0 INFINITE SERIES 0. Sequeces Preiary Questios. What is a 4 for the sequece a? solutio Substitutig 4 i the expressio for a gives a 4 4 4.. Which of the followig sequeces coverge to zero? a b + solutio a This sequece does ot coverge to zero: c + x x x + x + x + 0. b This sequece does ot coverge to zero: this is a geometric sequece with r > ; hece, the sequece diverges to. c Recall that if a coverges to 0, the a must also coverge to zero. Here,, which is a geometric sequece with 0 <r<; hece, coverges to zero. It therefore follows that coverges to zero. 3. Let a be the th decimal approximatio to. That is, a, a.4, a 3.4, etc. What is a? solutio a. 4. Which of the followig sequeces is defied recursively? a a 4 + b b 4 + b solutio a a ca be computed directly, sice it depeds o oly ad ot o precedig terms. Therefore a is defied explicitly ad ot recursively. b b is computed i terms of the precedig term b, hece the sequece {b } is defied recursively. 5. Theorem 5 says that every coverget sequece is bouded. Determie if the followig statemets are true or false ad if false, give a couterexample. a If {a } is bouded, the it coverges. b If {a } is ot bouded, the it diverges. c If {a } diverges, the it is ot bouded. solutio a This statemet is false. The sequece a cos π is bouded sice cos π for all, but it does ot coverge: sice a cos π, the terms assume the two values ad alterately, hece they do ot approach oe value. b By Theorem 5, a covergig sequece must be bouded. Therefore, if a sequece is ot bouded, it certaily does ot coverge. c The statemet is false. The sequece a is bouded, but it does ot approach oe it. 633 May 3, 0
634 CHAPTER 0 INFINITE SERIES Exercises. Match each sequece with its geeral term: a,a,a 3,a 4,... Geeral term a, 3, 3 4, 4 5,... i cos π b,,,,... ii! c,,,,... iii + d, 4, 6 8, 4 6... iv + solutio a The umerator of each term is the same as the idex of the term, ad the deomiator is oe more tha the umerator; hece a +,,, 3,... b The terms of this sequece are alteratig betwee ad so that the positive terms are i the eve places. Sice cos π for eve ad cos π for odd, we have a cos π,,,... c The terms a are for odd ad for eve. Hece, a +,,,... d The umerator of each term is!, ad the deomiator is ; hece, a!,,, 3,... I Exercises 3, calculate Let a the first four terms of the sequece, startig with. for,, 3,...Write out the first three terms of the followig sequeces. 3. c a 3 b! a + b c a +3 solutio c d Settig a,, 3, 4 i the formula for c gives d e a a + c 3! 3 3, c 3! 9, c 3 33 3! 7 6 9, c 4 34 4! 8 4 7 8. 5. a, a + a! 3 solutio b For!,, 3 we have: a a + a 3 4 3 5; a 3 a + a 3 5 3 47; a 4 a 3+ a3 3 09 3 445. The first four terms of {a } are, 5, 47, 445. 7. b 5 + cos π solutio b For, b, b, 3, 4+ we have b b 5 + cos π 4; b 5 + cos π 6; b 3 5 + cos 3π 4; b 4 5 + cos 4π 6. The first four terms of {b } are 4, 6, 4, 6. 9. c c + + + 3 + + solutio c ; c + 3 ; c 3 + + 3 3 + 3 6 ; c 4 + + 3 + 4 6 + 4 5. May 3, 0
SECTION 0. Sequeces 635. b, b a + 3, b b + + + + b + + solutio We eed to fid b 3 ad b 4. Settig 3 ad 4 ad usig the give values for b ad b we obtai: b 3 b 3 + b 3 b + b 3 + 8; b 4 b 4 + b 4 b 3 + b 8 + 3 9. The first four terms of the sequece {b } are, 3, 8, 9. 3. Fid a formula for the th term of each sequece. c -place decimal approximatio to e a, 8, 7,... b 6, 3 7, 4 8,... solutio a The deomiators are the third powers of the positive itegers startig with. Also, the sig of the terms is alteratig with the sig of the first term beig positive. Thus, a 3 + 3 ; a 3 + 3 ; a 3 3 3 3+ 3 3. This rule leads to the followig formula for the th term: a + 3. b Assumig a startig idex of, we see that each umerator is oe more tha the idex ad the deomiator is four more tha the umerator. Thus, the geeral term a is a + + 5. I Exercises Suppose 5 6, thatuse Theorem a 4 ad to determie b the7. it Determie: of the sequece or state that the sequece diverges. 5. aa a + b b a3 solutio We c cosπb have a f where fx ; thus, d a a b a fx. x x 7. b 5 a 0+ 9 4 solutio We have b f where fx 5x x + 9 ; thus, 9. c 5 + 9 x 5x x + 9 5. solutio a We 4 + have 3 4 c + f where fx x ; thus, x x x x 0.. c 9 solutio z We have c 3 f where fx 9 x ; thus, 9 x 9x Thus, the sequece 9 diverges. 3. a z 0 / + x solutio We have a f where fx x + ; thus, + x x x + x x x x + x x x + x x + x + 0. May 3, 0
636 CHAPTER 0 INFINITE SERIES + 5. a a l 9 3 + 4 + x + solutio We have a f where fx l ; thus, 9 + 4x l + 9 + 4 x l x + 9 + 4x l x x + l 3 9 + 4x I Exercises 7 30, use Theorem r l l 4 to determie the it of the sequece. + 7. a 4 + solutio We have 4 + x 4 + x 4 Sice x is a cotiuous fuctio for x>0, Theorem 4 tells us that 4 + 9. a a e 4/3+9 cos 3 3 + solutio We have 4 + 4 3 3 + Sice cos x is cotiuous for all x, Theorem 4 tells us that 3 cos 3 cos 3 + 3 cos / π + 3 3. Let a ta. + e Fid a umber M such that: a a 0.00 for M. b a 0.0000 for M. The use the it defiitio to prove that a. solutio a We have a + + + + +. Therefore a 0.00 provided + 0.00, that is, 999. It follows that we ca take M 999. b By part a, a 0.0000 provided + 0.0000, that is, 99999. It follows that we ca take M 99999. We ow prove formally that a. Usig part a, we kow that a + <ɛ, provided > ɛ. Thus, Let ɛ>0 ad take M ɛ. The, for >M, we have a + < M + ɛ. 33. Use the it Let b defiitio. to prove that 0. 3 solutio a FidWe a value see that of M such that b 0 5 for M. b Use the it defiitio to prove that b 0. 0 <ɛ May 3, 0
SECTION 0. Sequeces 637 provided > ɛ. Thus, let ɛ>0 ad take M. The, for >M, we have ɛ 0 < M ɛ. I Exercises 35 6, use the appropriate it laws ad theorems to determie the it of the sequece or show that it diverges. Use the it defiitio to prove that. + 35. a 0 + 9 solutio By the Limit Laws for Sequeces we have: 0 + 9 0 + 0 +. 9 9 Now, Because by the Limit Laws for Sequeces, Thus, we have ad 9 9 0, 9 9 0, 9. 9 0. 9 0 + 0 + 0 0. 9 37. c.0 d + 3 solutio Sice c f where fx.0 x, we have.0 x.0x so that the sequece diverges. 39. a / b e solutio Because x is a cotiuous fuctio, 4. c b 9! / solutio For 9, write / x /x x /x 0. The clearly c 9! 9 9 9 9 } {{ 9 } 0 9 9 9 } {{ } call this C Each factor is less tha 0 9! C 9 May 3, 0
638 CHAPTER 0 INFINITE SERIES sice each factor after the first ie is <. The squeeze theorem tells us that 0 9! C 9 C 9 C 0 0 so that c 0 as well. 43. a 3 + + a 8! 3 solutio 45. a cos a solutio Sice + 4 cos the followig holds: 3 + + 3x + x + 3 x x 3 3. cos. We ow apply the Squeeze Theorem for Sequeces ad the its 0 to coclude that cos 0. 47. d l 5 l! solutio c Note that so that e d 5! so d l 5! ed 5! 0 by the method of Exercise 4. If d coverged, we could, sice fx e x is cotiuous, the write which is impossible. Thus {d } diverges. 49. a d l + 4 l + 4 /3 ed e d 0 /3. solutio Let a + 4 Takig the atural logarithm of both sides of this expressio yields l a l + 4 /3 3 l + 4. Thus, l a 3 l + 4 /3 3 x l + 4 x 3 l + 4 x x 3 l + 0 3 l l /3. Because fx e x is a cotiuous fuctio, it follows that a el a e l a e l /3 /3. + b ta 5. c l 3 + 4 solutio Because fx l x is a cotiuous fuctio, it follows that x + c x l l 3x + 4 x x + 3x + 4 l 3. May 3, 0
SECTION 0. Sequeces 639 53. y e c + / solutio e e ad e >. By the Limit of Geometric Sequeces,we coclude that e. Thus, the give sequece diverges. 55. y a e + 3 5 solutio e + 3 e 3 5 + 5 5 assumig both its o the right-had side exist. But by the Limit of Geometric Sequeces, sice < 3 5 < 0 < e 5 < both its o the right-had side are 0, so that y coverges to 0. 57. a si π b 3 + solutio By 3 the 3 Theorem + 4 o Sequeces Defied by a Fuctio, we have Now, Thus, x x si π x x si π x x si π x x si π x. π x cos π x x cos π x π x x π cos 0 π π. π cos π x x si π π. 59. b 3 b! 4 + 7 π 4 solutio Divide the umerator ad deomiator by 4 to obtai Thus, a x 3 4 x 4 x + 7 a 3 4 3 + 7 4 4 4 4 4 + 7 4 4 3 4 4 + 7. 3 x 4 x 3 x 4 x x x 4 x + 7 x 4 x x 7 3 0 0 + 7 7. 6. a + a 3 4 + 7 3 solutio Takig the atural logarithm of both sides of this expressio yields Thus, l l a x + x x l a l + l + l x d dx l d dx Because fx e x is a cotiuous fuctio, it follows that + x x x + x +. x x a el a e l a e e. x + x + 0. a + May 3, 0
640 CHAPTER 0 INFINITE SERIES I Exercises 63 66, fid the it of the sequece usig L Hôpital s Rule. l 63. a solutio 65. c b + l + solutio l l x x x x d dx l x d dx x x d dx lx x x d dx x x x x 0 + x x x + x x x x x x + + x x + x x + x d dx lx x lx x x x x + x x + + x x + + x d dx x x + + x x + +/x I Exercises 67 70, d 3 use 3 the Squeeze + Theorem to evaluate a by verifyig the give iequality. 67. a 4 +, 8 4 a + x x + solutio That is, For all > we have 4 < 8, so the quotiet a < is smaller tha ad larger tha. 4 + 8 4 + 4 8 + 8 4 + 4 4 ; ad Now, sice 4 a > 8 + 8 8 4. 0, the Squeeze Theorem for Sequeces implies that a 0. 69. a + 3 /, 3 c + a 3 / + + + / 3 solutio Clearly + 3 3 for all. Therefore: +, + c + 3 / 3 / 3. + Also + 3 3 + 3 3,so Thus, Because + 3 / 3 / / 3. 3 + 3 / / 3. / 3 3 / 3 3 ad 3 3, the Squeeze Theorem for Sequeces guaratees + 3 / 3. 7. a Which + 0 of the / followig, 0 a statemets 0 is equivalet / to the assertio a L? Explai. a For every ɛ>0, the iterval L ɛ, L + ɛ cotais at least oe elemet of the sequece {a }. b For every ɛ>0, the iterval L ɛ, L + ɛ cotais all but at most fiitely may elemets of the sequece {a }. solutio Statemet b is equivalet to Defiitio of the it, sice the assertio a L <ɛfor all >M meas that L ɛ<a <L+ ɛ for all >M; that is, the iterval L ɛ, L + ɛ cotais all the elemets a except maybe the fiite umber of elemets a,a,...,a M. May 3, 0
SECTION 0. Sequeces 64 Statemet a is ot equivalet to the assertio a L. We show this, by cosiderig the followig sequece: for odd a + for eve Clearly for every ɛ>0, the iterval ɛ,ɛ L ɛ, L + ɛ for L 0 cotais at least oe elemet of {a }, but the sequece diverges rather tha coverges to L 0. Sice the terms i the odd places coverge to 0 ad the terms i the eve places coverge to. Hece, a does ot approach oe it. 73. Show Show that that a a 3 is icreasig. is decreasig. Fid a upper boud. + + solutio Let fx 3x x +. The f x 6xx + 3x x x x + x +. f x > 0 for x>0, hece f is icreasig o this iterval. It follows that a f is also icreasig. We ow show that M 3 is a upper boud for a, by writig: That is, a 3 for all. a 3 + 3 + 6 + 3 + 3. + 75. Give a example of a Show that a 3 diverget sequece {a } such that + is decreasig. a coverges. solutio Let a. The sequece {a } diverges because the terms alterate betwee + ad ; however, the sequece { a } coverges because it is a costat sequece, all of whose terms are equal to. 77. Usig the it defiitio, prove that if {a } coverges ad {b } diverges, the {a + b } diverges. Give a example of diverget sequeces {a } ad {b } such that {a + b } coverges. solutio We will prove this result by cotradictio. Suppose a L ad that {a + b } coverges to a it L. Now, let ɛ>0. Because {a } coverges to L ad {a + b } coverges to L, it follows that there exist umbers M ad M such that: a L < ɛ a + b L < ɛ for all >M, for all >M. Thus, for >M max{m,m }, a L < ɛ ad a + b L < ɛ. By the triagle iequality, Thus, for >M, b L L a + b a L L a + L + a + b L May 3, 0 L a + a + b L. b L L < ɛ + ɛ ɛ; that is, {b } coverges to L L, i cotradictio to the give data. Thus, {a + b } must diverge. 79. Theorem states that if Use the it defiitio x to prove fx that L, if the {a } the is a sequece coverget a f coverges ad sequece of itegers with it a L. Show that the L, the there exists a coverse umber is false. M such I other that awords, L for fidall a fuctio M. fxsuch that a f coverges but fxdoes ot exist. x solutio Let fx si πx ad a si π. The a f. Sice si πx is oscillatig betwee ad the it x fxdoes ot exist. However, the sequece {a } is the costat sequece i which a si π 0 for all, hece it coverges to zero. Use the it defiitio to prove that the it does ot chage if a fiite umber of terms are added or removed from a coverget sequece.
64 CHAPTER 0 INFINITE SERIES 8. Let b a +. Use the it defiitio to prove that if {a } coverges, the {b } also coverges ad a b. solutio Suppose {a } coverges to L. Let b a +, ad let ɛ>0. Because {a } coverges to L, there exists a M such that a L <ɛfor >M. Now, let M M. The, wheever >M, + >M+ M. Thus, for >M, b L a + L <ɛ. Hece, {b } coverges to L. Let {a } be a sequece such that a exists ad is ozero. Show that a exists if ad oly if there 83. Proceed as i Example to show that the sequece 3, 3 3, 3 3 3,...is icreasig ad bouded above by M exists 3. The a iteger prove that M such the it that exists the sig ad offid a does its value. ot chage for >M. solutio This sequece is defied recursively by the formula: Cosider the followig iequalities: a 3a I geeral, if we assume that a k >a k, the a + 3a, a 3. 3 3 > 3 a a >a ; a 3 3a > 3a a a 3 >a ; a 4 3a 3 > 3a a 3 a 4 >a 3. a k+ 3a k > 3a k a k. Hece, by mathematical iductio, a + >a for all ; that is, the sequece {a } is icreasig. Because a + 3a, it follows that a 0 for all. Now, a 3 < 3. If a k 3, the a k+ 3a k 3 3 3. Thus, by mathematical iductio, a 3 for all. Sice {a } is icreasig ad bouded, it follows by the Theorem o Bouded Mootoic Sequeces that this sequece is covergig. Deote the it by L a. Usig Exercise 8, it follows that L a + 3a 3 a 3L. Thus, L 3L, sol 0orL 3. Because the sequece is icreasig, we have a a 3 for all. Hece, the it also satisfies L 3. We coclude that the appropriate solutio is L 3; that is, a 3. Let {a } be the sequece defied recursively by Further Isights ad Challeges 85. Show that a 0 0, a +!. Hit: Verify that! / / by observig + a that half of the factors of! are greater tha or equal to /. Thus, a, a +, a 3 + + solutio We show that! /.,... For 4 eve, we have: a Show that if a <, the a + <. Coclude by iductio that a < for all. b Show that if a <, the! a a +. Coclude by iductio that {a } is icreasig. c Use a ad b to coclude that } L {{ } + } a exists. {{ The } +. compute } L{{ by showig } that L + L. factors factors factors Sice each oe of the factors is greater tha, we have: For 3 odd, we have:! + } {{ } factors } {{ } factors /.! + +. } {{ } } {{ } } {{ } + + factors factors factors May 3, 0
Sice each oe of the + factors is greater tha, we have:! + } {{ } + factors I either case we have! /. Thus, } {{ } + factors +/ / /. SECTION 0. Sequeces 643!. Sice, it follows that!. Thus, the sequece a! diverges. 87. Give positive umbers a <b, defie two sequeces recursively by! Let b. a + a b, b + a + b a Show that l b l k a Show that a. b for all Figure 3. b Show that {a k } is icreasig ad {b } is decreasig. c Show b Show that bthat l b coverges to l xdx, ad coclude that b e + a + b a.. 0 d Prove that both {a } ad {b } coverge ad have the same it. This it, deoted AGMa,b, is called the arithmetic-geometric mea of a ad b. e Estimate AGM, to three decimal places. Geometric mea Arithmetic mea a a + b + b AGMa, b FIGURE 3 x solutio a Examie the followig: b + a + a + b a b a + b a b a a b + b a b 0. We coclude that b + a + for all >. By the give iformatio b >a ; hece, b a for all. b By part a, b a for all, so a + a b a a a a for all. Hece, the sequece {a } is icreasig. Moreover, sice a b for all, b + a + b for all ; that is, the sequece {b } is decreasig. c Sice {a } is icreasig, a + a. Thus, b + b b b b + a + b + a a + b a a + b a b a. Now, by part a, a b for all. By part b, {b } is decreasig. Hece b b for all. Combiig the two iequalities we coclude that a b for all. That is, the sequece {a } is icreasig ad bouded 0 a b. By the Theorem o Bouded Mootoic Sequeces we coclude that {a } coverges. Similarly, sice {a } is icreasig, a a for all. We combie this iequality with b a to coclude that b a for all. Thus, {b } is decreasig ad bouded a b b ; hece this sequece coverges. To show that {a } ad {b } coverge to the same it, ote that Thus, b a b a b a b a b a b a. 0. May 3, 0
644 CHAPTER 0 INFINITE SERIES d We have a + a b, a ; b + a + b, b Computig the values of a ad b util the first three decimal digits are equal i successive terms, we obtai: a a b.89 b a + b +.07 a 3 a b.89.07.98 b 3 a + b a 4 a 3 b 3.98.89.07.98 b 4 a 3 + b 3.98 Thus, AGM,.98. 89. Let a Let c + H l + +, where + + + H is the th. harmoic umber a Calculate c,c,c 3,c 4. H + b Use a compariso of rectagles with the area uder + y 3 + + x over the iterval [, ] to prove that + dx x + dx a Show that a 0 for. Hit: Show that H c dx x + x. b Show that {a } is decreasig by iterpretig a a + as a area. c Prove c Use that the Squeeze Theorem to determie c. a exists. This it, deoted γ, is kow as Euler s Costat. It appears i may areas of mathematics, icludig aalysis ad umber theory, ad has bee calculated to more tha 00 millio decimal places, but it is still ot kow whether γ is a irratioal umber. The first 0 digits are γ 0.57756649. solutio a Sice the fuctio y x is decreasig, the left edpoit approximatio to the itegral + itegral; that is, dx x is greater tha this or + + 3 + + + + dx H x. dx x y 3 / 3 + x Moreover, sice the fuctio y x is positive for x>0, we have: + dx x dx x. May 3, 0
SECTION 0. Summig a Ifiite Series 645 Thus, dx H x l x l l l, ad a H l 0 for all. b To show that {a } is decreasig, we cosider the differece a a + : Now, l + l + a a + H l H + l + H H + + l + l + + + + + + + + l + l + + l + l. + dx x, whereas + is the right edpoit approximatio to the itegral + dx x. Recallig y x is decreasig, it follows that + dx x + y y x + + x so a a + 0. c By parts a ad b, {a } is decreasig ad 0 is a lower boud for this sequece. Hece 0 a a for all. A mootoic ad bouded sequece is coverget, so a exists. 0. Summig a Ifiite Series Preiary Questios. What role do partial sums play i defiig the sum of a ifiite series? solutio The sum of a ifiite series is defied as the it of the sequece of partial sums. If the it of this sequece does ot exist, the series is said to diverge.. What is the sum of the followig ifiite series? 4 + 8 + 6 + 3 + 64 + solutio This is a geometric series with c 4 ad r. The sum of the series is therefore 4 4. 3. What happes if you apply the formula for the sum of a geometric series to the followig series? Is the formula valid? + 3 + 3 + 3 3 + 3 4 + May 3, 0
646 CHAPTER 0 INFINITE SERIES solutio the gives This is a geometric series with c ad r 3. Applyig the formula for the sum of a geometric series 0 3 3. Clearly, this is ot valid: a series with all positive terms caot have a egative sum. The formula is ot valid i this case because a geometric series with r 3 diverges. 4. Arvid asserts that 0 because teds to zero. Is this valid reasoig? solutio Arvid s reasoig is ot valid. Though the terms i the series do ted to zero, the geeral term i the sequece of partial sums, S + + 3 + +, is clearly larger tha. The sum of the series therefore caot be zero. 5. Collee claims that coverges because Is this valid reasoig? 0 solutio Collee s reasoig is ot valid. Although the geeral term of a coverget series must ted to zero, a series whose geeral term teds to zero eed ot coverge. I the case of, the series diverges eve though its geeral term teds to zero. 6. Fid a N such that S N > 5 for the series. solutio The Nth partial sum of the series is: 7. Does there exist a N such that S N > 5 for the series solutio The series N S N + + N. } {{ } N? Explai. is a coverget geometric series with the commo ratio r. The sum of the series is: S. Notice that the sequece of partial sums {S N } is icreasig ad coverges to ; therefore S N for all N. Thus, there does ot exist a N such that S N > 5. 8. Give a example of a diverget ifiite series whose geeral term teds to zero. solutio Cosider the series. The geeral term teds to zero, sice 0 9 sum satisfies the followig iequality: 9 0 0. However, the Nth partial S N 9 0 + 9 0 + + N 9 0 N N 9 0 N 9 0 N 0. That is, S N N 0 for all N. Sice N 0, the sequece of partial sums S diverges; hece, the series N 0 9 diverges. May 3, 0
SECTION 0. Summig a Ifiite Series 647 Exercises. Fid a formula for the geeral term a ot the partial sum of the ifiite series. a 3 + 9 + 7 + 8 + b + 5 + 5 4 + 5 8 + c + 33 3 4 4 4 3 + d + + + + 3 + + 4 + + solutio a The deomiators of the terms are powers of 3, startig with the first power. Hece, the geeral term is: a 3. b The umerators are powers of 5, ad the deomiators are the same powers of. The first term is a so, c The geeral term of this series is, a 5. a +!. d Notice that the umerators of a equal for odd values of ad for eve values of. Thus, The formula ca also be rewritte as follows: a + + odd eve a + + +. + I Exercises Write3 6, i summatio compute the otatio: partial sums S, S 4, ad S 6. a + 4 + 9 + 6 + b 9 + 6 + 5 + 3. + 36 + + 3 + 4 + solutio c 3 + 5 7 + d 5 9 + 65 6 + 35 5 + 5,65 S + + 36 5 4 ; 5. + 3 + k k 3 4 + solutio k S 4 + + 3 + 4 05 44 ; S 6 + + 3 + 4 + 5 + 6 5369 3600. S + 3 + 6 4 6 3 ; S 4 S + a 3 + a 4 3 + 3 4 + 4 5 3 + + 0 4 5 ; S 6 S 4 + a 5 + a 6 4 5 + 5 6 + 6 7 4 5 + 30 + 4 6 7. j j! May 3, 0
648 CHAPTER 0 INFINITE SERIES 7. The series S + 5 + 5 + 5 3 + coverges to 5 4. Calculate S N for N,,... util you fid a S N that approximates 5 4 with a error less tha 0.000. solutio S S + 5 6 5. Note that S 3 + 5 + 5 3 5.4 S 3 + 5 + 5 + 5 56 5.48 S 4 + 5 + 5 + 5 + 65 78 65.496 S 5 + 5 + 5 + 5 + 65 + 35 3906 35.499.5 S 5.5.499 0.00008 < 0.000 I Exercises 9 ad 0, use The series S a computer 0!! + algebra! system to compute S 0, S 00, S 500, ad S 000 for the series. Do these values suggest covergece to the give value? 3! + is kow to coverge to e recall that 0!. Calculate S N for 9. N,,... util you fid a S N that approximates e with a error less tha 0.00. solutio The Write π 3 4 3 4 4 5 6 + 6 7 8 8 9 0 + a + + + Computig, we fid N S N a i π 3 0.035398635 4 S 0 0.035356796 S 00 0.035398074 S 500 0.035398690 S 000 0.0353986334 It appears that S N π 3 4.. Calculate S 3, S 4, ad S 5 ad the fid the sum of the telescopig series solutio S 3 S π 4 90 + + 4 + 3 4 + 4 4 + + 3 3 4 S 4 S 3 + 5 6 S 5 S 4 + 6 7 + 4 5 5 3 0 ; 6 3 ; 7 5 4. May 3, 0
SECTION 0. Summig a Ifiite Series 649 The geeral term i the sequece of partial sums is S N + 3 3 + 4 thus, 3 4 5 + + N + N + N + ; S S N N N N +. The sum of the telescopig series is therefore. 3. Calculate S 3, S 4, ad S 5 ad the fid the sum S 4 usig the idetity Write as a telescopig series ad fid its sum. solutio 4 S 3 + 3 S 4 S 3 + 7 9 S 5 S 4 + 9 The geeral term i the sequece of partial sums is thus, S N 3 + 3 5 + 3 5 + + 9 5 7 4 9 ; 5. 5 7 S S N N N + + 3 7 7 ; N N + N +. 5. Fid the sum of Use partial fractios 3 + to 3 rewrite 5 + 5 7 +. as a telescopig series ad fid its sum. solutio We may write this sum as + 3 +. + The geeral term i the sequece of partial sums is S N + 3 3 + 5 thus, ad 5 7 S N N N + + N N + N +, +. I Exercises 7, use Theorem 3 to prove that the followig series diverge. Fid a formula for the partial sum S N of ad show that the series diverges. 7. 0 + solutio The geeral term,, has it 0 + 0 + Sice the geeral term does ot ted to zero, the series diverges. 0 + / 0 ; N + ; N + May 3, 0
650 CHAPTER 0 INFINITE SERIES 0 9. + 3 3 4 + solutio The + geeral term a does ot ted to zero. I fact, because, a does ot exist. By Theorem 3, we coclude that the give series diverges.. cos + cos 3 + cos 4 + solutio The geeral term a cos + teds to, ot zero. By Theorem 3, we coclude that the give series diverges. I Exercises 3 36, use the 4 + formula for the sum of a geometric series to fid the sum or state that the series diverges. 3. 0 + 8 + 8 + solutio This is a geometric series with c ad r 8, so its sum is 7/8 8 7 8 4 3 3 5. 5 3 + 44 5 4 + 45 3 5 5 + solutio Rewrite this series as 3 3 This is a geometric series with r >, so it is diverget. 3 7. 4 7 9 3 4 5 solutio This is a geometric series with c ad r 4, startig at 4. Its sum is thus 9 cr 4 r c r 4 r 5 9 5 4 4 9 4 + 45 9 4 4 + 4 5 59,049 338 9 5 9. e π e 0 solutio Rewrite the series as e to recogize it as a geometric series with c e ad r e. Thus, 8 + 3. 5e 3 0 solutio Rewrite the series as e e e 8 5 + 5 8 0 0 0 e. + 5 0, 5 which is a sum of two geometric series. The first series has c 8 5 0 8 ad r 5 ; the secod has c 5 0 ad r 5. Thus, 8 8 5 0 5 8 0, 4 5 5 0 5 5 3 3, 5 May 3, 0
SECTION 0. Summig a Ifiite Series 65 ad 8 + 5 0 + 5 3 35 3. 0 33. 5 5 4 + 5 3 4 5 4 3 5 + solutio 0 8 This is a geometric series with c 5 ad r 4. Thus, 0 5 4 5 5 + 4 4 5 5 4 4. 7 35. 8 3 49 64 + 343 7 + 4 5 7 + 5 40 4096 7 3 + 6 + 7 4 + solutio This is a geometric series with c 7 8 ad r 7 8. Thus, 0 7 8 7 8 7 8 7 8 7 8 7 5 5. 8 37. Which of 5 9 + 5 the followig 3 + + 3 are 5 + 9 ot 5 + 7 geometric series? 7 a 5 + 9 b 0 3 4 c d π 0 5 solutio 7 a 9 7 : this is a geometric series with commo ratio r 7 9 9. 0 0 b The ratio betwee two successive terms is a + a + 4 4 4 4 + 4. + This ratio is ot costat sice it depeds o. Hece, the series is ot a geometric series. 4 3 c The ratio betwee two successive terms is a + a + + + + +. This ratio is ot costat sice it depeds o. Hece, the series is ot a geometric series. 0 d π : this is a geometric series with commo ratio r π π. 5 5 39. Prove that if a coverges ad b diverges, the a + b diverges. Hit: If ot, derive a cotradictio Use the method of Example 8 to show that diverges. k/3 by writig k b a + b a May 3, 0
65 CHAPTER 0 INFINITE SERIES solutio Suppose to the cotrary that a coverges, b diverges, but a + b coverges. The by the Liearity of Ifiite Series, we have so that b coverges, a cotradictio. b a + b a 4. Give a couterexample to 9 show + that each of the followig statemets is false. Prove the divergece of 5. a If the geeral term a teds0 to zero, the a 0. b The Nth partial sum of the ifiite series defied by {a } is a N. c If a teds to zero, the a coverges. d If a teds to L, the a L. solutio a Let a. The a 0, but a is a geometric series with c 0 ad r /, so its sum is /. b Let a. The the th partial sum is a + a + +a while a. c Let a. A example i the text shows that while a teds to zero, the sum a does ot coverge. d Let a. The clearly a teds to L, while the series a obviously diverges. 43. Compute the total area of the ifiitely may triagles i Figure 4. Suppose that S a is a ifiite series with partial sum S N 5 N. y 0 6 a What are the values of a ad a? b What is the value of a 3? c Fid a geeral formula for a. d Fid the sum a. 5 6 8 4 FIGURE 4 solutio The area of a triagle with base B ad height H is A BH. Because all of the triagles i Figure 4 have height, the area of each triagle equals oe-quarter of the base. Now, for 0, the th triagle has a base which exteds from x + to x. Thus, B + + ad A 4 B +3. The total area of the triagles is the give by the geometric series x 0 +3 0 8 8 4. 45. Fid the total legth of the ifiite zigzag path i Figure 5 each zag occurs at a agle of π The wier of a lottery receives m dollars at the ed of each year for N years. The preset 4. value PV of this prize N i today s dollars is PV m + r i, where r is the iterest rate. Calculate PV if m $50,000, r 0.06, ad i N 0. What is PV if N? π /4 π /4 FIGURE 5 May 3, 0
SECTION 0. Summig a Ifiite Series 653 solutio Because the agle at the lower left i Figure 5 has measure π 4 ad each zag i the path occurs at a agle of π 4, every triagle i the figure is a isosceles right triagle. Accordigly, the legth of each ew segmet i the path is times the legth of the previous segmet. Sice the first segmet has legth, the total legth of the path is 0 +. 47. Show that if a is a positive iteger, the Evaluate. Hit: Fid costats A, B, ad C such that + + + + A + B + + C + a + a + + a + solutio By partial fractio decompositio clearig the deomiators gives + a A + B + a ; A + a + B. Settig 0 the yields A a, while settig a yields B a. Thus, + a a a + a a, + a ad + a a. + a For N>a, the Nth partial sum is S N + a + 3 + + a a N + + N + + N + 3 + +. N + a Thus, + a S N + N a + 3 + +. a 49. Let A{bball } be dropped a sequece fromad a height let a of 0b ft begis b. toshow bouce. thateachatime coverges it strikesifthe adgroud, oly ifit returs b to exists. two-thirds of its previous height. What is the total distace traveled by the ball if it bouces ifiitely may times? solutio Let a b b. The geeral term i the sequece of partial sums for the series a is the S N b b 0 + b b + b 3 b + +b N b N b N b 0. Now, if b N exists, the so does S N ad a coverges. O the other had, if a coverges, the N N S N exists, which implies that b N also exists. Thus, a coverges if ad oly if N N b exists. Further Isights ad Challeges Assumptios Matter Show, by givig couterexamples, that the assertios of Theorem are ot valid if the Exercises 5 53 use the formula series a ad b are ot coverget. 0 0 + r + r + +r N rn 7 r 5. Professor GeorgeAdrews of Pesylvaia State Uiversity observed that we ca use Eq. 7 to calculate the derivative of fx x N for N 0. Assume that a 0 ad let x ra. Show that ad evaluate the it. f x N a N a x a x a a N r N r r May 3, 0
654 CHAPTER 0 INFINITE SERIES solutio Accordig to the defiitio of derivative of fxat x a Now, let x ra. The x a if ad oly if r, ad f x N a N a x a x a By Eq. 7 for a geometric sum, r N r f x N a N a x a x a. ra N a N r ra a r N an r a r rn r + r + r + +r N, a N r r N r. so r N r r + r + r + +r N + + + + N N. r Therefore, f a a N N Na N 53. Verify the Gregory Leibiz formula as follows. a Set Pierre r x de i Fermat Eq. 7 used ad geometric rearrage series to show to that compute the area uder the graph of fx x N over [0,A]. For 0 <r<, let Frbe the sum of the areas of the ifiitely may right-edpoit rectagles with edpoits Ar,as i Figure 6. As r teds to, the rectagles become arrower ad Frteds to the area uder the graph. a Show that Fr A N+ + x r x + x 4 + N x N + N x N r N+. + x b Show, by itegratig over [0, ], A that b Use Eq. 7 to evaluate x N dx Fr. π 0 r 4 3 + 5 N + + 7 N + x N dx N 0 + x c Use the Compariso Theorem for itegrals to prove that x N dx 0 0 + x N + Hit: Observe that the itegrad is x N. d Prove that π 4 3 + 5 7 + 9 Hit: Use b ad c to show that the partial sums S N of satisfy SN π 4 S N π N 4. solutio a Start with Eq. 7, ad substitute x for r: N+ + r + r + +r N rn r x + x 4 + + N x N N x N x x + x 4 + + N x N + x N x N + x + x x + x 4 + + N x N + N x N + x b The itegrals of both sides must be equal. Now, 0 + x dx ta x ta ta 0 π 0 4 while x + x 4 + + N x N + N x N 0 + x dx, ad thereby coclude that May 3, 0
SECTION 0. Summig a Ifiite Series 655 x 3 x3 + 5 x5 + + N N xn + N 0 3 + 5 + + N N + x N dx N 0 + x c Note that for x [0, ], we have + x, so that 0 xn + x xn By the Compariso Theorem for itegrals, we the see that x N dx 0 0 + x x N dx 0 N + xn+ 0 N + d Write a, ad let S N be the partial sums. The S N π 4 N 0 x N dx + x 0 x N dx + x N + x N dx + x Thus N S N π 4 so that π 4 3 + 5 7 + 9... 55. The Koch sowflake described i 904 by Swedish mathematicia Helge vo Koch is a ifiitely jagged fractal curve obtaied Cator s asdisappearig a it of polygoal Table followig curves it Larry is cotiuous Kop of but Hamilto has ocollege taget lie Take at aay table poit. of legth BegiL with Figure a equilateral 7. At stage triagle, remove stage 0 the ad sectio produce of legth stage L/4 by replacig cetered at the midpoit. Two sectios remai, each with legth less tha L/. At stage, remove sectios of legth L/4 each edge four edges of oe-third the legth, arraged as i Figure 8. Cotiue the process: At the th stage, replace fromeach eachedge of these withtwo foursectios edges of this oe-third stage removes the legth. L/8ofthe table. Now four sectios remai, each of legth less tha L/4. At stage 3, remove the four cetral sectios of legth a Show L/4 3 that the perimeter P of the polygo at the th stage satisfies P 4, etc. 3 P. Prove that P. The sowflake has ifiite legth. a Show that at the Nth stage, each remaiig sectio has legth less tha L/ N ad that the total amout of table b Let A removed 0 be the area of the origial equilateral triagle. Show that 34 is ew triagles are added at the th stage, each with area A 0 /9 for. Show that the total area of the Koch sowflake L 4 + 8 + 6 + + is 8 5 A 0. N+ b Show that i the it as N, precisely oe-half of the table remais. This result is curious, because there are o ozero itervals of table left at each stage, the remaiig sectios have a legth less tha L/ N. So the table has disappeared. Stage However, Stage we ca Stage place 3 ay object loger tha L/4 othe table. It will ot fall through because it will ot fitfigure through ay 8 of the removed sectios. solutio a Each edge of the polygo at the st stage is replaced by four edges of oe-third the legth; hece the perimeter of the polygo at the th stage is 4 3 times the perimeter of the polygo at the th stage. That is, P 4 3 P. Thus, P 4 3 P 0; P 4 3 P 4 P 0, P 3 4 3 3 P 4 3 P 0, 3 ad, i geeral, P 4 P0 3.As, it follows that 4 P P 0. 3 b Whe each edge is replaced by four edges of oe-third the legth, oe ew triagle is created. At the st stage, there are 3 4 edges i the sowflake, so 3 4 ew triagles are geerated at the th stage. Because the area of a equilateral triagle is proportioal to the square of its side legth ad the side legth for each ew triagle is oe-third the side legth of triagles from the previous stage, it follows that the area of the triagles added at each stage is reduced by a factor of 9 from the area of the triagles added at the previous stage. Thus, each triagle added at the th stage has a area of A 0 /9. This meas that the th stage cotributes 3 4 A0 9 3 4 4 A 0 9 May 3, 0
656 CHAPTER 0 INFINITE SERIES to the area of the sowflake. The total area is therefore A A 0 + 3 4 4 A 0 A 0 + 3 4 9 4 A 9 0 4 9 A 0 + 3 4 A 0 4 5 8 5 A 0. 0.3 Covergece of Series with Positive Terms Preiary Questios. Let S a. If the partial sums S N are icreasig, the choose the correct coclusio: a {a } is a icreasig sequece. b {a } is a positive sequece. solutio The correct respose is b. Recall that S N a + a + a 3 + +a N ; thus, S N S N a N.IfS N is icreasig, the S N S N 0. It the follows that a N 0; that is, {a } is a positive sequece.. What are the hypotheses of the Itegral Test? solutio The hypotheses for the Itegral Test are: A fuctio fxsuch that a f must be positive, decreasig, ad cotiuous for x. 3. Which test would you use to determie whether 3. coverges? solutio Because 3. 3., we see that the idicated series is a p-series with p 3. >. Therefore, the series coverges. 4. Which test would you use to determie whether + coverges? solutio Because ad + <, is a coverget geometric series, the compariso test would be a appropriate choice to establish that the give series coverges. e 5. Ralph hopes to ivestigate the covergece of by comparig it with. Is Ralph o the right track? solutio No, Ralph is ot o the right track. For, however, e < ; is a diverget series. The Compariso Test therefore does ot allow us to draw a coclusio about the covergece or divergece of the series e. Exercises I Exercises 4, use the Itegral Test to determie whether the ifiite series is coverget.. 4 solutio Let fx. This fuctio is cotiuous, positive ad decreasig o the iterval x, so the Itegral x4 Test applies. Moreover, May 3, 0
SECTION 0.3 Covergece of Series with Positive Terms 657 dx R x 4 x 4 dx R 3 R R 3 3. The itegral coverges; hece, the series also coverges. 4 3. /3 + 3 solutio Let fx x 3 3. This fuctio is cotiuous, positive ad decreasig o the iterval x, so the x Itegral Test applies. Moreover, R x /3 dx x /3 dx 3 R R /3. R The itegral diverges; hece, the series 5. 5 3 + 9 5/ 5 4 solutio Let fx /3 also diverges. x x 3 + 9. This fuctio is positive ad cotiuous for x 5. Moreover, because 5/ f x xx3 + 9 5/ x 5 x3 + 9 3/ 3x x 3 + 9 5 x36 x3 x 3 + 9 7/, we see that f x < 0 for x 5, so f is decreasig o the iterval x 5. The Itegral Test therefore applies. To evaluate the improper itegral, we use the substitutio u x 3 + 9,du 3x dx. We the fid x R x 5 x 3 dx + 9 5/ R 5 x 3 + 9 5/ dx R 3 +9 3 R 5634 9 R R 3 + 9 3/ 5634 3/ du u 5/ 9 5634 3/. The itegral coverges; hece, the series 5 3 + 9 also coverges. 5/ 7. + + 3/5 solutio Let fx x. This fuctio is positive, decreasig ad cotiuous o the iterval x, hece the + Itegral Test applies. Moreover, dx R x + dx R x + ta R π π R 4 π 4 π 4. The itegral coverges; hece, the series also coverges. + 9. + 4 solutio Let fx. This fuctio is positive, cotiuous ad decreasig o the iterval x, so the xx + Itegral Test applies. We compute the improper itegral usig partial fractios: dx R xx + R x dx x + l x R R R x + l R R + l l l l. The itegral coverges; hece, the series + coverges. May 3, 0
658 CHAPTER 0 INFINITE SERIES. le solutio Let fx. This fuctio is positive ad cotiuous for x. Moreover, xl x f x x l x 4 l x + x l x x x l x 4 l x + lx. Sice l x>0 for x>, f x is egative for x>; hece, f is decreasig for x. To compute the improper itegral, we make the substitutio u l x,du dx. We obtai: x R xl x dx l R R xl x dx du R l u R l R l l. The itegral coverges; hece, the series also coverges. l 3. l l solutio Note that Thus, l e l l e l l l. l l. Now, let fx. This fuctio is positive, cotiuous ad decreasig o the iterval x ; therefore, the Itegral xl Test applies. Moreover, dx R x l dx R x l l R R l, because l > 0. The itegral diverges; hece, the series also diverges. l 5. Show that 3 l 3 coverges by usig the Compariso Test with 3. + 8 solutio We compare the series with the p-series 3. For, 3 + 8 3. Sice coverges it is a p-series with p 3 >, the series 3 3 also coverges by the Compariso Test. + 8 7. Let S Show that 3 diverges by comparig with +. Verify that for,. +, + Ca either iequality be used to show that S diverges? Show that solutio + ad coclude that S diverges. For, + ad +. Takig the reciprocal of each of these iequalities yields + ad +. May 3, 0
These iequalities idicate that the series SECTION 0.3 Covergece of Series with Positive Terms 659 + is smaller tha both both diverge so either iequality allows us to show that S diverges. O the other had, for,,so + ad +. ad ; however, ad The series diverges, sice the harmoic series diverges. The Compariso Test the lets us coclude that the larger series + also diverges. I Exercises 9 30, use the Compariso Test to determie whether the ifiite series is coverget. Which of the followig iequalities ca be used to study the covergece of +? Explai. 9. +, + solutio We compare with the geometric series. For,. Sice coverges.. 3 /3 5 + + 4 + solutio For, coverges it is a geometric series with r, we coclude by the Compariso Test that /3 + also The series is a geometric series with r, so it coverges. By the Compariso test, so does /3 +. 4 3. m!+4 m m 3 + solutio For m, 4 m!+4 m 4 4 m m. 4 m The series is a geometric series with r, so it coverges. By the Compariso Test we ca therefore 4 4 m 4 coclude that the series m!+4m also coverges. m si k 5. k k 3 4 solutio For k, 0 si k, so 0 si k k k. The series is a p-series with p >, so it coverges. By the Compariso Test we ca therefore coclude that k k si k the series k k also coverges. May 3, 0
660 CHAPTER 0 INFINITE SERIES 7. k /3 3 + k k 5/4 3 k solutio Sice 3 > 0 for all, 3 + 3 3. 3 The series is a geometric series with r, so it coverges. By the Compariso Theorem we ca therefore 3 3 coclude that the series 3 + 3 also coverges. 9. + k! k solutio Note that for, so that +! 3 + } {{ } factors +! + +! + But is a geometric series with ratio r, so it coverges. By the compariso test, coverges as +! well. Exercise 3 36: For all a>0 ad b>, the iequalities! 3 l a, a <b are true for sufficietly large this ca be proved usig L Hopital s Rule. Use this, together with the Compariso Theorem, to determie whether the series coverges or diverges. 3. solutio l 3 For sufficietly large say k, although i this case suffices, we have l, so that k l 3 k 3 This is a p-series with p >, so it coverges. Thus l k 3 also coverges; addig back i the fiite umber of terms for k does ot affect this result. l 00 33.. l m m solutio Choose N so that l 0.0005 for N. The also for >N, l 00 0.0005 00 0.05. The k l 00 N. N 0.05. N.05 But N.05 is a p-series with p.05 >, so is coverget. It follows that l 00 N. is also coverget; l 00 addig back i the fiite umber of terms for,,...,n shows that. coverges as well. l 0 May 3, 0
SECTION 0.3 Covergece of Series with Positive Terms 66 35. 3 solutio Choose N such that for N. The N 3 N 3 The latter sum is a geometric series with r <, so it coverges. Thus the series o the left coverges as well. Addig 3 back i the fiite umber of terms for <Nshows that 3 coverges. 37. Show that 5 si coverges. Hit: Use si x x for x 0. solutio For, therefore, si > 0 for. Moreover, for, 0 <π; si. The series is a p-series with p >, so it coverges. By the Compariso Test we ca therefore coclude that the series si also coverges. I Exercises 39 48, si/ use the Limit Compariso Test to prove covergece or divergece of the ifiite series. Does coverge? l 39. 4 solutio Let a 4. For large, 4 4, so we apply the Limit Compariso Test with b. We fid a L 4 b 4 4. The series is a p-series with p >, so it coverges; hece, also coverges. Because L exists, by the Limit Compariso Test we ca coclude that the series 4 coverges. 4. 3 + solutio Let a. For large, 3 + 3 + 3, so we apply the Limit Compariso test with b. We fid The series a L b 3 + 3 3 +. is a p-series with p <, so it diverges; hece, also diverges. Because L>0, by the Limit Compariso Test we ca coclude that the series 3 + diverges. May 3, 0
66 CHAPTER 0 INFINITE SERIES 3 + 5 43. 3 3 7 + + 3 + 5 solutio Let a. For large, 3 + 5 3 3 3, so we apply the Limit Compariso Test with b. We fid 3+5 a ++ L b 3 3 + 5 + + 3. The series is a p-series with p >, so it coverges; hece, the series also coverges. Because L 3 3 + 5 exists, by the Limit Compariso Test we ca coclude that the series coverges. 3 45. e + + l e solutio Let a + l For large, + l, so apply the Compariso Test with b. We fid a L b + l + l The series is a p-series with p <, so it diverges. Because L exists, the Limit Compariso Test tells us the the origial series also diverges. 47. l cos+ 4 Hit: Compare with. 5/ solutio Let a cos, ad apply the Limit Compariso Test with b. We fid a cos L b As x, u x 0, so L cos x x x x si x x x si x x x x 3 si u u 0 u. The series is a p-series with p >, so it coverges. Because L exists, by the Limit Compariso Test we ca coclude that the series cos also coverges. I Exercises 49 78, determie covergece or divergece usig ay method covered so far. / Hit: Compare with the harmoic series. 49. 4 9 solutio Apply the Limit Compariso Test with a 9 ad b : a L 9 b 9. si x x. May 3, 0
SECTION 0.3 Covergece of Series with Positive Terms 663 Sice the p-series coverges, the series also coverges. Because L exists, by the Limit Compariso Test 4 we ca coclude that the series 4 9 coverges. 5. cos 4 + 9 solutio Apply the Limit Compariso Test with a 4 + 9 ad b : The series a 4+9 L b 4 + 9 4. is a diverget p-series. Because L>0, by the Limit Compariso Test we ca coclude that the series also diverges. 4 + 9 53. cos 5 + 3 solutio First rewrite a 5 + 4 + 4 ad observe + 4 + < 4 3 for. The series is a coverget p-series, so by the Compariso Test we ca coclude that the series 3 5 also coverges. + 55. 4/5 5 + si solutio 4 5 5 4 5 5 which is a geometric series startig at 5 with ratio r 5 >. Thus the series diverges. 4 57. 3/ l 3 solutio For 3, l >, so 3/ l > 3/ ad 3/ l < 3/. The series is a coverget p-series, so the series also coverges. By the Compariso Test we ca 3/ 3/ 3 therefore coclude that the series 3 3/ coverges. Hece, the series l 3/ also coverges. l 59. 4 /k l k 9/8 solutio a k k k 4/k 4 0 0; therefore, the series 4 /k diverges by the Divergece Test. k May 3, 0
664 CHAPTER 0 INFINITE SERIES 6. l 5 4 solutio By the commet precedig Exercise 3, we ca choose N so that for N, we have l < /8, so that l 4 < /. The N l 4 > which is a diverget p-series. Thus the series o the left diverges as well, ad addig back i the fiite umber of terms for <Ndoes ot affect the result. Thus l 4 diverges. 63. l 3 solutio For, l l ; therefore, N / l l. Now, let fx. For x, this fuctio is cotiuous, positive ad decreasig, so the Itegral Test applies. Usig x l x the substitutio u l x, du x dx,wefid dx R x l x dx l R R x l x du R l u ll R ll. R The itegral diverges; hece, the series the series l diverges. 65. l solutio For, ; therefore, also diverges. By the Compariso Test we ca therefore coclude that l. The series is a coverget geometric series, so also coverges. By the Compariso Test we ca therefore coclude that the series coverges. Hece, the series coverges. + 67. 4 3/ 3 solutio Let a + The a { 0 odd k k k eve Therefore, {a } cosists of 0s i the odd places ad the harmoic series i the eve places, so i a is just the sum of the harmoic series, which diverges. Thus i a diverges as well. + 3/ May 3, 0
SECTION 0.3 Covergece of Series with Positive Terms 665 69. si solutio Apply the Limit Compariso Test with a si ad b : si L si u u 0 u, where u. The harmoic series diverges. Because L>0, by the Limit Compariso Test we ca coclude that the series si also diverges. + 7. si/ 4 solutio For 3, + <,so The series + 4 < 4 is a coverget geometric series, so therefore coclude that the series 3. 3 + 4 coverges. Fially, the series also coverges. By the Compariso Test we ca + 4 coverges. l 73. 4 3 e 3 solutio By the commet precedig Exercise 3, we ca choose N 4 so that for N,l< /. The N l 3 N / 3 N 3/ 3 / To evaluate covergece of the latter series, let a 3/ 3 / ad b, ad apply the Limit Compariso 3/ Test: a L b 3/ 3 / 3/ 3 0 Thus a coverges if b does. But b is a coverget p-series. Thus a coverges ad, by the compariso test, so does the origial series. Addig back i the fiite umber of terms for <Ndoes ot affect covergece. 75. / 3 l l solutio By the commet precedig Exercise 3, we ca choose N so that for N,l< /4. The N / l > which is a diverget p-series. Thus the origial series diverges as well - as usual, addig back i the fiite umber of terms for <Ndoes ot affect covergece. 4 + 5 77. 3 4 3/ 5 l 4 7 solutio Apply the Limit Compariso Test with N 3/4 a 4 + 5 3 4 5 7, b 4 3 4 4 3 May 3, 0
666 CHAPTER 0 INFINITE SERIES We have a L b 4 + 5 3 4 5 7 3 4 4 + 45 3 4 0 68 + 45/ 0/ 68/ 4 Now, 4 + 5 b is a p-series with p >, so coverges. Sice L, we see that 3 4 5 coverges as 7 well. 79. For which a does 4 + 5 l a coverge? solutio First cosider the case a>0 but a. Let fx xl x a. This fuctio is cotiuous, positive ad decreasig for x, so the Itegral Test applies. Now, dx R xl x a dx l R R xl x a du R l u a a R l R a l a. Because { R l R a, 0 <a< 0, a > we coclude the itegral diverges whe 0 <a< ad coverges whe a>. Therefore l a coverges for a> ad diverges for 0 <a<. Next, cosider the case a. The series becomes. Let fx. For x, this fuctio is cotiuous, l x l x positive ad decreasig, so the Itegral Test applies. Usig the substitutio u l x, du x dx,wefid dx R x l x dx l R R x l x du R l u ll R ll. R The itegral diverges; hece, the series also diverges. l b Fially, cosider the case a<0. Let b a>0 so the series becomes. Sice l > for all 3, it follows that l b > so l b >. The series diverges, so by the Compariso Test we ca coclude that l b also diverges. Cosequetly, 3 3 l b diverges. Thus, To summarize: l a diverges for a<0. l a coverges if a> ad diverges if a. Approximatig Ifiite Sums I Exercises 8 83, let a f, where fxis a cotiuous, decreasig fuctio such that fx For which 0 ada does fxdx a coverges. l coverge? 8. Show that fxdx a a + fxdx 3 May 3, 0
SECTION 0.3 Covergece of Series with Positive Terms 667 solutio From the proof of the Itegral Test, we kow that N a + a 3 + a 4 + +a N fxdx fxdx; that is, S N a fxdx or S N a + fxdx. Also from the proof of the Itegral test, we kow that N fxdx a + a + a 3 + +a N S N a N S N. Thus, N fxdx S N a + fxdx. Takig the it as N yields Eq. 3, as desired. 83. Let S Usig Eq. a.3, Arguig showas that i Exercise 8, show that M 5 M+. 6 a + fxdx S a + fxdx 4 M+ M+ This series coverges slowly. Use a computer algebra system to verify that S N < 5 for N 43,8 ad S 43,9 Coclude 5.000000. that M 0 S a + fxdx a M+ 5 M+ This provides a method for approximatig S with a error of at most a M+. solutio Followig the proof of the Itegral Test ad the argumet i Exercise 8, but startig with M + rather tha, we obtai fxdx a a M+ + fxdx. M+ M+ M+ Addig M a to each part of this iequality yields Subtractig M M a + fxdx M+ M+ a S a + fxdx. M+ a + fxdx from each part of this last iequality the gives us M+ M 0 S a + fxdx a M+. M+ 85. Apply Eq. 4 with M 40,000 to show that Use Eq. 4 with M 43,9 to prove that.644934066 5.59580 5.595839.644934068. Is this cosistet with Euler s result, accordig to which this ifiite series has sum π /6? solutio Usig Eq. 4 with fx x, a ad M 40,000, we fid dx S 40,000 + 40,00 x S dx 40,00 + 40,00 x. May 3, 0
668 CHAPTER 0 INFINITE SERIES Now, S 40,000.644909067; S 40,00 S 40,000 + 40,00.6449090678; ad dx R 40,00 x dx R 40,00 x R R 40,00 40,00 0.000049994. Thus, or.644909067 + 0.000049994.6449090678 + 0.000049994,.6449340665.644934067. Sice π.6449340668, our approximatio is cosistet with Euler s result. 6 87. Usig a CAS ad Eq. 5, determie the value of 5 to withi a error less tha 0 4. Usig a CAS ad Eq. 5, determie the value of 6 to withi a error less tha 0 4. Check that your solutio Usig Eq. 5 with fx x 5 ad a 5, we have result is cosistet with that of Euler, who proved that the sum is equal to π 6 /945. M+ 0 5 5 + x 5 dx M + 5. M+ To guaratee a error less tha 0 4, we eed M + 5 0 4. This yields M 0 4/5 5.3, so we choose M 6. Now, 7 5.0368498887, ad Thus, R x 5 dx x 5 dx 7 R 7 4 R 4 7 4 R 4 7 4 0.0000433. 7 5 5 + x 5 dx.0368498887 + 0.0000433.03695400. 7 89. TheHow followig far caargumet a stack of proves idetical the divergece books of mass of the mharmoic ad uit legth series S exted without / without tippig usig over? thethe Itegral stacktest. will ot tip over if the + st book is placed at the bottom of the stack with its right edge located at the ceter of mass Let of the first books Figure 5. Let c be the ceter of mass of the first books, measured alog the x-axis, where we take the positive x-axis to the left of the origi as i Figure 6. Recall that if a object of mass m has ceter of mass at x ad a secod object of ms has + ceter of mass x, the the ceter of mass of the system has x-coordiate 3 + 5 +, S + 4 + 6 + m x + m x Show that if S coverges, the m + m a S ad S also coverge ad S S + S. b Sa Show that if the + st book is placed with its right edge at c, the its ceter of mass is located at c +. >S ad S S. Observe b Cosider that b cotradicts the first books a, ad as acoclude sigle object that Sofdiverges. mass m with ceter of mass at c ad the + st book as a secod solutio object of Assume mass m. throughout Show that if that the S coverges; + st book we is will placed derive with a cotradictio. its right edge Write at c, the c + c + +. c Prove that c. Thus, by usig eough books, the stack ca be exteded as far as desired without tippig over. a, b, c May 3, 0