n n 2 n n + 1 +
|
|
- Cameron Doyle
- 6 years ago
- Views:
Transcription
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 is equivalet with < ε. Note that the previous expressio is positive so we do ot eed the absolute value bars. Sice 1 < if we choose N = (1/(2ε)) 2, it follows that a +1 a < ε if N. (5pts) Is a a Cauchy sequece? Why? Solutio: a is ot a Cauchy sequece. To prove this, suppose by cotradictio that a is Cauchy. The a must be coverget by Theorem 6.4. The sequece a is, however, diverget. Thus it ca ot be Cauchy.
2 (7pts) 2. (a) Show that if a coverges absolutely, the so does a 2. Is this true without the hypothesis of absolute covergece (prove or give a couterexample)? Solutio: If the series a coverges absolutely it coverges. The lim a = 0. The there is N 1 such that a < 1 for all N. It follows that a 2 a for all N. The compariso test for positive series implies that =N a2 coverges. From the tail theorem we coclude that a 2 coverges. The coclusio fails without the hypothesis of absolute covergece. Cosider ( 1). This series is coverget by the Cauchy test. The series 1, however, diverges (it is a p-series with p = 1). (3pts) If a coverges ad a 0, does it follow that a coverges? Prove or give a couterexample. Solutio: If a coverges ad a 0, it does ot follow that a coverges. For example, 1 coverges (p-series with p = 2) but 1 2 diverges. Page 2 of 8 Please go o to the ext page...
3 (20pts) 3. Fid the radius of covergece of =0 (si )x (with proof). Solutio: The mai poit about this problem is that we ca ot use the root or the ratio test to determie the radius of covergece of this series (because the sequeces (si()) (1/) ad si(+1) si() are diverget). Notice that si()x x for all 1. If x < 1 we kow that x coverges absolutely. The compariso theorem implies that si()x coverges absolutely. For x = 1, si() diverges sice { si() } does ot coverge to 0. Thus R = by Theorem-Defiitio 8.1. Page 3 of 8 Please go o to the ext page...
4 (10pts) 4. Determie if the series is coverget. If yes, fid it s value. l + 2 =1 Solutio: Sice l +2 = l() l(+2) we see that the series is a telescopig series. We compute the th partial sum as follows: The s = ( ) l(k) l(k + 2) = l(1) + l(2) l( + 1) l( + 2). k=1 Thus the series is diverget. lim s ( ) = l(2) lim l( + 1) + l( + 2) =. Page 4 of 8 Please go o to the ext page...
5 (5pts) 5. Let b be a decreasig sequece with lim b = 0. (a) Prove that =1 (b b +1 ) coverges. Solutio: We otice this is a telescopig series. The partial sum equals s = (b 1 b 2 ) + (b 2 b 3 ) + + (b b +1 ) = b 1 b +1. Sice lim b = 0 it follows that the sequece of partial sums is coverget ad equals b 1. Thus (b b +1 ) coverges. (10pts) Let a be a bouded sequece. Prove that =1 a (b b +1 ) coverges. Solutio: By hypothesis, there exists M > 0 such that a M for all 1. Sice b b +1 0 we have that a (b b +1 ) M(b b +1 ). By the liearity theorem M(b b +1 ) coverges. By the compariso theorem, a (b b +1 ) coverges. Thus =1 a (b b +1 ) coverges. Page 5 of 8 Please go o to the ext page...
6 (10pts) 6. Prove the followig two statemets: (a) Every real umber is a cluster poit of some sequece of ratioal umbers. Solutio: Let r be a real umber. For ay 1, Theorem 2.5 o page 25 implies that there is a ratioal umber a such that r < a < r + 1. By the squeeze theorem, lim a = r. Theorem 6.2 implies that r is a cluster poit for the sequece a. (5pts) Every real umber is a cluster poit of some sequece of irratioal umbers. Solutio: The proof is similar with the previous part. If r is ay real umber, Theorem 2.5 implies that there is a irratioal umber a with r < a < r + 1 for all 1. Thus r = lim a ad r is a cluster poit of a sequece of irratioal umbers. Page 6 of 8 Please go o to the ext page...
7 (3pts) 7. Give examples for the followig or explai why o example exists. (a) A series that has bouded partial sums but does ot coverge. Solutio: Cosider the sequece a = ( 1) for all 1. The s 2k+1 = 1 ad s 2k = 0 for all k 1. Thus { s } is a bouded sequece which diverges. (4pts) A sequece which has a ifiite umber of cluster poits. Solutio: Example 6.2A a) i the textbook says that for the sequece 1; 1, 2; 1, 2, 3;... every iteger is a cluster poit. (3pts) (c) A power series whose radius of covergece is 2. Solutio: Cosider the power series x 2. We compute the radius of covergece usig the ratio test: lim a +1 a = x 2. Thus the series is absolutely coverget if x < 2 ad diverget for x > 2. Thus R = 2. Page 7 of 8 Please go o to the ext page...
8 (10pts) 8. Prove, usig the defiitio, that if { a } ad { b } are Cauchy sequeces the { a b } is Cauchy. Solutio: We proved i class that a Cauchy sequece is bouded (see also part (A) of the proof of Theorem 6.4). Thus there are M 1 > 0 ad M 2 > 0 such that a M 1 ad b M 2 for all 1. Let M = max{m 1, M 2 }. Thus a M ad b M for all 1. Let ε > 0. Sice the sequece a is Cauchy there exists N 1 1 such that a a m < 2M for all m, N 1. Similarly there exists N 2 1 such that b b m < for all m, N 2. Let N = max{n 1, N 2 }. If, m N we have ε 2M a b a m b m = a b a b m + a b m a m b m Sice ε was arbitrary, {a b } is a Cauchy sequece. ε a b a b m + a b m a m b m = a b b m + b m a a m ε < M 2M + M ε 2M = ε. Page 8 of 8 Ed of exam.
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 informationMath 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 informationPlease 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 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 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 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 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 informationSCORE. 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 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 informationPLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)
Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages
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 informationMath 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 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 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 information1 Introduction to Sequences and Series, Part V
MTH 22 Calculus II Essex Couty College Divisio of Mathematics ad Physics Lecture Notes #8 Sakai Web Project Material Itroductio to Sequeces ad Series, Part V. The compariso test that we used prior, relies
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 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 informationAre 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 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 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 informationLECTURE SERIES WITH NONNEGATIVE TERMS (II). SERIES WITH ARBITRARY TERMS
LECTURE 4 SERIES WITH NONNEGATIVE TERMS II). SERIES WITH ARBITRARY TERMS Series with oegative terms II) Theorem 4.1 Kummer s Test) Let x be a series with positive terms. 1 If c ) N i 0, + ), r > 0 ad 0
More informationSCORE. 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 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 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 informationThe Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1
460_0906.qxd //04 :8 PM Page 69 SECTION 9.6 The Ratio ad Root Tests 69 Sectio 9.6 EXPLORATION Writig a Series Oe of the followig coditios guaratees that a series will diverge, two coditios guaratee that
More informationSolutions to Practice Midterms. Practice Midterm 1
Solutios to Practice Midterms Practice Midterm. a False. Couterexample: a =, b = b False. Couterexample: a =, b = c False. Couterexample: c = Y cos. = cos. + 5 = 0 sice both its exist. + 5 cos π. 5 + 5
More informationArkansas 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 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 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 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 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 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 information5.6 Absolute Convergence and The Ratio and Root Tests
5.6 Absolute Covergece ad The Ratio ad Root Tests Bria E. Veitch 5.6 Absolute Covergece ad The Ratio ad Root Tests Recall from our previous sectio that diverged but ( ) coverged. Both of these sequeces
More informationBC: Q401.CH9A Convergent and Divergent Series (LESSON 1)
BC: Q40.CH9A Coverget ad Diverget Series (LESSON ) INTRODUCTION Sequece Notatio: a, a 3, a,, a, Defiitio: A sequece is a fuctio f whose domai is the set of positive itegers. Defiitio: A ifiite series (or
More information9.3 The INTEGRAL TEST; p-series
Lecture 9.3 & 9.4 Math 0B Nguye of 6 Istructor s Versio 9.3 The INTEGRAL TEST; p-series I this ad the followig sectio, you will study several covergece tests that apply to series with positive terms. Note
More information4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3
Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x
More informationREVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.
REVIEW, MATH 00. Let a = +. a) Determie whether the sequece a ) is coverget. b) Determie whether a is coverget.. Determie whether the series is coverget or diverget. If it is coverget, fid its sum. a)
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 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 information10.6 ALTERNATING SERIES
0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose
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 informationJANE PROFESSOR WW Prob Lib1 Summer 2000
JANE PROFESSOR WW Prob Lib Summer 000 Sample WeBWorK problems. WeBWorK assigmet Series6CompTests due /6/06 at :00 AM..( pt) Test each of the followig series for covergece by either the Compariso Test or
More informationConvergence: nth-term Test, Comparing Non-negative Series, Ratio Test
Covergece: th-term Test, Comparig No-egative Series, Ratio Test Power Series ad Covergece We have writte statemets like: l + x = x x2 + x3 2 3 + x + But we have ot talked i depth about what values of x
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 informationSequences 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 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 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 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 informationMA541 : 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 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 informationCalculus 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 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 information11.6 Absolute Convergence and the Ratio and Root Tests
.6 Absolute Covergece ad the Ratio ad Root Tests The most commo way to test for covergece is to igore ay positive or egative sigs i a series, ad simply test the correspodig series of positive terms. Does
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 informationSection 1.4. Power Series
Sectio.4. Power Series De itio. The fuctio de ed by f (x) (x a) () c 0 + c (x a) + c 2 (x a) 2 + c (x a) + ::: is called a power series cetered at x a with coe ciet sequece f g :The domai of this fuctio
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 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 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 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 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 informationMTH 246 TEST 3 April 4, 2014
MTH 26 TEST April, 20 (PLEASE PRINT YOUR NAME!!) Name:. (6 poits each) Evaluate lim! a for the give sequece fa g. (a) a = 2 2 5 2 5 (b) a = 2 7 2. (6 poits) Fid the sum of the telescopig series p p 2.
More informationInfinite Sequence and Series
Chapter 7 Ifiite Sequece ad Series 7. Sequeces A sequece ca be thought of as a list of umbers writte i a defiite order: a,a 2,a 3,a 4,...,a,... The umber a is called the first term, a 2 is the secod term,
More informationReview for Test 3 Math 1552, Integral Calculus Sections 8.8,
Review for Test 3 Math 55, Itegral Calculus Sectios 8.8, 0.-0.5. Termiology review: complete the followig statemets. (a) A geometric series has the geeral form k=0 rk.theseriescovergeswhe r is less tha
More informationSection 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 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 informationSequence 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 informationMath 113 Exam 4 Practice
Math Exam 4 Practice Exam 4 will cover.-.. 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
More informationNATIONAL 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 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 informationMIDTERM 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 informationMTH 133 Solutions to Exam 2 November 16th, Without fully opening the exam, check that you have pages 1 through 12.
Name: Sectio: Recitatio Istructor: INSTRUCTIONS Fill i your ame, etc. o this first page. Without fully opeig the exam, check that you have pages through. Show all your work o the stadard respose questios.
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 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 I: Covers Sequence through Series Comparison Tests
Part I: Covers Sequece through Series Compariso Tests. Give a example of each of the followig: (a) A geometric sequece: (b) A alteratig sequece: (c) A sequece that is bouded, but ot coverget: (d) A sequece
More information11.5 Alternating Series, Absolute and Conditional Convergence
.5.5 Alteratig Series, Absolute ad Coditioal Covergece We have see that the harmoic series diverges. It may come as a surprise the to lear that ) 2 + 3 4 + + )+ + = ) + coverges. To see this, let s be
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 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 informationANSWERS 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 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 informationMath 113 (Calculus 2) Section 12 Exam 4
Name: Row: Math Calculus ) Sectio Exam 4 8 0 November 00 Istructios:. Work o scratch paper will ot be graded.. For questio ad questios 0 through 5, show all your work i the space provided. Full credit
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 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 informationMath 163 REVIEW EXAM 3: SOLUTIONS
Math 63 REVIEW EXAM 3: SOLUTIONS These otes do ot iclude solutios to the Cocept Check o p8. They also do t cotai complete solutios to the True-False problems o those pages. Please go over these problems
More informationPractice Test Problems for Test IV, with Solutions
Practice Test Problems for Test IV, with Solutios Dr. Holmes May, 2008 The exam will cover sectios 8.2 (revisited) to 8.8. The Taylor remaider formula from 8.9 will ot be o this test. The fact that sums,
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 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 informationMTH 133 Solutions to Exam 2 Nov. 18th 2015
Name: Sectio: Recitatio Istructor: READ THE FOLLOWING INSTRUCTIONS. Do ot ope your exam util told to do so. No calculators, cell phoes or ay other electroic devices ca be used o this exam. Clear your desk
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 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 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 information1. Do the following sequences converge or diverge? If convergent, give the limit. Explicitly show your reasoning. 2n + 1 n ( 1) n+1.
Solutio: APPM 36 Review #3 Summer 4. Do the followig sequeces coverge or iverge? If coverget, give the limit. Eplicitly show your reasoig. a a = si b a = { } + + + 6 c a = e Solutio: a Note si a so, si
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 informationTHE INTEGRAL TEST AND ESTIMATES OF SUMS
THE INTEGRAL TEST AND ESTIMATES OF SUMS. Itroductio Determiig the exact sum of a series is i geeral ot a easy task. I the case of the geometric series ad the telescoig series it was ossible to fid a simle
More informationSUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11
UTCLIFFE NOTE: CALCULU WOKOWKI CHAPTER Ifiite eries Coverget or Diverget eries Cosider the sequece If we form the ifiite sum 0, 00, 000, 0 00 000, we hve wht is clled ifiite series We wt to fid the sum
More informationA detailed proof of the irrationality of π
Matthew Straugh Math 4 Midterm A detailed proof of the irratioality of The proof is due to Iva Nive (1947) ad essetial to the proof are Lemmas ad 3 due to Charles Hermite (18 s) First let us itroduce some
More informationTesting for Convergence
9.5 Testig for Covergece Remember: The Ratio Test: lim + If a is a series with positive terms ad the: The series coverges if L . The test is icoclusive if L =. a a = L This
More informationSection 5.5. Infinite Series: The Ratio Test
Differece Equatios to Differetial Equatios Sectio 5.5 Ifiite Series: The Ratio Test I the last sectio we saw that we could demostrate the covergece of a series a, where a 0 for all, by showig that a approaches
More informationReal 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 information11.6 Absolute Convrg. (Ratio & Root Tests) & 11.7 Strategy for Testing Series
11.6 Absolute Covrg. (Ratio & Root Tests) & 11.7 Strategy for Testig Series http://screecast.com/t/ri3unwu84 Give ay series Σ a, we ca cosider the correspodig series 1 a a a a 1 2 3 whose terms are the
More informationInfinite Series. Definition. An infinite series is an expression of the form. Where the numbers u k are called the terms of the series.
Ifiite Series Defiitio. A ifiite series is a expressio of the form uk = u + u + u + + u + () 2 3 k Where the umbers u k are called the terms of the series. Such a expressio is meat to be the result of
More information