Sequence A sequence is a function whose domain of definition is the set of natural numbers.
|
|
- Margaret Barton
- 5 years ago
- Views:
Transcription
1 Chapter Sequeces Course Title: Real Aalysis Course Code: MTH3 Course istructor: Dr Atiq ur Rehma Class: MSc-I Course URL: wwwmathcityorg/atiq/fa8-mth3 Sequeces form a importat compoet of Mathematical Aalysis ad arise i may situatios The first rigorous treatmet of sequeces was made by George Cator (845-98) ad A Cauchy ( ) A sequece (of real umbers, of sets, of fuctios, of aythig) is simply a list There is a first elemet i the list, a secod elemet, a third elemet, ad so o cotiuig i a order forever I mathematics a fiite list is ot called a sequece (some authors cosidered it fiite sequece); a sequece must cotiue without iterruptio Formally it is defied as follows: Sequece A sequece is a fuctio whose domai of defiitio is the set of atural umbers Notatio: A ifiite sequece is usually deoted as { } s = or { s : N } or { s, s, s,} or simply as { } 3 But it is ot limited to above otatios oly The values eg i) { } = {,,3, } ii) s or by ( s ) s are called the terms or the elemets of the sequece { } =,,, 3 + ( ) =,,,, iii) { } { } iv) {,3,5,7,, }, a sequece of positive prime umbers Subsequece It is a sequece whose terms are cotaied i give sequece A subsequece of { s } is usually writte as { s } Icreasig Sequece s is said to be a icreasig sequece if s + s A sequece { } Decreasig Sequece s is said to be a decreasig sequece if s + s A sequece { } Mootoic Sequece s is said to be mootoic sequece if it is either icreasig or decreasig A sequece { } s
2 Ch : Sequeces - - Remars: A sequece { } s is mootoically icreasig if s s + s s is mootoically icreasig if + s, A positive term sequece { } A sequece { } s is mootoically decreasig if s s + A positive term sequece { s } is mootoically decreasig if s, s + Strictly Icreasig or Decreasig A sequece { s } is called strictly icreasig or decreasig accordig as s > + s or s < + s Examples: { } = {,,3,} is a icreasig sequece is a decreasig sequece { cos π } = {,,,,} is either icreasig or decreasig Questios: ) Prove that + is a decreasig sequece + ) Is is icreasig or decreasig sequece? + Bouded Sequece s is said to be bouded if there is a umber λ A sequece { } s < λ N For such a sequece, every term belogs to the iterval [ λ, λ] It ca be oted that if the sequece is bouded the its supremum ad ifimum exist If S ad s are the supremum ad ifimum of the bouded sequece { s }, the we write S = sup s ad s = if s
3 Ch : Sequeces Examples ( ) (i) { } u = is a bouded sequece v = si x is also bouded sequece Its supremum is ad ifimum is (ii) { } { } (iii) The geometric sequece { } bouded below by a π (iv) ta is a ubouded sequece ar, r > is a ubouded above sequece It is Covergece of the sequece The sequece,,,,,, is gettig closer ad closer to the umber We say that this sequece coverges to or that the limit of the sequece is the umber How should this idea be properly defied? The study of coverget sequeces was udertae ad developed i the eighteeth cetury without ay precise defiitio The closest oe might fid to a defiitio i the early literature would have bee somethig lie A sequece { s } coverges to a umber L if the terms of the sequece get closer ad closer to L However, this is too vague ad too wea to serve as defiitio but a rough guide for the ituitio, this is misleadig i other respects What about the sequece,,,,,,,,,? Surely this should coverge to but the terms do ot get steadily closer ad closer but bac off a bit at each secod step The defiitio that captured the idea i the best way was give by Augusti Cauchy i the 8s He foud a formulatio that expressed the idea of arbitrarily close usig iequalities Defiitio A sequece { s } of real umbers is said to coverget to limit s as, if for every real umber >, there exists a positive iteger, depedig o, so that s s < wheever > A sequece that coverges is said to be coverget A sequece that fails to coverge is said to diverget We will try to uderstad it by graph of some sequece Graphs of ay four sequeces is draw i the picture below
4 Ch : Sequeces Examples a) Prove that lim = (or coverges to ) Solutio: Let > be give By the Archimedea Property, there is a positive iteger = ( ) >, that is, < The, if, we have < < Thus we proved that for all >, there exists, depedig upo, = < wheever Hece coverges to poit b) Prove that lim = (by defiitio) + Solutio: Let > be give Now cosider = < < ( + > > ) + + Now if we choose <, the the above expressio gives us < wheever + Hece, we coclude that, lim = +
5 Ch : Sequeces c) Prove that lim = 3 (by defiitio) + Solutio: Let > be give Now cosider = + + = < < ( + > > ) + + Now if we choose <, the the above expressio gives us < wheever Hece, we coclude that lim = 3 + Questios: Use defiitio of the limits to prove the followigs: a) lim = b) lim = Review Triagular iequality: If a, b R, the a b a ± b a + b If a < for all >, the a = Theorem A coverget sequece of real umber has oe ad oly oe limit (ie limit of the sequece is uique) s coverges to two limits s ad t, where s t The for all >, there exists two positive itegers ad s s < > () ad s t < > () As () ad () hold simultaeously for all > max(, ) Thus for all > max(, ) we have s t = s s + s t s s + s t Suppose { } < + = As is arbitrary, we get s = t, that is, the limit of the sequece is uique
6 Ch : Sequeces Questio Prove that if lim s = t, the lim s = t but coverse is ot true i geeral Review: For all a, b, c R, a b < c c b < a < c + b or c a < b < c + a Theorem (Sadwich Theorem or Squeeze Theorem) t be two coverget sequeces lim s Suppose that { } s ad { }, the the sequece { } = limt If s < u < t u also coverges to s Sice the sequece { s } ad { t } coverge to the same limit s (say), therefore for give > there exists two positive itegers, > s s < >, t s < > ie s < s < s + >, s < t < s + > Sice we have give s < u < t > s < s < u < t < s + > max(,, ) s < u < s + > max(,, ) ie u s < > max(,, ) ie limu = s Example Show that lim = ( + ) ( + ) ( ) Solutio Cosider s = ( + ) ( + ) ( ) As s < + + +, ( ) ( ) ( ) that is, times s < ( ) s < 4 times
7 Ch : Sequeces lim lim s < lim 4 lim s < lim s = Cauchy Sequece A sequece { s } of real umber is said to be a Cauchy sequece if for give umber >, there exists a positive iteger ( ) s sm < m, > Example The sequece is a Cauchy sequece Suppose s = ad > be give We choose a positive iteger = ( ) such that > The if m,, we have < ad similarly < Therefore, it m follows that if m,, the s sm = + < + = m m Sice > is arbitrary, we coclude that is Cauchy sequece Theorem A Cauchy sequece of real umbers is bouded s be a Cauchy sequece The for give umber >, there exists a positive iteger s sm < m, > Tae =, the we have Let { } Fix m = + the + + s s = s s + s s s + s + + sm < m, > < + s + >
8 Ch : Sequeces = λ >, ad λ s + = + ( chages as chages) Hece we coclude that { s } is a Cauchy sequece, which is bouded oe Note: (i) Coverget sequece is bouded (ii) The coverse of the above theorem does ot hold ie every bouded sequece is ot Cauchy s where s = ( ), It is bouded sequece because Cosider the sequece { } ( ) = < But it is ot a Cauchy sequece if it is the for = we should be able to fid a positive iteger s sm < for all m, > But with m = +, = + whe + >, we arrive at s s = ( ) ( ) m + + = + = < is absurd Hece { s } is ot a Cauchy sequece Also this sequece is ot a coverget sequece (it is a oscillatory sequece) Questio: Prove that every Cauchy sequece of real umber is bouded but coverse is ot true Theorem If the sequece { s } coverges to s the a positive iteger s > s for all > We fix = s > a positive iteger s s < for > s s < s Now s = s s s s s s + s s < ( ) s < s
9 Ch : Sequeces Theorem Let a ad b be fixed real umbers if { s } ad { } respectively, the as + bt coverges to as + bt (i) { } (ii) { } s t coverges to st t coverge to s ad t s (iii) coverges to s t t, provided t ad t Sice { s } ad { } t coverge to s ad t respectively, < s s > N t t < > N Also λ > s < λ (i) We have as + bt as + bt = a s s + b t t > ( { } ( ) ( ) ( ) ( ) a( s s) + b( t t) s is bouded ) < a + b > max (, ) =, where = a + b a certai umber This implies { as bt } (ii) (iii) + coverges to as + bt s t st = s t s t + s t st = s ( t t) + t ( s s ) s ( t t) + t ( s s ) < λ + t > max (, ) =, where = λ + t a certai umber This implies { } t t = t t t t s t coverges to st t t t t t t = < > max (, ) = = 3, where 3 t t t > t = a certai umber This implies t coverges to t s Hece s t = t coverges to s s t t = ( from (ii) )
10 Ch : Sequeces - - Theorem For each irratioal umber x, there exists a sequece { } umbers lim r = x Sice x ad x + are two differet real umbers a ratioal umber r x < r < x + Similarly a ratioal umber r r x < r < mi r, x + < x + Cotiuig i this maer we have x < r3 < mi r, x + < x + 3 x < r4 < mi r3, x + < x + 4 r of distict ratioal x < r < mi r, x + < x + r of the distict ratioal umber This implies that there is a sequece { } Sice Therefore Theorem lim x < r < x + x = lim x + = x ( ) lim r = x Let a sequece { } (i) If { } (ii) If { } s be a bouded sequece s is mootoically icreasig the it coverges to its supremum s is mootoically decreasig the it coverges to its ifimum Proof (i) Let S = sup s ad tae > Sice there exists s Sice { } S < s s is mootoically icreasig, therefore
11 Ch : Sequeces - - S < s < s < S < S + for > S < s < S + for > s S < for > lim s = S (ii) Let s = if s ad tae > Sice there exists s Sice { } s < s + s is mootoically decreasig, therefore s < s < s < s < s + for > s < s < s + for > s s < for > Thus lim s = s Questio: Let { } s be a sequece ad lim s = s The prove that lim s + = s Prove that a bouded icreasig sequece coverges to its supremum 3 Prove that a bouded decreasig sequece coverges to its ifimum Recurrece Relatio A sequece is said to be defied recursively or by recurrece relatio if the geeral term is give as a relatio of its precedig ad succeedig terms i the sequece together with some iitial coditio Example: Let t > ad let { } t be defied by t (i) Show that { t } is decreasig sequece (ii) It is bouded below (iii) Fid the limit of the sequece Sice t > ad { } t is defied by t t > Also t t+ = t + t ( ) = + t for = + t ; t t + t = = > t t t > t + This implies that t is mootoically decreasig
12 Ch : Sequeces - - Sice t >, t is bouded below Sice t is decreasig ad bouded below therefore t is coverget Let us suppose limt = t The lim t+ = lim t lim = limt t t = t = t t = t t t ( t ) = t = t t + = Questio: Let { t } be a positive term sequece Fid the limit of the sequece if 4t+ = 3t for all 5 u be a sequece of positive umbers The fid the limit of the sequece Let { } = + for if u+ u u 4 The Fiboacci umbers are: F = F =, ad for every 3, F is defied by F the recurrece relatio F = F + F Fid the lim (this limit is ow as F golde umber) Theorem Every Cauchy sequece of real umbers has a coverget subsequece s is a Cauchy sequece Let > the a positive iteger s s <,, =,,3, Suppose { } Put b = ( s ) ( ) ( ) + s s + + s s b = ( s ) ( ) ( ) + s s + + s s ( ) + ( ) + + ( ) s s s s s s < = ( ) = =
13 Ch : Sequeces b < This gives { } b is bouded Sice b = b + ( s + s ) { } where + b is coverget b = s s s = b + s, coverget s is a certai fix umber therefore { } s which is a subsequece of { s } is Theorem (Cauchy s Geeral Priciple for Covergece) A sequece of real umber is coverget if ad oly if it is a Cauchy sequece s be a coverget sequece, which coverges to s The for give > a positive iteger, s s < > Now for > m > s sm = s s + s sm s s s s = s s + s s Let { } + m m < + = s is a Cauchy sequece This shows that { } s is a Cauchy sequece the for >, there exists a positive iteger m s sm <, m > m (i) s is a Cauchy sequece, Coversely, suppose that { } Sice { } therefore it has a subsequece { s } covergig to s (say) This implies there exists a positive iteger m s s < > m (ii) Now s s = s s + s s this shows that { } s s + s s < + = > max( m, m), s is a coverget sequece
14 Ch : Sequeces Example Prove that is diverget sequece 3 t be defied by Let { } For m, N, t = > m we have t tm = m + m + > ( m times) = ( m) = m I particular if = m the t tm > t is ot a Cauchy sequece therefore it is diverget This implies that { } Theorem (ested itervals) I Suppose that { I } is a sequece of the closed iterval I [ a, b ] I, ad ( b a ) I + poit as the Sice I + I a < a < a < < a < a < b < b < < b < b < b { } 3 3 =, cotais oe ad oly oe a is icreasig sequece, bouded above by b ad bouded below by a b is decreasig sequece bouded below by a ad bouded above by b b both are coverget Ad { } { } Suppose { } a ad { } a coverges to a ad { } a b = a a + a b + b b But b coverges to b a a + a b + b b as a = b ad a < a < b
15 Ch : Sequeces Limit Iferior of the sequece Suppose { s } is bouded below the we defie limit iferior of { } limif s = limu, where u = if { s : } If s is ot bouded below the limif s = Limit Superior of the sequece Suppose { s } is bouded above the we defie limit superior of { } limsup s = limv, where v = sup { s : } If s is ot bouded above the we have limsup s = + s as follow s as follow Note: (i) A bouded sequece has uique limit iferior ad superior (ii) Let { s } cotais all the ratioal umbers, the every real umber is a subsequecial limit the limit superior of s is + ad limit iferior of s is (iii) Let { s } = ( ) + the limit superior of s is ad limit iferior of s is (iv) Let s = + cosπ u = if s : The { } = if + cos π, + cos( + ) π, + cos( + ) π, cosπ if is odd = + cos( + ) π if is eve + lim if s = limu = ( ) Also v = sup { s : } + cos( + ) π if is odd + = + cosπ if is eve lim sup s = limv = ( )
16 Ch : Sequeces Theorem s is a coverget sequece the If { } ( ) ( ) lim s = lim if s = lim sup s Let lim s = s the for a real umber >, a positive iteger s s < (i) ie s < s < s + v = sup s : s < v < s + s < limv < s + (ii) If { } The from (i) ad (ii) we have s = lim sup { s } We ca have the same result for limit iferior of { } u if { s : } s by taig = Refereces: W Rudi, Priciple of Mathematical Aalysis, 3 rd Editio, McGraw-Hill, Ic, 976 RG Bartle ad DR Sherbert, Itroductio to Real Aalysis, 4 th Editio, Joh Wiley & Sos, Ic, 3 BS Thomso, JB Brucer ad AM Brucer, Elemetary Real Aalysis, Pretice Hall (Pearso), URL:
MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions
MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationAssignment 5: Solutions
McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece
More informationInfinite 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
More information1 Introduction. 1.1 Notation and Terminology
1 Itroductio You have already leared some cocepts of calculus such as limit of a sequece, limit, cotiuity, derivative, ad itegral of a fuctio etc. Real Aalysis studies them more rigorously usig a laguage
More informationM17 MAT25-21 HOMEWORK 5 SOLUTIONS
M17 MAT5-1 HOMEWORK 5 SOLUTIONS 1. To Had I Cauchy Codesatio Test. Exercise 1: Applicatio of the Cauchy Codesatio Test Use the Cauchy Codesatio Test to prove that 1 diverges. Solutio 1. Give the series
More informationRead carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.
THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More information10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term.
0. Sequeces A sequece is a list of umbers writte i a defiite order: a, a,, a, a is called the first term, a is the secod term, ad i geeral eclusively with ifiite sequeces ad so each term Notatio: the sequece
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationSolutions to Tutorial 3 (Week 4)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial Week 4 MATH2962: Real ad Complex Aalysis Advaced Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
More information1 Lecture 2: Sequence, Series and power series (8/14/2012)
Summer Jump-Start Program for Aalysis, 202 Sog-Yig Li Lecture 2: Sequece, Series ad power series (8/4/202). More o sequeces Example.. Let {x } ad {y } be two bouded sequeces. Show lim sup (x + y ) lim
More informationTopics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.
MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem 4.1.8. 4. Theorems 4.1.13
More informationn=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n
Series. Defiitios ad first properties A series is a ifiite sum a + a + a +..., deoted i short by a. The sequece of partial sums of the series a is the sequece s ) defied by s = a k = a +... + a,. k= Defiitio
More informationLecture 17Section 10.1 Least Upper Bound Axiom
Lecture 7Sectio 0. Least Upper Boud Axiom Sectio 0.2 Sequeces of Real Numbers Jiwe He Real Numbers. Review Basic Properties of R: R beig Ordered Classificatio N = {0,, 2,...} = {atural umbers} Z = {...,
More informationCHAPTER 1 SEQUENCES AND INFINITE SERIES
CHAPTER SEQUENCES AND INFINITE SERIES SEQUENCES AND INFINITE SERIES (0 meetigs) Sequeces ad limit of a sequece Mootoic ad bouded sequece Ifiite series of costat terms Ifiite series of positive terms Alteratig
More informationsin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =
60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece
More informationn n 2 n n + 1 +
Istructor: Marius Ioescu 1. Let a =. (5pts) (a) Prove that for every ε > 0 there is N 1 such that a +1 a < ε if N. Solutio: Let ε > 0. The a +1 a < ε is equivalet with + 1 < ε. Simplifyig, this iequality
More informationDefinition An infinite sequence of numbers is an ordered set of real numbers.
Ifiite sequeces (Sect. 0. Today s Lecture: Review: Ifiite sequeces. The Cotiuous Fuctio Theorem for sequeces. Usig L Hôpital s rule o sequeces. Table of useful its. Bouded ad mootoic sequeces. Previous
More informationSequences. A Sequence is a list of numbers written in order.
Sequeces A Sequece is a list of umbers writte i order. {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 d term,
More informationMAS111 Convergence and Continuity
MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece
More informationNotes #3 Sequences Limit Theorems Monotone and Subsequences Bolzano-WeierstraßTheorem Limsup & Liminf of Sequences Cauchy Sequences and Completeness
Notes #3 Sequeces Limit Theorems Mootoe ad Subsequeces Bolzao-WeierstraßTheorem Limsup & Limif of Sequeces Cauchy Sequeces ad Completeess This sectio of otes focuses o some of the basics of sequeces of
More informationPart A, for both Section 200 and Section 501
Istructios Please write your solutios o your ow paper. These problems should be treated as essay questios. A problem that says give a example or determie requires a supportig explaatio. I all problems,
More informationMA131 - Analysis 1. Workbook 9 Series III
MA3 - Aalysis Workbook 9 Series III Autum 004 Cotets 4.4 Series with Positive ad Negative Terms.............. 4.5 Alteratig Series.......................... 4.6 Geeral Series.............................
More informationSequences and Series
Sequeces ad Series Sequeces of real umbers. Real umber system We are familiar with atural umbers ad to some extet the ratioal umbers. While fidig roots of algebraic equatios we see that ratioal umbers
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More information7 Sequences of real numbers
40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are
More information2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.
CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationMath 299 Supplement: Real Analysis Nov 2013
Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality
More informationMA131 - Analysis 1. Workbook 3 Sequences II
MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................
More informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you
More informationMATH 147 Homework 4. ( = lim. n n)( n + 1 n) n n n. 1 = lim
MATH 147 Homework 4 1. Defie the sequece {a } by a =. a) Prove that a +1 a = 0. b) Prove that {a } is ot a Cauchy sequece. Solutio: a) We have: ad so we re doe. a +1 a = + 1 = + 1 + ) + 1 ) + 1 + 1 = +
More information{ } { S n } is monotonically decreasing if Sn
Sequece A sequece is fuctio whose domi of defiitio is the set of turl umers. Or it c lso e defied s ordered set. Nottio: A ifiite sequece is deoted s { } S or { S : N } or { S, S, S,...} or simply s {
More informationMath 140A Elementary Analysis Homework Questions 3-1
Math 0A Elemetary Aalysis Homework Questios -.9 Limits Theorems for Sequeces Suppose that lim x =, lim y = 7 ad that all y are o-zero. Detarime the followig limits: (a) lim(x + y ) (b) lim y x y Let s
More informationSOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.
SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad
More informationSolutions to Tutorial 5 (Week 6)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial 5 (Wee 6 MATH2962: Real ad Complex Aalysis (Advaced Semester, 207 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
More informationLimit superior and limit inferior c Prof. Philip Pennance 1 -Draft: April 17, 2017
Limit erior ad limit iferior c Prof. Philip Peace -Draft: April 7, 207. Defiitio. The limit erior of a sequece a is the exteded real umber defied by lim a = lim a k k Similarly, the limit iferior of a
More informationREAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS
REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai
More informationSequences. Notation. Convergence of a Sequence
Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationSUMMARY OF SEQUENCES AND SERIES
SUMMARY OF SEQUENCES AND SERIES Importat Defiitios, Results ad Theorems for Sequeces ad Series Defiitio. A sequece {a } has a limit L ad we write lim a = L if for every ɛ > 0, there is a correspodig iteger
More information} is said to be a Cauchy sequence provided the following condition is true.
Math 4200, Fial Exam Review I. Itroductio to Proofs 1. Prove the Pythagorea theorem. 2. Show that 43 is a irratioal umber. II. Itroductio to Logic 1. Costruct a truth table for the statemet ( p ad ~ r
More informationChapter 3. Strong convergence. 3.1 Definition of almost sure convergence
Chapter 3 Strog covergece As poited out i the Chapter 2, there are multiple ways to defie the otio of covergece of a sequece of radom variables. That chapter defied covergece i probability, covergece i
More information2.4 Sequences, Sequences of Sets
72 CHAPTER 2. IMPORTANT PROPERTIES OF R 2.4 Sequeces, Sequeces of Sets 2.4.1 Sequeces Defiitio 2.4.1 (sequece Let S R. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For each
More information3. Sequences. 3.1 Basic definitions
3. Sequeces 3.1 Basic defiitios Defiitio 3.1 A (ifiite) sequece is a fuctio from the aturals to the real umbers. That is, it is a assigmet of a real umber to every atural umber. Commet 3.1 This is the
More informationSection 7 Fundamentals of Sequences and Series
ectio Fudametals of equeces ad eries. Defiitio ad examples of sequeces A sequece ca be thought of as a ifiite list of umbers. 0, -, -0, -, -0...,,,,,,. (iii),,,,... Defiitio: A sequece is a fuctio which
More informationSeries III. Chapter Alternating Series
Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with
More informationMetric Space Properties
Metric Space Properties Math 40 Fial Project Preseted by: Michael Brow, Alex Cordova, ad Alyssa Sachez We have already poited out ad will recogize throughout this book the importace of compact sets. All
More informationIt is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.
Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable
More informationd) If the sequence of partial sums converges to a limit L, we say that the series converges and its
Ifiite Series. Defiitios & covergece Defiitio... Let {a } be a sequece of real umbers. a) A expressio of the form a + a +... + a +... is called a ifiite series. b) The umber a is called as the th term
More informationMa 530 Infinite Series I
Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li
More informationINFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationSequences, Series, and All That
Chapter Te Sequeces, Series, ad All That. Itroductio Suppose we wat to compute a approximatio of the umber e by usig the Taylor polyomial p for f ( x) = e x at a =. This polyomial is easily see to be 3
More informationlim za n n = z lim a n n.
Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget
More informationIf a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?
2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a
More informationAxioms of Measure Theory
MATH 532 Axioms of Measure Theory Dr. Neal, WKU I. The Space Throughout the course, we shall let X deote a geeric o-empty set. I geeral, we shall ot assume that ay algebraic structure exists o X so that
More informationMathematical Methods for Physics and Engineering
Mathematical Methods for Physics ad Egieerig Lecture otes Sergei V. Shabaov Departmet of Mathematics, Uiversity of Florida, Gaiesville, FL 326 USA CHAPTER The theory of covergece. Numerical sequeces..
More informationChapter 3 Selected Exercises (Rudin)
W. Rudi, Priciples of Mathematical Aalysis, rd Editio. Prove that covergece of {s } implies covergece of { s }. Is the coverse true? (i) Suppose {s } coverges to some s, the for ε > 0, there exists a iteger
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationIn this section, we show how to use the integral test to decide whether a series
Itegral Test Itegral Test Example Itegral Test Example p-series Compariso Test Example Example 2 Example 3 Example 4 Example 5 Exa Itegral Test I this sectio, we show how to use the itegral test to decide
More informationCouncil for Innovative Research
ABSTRACT ON ABEL CONVERGENT SERIES OF FUNCTIONS ERDAL GÜL AND MEHMET ALBAYRAK Yildiz Techical Uiversity, Departmet of Mathematics, 34210 Eseler, Istabul egul34@gmail.com mehmetalbayrak12@gmail.com I this
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationE. Incorrect! Plug n = 1, 2, 3, & 4 into the general term formula. n =, then the first four terms are found by
Calculus II - Problem Solvig Drill 8: Sequeces, Series, ad Covergece Questio No. of 0. Fid the first four terms of the sequece whose geeral term is give by a ( ) : Questio #0 (A) (B) (C) (D) (E) 8,,, 4
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisher-provided material Overview May mathematical
More informationMeasure and Measurable Functions
3 Measure ad Measurable Fuctios 3.1 Measure o a Arbitrary σ-algebra Recall from Chapter 2 that the set M of all Lebesgue measurable sets has the followig properties: R M, E M implies E c M, E M for N implies
More informationReal Variables II Homework Set #5
Real Variables II Homework Set #5 Name: Due Friday /0 by 4pm (at GOS-4) Istructios: () Attach this page to the frot of your homework assigmet you tur i (or write each problem before your solutio). () Please
More information3 Gauss map and continued fractions
ICTP, Trieste, July 08 Gauss map ad cotiued fractios I this lecture we will itroduce the Gauss map, which is very importat for its coectio with cotiued fractios i umber theory. The Gauss map G : [0, ]
More informationLecture Notes for Analysis Class
Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios
More informationTRUE/FALSE QUESTIONS FOR SEQUENCES
MAT1026 CALCULUS II 21.02.2012 Dokuz Eylül Üiversitesi Fe Fakültesi Matematik Bölümü Istructor: Egi Mermut web: http://kisi.deu.edu.tr/egi.mermut/ TRUE/FALSE QUESTIONS FOR SEQUENCES Write TRUE or FALSE
More informationSCORE. Exam 2. MA 114 Exam 2 Fall 2016
MA 4 Exam Fall 06 Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use
More informationMATH 312 Midterm I(Spring 2015)
MATH 3 Midterm I(Sprig 05) Istructor: Xiaowei Wag Feb 3rd, :30pm-3:50pm, 05 Problem (0 poits). Test for covergece:.. 3.. p, p 0. (coverges for p < ad diverges for p by ratio test.). ( coverges, sice (log
More informationINFINITE SEQUENCES AND SERIES
11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More informationFINALTERM EXAMINATION Fall 9 Calculus & Aalytical Geometry-I Questio No: ( Mars: ) - Please choose oe Let f ( x) is a fuctio such that as x approaches a real umber a, either from left or right-had-side,
More informationFUNDAMENTALS OF REAL ANALYSIS by
FUNDAMENTALS OF REAL ANALYSIS by Doğa Çömez Backgroud: All of Math 450/1 material. Namely: basic set theory, relatios ad PMI, structure of N, Z, Q ad R, basic properties of (cotiuous ad differetiable)
More informationSeries Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1
Defiitio: We say that the series S = Series Review i= a i is the sum of the first terms. i= a i coverges if lim S exists ad is fiite, where The above is the defiitio of covergece for series. order to see
More informationSequences and Limits
Chapter Sequeces ad Limits Let { a } be a sequece of real or complex umbers A ecessary ad sufficiet coditio for the sequece to coverge is that for ay ɛ > 0 there exists a iteger N > 0 such that a p a q
More informationMATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and
MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f
More informationChapter 0. Review of set theory. 0.1 Sets
Chapter 0 Review of set theory Set theory plays a cetral role i the theory of probability. Thus, we will ope this course with a quick review of those otios of set theory which will be used repeatedly.
More information1+x 1 + α+x. x = 2(α x2 ) 1+x
Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem
More informationQuiz No. 1. ln n n. 1. Define: an infinite sequence A function whose domain is N 2. Define: a convergent sequence A sequence that has a limit
Quiz No.. Defie: a ifiite sequece A fuctio whose domai is N 2. Defie: a coverget sequece A sequece that has a limit 3. Is this sequece coverget? Why or why ot? l Yes, it is coverget sice L=0 by LHR. INFINITE
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
More information2.4.2 A Theorem About Absolutely Convergent Series
0 Versio of August 27, 200 CHAPTER 2. INFINITE SERIES Add these two series: + 3 2 + 5 + 7 4 + 9 + 6 +... = 3 l 2. (2.20) 2 Sice the reciprocal of each iteger occurs exactly oce i the last series, we would
More informationNotice that this test does not say anything about divergence of an alternating series.
MATH 572H Sprig 20 Worksheet 7 Topics: absolute ad coditioal covergece; power series. Defiitio. A series b is called absolutely coverget if the series b is coverget. If the series b coverges, while b diverges,
More informationSingular Continuous Measures by Michael Pejic 5/14/10
Sigular Cotiuous Measures by Michael Peic 5/4/0 Prelimiaries Give a set X, a σ-algebra o X is a collectio of subsets of X that cotais X ad ad is closed uder complemetatio ad coutable uios hece, coutable
More informationLecture 3 : Random variables and their distributions
Lecture 3 : Radom variables ad their distributios 3.1 Radom variables Let (Ω, F) ad (S, S) be two measurable spaces. A map X : Ω S is measurable or a radom variable (deoted r.v.) if X 1 (A) {ω : X(ω) A}
More informationContinued Fractions and Pell s Equation
Max Lah Joatha Spiegel May, 06 Abstract Cotiued fractios provide a useful, ad arguably more atural, way to uderstad ad represet real umbers as a alterative to decimal expasios I this paper, we eumerate
More informationAbstract Vector Spaces. Abstract Vector Spaces
Astract Vector Spaces The process of astractio is critical i egieerig! Physical Device Data Storage Vector Space MRI machie Optical receiver 0 0 1 0 1 0 0 1 Icreasig astractio 6.1 Astract Vector Spaces
More informationUniversity of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!
Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad
More informationMath 341 Lecture #31 6.5: Power Series
Math 341 Lecture #31 6.5: Power Series We ow tur our attetio to a particular kid of series of fuctios, amely, power series, f(x = a x = a 0 + a 1 x + a 2 x 2 + where a R for all N. I terms of a series
More informationMATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1
MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The
More informationFall 2013 MTH431/531 Real analysis Section Notes
Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters
More informationINFINITE SERIES PROBLEMS-SOLUTIONS. 3 n and 1. converges by the Comparison Test. and. ( 8 ) 2 n. 4 n + 2. n n = 4 lim 1
MAC 3 ) Note that 6 3 + INFINITE SERIES PROBLEMS-SOLUTIONS 6 3 +
More informationAlternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.
0_0905.qxd //0 :7 PM Page SECTION 9.5 Alteratig Series Sectio 9.5 Alteratig Series Use the Alteratig Series Test to determie whether a ifiite series coverges. Use the Alteratig Series Remaider to approximate
More information5 Sequences and Series
Bria E. Veitch 5 Sequeces ad Series 5. Sequeces A sequece is a list of umbers i a defiite order. a is the first term a 2 is the secod term a is the -th term The sequece {a, a 2, a 3,..., a,..., } is a
More information