 Examples
More informationMATH 140A  HW 3 SOLUTIONS
MATH 40A  HW 3 SOLUTIONS Problem (WR Ch 2 #2). Let K R cosist of 0 ad the umbers / for = 23... Prove that K is compact directly from the defiitio (without usig the HeieBorel theorem). Solutio. Let {G
More informationWinter Camp 2012 Sequences Alexander Remorov. Sequences. Alexander Remorov
Witer Camp 202 Sequeces Alexader Remorov Sequeces Alexader Remorov alexaderrem@gmail.com Warmup Problem : Give a positive iteger, cosider a sequece of real umbers a 0, a,..., a defied as a 0 = 2 ad =
More informationORDERS OF GROWTH KEITH CONRAD
ORDERS OF GROWTH KEITH CONRAD Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really wat to uderstad their behavior It also helps you better grasp topics i calculus
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More information6.10 The Binomial Series
6 THE BINOMIAL SERIES 375 6 The Biomial Series 6 Itroductio This sectio focuses o derivig a Maclauri series for fuctios of the form f x) + x) for ay umber We use the results we obtaied i the sectio o Taylor
More information5.3 The Integral Test and Estimates of Sums
5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch 5.3 The Itegral Test ad Estimates of Sums The ext few sectios we lear techiques that help determie if a series coverges. I the last sectio we were
More informationSection 8.2 Markov and Chebyshev Inequalities and the Weak Law of Large Numbers
Sectio 8. Markov ad Chebyshev Iequalities ad the Weak Law of Large Numbers THEOREM (Markov s Iequality): Suppose that X is a radom variable takig oly oegative values. The, for ay a > 0 we have X a} E[X]
More informationSolutions for problems in the 9 th International Mathematics Competition for University Students
Solutios for problems i the 9 th Iteratioal Mathematics Competitio for Uiversity Studets Warsaw, July 9  July 25, 2002 First Day Problem. A stadard parabola is the graph of a quadratic polyomial y = x
More information1 The Binomial Theorem: Another Approach
The Biomial Theorem: Aother Approach Pascal s Triagle I class (ad i our text we saw that, for iteger, the biomial theorem ca be stated (a + b = c a + c a b + c a b + + c ab + c b, where the coefficiets
More informationSection 9.2 Series and Convergence
Sectio 9. Series ad Covergece Goals of Chapter 9 Approximate Pi Prove ifiite series are aother importat applicatio of limits, derivatives, approximatio, slope, ad cocavity of fuctios. Fid challegig atiderivatives
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationNewton s method is a iterative scheme for solving equations by finding zeros of a function. The
Usig Iterative Methods to Solve Equatios Newto s method is a iterative scheme for solvig equatios by fidig zeros of a fuctio. The f Newto iteratio 1 f fids the zeros of f by fidig the fied poit of g f
More informationMath 230a Homework 4
Math 30a Homework 4 Erik Lewis December 0, 006 Rudi. Prove that the covergece of {s } implies the covergece of { s }. Is the coverse true? Proof: Suppose {s } coverges to s. So we kow that s s < ɛ. We
More informationChapter Suppose you wish to use the Principle of Mathematical Induction to prove that 1 1! + 2 2! + 3 3! n n! = (n + 1)! 1 for all n 1.
Chapter 4. Suppose you wish to prove that the followig is true for all positive itegers by usig the Priciple of Mathematical Iductio: + 3 + 5 +... + ( ) =. (a) Write P() (b) Write P(7) (c) Write P(73)
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationChapter Eleven. Taylor Series. (x a) k. c k. k= 0
Chapter Eleve Taylor Series 111 Power Series Now that we are kowledgeable about series, we ca retur to the problem of ivestigatig the approximatio of fuctios by Taylor polyomials of higher ad higher degree
More informationContinued Fractions continued. 3. Best rational approximations
Cotiued Fractios cotiued 3. Best ratioal approximatios We hear so much about π beig approximated by 22/7 because o other ratioal umber with deomiator < 7 is closer to π. Evetually 22/7 is defeated by 333/06
More informationUniform convergence and power series
Uiform covergece ad power series Review of what we already kow about power series A power series cetered at a R is a expressio of the form c (x a) The umbers c are called the coefficiets of the power series
More informationWhen π(n) does not divide n
Whe π() does ot divide Germá Adrés Paz Jauary 26, 2015 Abstract Let π() deote the primecoutig fuctio ad let / log 1 log 1 f() = log log 0.1. I this paper we prove that if is a iteger 60184 ad f() = 0,
More informationMATH3283W LECTURE NOTES: WEEK 3
MATH3283W LECTURE NOTES: WEEK 3 2//200 Proof without words: picture depicts What is beig proved from Fig.3.? Addig more ad more dots gives bigger ad bigger squares. It is too vague ad it is ot actually
More information0,1 is an accumulation
Sectio 5.4 1 Accumulatio Poits Sectio 5.4 BolzaoWeierstrass ad HeieBorel Theorems Purpose of Sectio: To itroduce the cocept of a accumulatio poit of a set, ad state ad prove two major theorems of real
More informationFrom compound interest to the natural exponential function
From compoud iterest to the atural expoetial fuctio Abstract The most commo itroductio to the atural expoetial fuctio occurs i coectio with compoud iterest. I this paper we will show how to build o this
More informationElementary Results on the Distribution of Primes
Chapter 4 Elemetary Results o the Distributio of Primes 4. Itroductio Defiitio 4.. For real umber > 0, let π deote the umber of primes ot eceedig. The behavior of π as the fuctio of has bee studied by
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationMeasurable Functions
Measurable Fuctios Dug Le 1 1 Defiitio It is ecessary to determie the class of fuctios that will be cosidered for the Lebesgue itegratio. We wat to guaratee that the sets which arise whe workig with these
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationYour grandmother and her financial counselor
Sectio 10. Arithmetic Sequeces 963 Objectives Sectio 10. Fid the commo differece for a arithmetic sequece. Write s of a arithmetic sequece. Use the formula for the geeral of a arithmetic sequece. Use the
More information3.3 Convergence Tests for Infinite Series
3.3 Covergece Tests for Ifiite Series 3.3. The itegral test We may plot the sequece a i the Cartesia plae, with idepedet variable ad depedet variable a. The sum a ca the be represeted geometrically as
More informationRandomized Algorithms I, Spring 2016, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 29, 2016)
Radomized Algorithms I, Sprig 0, Departmet of Computer Sciece, Uiversity of Helsiki Homework : Solutios Discussed Jauary 9, 0). Exercise.: Cosider the followig ballsadbi game. We start with oe black
More informationFirst, it is important to recall some key definitions and notations before discussing the individual series tests.
Ifiite Series Tests Coverget or Diverget? First, it is importat to recall some key defiitios ad otatios before discussig the idividual series tests. sequece  a fuctio that takes o real umber values a
More information8.5 Alternating infinite series
65 8.5 Alteratig ifiite series I the previous two sectios we cosidered oly series with positive terms. I this sectio we cosider series with both positive ad egative terms which alterate: positive, egative,
More informationTILE PATTERNS & GRAPHING
TILE PATTERNS & GRAPHING LESSON 1 THE BIG IDEA Tile patters provide a meaigful cotext i which to geerate equivalet algebraic expressios ad develop uderstadig of the cocept of a variable. Such patters are
More informationBINOMIAL THEOREM INTRODUCTION. Development of Binomial Theorem
8 BINOMIAL THEOREM INTRODUCTION I previous classes, you have leart the squares ad cubes of biomial epressios like a + b, a b ad used these to fid the values of umbers like (10), (998) by epressig these
More informationGAMMA FUNCTION. n! = 1 ( log 0 s)n ds for n > 0 For the time being, we will refer to these as the product and integral definitions.
GAMMA FUNCTION Abstract. I this paper we eplore the history ad properties of the Gamma fuctio i a aalytic umber theoretical cotet. We aalyze the behavior of the Gamma fuctio at its critical poits ad poits
More informationSome review problems for Midterm 1
Some review problems for Midterm February 6, 05 Problems Here are some review problems to practice what we ve leared so far. Problems a,b,c,d ad p,q,r,s are t from the book, ad are a little tricky. You
More informationDiscrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 14
CS 70 Discrete Mathematics ad Probability Theory Fall 2009 Satish Rao, David Tse Note 14 Some Importat Distributios The first importat distributio we leart i the last lecture ote is the biomial distributio
More informationFIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix
FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if
More informationStirling s Formula and DeMoivreLaplace Central Limit Theorem
Stirlig s Formula ad DeMoivreLaplace Cetral Limit Theorem Márto Balázs ad Bálit Tóth October 6, 04 Uiversity of Bristol / Budapest Uiversity of Techology ad Ecoomics Usig Stirlig s formula we prove oe
More informationSection 1.1 1, 1 2, 1 4, 1 8,...
Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?
More informationSection IV.5: Recurrence Relations from Algorithms
Sectio IV.5: Recurrece Relatios from Algorithms Give a recursive algorithm with iput size, we wish to fid a Θ (best big O) estimate for its ru time T() either by obtaiig a explicit formula for T() or by
More informationBasic Elements of Arithmetic Sequences and Series
MA40S PRECALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic
More informationSERIES AND ERROR n 1. for all n.
SERIES AND ERROR The AP Calculus BC course descriptio icludes two kids of error bouds: Alteratig series with error boud Lagrage error boud for Taylor polyomials Both types of error have bee tested o the
More information14 Dihedral Groups. In this paragraph, we examine the symmetry groups of regular polygons.
Dihedral Groups I this paragraph, we examie the symmetry groups of regular polygos. Let F be ay oempty subset of the Euclidea plae E.. Here F might be a set with a sigle poit, a lie, a geometric figure.
More informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More informationA curious result related to Kempner s series
Amer. Math. Mothly, 5 (28), p. 933938. A curious result related to Kemper s series Bakir FARHI bakir.farhi@gmail.com Abstract It is well kow sice A. J. Kemper s work that the series of the reciprocals
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationMath 115 HW #4 Solutions
Math 5 HW #4 Solutios From 2.5 8. Does the series coverge or diverge? ( ) 3 + 2 = Aswer: This is a alteratig series, so we eed to check that the terms satisfy the hypotheses of the Alteratig Series Test.
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More informationExam3 Solutions, Math n + 2. Solution: First note that this series is a telescoping series. Let S n be the nth partial sum of the series.
Exam3 Solutios, Math 0560. Fid the sum of the followig series. [ l( + + ] l( + + 3 Solutio: First ote that this series is a telescopig series. Let S be the th partial sum of the series. S l 3 l 3 4 S
More information1 n. n > dt. t < n 1 + n=1
Math 05 otes C. Pomerace The harmoic sum The harmoic sum is the sum of recirocals of the ositive itegers. We kow from calculus that it diverges, this is usually doe by the itegral test. There s a more
More informationLecture Notes 4 Convergence (Chapter 5)
Radom Samples Lecture Notes 4 Covergece Chapter 5 Let X,..., X F. A statistic is ay fuctio T = gx,..., X. Recall that the sample mea is X = X i ad sample variace is S 2 = Let µ = EX i ad σ 2 = VarX i.
More informationThe geometric series and the ratio test
The geometric series ad the ratio test Today we are goig to develop aother test for covergece based o the iterplay betwee the it compariso test we developed last time ad the geometric series. A ote about
More informationn Find all pairs of integers x, y which satisfy the diophantine equation 50x 65y = 75. a i. G n = a i a 1 a 2 a k
Math 00. Treibergs σιι Fial Exam Name: Sample March 7, 0 Sample Fial Questios.. Show usig iductio that 5 (7 ) for every positive iteger.. Prove that! > for all itegers 4. [From miterm for Math 30, Sept.
More informationif, for every ǫ > 0, no matter how small, there exists a number N for which
Chapter 2 Ifiite Series 2. Sequeces A sequece of complex umbers {z } is a coutably ifiite set of umbers, z, z 2, z 3,...,z,... (2.) That is, for every positive iteger k, there is a umber, the kth term
More information5. SEQUENCES AND SERIES
5. SEQUENCES AND SERIES 5.. Limits of Sequeces Let N = {0,,,... } be the set of atural umbers ad let R be the set of real umbers. A ifiite real sequece u 0, u, u, is a fuctio from N to R, where we write
More informationABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS. Absolute Covergece Give ay series a we ca cosider a related series a a + a 2 +... + a +... whose terms are the absolute values of the terms of the origial
More informationHW 3 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson
HW 3 Solutios Math 5, Witer 009, Prof. Yitzhak Katzelso 7.4 a) Let x =. The lim x = 0 (as you may easily check  for ǫ > 0, just let N be ǫ ), which is ratioal, eve though all the x are irratioal. b) Cosider
More informationSection 1.6: Proof by Mathematical Induction
Sectio.6 Proof by Iductio Sectio.6: Proof by Mathematical Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both weak ad the strog versios, ad show how certai types of theorems
More information2. Introduction to Statistics and Sampling
. Defiitios. Itroductio to Statistics ad Samplig.. Populatio.. Sample..3 Probability..4 Cotiuous vs. discrete variables we will cocetrate o cotiuous variables. Graphical Represetatio of a Fiite Sample
More informationMath 211 Solutions To Sample Exam 2 Problems
Solutios to Sample Exam Problems Math Math Solutios To Sample Exam Problems. Solve each of the followig differetial equatios / iitial value problems. (a) (+x) dy dx 4y (b) y dy dx x x 6 Solutio: (a) Usig
More information. This is read n choose k. To refresh your memory, the binomial coefficients have the following formula and recurrence. n 1. n k. k 1.
. WEEK.. LECTURE : BINOMIAL COEFFICIENTS,UNIMODAL ITY, LOGCONCAVITY Recall that the umber of elemet subsets of a elemet set is deoted by. This is read choose. To refresh your memory, the biomial coefficiets
More informationMath Discrete Math Combinatorics MULTIPLICATION PRINCIPLE:
Math 355  Discrete Math 4.14.4 Combiatorics Notes MULTIPLICATION PRINCIPLE: If there m ways to do somethig ad ways to do aother thig the there are m ways to do both. I the laguage of set theory: Let
More informationTHE UNIVERSITY OF TEXAS AT AUSTIN Department of Information, Risk, and Operations Management
THE UNIVERSITY OF TEAS AT AUSTIN Departmet of Iformatio, Risk, ad Operatios Maagemet BA 386T Tom Shively ESTIMATION AND SAMPLING DISTRIBUTIONS The purpose of these otes is to summarize the cocepts regardig
More informationThe Integral Test. and. n 1. either both converge or both diverge. f i f 2 f 3... f n. f i f 1 f 2... f n 1. f i f x dx n 1. f i. i 1.
0_090.qd //0 :5 PM Page 7 SECTIO 9. The Itegral Test ad pseries 7 Sectio 9. The Itegral Test ad pseries Use the Itegral Test to determie whether a ifiite series coverges or diverges. Use properties of
More informationMATH 289 WINTER 2011 PROBLEM SET 3: INEQUALITIES
MATH 89 WINTER 0 PROBLEM SET 3: INEQUALITIES. Elemetary iequalities Perhaps the most fudametal iequality for real umbers is x 0, x R. Usig this iequality oe ca deduce may more iequalities. For example,
More information