Sectio 85 Alteratig Series ad Absolute Covergece AP E XC II Versio 20 Authors Gregory Hartma, PhD Departmet of Applied Mathema cs Virgiia Military Is tute Bria Heiold, PhD Departmet of Mathema cs ad Computer Sciece Mout Sait Mary s Uiversity Troy Siemers, PhD Departmet of Applied Mathema cs Virgiia Military Is tute Dimplekumar Chalishajar, PhD Departmet of Applied Mathema cs Virgiia Military Is tute Editor Jeifer Bowe, PhD Departmet of Mathema cs ad Computer Sciece The College of Wooster
Copyright 2014 Gregory Hartma Licesed to the public uder Crea ve Commos A ribu o-nocommercial 30 Uited States Licese
Cotets Preface Table of Cotets iii v 5 Itegra o 185 51 A deriva ves ad Idefiite Itegra o 185 52 The Defiite Itegral 194 53 Riema Sums 204 54 The Fudametal Theorem of Calculus 221 55 Numerical Itegra o 233 6 Techiques of A differe a o 247 61 Subs tu o 247 62 Itegra o by Parts 266 63 Trigoometric Itegrals 276 64 Trigoometric Subs tu o 286 65 Par al Frac o Decomposi o 295 66 Hyperbolic Fuc os 303 67 L Hôpital s Rule 313 68 Improper Itegra o 321 7 Applica os of Itegra o 333 71 Area Betwee Curves 334 72 Volume by Cross-Sec oal Area; Disk ad Washer Methods 341 73 The Shell Method 348 74 Arc Legth ad Surface Area 356 75 Work 365 76 Fluid Forces 375 8 Sequeces ad Series 383 81 Sequeces 383 82 Ifiite Series 395 83 Itegral ad Compariso Tests 410 84 Ra o ad Root Tests 419
85 Altera g Series ad Absolute Covergece 424 86 Power Series 434 87 Taylor Polyomials 446 88 Taylor Series 457 A Solu os To Selected Problems A1 Idex A11
Chapter 8 Sequeces ad Series 85 Altera g Series ad Absolute Covergece All of the series covergece tests we have used require that the uderlyig sequece {a } be a posi ve sequece (We ca relax this with Theorem 64 ad state that there must be a N > 0 such that a > 0 for all > N; that is, {a } is posi ve for all but a fiite umber of values of ) I this sec o we explore series whose summa o icludes ega ve terms We start with a very specific form of series, where the terms of the summa o alterate betwee beig posi ve ad ega ve Defii o 34 Altera g Series Let {a } be a posi ve sequece A altera g series is a series of either the form ( 1) a or ( 1) +1 a Recall the terms of Harmoic Series come from the Harmoic Sequece {a } = {1/} A importat altera g series is the Altera g Harmoic Series: ( 1) +1 1 = 1 1 2 + 1 3 1 4 + 1 5 1 6 + Geometric Series ca also be altera g series whe r < 0 For istace, if r = 1/2, the geometric series is ( ) 1 = 1 1 2 2 + 1 4 1 8 + 1 16 1 32 + Theorem 60 states that geometric series coverge whe r < 1 ad gives the sum: r = 1 Whe r = 1/2 as above, we fid 1 r ( ) 1 1 = 2 1 ( 1/2) = 1 3/2 = 2 3 A powerful covergece theorem exists for other altera g series that meet a few codi os 424
85 Altera g Series ad Absolute Covergece Theorem 70 Altera g Series Test Let {a } be a posi ve, decreasig sequece where lim a = 0 The coverge ( 1) a ad ( 1) +1 a 1 y The basic idea behid Theorem 70 is illustrated i Figure 817 A posi ve, decreasig sequece {a } is show alog with the par al sums L 05 S = ( 1) i+1 a i = a 1 a 2 + a 3 a 4 + + ( 1) a i=1 Because {a } is decreasig, the amout by which S bouces up/dow decreases Moreover, the odd terms of S form a decreasig, bouded sequece, while the eve terms of S form a icreasig, bouded sequece Sice bouded, mootoic sequeces coverge (see Theorem 59) ad the terms of {a } approach 0, oe ca show the odd ad eve terms of S coverge to the same commo limit L, the sum of the series 2 4 6 a S Figure 817: Illustra g covergece with the Altera g Series Test 8 10 Example 251 Applyig the Altera g Series Test Determie if the Altera g Series Test applies to each of the followig series 1 ( 1) +1 1 2 ( 1) l 3 +1 si ( 1) 2 S 1 This is the Altera g Harmoic Series as see previously The uderlyig sequece is {a } = {1/}, which is posi ve, decreasig, ad approaches 0 as Therefore we ca apply the Altera g Series Test ad coclude this series coverges While the test does ot state what the series coverges to, we will see later that ( 1) +1 1 = l 2 2 The uderlyig sequece is {a } = {l /} This is posi ve ad approaches 0 as (use L Hôpital s Rule) However, the sequece is ot decreasig for all It is straigh orward to compute a 1 = 0, a 2 0347, 425
Chapter 8 Sequeces ad Series a 3 0366, ad a 4 0347: the sequece is icreasig for at least the first 3 terms We do ot immediately coclude that we caot apply the Altera g Series Test Rather, cosider the log term behavior of {a } Trea g a = a() as a co uous fuc o of defied o (1, ), we ca take its deriva ve: a () = 1 l 2 The deriva ve is ega ve for all 3 (actually, for all > e), meaig a() = a is decreasig o (3, ) We ca apply the Altera g Series Test to the series whe we start with = 3 ad coclude that ( 1) l coverges; addig the terms with = 1 ad = 2 do ot =3 chage the covergece (ie, we apply Theorem 64) The importat lesso here is that as before, if a series fails to meet the criteria of the Altera g Series Test o oly a fiite umber of terms, we ca s ll apply the test 3 The uderlyig sequece is {a } = si / This sequece is posi ve ad approaches 0 as However, it is ot a decreasig sequece; the value of si oscillates betwee 0 ad 1 as We caot remove a fiite umber of terms to make {a } decreasig, therefore we caot apply the Altera g Series Test Keep i mid that this does ot mea we coclude the series diverges; i fact, it does coverge We are just uable to coclude this based o Theorem 70 Key Idea 31 gives the sum of some importat series Two of these are 1 2 = π2 6 164493 ad ( 1) +1 2 = π2 12 082247 These two series coverge to their sums at differet rates To be accurate to two places a er the decimal, we eed 202 terms of the first series though oly 13 of the secod To get 3 places of accuracy, we eed 1069 terms of the first series though oly 33 of the secod Why is it that the secod series coverges so much faster tha the first? While there are may factors ivolved whe studyig rates of covergece, the altera g structure of a altera g series gives us a powerful tool whe approxima g the sum of a coverget series 426
85 Altera g Series ad Absolute Covergece Theorem 71 The Altera g Series Approxima o Theorem Let {a } be a sequece that sa sfies the hypotheses of the Altera g Series Test, ad let S ad L be the th par al sums ad sum, respec vely, of either ( 1) a or ( 1) +1 a The 1 S L < a +1, ad 2 L is betwee S ad S +1 Part 1 of Theorem 71 states that the th par al sum of a coverget alteratig series will be withi a +1 of its total sum Cosider the altera g series ( 1) +1 we looked at before the statemet of the theorem, 2 Sice a 14 = 1/14 2 00051, we kow that S 13 is withi 00051 of the total sum That is, we kow S 13 is accurate to at least 1 place a er the decimal (The 5 i the third place a er the decimal could cause a carry meaig S 13 is t accurate to two places a er the decimal; i this par cular case, that does t happe) Moreover, Part 2 of the theorem states that sice S 13 08252 ad S 14 08201, we kow the sum L lies betwee 08201 ad 08252, assurig us that S 13 is ideed accurate to two decimal places Some altera g series coverge slowly I Example 251 we determied the +1 l series ( 1) coverged With = 1001, we fid l / 00069, meaig that S 1000 01633 is accurate to oe, maybe two, places a er the decimal Sice S 1001 01564, we kow the sum L is 01564 L 01633 Example 252 Approxima g the sum of coverget altera g series Approximate the sum of the followig series, accurate to two places a er the decimal 1 ( 1) +1 1 3 2 +1 l ( 1) S 1 To be esure accuracy to two places a er the decimal, we eed a < 427
Chapter 8 Sequeces ad Series 00001: 1 3 < 00001 3 > 10, 000 > 3 10000 215 With = 22, we are assured accuracy to two places a er the decimal With S 21 09015, we are cofidet that the sum L of the series is about 090 We ca arrive at this approxima o aother way Part 2 of Theorem 71 states that the sum L lies betwee successive par al sums It is straightforward to compute S 6 0899782, S 7 09027 ad S 8 09007 We kow the sum must lie betwee these last two par al sums; sice they agree to two places a er the decimal, we kow L 090 2 We agai solve for such that a < 00001; that is, we wat such that l()/ < 00001 This caot be solved algebraically, so we approximate the solu o usig Newto s Method Let f(x) = l(x)/x 00001 We wat to fid where f(x) = 0 Assumig that x must be large, we let x 1 = 1000 Recall that x +1 = x f(x )/f (x ); we compute f (x) = ( 1 l(x) ) /x 2 Thus: x 2 = 1000 = 215234 l(1000)/1000 00001 ( 1 l(1000) ) /1000 2 Usig a computer, we fid that a er 12 itera os we fid x 116, 671 With S 116,671 01598 ad S 116,672 01599, we kow that the sum L is betwee these two values Simply sta g that L 015 is misleadig, as L is very, very close to 016 1 Oe of the famous results of mathema cs is that the Harmoic Series, diverges, yet the Altera g Harmoic Series, ( 1) +1 1, coverges The o o that altera g the sigs of the terms i a series ca make a series coverge leads us to the followig defii os 428
85 Altera g Series ad Absolute Covergece Defii o 35 Absolute ad Codi oal Covergece 1 A series a coverges absolutely if a coverges 2 A series a coverges codi oally if a diverges a coverges but Note: I Defii o 35, a is ot ecessarily a altera g series; it just may have some ega ve terms Thus we say the Altera g Harmoic Series coverges codi oally Example 253 Determiig absolute ad codi oal covergece Determie if the followig series coverges absolutely, codi oally, or diverges 1 ( 1) + 3 2 + 2 + 5 2 ( 1) 2 + 2 + 5 2 3 =3 ( 1) 3 3 5 10 S 1 We ca show the series + 3 ( 1) 2 + 2 + 5 = + 3 2 + 2 + 5 diverges usig the Limit Compariso Test, comparig with 1/ The series ( 1) + 3 2 coverges usig the Altera g Series + 2 + 5 Test; we coclude it coverges codi oally 2 We ca show the series ( 1) 2 + 2 + 5 = 2 2 + 2 + 5 2 coverges usig the Ra o Test Therefore we coclude ( 1) 2 + 2 + 5 2 coverges absolutely 429
Chapter 8 Sequeces ad Series 3 The series 3 3 ( 1) 5 10 = 3 3 5 10 =3 =3 diverges usig the th Term Test, so it does ot coverge absolutely The series ( 1) 3 3 fails the codi os of the Altera g Series 5 10 =3 Test as (3 3)/(5 10) does ot approach 0 as We ca state further that this series diverges; as, the series effec vely adds ad subtracts 3/5 over ad over This causes the sequece of par al sums to oscillate ad ot coverge Therefore the series ( 1) 3 3 5 10 diverges Kowig that a series coverges absolutely allows us to make two importat statemets, give i the followig theorem The first is that absolute covergece is stroger tha regular covergece That is, just because coverges, we caot coclude that coverges absolutely tells us that a a will coverge, but kowig a series a will coverge Oe reaso this is importat is that our covergece tests all require that the uderlyig sequece of terms be posi ve By takig the absolute value of the terms of a series where ot all terms are posi ve, we are o e able to apply a appropriate test ad determie absolute covergece This, i tur, determies that the series we are give also coverges The secod statemet relates to rearragemets of series Whe dealig with a fiite set of umbers, the sum of the umbers does ot deped o the order which they are added (So 1+2+3 = 3+1+2) Oe may be surprised to fid out that whe dealig with a ifiite set of umbers, the same statemet does ot always hold true: some ifiite lists of umbers may be rearraged i differet orders to achieve differet sums The theorem states that the terms of a absolutely coverget series ca be rearraged i ay way without affec g the sum 430
85 Altera g Series ad Absolute Covergece Theorem 72 Absolute Covergece Theorem Let a be a series that coverges absolutely 1 a coverges 2 Let {b } be ay rearragemet of the sequece {a } The b = a I Example 253, we determied the series i part 2 coverges absolutely Theorem 72 tells us the series coverges (which we could also determie usig the Altera g Series Test) The theorem states that rearragig the terms of a absolutely coverget series does ot affect its sum This implies that perhaps the sum of a codi oally coverget series ca chage based o the arragemet of terms Ideed, it ca The Riema Rearragemet Theorem (amed a er Berhard Riema) states that ay codi oally coverget series ca have its terms rearraged so that the sum is ay desired value, icludig! As a example, cosider the Altera g Harmoic Series oce more We have stated that ( 1) +1 1 = 1 1 2 + 1 3 1 4 + 1 5 1 6 + 1 = l 2, 7 (see Key Idea 31 or Example 251) Cosider the rearragemet where every posi ve term is followed by two ega ve terms: 1 1 2 1 4 + 1 3 1 6 1 8 + 1 5 1 10 1 12 (Covice yourself that these are exactly the same umbers as appear i the Altera g Harmoic Series, just i a differet order) Now group some terms 431
Chapter 8 Sequeces ad Series ad simplify: ( 1 1 2 ) 1 ( 1 4 + 3 1 6 1 ) 1 ( 1 8 + 5 1 ) 1 10 12 + = 2 1 4 + 1 6 1 8 + 1 10 1 12 + = 1 (1 12 2 + 13 14 + 15 16 ) + = 1 l 2 2 By rearragig the terms of the series, we have arrived at a differet sum! (Oe could try to argue that the Altera g Harmoic Series does ot actually coverge to l 2, because rearragig the terms of the series should t chage the sum However, the Altera g Series Test proves this series coverges to L, for some umber L, ad if the rearragemet does ot chage the sum, the L = L/2, implyig L = 0 But the Altera g Series Approxima o Theorem quickly shows that L > 0 The oly coclusio is that the rearragemet did chage the sum) This is a icredible result We ed here our study of tests to determie covergece The back cover of this text cotais a table summarizig the tests that oe may fid useful While series are worthy of study i ad of themselves, our ul mate goal withi calculus is the study of Power Series, which we will cosider i the ext sec o We will use power series to create fuc os where the output is the result of a ifiite summa o 432
Exercises 85 Terms ad Cocepts 1 Why is 2 A series si ot a altera g series? ( 1) a coverges whe {a } is, ad lim a = 3 Give a example of a series where a does ot a coverges but 4 The sum of a coverget series ca be chaged by rearragig the order of its terms Problems I Exercises 5 20, a altera g series 5 6 7 8 9 10 11 12 13 a is give =i (a) Determie if the series coverges or diverges (b) Determie if a coverges or diverges (c) If a coverges, determie if the covergece is codi oal or absolute ( 1) +1 2 ( 1) +1! ( 1) + 5 3 5 ( 1) 2 2 ( 1) +1 3 + 5 2 3 + 1 ( 1) l + 1 ( 1) l =2 ( 1) +1 1 + 3 + 5 + + (2 1) cos ( π ) 14 15 16 17 18 19 20 si ( ( + 1/2)π ) l ( 2 ) 3 ( e) ( 1) 2! ( 1) 2 2 ( 1) ( 1000)! Let S be the th par al sum of a series I Exercises 21 24, a coverget altera g series is give ad a value of Compute S ad S +1 ad use these values to fid bouds o the sum of the series 21 22 23 24 ( 1) l( + 1), = 5 ( 1) +1, = 4 4 ( 1), = 6! ( 1 2 ), = 9 I Exercises 25 28, a coverget altera g series is give alog with its sum ad a value of ε Use Theorem 71 to fid such that the th par al sum of the series is withi ε of the sum of the series 25 26 27 28 ( 1) +1 4 ( 1)! = 7π4 720, ε = 0001 = 1 e, ε = 00001 ( 1) 2 + 1 = π 4, ε = 0001 ( 1) (2)! = cos 1, ε = 10 8
Solutios to Odd Exercises 1 si 29 Diverges; compare to Just as lim = 1, 0 si(1/) lim = 1 1/ 1 31 Coverges; compare to 3/2 33 Coverges; Itegral Test 35 Diverges; the th Term Test ad Direct Compariso Test ca be used 37 Coverges; the Direct Compariso Test ca be used with sequece 1/3 39 Diverges; the th Term Test ca be used, alog with the Itegral Test 41 (a) Coverges; use Direct Compariso Test as a < (b) Coverges; sice origial series coverges, we kow lim a = 0 Thus for large, a a +1 < a (c) Coverges; similar logic to part (b) so (a ) 2 < a (d) May coverge; certaily a > a but that does ot mea it does ot coverge (e) Does ot coverge, usig logic from (b) ad th Term Test Sec o 84 1 algebraic, or polyomial 3 Itegral Test, Limit Compariso Test, ad Root Test 5 Coverges 7 Coverges 9 The Ra o Test is icoclusive; the p-series Test states it diverges 11 Coverges 2! 13 Coverges; ote the summa o ca be rewri e as 3!, from which the Ra o Test ca be applied 15 Coverges 17 Coverges 19 Diverges 21 Diverges The Root Test is icoclusive, but the th -Term Test shows divergece (The terms of the sequece approach e 2, ot 0, as ) 23 Coverges 25 Diverges; Limit Compariso Test 27 Coverges; Ra o Test or Limit Compariso Test with 1/3 29 Diverges; th -Term Test or Limit Compariso Test with 1 31 Diverges; Direct Compariso Test with 1/ 33 Coverges; Root Test Sec o 85 1 The sigs of the terms do ot alterate; i the give series, some terms are ega ve ad the others posi ve, but they do ot ecessarily alterate 3 May examples exist; oe commo example is a = ( 1) / 5 (a) coverges (b) coverges (p-series) (c) absolute 7 (a) diverges (limit of terms is ot 0) (b) diverges (c) /a; diverges 9 (a) coverges (b) diverges (Limit Compariso Test with 1/) (c) codi oal 11 (a) diverges (limit of terms is ot 0) (b) diverges (c) /a; diverges 13 (a) diverges (terms oscillate betwee ±1) (b) diverges (c) /a; diverges 15 (a) coverges (b) coverges (Geometric Series with r = 2/3) (c) absolute 17 (a) coverges (b) coverges (Ra o Test) (c) absolute 19 (a) coverges (b) diverges (p-series Test with p = 1/2) (c) codi oal 21 S 5 = 11906; S 6 = 06767; ( 1) 11906 l( + 1) 06767 23 S 6 = 03681; S 7 = 03679; ( 1) 03681 03679! 25 = 5 27 Usig the theorem, we fid = 499 guaratees the sum is withi 0001 of π/4 (Covergece is actually faster, as the sum is withi ε of π/24 whe 249) Sec o 86 1 1 3 5 5 1 + 2x + 4x 2 + 8x 3 + 16x 4 7 1 + x + x2 2 + x3 6 + x4 24 9 (a) R = (b) (, ) 11 (a) R = 1 (b) (2, 4] 13 (a) R = 2 (b) ( 2, 2) 15 (a) R = 1/5 (b) (4/5, 6/5) 17 (a) R = 1 (b) ( 1, 1) 19 (a) R = (b) (, )