Math 140A Elementary Analysis Homework Questions 3-1

Size: px
Start display at page:

Download "Math 140A Elementary Analysis Homework Questions 3-1"

Transcription

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 = ad for let s + = s +. (a) List the first four terms of (s ). (b) It turs out that (s ) coverges. Assume this fact ad prove that the limit is ( + ). 6 Let x = ad x + = x for. (a) Show that if a = lim x, the a = or a = 0. (b) Does lim x exist? Explai. (c) Discuss the apparet cotradictio betwee parts (a) ad (b). Assume all s = 0 ad that the limit L = lim exists. s + s (a) Show that if L <, the lim s = 0. Hit: Select a so that L < a < ad obtai N so that s + < a s for N. The show that s < a N s N for > N. (b) Show that if L >, the lim s = +. Hit: Apply (a) to the sequece t = s. Let p > 0. Use questio to show that a lim p = 0 if a + if a > does ot exist if a <.0 Mootoe Sequeces ad Cauchy Sequeces 6 (a) Let (s ) be a sequece such that s + s < N. Prove that (s ) is a Cauchy sequece ad hece a coverget sequece. (b) Is the result i (a) true if we oly assume that s + s < for all N. 7 Let S be a bouded oempty subset of R ad suppose sup S / S. Prove that there is a odecreasig sequece (s ) of poits i S such that lim s = sup S. 8 Let (s ) be a odecreasig sequece of positive umbers ad defie σ = (s + s + + s ). Prove that (σ ) is a odecreasig sequece. 9 Let s = ad s + = + s for. (a) Fid s, s ad s.

2 (b) Show that lim s exists. (c) Prove that lim s = 0. 0 Let s = ad s + = (s + ) for. (a) Fid s, s ad s. (b) Use iductio to show that s > for all (c) Show that (s ) is a oicreasig sequece. (d) Show that lim s exists ad fid lim s.. Subsequeces Cosider the sequeces defied as follows: a = ( ) b = c = d = (a) For each sequece, give a example of a mootoe subsequece. (b) For each sequece, give its set of subsequetial limits. (c) For each sequece, give its lim sup ad lim if. (d) Which of the sequeces coverges? diverges to +? diverges to? (e) Which of the sequeces is bouded? 6 Show that every subsequece of a subsequece of a give sequece is itself a subsequece of the give sequece. 7 Let (r ) be a eumeratio of the set Q of all ratioal umbers. Show that there exists a subsequece (r k ) such that lim k r k = +. 9 (a) Show that the closed iterval [a, b] is a closed set. Hit: a set is closed if it cotais all subsequetial limits of all sequeces i itself. (b) Is there a sequece (s ) such that (0, ) is its set of subsequetial limits? 0 Let (s ) be the sequece of umbers i the followig figure, listed i the idicated order. (a) Fid the set S of subsequetial limits of (s ). (b) Determie lim sup s ad lim if s......

3 Math 0A Elemetary Aalysis Homework Aswers -.9 Limits Theorems for Sequeces (a) lim(x + y ) = lim x + lim y = + 7 = 0. (b) lim y x y = lim lim y lim x (lim y ) = 7 7 = 8 9. Let s = ad for let s + = s +. (a) (s ) = (,, +, + +,...). (b) If lim s = s the takig the limit of both sides of s + = s + results i s = s + = s = s + = s = ( ± ) Give the s is clearly a positive sequece, it follows that s 0 ad so s = ( + ). 6 Let x = ad x + = x for. (a) If a = lim x, the takig limits of x + = x results i a = a, which has solutios a = ad a = 0. (b) Give that x = ad x + = x, we see that x is always. It is therefore impossible for lim x to equal either or 0. lim x does ot exist: ideed x +. For a proof, we use iductio to see that x for all : Clearly true for = sice x =. Suppose x for some. The x + = x +. Hece result. (c) There is o cotradictio. We eeded to assume the existece of a limit i (a) to the calculate what it would be. If we do t assume there is a limit, (a) is meaigless. (a) If L < the let a = +L L. Clearly L < a <. Now let ɛ =. Sice L = lim s +, s there exists N N such that N = s + s L < ɛ = ɛ + L < s + s < ɛ + L = s + < (ɛ + L) s = + L s = a s Iteratig this we immediately obtai s < a N s N. Sice a 0 for 0 < a <, we coclude that lim s = 0. (b) Writig t = s, we have L = lim t + t existig ad 0 < L <. Applyig part(a) to (t ) we obtai lim t 0, whece lim s = +. If s = a, the p s + s = a + p ( (+) p a = a p +) a. This is the limit L from questio. We immediately have lim a { p = 0 if a < + if a >

4 Additioally, if a =, we have the sequece or ( ) p both of which coverge to zero. If p a > the a = p a which therefore diverges to +. If a <, the a p = ( ) a p which p diverges by oscillatio. Hece result..0 Mootoe Sequeces ad Cauchy Sequeces 6 (a) Suppose m >, the s m s = s m s m + s m + s + s s m s m + + s + s < m + + j = j=0 Now let ɛ > 0 be give ad let N = l ɛ l. The m > > N = s m s < < N = ɛ. (s ) is therefore Cauchy. (b) If we oly assume that s + s < for all N, the the best we ca do is to coclude that s m s < m + + m. We caot coclude that this is bouded by ɛ. As a couterexample, cosider s = l. By the Mea Value Theorem applied to f (x) = l x o the iterval [, + ], l( + ) l + = x some x (, + ) Thus s + s <. However s +, so (s ) is ot Cauchy. 7 Let t = sup S. Sice t < sup S, the set S (t, sup S) is o-empty. Defie s to be the ifimum of this set. This is a odecreasig sequece (sice S (t +, sup S) is a subset of S (t, sup S) ad A B = if B if A). Moreover, 0 < sup S s sup S t = 0 hece lim s = sup S. 8 Cosider σ + σ. Sice s + s for all, we have that σ + σ = + (s + s + + s + s + ) (s + s + + s ) = ( + ) (s + + ( ( + ))(s + s + + s )) = ( + ) (s + (s + s + + s ))

5 ( + ) (s + (s + + s s + )) = 0 Thus (σ ) is a odecreasig sequece. 9 (a) s =, s = ( ) = 6 ad s = ( ) 6 = 8. (b) We claim that (s ) is oicreasig. Ideed, s + s = + s s = s + (s ). Sice s it follows that s s 0 = s. Cotiuig by iductio, we see that s + s 0 for all. (s ) is also a positive sequece ad is thus bouded below. (s ) thus coverges. (c) By (b), we kow that lim s = s for some s. Give this, s must satisfy s = lim s + = lim + s = s = s = 0,. Sice s = 6 choice. < ad (s ) is o-icreasig, it follows that s = 0 is the oly possible 0 (a) s =, s = 9 ad s = 7. (b) Certaily s = >. Now suppose that s > for some. The s + = (s + ) > ( ) + = Hece, by iductio, s > for all. ( ) = 0, thus (s ) is oicreasig. (c) s + s = ( s ) < (d) (s ) is oicreasig ad bouded below, thus lim s exists. Call the limit s. Clearly s satisfies s = lim s + = lim (s + ) = (s + ) = s = = s =.. Subsequeces (a) a k =. Here k = k. (b ), (c ) ad (d ) are already mootoe. For the last, observe that d = is decreasig (7 ) = (7 ) (b) (a ) has subsequetial limit set {, }. (b ) has subsequetial limit set {0}. (c ) has subsequetial limit set {+ }. (d ) has subsequetial limit set { 6 7 }.

6 (c) lim sup a =, lim if a =. lim sup b = 0 = lim if b. lim sup c = + = lim if c. lim sup d = 6 7 = lim if a. (d) (b ) ad (d ) coverge, (c ) diverges to +, ad (a ) diverges by oscillatio. (e) All but (c ) are bouded. 6 Here are two possible argumets. Use the defiitio of the subsequece (s k ) as a fuctio t = s σ, where σ(k) = k is a icreasig fuctio σ : N N. Thus a subsequece of (s k ) is a fuctio u = t µ, where µ(l) = k l is a icreasig fuctio µ : N N. But the, by the associativity of fuctioal compositio, we have u = (s σ) µ = s (σ µ), which defies (s kl ) as a subsequece of (s ). It should be clear that σ µ : N N is icreasig. View a subsequece as a ordered subset of a sequece usig the orderig s m s = m. If (s kl ) is a subsequece of (s k ), which is a subsequece of (s ), the we have the iclusio (s kl ) (s k ) (s ) as ordered sets. Sice the subset relatio preserves orderig ad set iclusio is trasitive, it is immediate that (s kl ) is a subsequece of (s ). 7 Let (r ) be a eumeratio of the set Q of all ratioal umbers. As observed i lectures, every real umber ad {+, } are subsequetial limits if (r ). Thus lim sup r = + whece, (Corollary.) there exists a mootoe subsequece (r k ) such that lim k r k = lim sup r = +. 9 (a) We must prove that the set of all limits of all sequeces takig values i [a, b] is itself [a, b]. Firstly, trivially ote that every t [a, b] is the limit of the sequece (t, t, t, t, t,...). Now let a s b. The ay subsequece of s with a limit s must satisfy a s b. Hece s [a, b], ad so [a, b] is closed. (b) First a cheatig aswer: Ay set of subsequetial limits must be closed. (0, ) is ope, hece there is o such sequece. We eed to do better tha this, for we have ot yet got a defiitio of ope, ad so caot categorize it as somehow (at least for bouded sets), a opposite cocept to closed. Suppose there is a sequece (s ) such that S = (0, ) is its set of subsequetial limits. The (Corollary. from book) there exists a mootoe subsequece whose limit is lim sup s. However (Theorem.7) we kow that lim sup s = sup S =. Thus is the limit of some subsequece of (s ). Cotradictio. 0 (a) Sice each term is repeated ifiitely ofte, the set S of subsequetial limits of (s ) is precisely { : N}. (b) lim sup s = ad lim if s = 0 as these are the supremum ad ifimum of the above set S. 6

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

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

Solutions to Tutorial 3 (Week 4)

Solutions 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 information

Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.

Read 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 information

M17 MAT25-21 HOMEWORK 5 SOLUTIONS

M17 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 information

Sequences and Series

Sequences 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 information

Topics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.

Topics. 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 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

(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 information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1

MATH 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 information

Real Analysis Fall 2004 Take Home Test 1 SOLUTIONS. < ε. Hence lim

Real Analysis Fall 2004 Take Home Test 1 SOLUTIONS. < ε. Hence lim Real Aalysis Fall 004 Take Home Test SOLUTIONS. Use the defiitio of a limit to show that (a) lim si = 0 (b) Proof. Let ε > 0 be give. Defie N >, where N is a positive iteger. The for ε > N, si 0 < si

More information

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

2.1. 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 information

MAT1026 Calculus II Basic Convergence Tests for Series

MAT1026 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 information

Part A, for both Section 200 and Section 501

Part 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 information

Solutions to Math 347 Practice Problems for the final

Solutions 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 information

n n 2 n n + 1 +

n 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 information

ANSWERS TO MIDTERM EXAM # 2

ANSWERS TO MIDTERM EXAM # 2 MATH 03, FALL 003 ANSWERS TO MIDTERM EXAM # PENN STATE UNIVERSITY Problem 1 (18 pts). State ad prove the Itermediate Value Theorem. Solutio See class otes or Theorem 5.6.1 from our textbook. Problem (18

More information

TRUE/FALSE QUESTIONS FOR SEQUENCES

TRUE/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 information

Math 299 Supplement: Real Analysis Nov 2013

Math 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 information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

1 Introduction. 1.1 Notation and Terminology

1 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 information

Definition An infinite sequence of numbers is an ordered set of real numbers.

Definition 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 information

1 Lecture 2: Sequence, Series and power series (8/14/2012)

1 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 information

Please do NOT write in this box. Multiple Choice. Total

Please do NOT write in this box. Multiple Choice. Total Istructor: Math 0560, Worksheet Alteratig Series Jauary, 3000 For realistic exam practice solve these problems without lookig at your book ad without usig a calculator. Multiple choice questios should

More information

62. 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 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 information

Limit superior and limit inferior c Prof. Philip Pennance 1 -Draft: April 17, 2017

Limit 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 information

MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and

MATH301 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 information

NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION ADVANCED CALCULUS II. November 2003 Time allowed :

NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION ADVANCED CALCULUS II. November 2003 Time allowed : NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER EXAMINATION 003-004 MA08 ADVANCED CALCULUS II November 003 Time allowed : hours INSTRUCTIONS TO CANDIDATES This examiatio paper cosists of TWO

More information

Math Solutions to homework 6

Math Solutions to homework 6 Math 175 - Solutios to homework 6 Cédric De Groote November 16, 2017 Problem 1 (8.11 i the book): Let K be a compact Hermitia operator o a Hilbert space H ad let the kerel of K be {0}. Show that there

More information

Sequences. A Sequence is a list of numbers written in order.

Sequences. 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 information

MATH 312 Midterm I(Spring 2015)

MATH 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 information

Notes #3 Sequences Limit Theorems Monotone and Subsequences Bolzano-WeierstraßTheorem Limsup & Liminf of Sequences Cauchy Sequences and Completeness

Notes #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 information

page Suppose that S 0, 1 1, 2.

page Suppose that S 0, 1 1, 2. page 10 1. Suppose that S 0, 1 1,. a. What is the set of iterior poits of S? The set of iterior poits of S is 0, 1 1,. b. Give that U is the set of iterior poits of S, evaluate U. 0, 1 1, 0, 1 1, S. The

More information

Math 341 Lecture #31 6.5: Power Series

Math 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 information

Math 132, Fall 2009 Exam 2: Solutions

Math 132, Fall 2009 Exam 2: Solutions Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of

More information

HOMEWORK #4 - MA 504

HOMEWORK #4 - MA 504 HOMEWORK #4 - MA 504 PAULINHO TCHATCHATCHA Chapter 2, problem 19. (a) If A ad B are disjoit closed sets i some metric space X, prove that they are separated. (b) Prove the same for disjoit ope set. (c)

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

SCORE. 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 information

Math 104: Homework 2 solutions

Math 104: Homework 2 solutions Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does

More information

MAS111 Convergence and Continuity

MAS111 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 information

10.1 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.

10.1 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 information

d) If the sequence of partial sums converges to a limit L, we say that the series converges and its

d) 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 information

2 Banach spaces and Hilbert spaces

2 Banach spaces and Hilbert spaces 2 Baach spaces ad Hilbert spaces Tryig to do aalysis i the ratioal umbers is difficult for example cosider the set {x Q : x 2 2}. This set is o-empty ad bouded above but does ot have a least upper boud

More information

University of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!

University 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 information

7 Sequences of real numbers

7 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 information

1+x 1 + α+x. x = 2(α x2 ) 1+x

1+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 information

Chapter 6 Infinite Series

Chapter 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 information

FUNDAMENTALS OF REAL ANALYSIS by

FUNDAMENTALS 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 information

In this section, we show how to use the integral test to decide whether a series

In 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 information

If 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?

If 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 information

6.3 Testing Series With Positive Terms

6.3 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 information

MA131 - Analysis 1. Workbook 9 Series III

MA131 - 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 information

Mathematical Methods for Physics and Engineering

Mathematical 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 information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

Math 25 Solutions to practice problems

Math 25 Solutions to practice problems Math 5: Advaced Calculus UC Davis, Sprig 0 Math 5 Solutios to practice problems Questio For = 0,,, 3,... ad 0 k defie umbers C k C k =! k!( k)! (for k = 0 ad k = we defie C 0 = C = ). by = ( )... ( k +

More information

Lecture Notes for Analysis Class

Lecture 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 information

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS

REAL 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 information

MATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006

MATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006 MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the

More information

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6 Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is

More information

Assignment 5: Solutions

Assignment 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 information

MA131 - Analysis 1. Workbook 3 Sequences II

MA131 - 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 information

Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan

Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan Arkasas Tech Uiversity MATH 94: Calculus II Dr Marcel B Fia 85 Power Series Let {a } =0 be a sequece of umbers The a power series about x = a is a series of the form a (x a) = a 0 + a (x a) + a (x a) +

More information

Math 140A Elementary Analysis Homework Questions 1

Math 140A Elementary Analysis Homework Questions 1 Math 14A Elemetary Aalysis Homewor Questios 1 1 Itroductio 1.1 The Set N of Natural Numbers 1 Prove that 1 2 2 2 2 1 ( 1(2 1 for all atural umbers. 2 Prove that 3 11 (8 5 4 2 for all N. 4 (a Guess a formula

More information

Lecture 17Section 10.1 Least Upper Bound Axiom

Lecture 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 information

CHAPTER 1 SEQUENCES AND INFINITE SERIES

CHAPTER 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 information

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing

Physics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals

More information

Real Variables II Homework Set #5

Real 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 information

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

Alternating 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 information

Convergence of random variables. (telegram style notes) P.J.C. Spreij

Convergence of random variables. (telegram style notes) P.J.C. Spreij Covergece of radom variables (telegram style otes).j.c. Spreij this versio: September 6, 2005 Itroductio As we kow, radom variables are by defiitio measurable fuctios o some uderlyig measurable space

More information

Calculus II Homework: The Comparison Tests Page 1. a n. 1 n 2 + n + 1. n= n. n=1

Calculus II Homework: The Comparison Tests Page 1. a n. 1 n 2 + n + 1. n= n. n=1 Calculus II Homework: The Compariso Tests Page Questios l coverget or diverget? + 7 3 + 5 coverget or diverget? Example Suppose a ad b are series with positive terms ad b is kow to be coverget. a If a

More information

Math 61CM - Solutions to homework 3

Math 61CM - Solutions to homework 3 Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig

More information

Fall 2013 MTH431/531 Real analysis Section Notes

Fall 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 information

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

sin(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 information

HOMEWORK #10 SOLUTIONS

HOMEWORK #10 SOLUTIONS Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2017

SCORE. Exam 2. MA 114 Exam 2 Fall 2017 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 a graphig calculator

More information

Series III. Chapter Alternating Series

Series 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 information

Chapter 3 Selected Exercises (Rudin)

Chapter 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 information

Exponential Functions and Taylor Series

Exponential Functions and Taylor Series MATH 4530: Aalysis Oe Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 2017 MATH 4530: Aalysis Oe Outlie

More information

Metric Space Properties

Metric 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 information

Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence

Chapter 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 information

Basic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S.

Basic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S. Basic Sets Example 1. Let S = {1, {2, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 (b) {1} S (c) {2, 3} S (d) {1, 4} S (e) 2 S. (f) S = {1, 4, {2, 3}} (g) S Example 2. Compute the

More information

Definition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.

Definition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4. 4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad

More information

LECTURES 5 AND 6: CONVERGENCE TESTS

LECTURES 5 AND 6: CONVERGENCE TESTS LECTURES 5 AND 6: CONVERGENCE TESTS I the previous lecture we saw a very basic test for covergece - the compariso test. Here we develop slightly more sophisticated machiery for determiig the covergece

More information

Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology

Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009 Cotets Limit ad Cotiuity 7 Limit of Sequece 8 Defiitio 8 Property 3 3 Ifiity ad Ifiitesimal 8 4

More information

Sequences and Limits

Sequences 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 information

f n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that

f n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that Lecture 15 We have see that a sequece of cotiuous fuctios which is uiformly coverget produces a limit fuctio which is also cotiuous. We shall stregthe this result ow. Theorem 1 Let f : X R or (C) be a

More information

Are the following series absolutely convergent? n=1. n 3. n=1 n. ( 1) n. n=1 n=1

Are the following series absolutely convergent? n=1. n 3. n=1 n. ( 1) n. n=1 n=1 Absolute covergece Defiitio A series P a is called absolutely coverget if the series of absolute values P a is coverget. If the terms of the series a are positive, absolute covergece is the same as covergece.

More information

Solutions to Tutorial 5 (Week 6)

Solutions 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 information

Sequence A sequence is a function whose domain of definition is the set of natural numbers.

Sequence A sequence is a function whose domain of definition is the set of natural numbers. 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

More information

Math 128A: Homework 1 Solutions

Math 128A: Homework 1 Solutions Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that

More information

SCORE. Exam 2. MA 114 Exam 2 Fall 2016

SCORE. Exam 2. MA 114 Exam 2 Fall 2016 Exam 2 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 a graphig calculator

More information

Solutions to Homework 1

Solutions to Homework 1 Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2.

More information

Integrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number

Integrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios

More information

and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.

and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime. MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,

More information

SOLUTIONS 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. 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 information

Final Solutions. 1. (25pts) Define the following terms. Be as precise as you can.

Final Solutions. 1. (25pts) Define the following terms. Be as precise as you can. Mathematics H104 A. Ogus Fall, 004 Fial Solutios 1. (5ts) Defie the followig terms. Be as recise as you ca. (a) (3ts) A ucoutable set. A ucoutable set is a set which ca ot be ut ito bijectio with a fiite

More information

n=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

n=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 information

Math 113 Exam 3 Practice

Math 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 information

Section 11.8: Power Series

Section 11.8: Power Series Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i

More information

Math 210A Homework 1

Math 210A Homework 1 Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the Cauchy-Riema equatios? Solutio. a) A fuctio f : G C is called

More information

INFINITE SEQUENCES AND SERIES

INFINITE 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 information

Math 220A Fall 2007 Homework #2. Will Garner A

Math 220A Fall 2007 Homework #2. Will Garner A Math 0A Fall 007 Homewor # Will Garer Pg 3 #: Show that {cis : a o-egative iteger} is dese i T = {z œ : z = }. For which values of q is {cis(q): a o-egative iteger} dese i T? To show that {cis : a o-egative

More information