This page intentionally left blank Students Solutions Manualto accompany Jon RogawskisMultivariableCALCULUSSECOND EDITIONGREGORY P. DRESDENWashington and Lee UniversityJENNIFER BOWENThe College of WoosterRANDALL PAULOregon Institute of TechnologyW. H. FREEMAN AND COMPANYNEW YORK 2012 by W. H. Freeman and CompanyISBN-13: 978-1-4292-5508-0ISBN-10: 1-4292-5508-0All rights reservedPrinted in the United States of AmericaFirst PrintingW. H. Freeman and Company, 41 Madison Avenue, New York, NY 10010Houndmills, Basingstoke RG21 6XS, Englandwww.whfreeman.comCONTENTSChapter 10 INFINITE SERIES (LT Chapter 11) 110.1 Sequences (LT Section 11.1) 110.2 Summing an Innite Series (LT Section 11.2) 1310.3 Convergence of Series with Positive Terms(LT Section 11.3) 2310.4 Absolute and Conditional Convergence(LT Section 11.4) 3810.5 The Ratio and Root Tests (LT Section 11.5) 4410.6 Power Series (LT Section 11.6) 5210.7 Taylor Series (LT Section 11.7) 64Chapter Review Exercises 81Chapter 11 PARAMETRIC EQUATIONS,POLAR COORDINATES,AND CONIC SECTIONS(LT Chapter 12) 9611.1 Parametric Equations (LT Section 12.1) 9611.2 Arc Length and Speed (LT Section 12.2) 11211.3 Polar Coordinates (LT Section 12.3) 12011.4 Area and Arc Length in Polar Coordinates(LT Section 12.4) 13311.5 Conic Sections (LT Section 12.5) 143Chapter Review Exercises 154Chapter 12 VECTOR GEOMETRY(LT Chapter 13) 16612.1 Vectors in the Plane (LT Section 13.1) 16612.2 Vectors in Three Dimensions (LT Section 13.2) 17612.3 Dot Product and the Angle between Two Vectors(LT Section 13.3) 18412.4 The Cross Product (LT Section 13.4) 19812.5 Planes in Three-Space (LT Section 13.5) 21212.6 A Survey of Quadric Surfaces (LT Section 13.6) 22312.7 Cylindrical and Spherical Coordinates(LT Section 13.7) 230Chapter Review Exercises 240Chapter 13 CALCULUS OF VECTOR-VALUEDFUNCTIONS (LT Chapter 14) 25013.1 Vector-Valued Functions (LT Section 14.1) 25013.2 Calculus of Vector-Valued Functions(LT Section 14.2) 26113.3 Arc Length and Speed (LT Section 14.3) 27313.4 Curvature (LT Section 14.4) 28213.5 Motion in Three-Space (LT Section 14.5) 30313.6 Planetary Motion According to Kepler and Newton(LT Section 14.6) 319Chapter Review Exercises 326Chapter 14 DIFFERENTIATION IN SEVERALVARIABLES (LT Chapter 15) 33614.1 Functions of Two or More Variables(LT Section 15.1) 33614.2 Limits and Continuity in Several Variables(LT Section 15.2) 34514.3 Partial Derivatives (LT Section 15.3) 35114.4 Differentiability and Tangent Planes(LT Section 15.4) 36314.5 The Gradient and Directional Derivatives(LT Section 15.5) 37414.6 The Chain Rule (LT Section 15.6) 38614.7 Optimization in Several Variables(LT Section 15.7) 39914.8 Lagrange Multipliers: Optimizing with a Constraint(LT Section 15.8) 419Chapter Review Exercises 442Chapter 15 MULTIPLE INTEGRATION(LT Chapter 16) 45815.1 Integration in Two Variables (LT Section 16.1) 45815.2 Double Integrals over More General Regions(LT Section 16.2) 47115.3 Triple Integrals (LT Section 16.3) 48915.4 Integration in Polar, Cylindrical, and SphericalCoordinates (LT Section 16.4) 50215.5 Applications of Multiple Integrals(LT Section 16.5) 52115.6 Change of Variables (LT Section 16.6) 537Chapter Review Exercises 554Chapter 16 LINE AND SURFACE INTEGRALS(LT Chapter 17) 57616.1 Vector Fields (LT Section 17.1) 57616.2 Line Integrals (LT Section 17.2) 58116.3 Conservative Vector Fields (LT Section 17.3) 59916.4 Parametrized Surfaces and Surface Integrals(LT Section 17.4) 60616.5 Surface Integrals of Vector Fields(LT Section 17.5) 622Chapter Review Exercises 636Chapter 17 FUNDAMENTAL THEOREMS OFVECTOR ANALYSIS(LT Chapter 18) 64917.1 Greens Theorem (LT Section 18.1) 64917.2 Stokes Theorem (LT Section 18.2) 66517.3 Divergence Theorem (LT Section 18.3) 678Chapter Review Exercises 69312345678910111213141516171819202122232425262728293031323334353637383940414243444546474849S 50R 511st Pass Pages1019763_FM_VOL-I.qxp 9/17/07 4:22 PM Page viiiThis page was intentionally left blankMay 18, 201110 INFINITE SERIES10.1 Sequences (LT Section 11.1)Preliminary Questions1. What is a4 for the sequence an = n2n?solution Substituting n = 4 in the expression for an givesa4 = 424 = 12.2. Which of the following sequences converge to zero?(a) n2n2+1 (b) 2n(c)_12_nsolution(a) This sequence does not converge to zero:limnn2n2+1 = limxx2x2+1 = limx11 + 1x2= 11 +0 = 1.(b) This sequence does not converge to zero: this is a geometric sequence with r = 2 > 1; hence, the sequence divergesto .(c) Recall that if |an| converges to 0, then an must also converge to zero. Here,_12_n=_12_n,which is a geometric sequence with 0 < r < 1; hence, (12)nconverges to zero. It therefore follows that (12)nconvergesto zero.3. Let an be the nth decimal approximation to2. That is, a1 = 1, a2 = 1.4, a3 = 1.41, etc. What is limnan?solution limnan =2.4. Which of the following sequences is dened recursively?(a) an =4 +n (b) bn =_4 +bn1solution(a) an can be computed directly, since it depends on n only and not on preceding terms. Therefore an is dened explicitlyand not recursively.(b) bn is computed in terms of the preceding term bn1, hence the sequence {bn} is dened recursively.5. Theorem 5 says that every convergent sequence is bounded. Determine if the following statements are true or falseand if false, give a counterexample.(a) If {an} is bounded, then it converges.(b) If {an} is not bounded, then it diverges.(c) If {an} diverges, then it is not bounded.solution(a) This statement is false. The sequence an = cos n is bounded since 1 cos n 1 for all n, but it does notconverge: since an = cos n = (1)n, the terms assume the two values 1 and 1 alternately, hence they do not approachone value.(b) By Theorem 5, a converging sequence must be bounded. Therefore, if a sequence is not bounded, it certainly doesnot converge.(c) The statement is false. The sequence an = (1)nis bounded, but it does not approach one limit.1May 18, 20112 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)Exercises1. Match each sequence with its general term:a1, a2, a3, a4, . . . General term(a) 12, 23, 34, 45, . . . (i) cos n(b) 1, 1, 1, 1, . . . (ii) n!2n(c) 1, 1, 1, 1, . . . (iii) (1)n+1(d) 12, 24, 68, 2416 . . . (iv) nn +1solution(a) The numerator of each term is the same as the index of the term, and the denominator is one more than the numerator;hence an = nn+1, n = 1, 2, 3, . . . .(b) The terms of this sequence are alternating between 1 and 1 so that the positive terms are in the even places. Sincecos n = 1 for even n and cos n = 1 for odd n, we have an = cos n, n = 1, 2, . . . .(c) The terms an are 1 for odd n and 1 for even n. Hence, an = (1)n+1, n = 1, 2, . . .(d) The numerator of each term is n!, and the denominator is 2n; hence, an = n!2n, n = 1, 2, 3, . . . .Let an = 12n 1 for n = 1, 2, 3, . . . . Write out the rst three terms of the following sequences.(a) bn = an+1 (b) cn = an+3(c) dn = a2n (d) en = 2anan+1In Exercises 312, calculate the rst four terms of the sequence, starting with n = 1.3. cn = 3nn!solution Setting n = 1, 2, 3, 4 in the formula for cn givesc1 = 311! = 31 = 3, c2 = 322! = 92,c3 = 333! = 276 = 92, c4 = 344! = 8124 = 278 .bn = (2n 1)!n!5. a1 = 2, an+1 = 2a2n3solution For n = 1, 2, 3 we have:a2 = a1+1 = 2a21 3 = 2 4 3 = 5;a3 = a2+1 = 2a22 3 = 2 25 3 = 47;a4 = a3+1 = 2a23 3 = 2 2209 3 = 4415.The rst four terms of {an} are 2, 5, 47, 4415.b1 = 1, bn = bn1+ 1bn17. bn = 5 +cos nsolution For n = 1, 2, 3, 4 we haveb1 = 5 +cos = 4;b2 = 5 +cos 2 = 6;b3 = 5 +cos 3 = 4;b4 = 5 +cos 4 = 6.The rst four terms of {bn} are 4, 6, 4, 6.cn = (1)2n+1 9. cn = 1 + 12 + 13 + + 1nsolutionc1 = 1;c2 = 1 + 12 = 32;c3 = 1 + 12 + 13 = 32 + 13 = 116 ;c4 = 1 + 12 + 13 + 14 = 116 + 14 = 2512.May 18, 2011S E C T I ON 10.1 Sequences (LT SECTION 11.1) 3an = n +(n +1) +(n +2) + +(2n)11. b1 = 2, b2 = 3, bn = 2bn1+bn2solution We need to nd b3 and b4. Setting n = 3 and n = 4 and using the given values for b1 and b2 we obtain:b3 = 2b31+b32 = 2b2+b1 = 2 3 +2 = 8;b4 = 2b41+b42 = 2b3+b2 = 2 8 +3 = 19.The rst four terms of the sequence {bn} are 2, 3, 8, 19.cn = n-place decimal approximation to e13. Find a formula for the nth term of each sequence.(a) 11, 18 , 127, . . . (b) 26, 37, 48, . . .solution(a) The denominators are the third powers of the positive integers starting with n = 1. Also, the sign of the terms isalternating with the sign of the rst term being positive. Thus,a1 = 113 = (1)1+113 ; a2 = 123 = (1)2+123 ; a3 = 133 = (1)3+133 .This rule leads to the following formula for the nth term:an = (1)n+1n3 .(b) Assuming a starting index of n = 1, we see that each numerator is one more than the index and the denominator isfour more than the numerator. Thus, the general term an isan = n +1n +5.Suppose that limnan = 4 and limnbn = 7. Determine:(a) limn(an+bn) (b) limna3n(c) limncos(bn) (d) limn(a2n2anbn)In Exercises 1526, use Theorem 1 to determine the limit of the sequence or state that the sequence diverges.15. an = 12solution We have an = f (n) where f (x) = 12; thus,limnan = limxf (x) = limx12 = 12.an = 20 4n217. bn = 5n 112n +9solution We have bn = f (n) where f (x) = 5x 112x +9; thus,limn5n 112n +9 = limx5x 112x +9 = 512.an = 4 +n 3n24n2+119. cn = 2nsolution We have cn = f (n) where f (x) = 2x; thus,limn_2n_= limx2x= limx 12x = 0.zn =_13_n21. cn = 9nsolution We have cn = f (n) where f (x) = 9x; thus,limn9n= limx9x= Thus, the sequence 9ndiverges.zn = 101/n23. an = n_n2+1solution We have an = f (n) where f (x) = x_x2+1; thus,limnn_n2+1= limxx_x2+1= limxxxx2+1x= limx1_x2+1x2= limx1_1 + 1x2= 11 +0 = 1.May 18, 20114 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)an = n_n3+125. an = ln_ 12n +29 +4n_solution We have an = f (n) where f (x) = ln_ 12x +29 +4x_; thus,limnln_ 12n +29 +4n_= limxln_ 12x +29 +4x_= ln limx_ 12x +29 +4x_= ln 3rn = ln n ln(n2+1)In Exercises 2730, use Theorem 4 to determine the limit of the sequence.27. an =_4 + 1nsolution We havelimn4 + 1n= limx4 + 1x= 4Sincex is a continuous function for x > 0, Theorem 4 tells us thatlimn_4 + 1n=_ limn4 + 1n=4 = 2an = e4n/(3n+9)29. an = cos1_ n32n3+1_solution We havelimnn32n3+1 = 12Since cos1(x) is continuous for all x, Theorem 4 tells us thatlimncos1_ n32n3+1_= cos1_ limnn32n3+1_= cos1(1/2) = 3an = tan1(en)31. Let an = nn +1. Find a number M such that:(a) |an1| 0.001 for n M.(b) |an1| 0.00001 for n M.Then use the limit denition to prove that limnan = 1.solution(a) We have|an1| =nn +1 1=n (n +1)n +1=1n +1= 1n +1.Therefore |an1| 0.001 provided 1n+1 0.001, that is, n 999. It follows that we can take M = 999.(b) By part (a), |an1| 0.00001 provided 1n+1 0.00001, that is, n 99999. It follows that we can take M = 99999.We now prove formally that limnan = 1. Using part (a), we know that|an1| = 1n +1 < ,provided n > 1

1. Thus, Let > 0 and take M = 1

1. Then, for n > M, we have|an1| = 1n +1 1

.May 18, 2011S E C T I ON 10.1 Sequences (LT SECTION 11.1) 5Thus, let > 0 and take M = 1

. Then, for n > M, we have|n20| =1n2= 1n2 1. By the Limit of Geometric Sequences,we conclude that limn_e2_n= . Thus,the given sequence diverges.May 18, 2011S E C T I ON 10.1 Sequences (LT SECTION 11.1) 7an = n2n55. yn = en+(3)n5nsolutionlimnen+(3)n5n = limn_e5_n+ limn_35_nassuming both limits on the right-hand side exist. But by the Limit of Geometric Sequences, since1 0, the interval (L , L +) contains at least one element of the sequence {an}.(b) For every > 0, the interval (L , L +) contains all but at most nitely many elements of the sequence {an}.May 18, 2011S E C T I ON 10.1 Sequences (LT SECTION 11.1) 9solution Statement (b) is equivalent to Denition 1 of the limit, since the assertion |an L| < for all n > Mmeans that L < an < L + for all n > M; that is, the interval (L , L + ) contains all the elements an except(maybe) the nite number of elements a1, a2, . . . , aM.Statement (a) is not equivalent to the assertion limnan = L. We show this, by considering the following sequence:an =1nfor odd n1 + 1nfor even nClearly for every > 0, the interval (, ) = (L , L + ) for L = 0 contains at least one element of {an}, but thesequence diverges (rather than converges to L = 0). Since the terms in the odd places converge to 0 and the terms in theeven places converge to 1. Hence, an does not approach one limit.Show that an = 12n +1 is decreasing.73. Show that an = 3n2n2+2 is increasing. Find an upper bound.solution Let f (x) = 3x2x2+2. Thenf

(x) = 6x(x2+2) 3x2 2x(x2+2)2 = 12x(x2+2)2.f

(x) > 0 for x > 0, hence f is increasing on this interval. It follows that an = f (n) is also increasing. We now showthat M = 3 is an upper bound for an, by writing:an = 3n2n2+2 3n2+6n2+2 = 3(n2+2)n2+2 = 3.That is, an 3 for all n.Show that an = 3n +1 n is decreasing.75. Give an example of a divergent sequence {an} such that limn|an| converges.solution Let an = (1)n. The sequence {an} diverges because the terms alternate between +1 and 1; however, thesequence {|an|} converges because it is a constant sequence, all of whose terms are equal to 1.Give an example of divergent sequences {an} and {bn} such that {an+bn} converges.77. Using the limit denition, prove that if {an} converges and {bn} diverges, then {an+bn} diverges.solution We will prove this result by contradiction. Suppose limnan = L1 and that {an + bn} converges to alimit L2. Now, let > 0. Because {an} converges to L1 and {an+bn} converges to L2, it follows that there exist numbersM1 and M2 such that:|anL1| M1,| (an+bn) L2| M2.Thus, for n > M = max{M1, M2},|anL1| M

. Now, let M = M

1. Then, whenever n > M, n + 1 > M + 1 = M

. Thus,for n > M,|bnL| = |an+1L| < .Hence, {bn} converges to L.Let {an} be a sequence such that limn|an| exists and is nonzero. Show that limnan exists if and only if thereexists an integer M such that the sign of an does not change for n > M.83. Proceed as in Example 12 to show that the sequence3,_33,_3_33, . . . is increasing and bounded above byM = 3. Then prove that the limit exists and nd its value.solution This sequence is dened recursively by the formula:an+1 =_3an, a1 =3.Consider the following inequalities:a2 =3a1 =_33 >3 = a1 a2 > a1;a3 =3a2 >3a1 = a2 a3 > a2;a4 =3a3 >3a2 = a3 a4 > a3.In general, if we assume that ak > ak1, thenak+1 =_3ak >_3ak1 = ak.Hence, by mathematical induction, an+1 > an for all n; that is, the sequence {an} is increasing.Because an+1 =3an, it follows that an 0 for all n. Now, a1 =3 < 3. If ak 3, thenak+1 =_3ak 3 3 = 3.Thus, by mathematical induction, an 3 for all n.Since {an} is increasing and bounded, it follows by the Theorem on Bounded Monotonic Sequences that this sequenceis converging. Denote the limit by L = limnan. Using Exercise 81, it follows thatL = limnan+1 = limn_3an =_3 limnan =3L.Thus, L2= 3L, so L = 0 or L = 3. Because the sequence is increasing, we have an a1 =3 for all n. Hence, thelimit also satises L 3. We conclude that the appropriate solution is L = 3; that is, limnan = 3.Let {an} be the sequence dened recursively bya0 = 0, an+1 =_2 +anThus, a1 =2, a2 =_2 +2, a3 =_2 +_2 +2, . . . .(a) Show that if an < 2, then an+1 < 2. Conclude by induction that an < 2 for all n.(b) Show that if an < 2, then an an+1. Conclude by induction that {an} is increasing.(c) Use (a) and (b) to conclude that L = limnan exists. Then compute L by showing that L =2 +L.Further Insights and Challenges85. Show that limnnn! = . Hint: Verify that n! (n/2)n/2by observing that half of the factors of n! are greaterthan or equal to n/2.solution We show that n! _n2_n/2. For n 4 even, we have:n! = 1 n2. ,, .n2 factors_n2 +1_ n. ,, .n2 factors_n2 +1_ n. ,, .n2 factors.Since each one of the n2 factors is greater than n2, we have:n! _n2 +1_ n. ,, .n2 factors n2 n2. ,, .n2 factors=_n2_n/2.For n 3 odd, we have:n! = 1 n 12. ,, .n12 factors n +12 n. ,, .n+12 factors n +12 n. ,, .n+12 factors.May 18, 2011S E C T I ON 10.1 Sequences (LT SECTION 11.1) 11Since each one of the n+12 factors is greater than n2, we have:n! n +12 n. ,, .n+12 factors n2 n2. ,, .n+12 factors=_n2_(n+1)/2=_n2_n/2_n2 _n2_n/2.In either case we have n! _n2_n/2. Thus,nn! _n2.Since limn_n2 = , it follows that limnnn! = . Thus, the sequence an = nn! diverges.Let bn =nn!n.(a) Show that ln bn = 1nn

k=1ln kn.(b) Show that ln bn converges to_ 10ln x dx, and conclude that bn e1.87. Given positive numbers a1 < b1, dene two sequences recursively byan+1 =_anbn, bn+1 = an+bn2(a) Show that an bn for all n (Figure 13).(b) Show that {an} is increasing and {bn} is decreasing.(c) Show that bn+1an+1 bnan2 .(d) Prove that both {an} and {bn} converge and have the same limit. This limit, denoted AGM(a1, b1), is called thearithmetic-geometric mean of a1 and b1.(e) Estimate AGM(1,2) to three decimal places.xan an+1 bn+1 bnGeometricmeanAGM(a1, b1)ArithmeticmeanFIGURE 13solution(a) Examine the following:bn+1an+1 = an+bn2 _anbn = an+bn2anbn2 =_an_22anbn+_bn_22=_anbn_22 0.We conclude that bn+1 an+1 for all n > 1. By the given information b1 > a1; hence, bn an for all n.(b) By part (a), bn an for all n, soan+1 =_anbn an an =_a2n = anfor all n. Hence, the sequence {an} is increasing. Moreover, since an bn for all n,bn+1 = an+bn2 bn+bn2 = 2bn2 = bnfor all n; that is, the sequence {bn} is decreasing.(c) Since {an} is increasing, an+1 an. Thus,bn+1an+1 bn+1an = an+bn2 an = an+bn2an2 = bnan2 .Now, by part (a), an bn for all n. By part (b), {bn} is decreasing. Hence bn b1 for all n. Combining the two inequalitieswe conclude that an b1 for all n. That is, the sequence {an} is increasing and bounded (0 an b1). By the Theoremon Bounded Monotonic Sequences we conclude that {an} converges. Similarly, since {an} is increasing, an a1 for alln. We combine this inequality with bn an to conclude that bn a1 for all n. Thus, {bn} is decreasing and bounded(a1 bn b1); hence this sequence converges.To show that {an} and {bn} converge to the same limit, note thatbnan bn1an12 bn2an222 b1a12n1 .Thus,limn(bnan) = (b1a1) limn12n1 = 0.May 18, 201112 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)(d) We havean+1 =_anbn, a1 = 1; bn+1 = an+bn2 , b1 =2Computing the values of an and bn until the rst three decimal digits are equal in successive terms, we obtain:a2 =_a1b1 =_1 2 = 1.1892b2 = a1+b12 = 1 +22 = 1.2071a3 =_a2b2 =1.1892 1.2071 = 1.1981b3 = a2+b22 = 1.1892 1.20712 = 1.1981a4 =_a3b3 = 1.1981b4 = a3+b32 = 1.1981Thus,AGM_1,2_ 1.198.Let cn = 1n+ 1n +1 + 1n +2 + + 12n.(a) Calculate c1, c2, c3, c4.(b) Use a comparison of rectangles with the area under y = x1over the interval [n, 2n] to prove that_ 2nndxx+ 12n cn _ 2nndxx+ 1n(c) Use the Squeeze Theorem to determine limncn.89. Let an = Hnln n, where Hn is the nth harmonic numberHn = 1 + 12 + 13 + + 1n(a) Show that an 0 for n 1. Hint: Show that Hn _ n+11dxx.(b) Show that {an} is decreasing by interpreting anan+1 as an area.(c) Prove that limnan exists.This limit, denoted , is known as Eulers Constant. It appears in many areas of mathematics, including analysis andnumber theory, and has been calculated to more than 100 million decimal places, but it is still not known whether is anirrational number. The rst 10 digits are 0.5772156649.solution(a) Since the function y = 1x is decreasing, the left endpoint approximation to the integral_n+11 dxx is greater than thisintegral; that is,1 1 + 12 1 + 13 1 + + 1n 1 _ n+11dxxorHn _ n+11dxx.11yx2 3 n n + 11/n12 13Moreover, since the function y = 1x is positive for x > 0, we have:_ n+11dxx_ n1dxx.Thus,Hn _ n1dxx= ln xn1 = ln n ln 1 = ln n,May 18, 2011S E C T I ON 10.2 Summing an Infinite Series (LT SECTION 11.2) 13andan = Hnln n 0 for all n 1.(b) To show that {an} is decreasing, we consider the difference anan+1:anan+1 = Hnln n _Hn+1ln(n +1)_= HnHn+1+ln(n +1) ln n= 1 + 12 + + 1n_1 + 12 + + 1n+ 1n +1_+ln(n +1) ln n= 1n +1 +ln(n +1) ln n.Now, ln(n +1) ln n =_n+1ndxx , whereas 1n+1 is the right endpoint approximation to the integral_n+1ndxx . Recallingy = 1x is decreasing, it follows that_ n+1ndxx 1n +1yxn n + 1y = 1x1n + 1soanan+1 0.(c) By parts (a) and (b), {an} is decreasing and 0 is a lower bound for this sequence. Hence 0 an a1 for all n. Amonotonic and bounded sequence is convergent, so limnan exists.10.2 Summing an Infinite Series (LT Section 11.2)Preliminary Questions1. What role do partial sums play in dening the sum of an innite series?solution The sumof an innite series is dened as the limit of the sequence of partial sums. If the limit of this sequencedoes not exist, the series is said to diverge.2. What is the sum of the following innite series?14 + 18 + 116 + 132 + 164 + solution This is a geometric series with c = 14 and r = 12. The sum of the series is therefore141 12=1412= 12.3. What happens if you apply the formula for the sum of a geometric series to the following series? Is the formula valid?1 +3 +32+33+34+ solution This is a geometric series with c = 1 and r = 3. Applying the formula for the sum of a geometric seriesthen gives

n=03n= 11 3 = 12.Clearly, this is not valid: a series with all positive terms cannot have a negative sum. The formula is not valid in this casebecause a geometric series with r = 3 diverges.May 18, 201114 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)4. Arvind asserts that

n=11n2 = 0 because 1n2 tends to zero. Is this valid reasoning?solution Arvinds reasoning is not valid. Though the terms in the series do tend to zero, the general term in thesequence of partial sums,Sn = 1 + 122 + 132 + + 1n2,is clearly larger than 1. The sum of the series therefore cannot be zero.5. Colleen claims that

n=11nconverges becauselimn1n= 0Is this valid reasoning?solution Colleens reasoning is not valid. Although the general term of a convergent series must tend to zero, a serieswhose general term tends to zero need not converge. In the case of

n=11n, the series diverges even though its generalterm tends to zero.6. Find an N such that SN > 25 for the series

n=12.solution The Nth partial sum of the series is:SN =N

n=12 = 2 + +2. ,, .N= 2N.7. Does there exist an N such that SN > 25 for the series

n=12n? Explain.solution The series

n=12nis a convergent geometric series with the common ratio r = 12. The sum of the series is:S =121 12= 1.Notice that the sequence of partial sums {SN} is increasing and converges to 1; therefore SN 1 for all N. Thus, theredoes not exist an N such that SN > 25.8. Give an example of a divergent innite series whose general term tends to zero.solution Consider the series

n=11n910. The general term tends to zero, since limn1n910= 0. However, the Nth partialsum satises the following inequality:SN = 11 910+ 12 910+ + 1N910 NN910= N1 910 = N110 .That is, SN N110 for all N. Since limNN110 = , the sequence of partial sums Sn diverges; hence, the series

n=11n910diverges.Exercises1. Find a formula for the general term an (not the partial sum) of the innite series.(a) 13 + 19 + 127 + 181 + (b) 11 + 52 + 254 + 1258 + (c) 11 222 1 + 333 2 1 444 3 2 1 + (d) 212+1 + 122+1 + 232+1 + 142+1 + solution(a) The denominators of the terms are powers of 3, starting with the rst power. Hence, the general term is:an = 13n.May 18, 2011S E C T I ON 10.2 Summing an Infinite Series (LT SECTION 11.2) 15(b) The numerators are powers of 5, and the denominators are the same powers of 2. The rst term is a1 = 1 so,an =_52_n1.(c) The general term of this series is,an = (1)n+1nnn!.(d) Notice that the numerators of an equal 2 for odd values of n and 1 for even values of n. Thus,an =2n2+1 odd n1n2+1 even nThe formula can also be rewritten as follows:an = 1 + (1)n+1+12n2+1 .Write in summation notation:(a) 1 + 14 + 19 + 116 + (b) 19 + 116 + 125 + 136 + (c) 1 13 + 15 17 + (d) 1259 + 62516 + 312525 + 15,62536 + In Exercises 36, compute the partial sums S2, S4, and S6.3. 1 + 122 + 132 + 142 + solutionS2 = 1 + 122 = 54;S4 = 1 + 122 + 132 + 142 = 205144;S6 = 1 + 122 + 132 + 142 + 152 + 162 = 53693600.

k=1(1)kk15. 11 2 + 12 3 + 13 4 + solutionS2 = 11 2 + 12 3 = 12 + 16 = 46 = 23;S4 = S2+a3+a4 = 23 + 13 4 + 14 5 = 23 + 112 + 120 = 45;S6 = S4+a5+a6 = 45 + 15 6 + 16 7 = 45 + 130 + 142 = 67.

j=11j!7. The series S = 1 +_15_+_15_2+_15_3+ converges to 54. Calculate SN for N = 1, 2, . . . until you nd an SNthat approximates 54 with an error less than 0.0001.solutionS1 = 1S2 = 1 + 15 = 65 = 1.2S3 = 1 + 15 + 125 = 3125 = 1.24S3 = 1 + 15 + 125 + 1125 = 156125 = 1.248S4 = 1 + 15 + 125 + 1125 + 1625 = 781625 = 1.2496S5 = 1 + 15 + 125 + 1125 + 1625 + 13125 = 39063125 = 1.24992Note that1.25 S5 = 1.25 1.24992 = 0.00008 < 0.0001May 18, 201116 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)The series S = 10! 11! + 12! 13! + is known to converge to e1(recall that 0! = 1). Calculate SN forN = 1, 2, . . . until you nd an SN that approximates e1with an error less than 0.001.In Exercises 9 and 10, use a computer algebra system to compute S10, S100, S500, and S1000 for the series. Do thesevalues suggest convergence to the given value?9. 34 = 12 3 4 14 5 6 + 16 7 8 18 9 10 + solution Writean = (1)n+12n (2n +1) (2n +2)ThenSN =N

i=1anComputing, we nd 34 0.0353981635S10 0.03535167962S100 0.03539810274S500 0.03539816290S1000 0.03539816334It appears that SN 34 .490 = 1 + 124 + 134 + 144 + 11. Calculate S3, S4, and S5 and then nd the sum of the telescoping seriesS =

n=1_ 1n +1 1n +2_solutionS3 =_12 13_+_13 14_+_14 15_= 12 15 = 310;S4 = S3+_15 16_= 12 16 = 13;S5 = S4+_16 17_= 12 17 = 514.The general term in the sequence of partial sums isSN =_12 13_+_13 14_+_14 15_+ +_ 1N +1 1N +2_= 12 1N +2;thus,S = limNSN = limN_12 1N +2_= 12.The sum of the telescoping series is therefore 12.Write

n=31n(n 1)as a telescoping series and nd its sum.13. Calculate S3, S4, and S5 and then nd the sum S =

n=114n21 using the identity14n21 = 12_ 12n 1 12n +1_solutionS3 = 12_11 13_+ 12_13 15_+ 12_15 17_= 12_1 17_= 37;S4 = S3+ 12_17 19_= 12_1 19_= 49;S5 = S4+ 12_19 111_= 12_1 111_= 511.May 18, 2011S E C T I ON 10.2 Summing an Infinite Series (LT SECTION 11.2) 17The general term in the sequence of partial sums isSN = 12_11 13_+ 12_13 15_+ 12_15 17_+ + 12_ 12N 1 12N +1_= 12_1 12N +1_;thus,S = limNSN = limN12_1 12N +1_= 12.Use partial fractions to rewrite

n=11n(n +3)as a telescoping series and nd its sum.15. Find the sum of 11 3 + 13 5 + 15 7 + .solution We may write this sum as

n=11(2n 1)(2n +1)=

n=112_ 12n 1 12n +1_.The general term in the sequence of partial sums isSN = 12_11 13_+ 12_13 15_+ 12_15 17_+ + 12_ 12N 1 12N +1_= 12_1 12N +1_;thus,limNSN = limN12_1 12N +1_= 12,and

n=11(2n 1)(2n +1)= 12.Find a formula for the partial sum SN of

n=1(1)n1and show that the series diverges.In Exercises 1722, use Theorem 3 to prove that the following series diverge.17.

n=1n10n +12solution The general term, n10n +12, has limitlimnn10n +12 = limn110 +(12/n)= 110Since the general term does not tend to zero, the series diverges.

n=1n_n2+119. 01 12 + 23 34 + solution The general term an = (1)n1 n1n does not tend to zero. In fact, because limn n1n = 1, limnandoes not exist. By Theorem 3, we conclude that the given series diverges.

n=1(1)nn221. cos 12 +cos 13 +cos 14 + solution The general term an = cos 1n+1 tends to 1, not zero. By Theorem 3, we conclude that the given seriesdiverges.

n=0__4n2+1 n_In Exercises 2336, use the formula for the sum of a geometric series to nd the sum or state that the series diverges.23. 11 + 18 + 182 + solution This is a geometric series with c = 1 and r = 18, so its sum is11 18= 17/8 = 874353 + 4454 + 4555 + 25.

n=3_ 311_nsolution Rewrite this series as

n=3_113_nThis is a geometric series with r = 113 > 1, so it is divergent.May 18, 201118 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=27 (3)n5n27.

n=4_49_nsolution This is a geometric series with c = 1 and r = 49, starting at n = 4. Its sum is thuscr41 r= cr4r5 = 14494 + 4595= 959 44+45 = 59,0493328

n=0_e_n 29.

n=1ensolution Rewrite the series as

n=1_1e_nto recognize it as a geometric series with c = 1e and r = 1e. Thus,

n=1en=1e1 1e= 1e 1.

n=2e32n31.

n=08 +2n5nsolution Rewrite the series as

n=085n +

n=02n5n =

n=08 _15_n+

n=0_25_n,which is a sum of two geometric series. The rst series has c = 8_15_0= 8 and r = 15; the second has c =_25_0= 1and r = 25. Thus,

n=08 _15_n= 81 15= 845= 10,

n=0_25_n= 11 25= 135= 53,and

n=08 +2n5n = 10 + 53 = 353 .

n=03(2)n5n8n33. 5 54 + 542 543 + solution This is a geometric series with c = 5 and r = 14. Thus,

n=05 _14_n= 51 _14_ = 51 + 14= 554= 4.237 + 2472 + 2573 + 2674 + 35. 78 4964 + 343512 24014096 + solution This is a geometric series with c = 78 and r = 78. Thus,

n=078 _78_n=781 _78_ =78158= 715.May 18, 2011S E C T I ON 10.2 Summing an Infinite Series (LT SECTION 11.2) 19259 + 53 +1 + 35 + 925 + 27125 + 37. Which of the following are not geometric series?(a)

n=07n29n (b)

n=31n4(c)

n=0n22n (d)

n=5nsolution(a)

n=07n29n =

n=0_ 729_n: this is a geometric series with common ratio r = 729.(b) The ratio between two successive terms isan+1an=1(n+1)41n4= n4(n +1)4 =_ nn +1_4.This ratio is not constant since it depends on n. Hence, the series

n=31n4 is not a geometric series.(c) The ratio between two successive terms isan+1an=(n+1)22n+1n22n= (n +1)2n2 2n2n+1 =_1 + 1n_2 12.This ratio is not constant since it depends on n. Hence, the series

n=0n22n is not a geometric series.(d)

n=5n=

n=5_1_n: this is a geometric series with common ratio r = 1.Use the method of Example 8 to show that

k=11k1/3 diverges.39. Prove that if

n=1an converges and

n=1bn diverges, then

n=1(an+bn) diverges. Hint: If not, derive a contradictionby writing

n=1bn =

n=1(an+bn)

n=1ansolution Suppose to the contrary that

n=1 an converges,

n=1 bn diverges, but

n=1(an+bn) converges. Thenby the Linearity of Innite Series, we have

n=1bn =

n=1(an+bn)

n=1anso that

n=1 bn converges, a contradiction.Prove the divergence of

n=09n+2n5n .41. Give a counterexample to show that each of the following statements is false.(a) If the general term an tends to zero, then

n=1an = 0.(b) The Nth partial sum of the innite series dened by {an} is aN.(c) If an tends to zero, then

n=1an converges.(d) If an tends to L, then

n=1an = L.solution(a) Let an = 2n. Then limnan = 0, but an is a geometric series with c = 20= 1 and r = 1/2, so its sum is11 (1/2)= 2.(b) Let an = 1. Then the nthpartial sum is a1+a2+ +an = n while an = 1.May 18, 201120 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)(c) Let an = 1n. An example in the text shows that while an tends to zero, the sum

n=1an does not converge.(d) Let an = 1. Then clearly an tends to L = 1, while the series

n=1 an obviously diverges.Suppose that S =

n=1an is an innite series with partial sum SN = 5 2N2.(a) What are the values of10

n=1an and16

n=5an?(b) What is the value of a3?(c) Find a general formula for an.(d) Find the sum

n=1an.43. Compute the total area of the (innitely many) triangles in Figure 4.18141211612yx1FIGURE 4solution The area of a triangle with base B and height H is A = 12BH. Because all of the triangles in Figure 4 haveheight 12, the area of each triangle equals one-quarter of the base. Now, for n 0, the nth triangle has a base whichextends from x = 12n+1 to x = 12n. Thus,B = 12n 12n+1 = 12n+1 and A = 14B = 12n+3.The total area of the triangles is then given by the geometric series

n=012n+3 =

n=018_12_n=181 12= 14.The winner of a lottery receives m dollars at the end of each year for N years. The present value (PV) of this prizein todays dollars is PV =N

i=1m(1 +r)i, where r is the interest rate. Calculate PV if m = $50,000, r = 0.06, andN = 20. What is PV if N = ?45. Find the total length of the innite zigzag path in Figure 5 (each zag occurs at an angle of 4 ).1/4 /4FIGURE 5solution Because the angle at the lower left in Figure 5 has measure 4 and each zag in the path occurs at an angle of4 , every triangle in the gure is an isosceles right triangle. Accordingly, the length of each new segment in the path is12 times the length of the previous segment. Since the rst segment has length 1, the total length of the path is

n=0_ 12_n= 11 12=22 1 = 2 +2.Evaluate

n=11n(n +1)(n +2). Hint: Find constants A, B, and C such that1n(n +1)(n +2)= An+ Bn +1 + Cn +247. Show that if a is a positive integer, then

n=11n(n +a)= 1a_1 + 12 + + 1a_solution By partial fraction decomposition1n (n +a)= An+ Bn +a;clearing the denominators gives1 = A(n +a) +Bn.Setting n = 0 then yields A = 1a, while setting n = a yields B = 1a. Thus,1n (n +a)=1an1an +a= 1a_1n 1n +a_,May 18, 2011S E C T I ON 10.2 Summing an Infinite Series (LT SECTION 11.2) 21and

n=11n(n +a)=

n=11a_1n 1n +a_.For N > a, the Nth partial sum isSN = 1a_1 + 12 + 13 + + 1a_ 1a_ 1N +1 + 1N +2 + 1N +3 + + 1N +a_.Thus,

n=11n(n +a)= limNSN = 1a_1 + 12 + 13 + + 1a_.Aball dropped from a height of 10 ft begins to bounce. Each time it strikes the ground, it returns to two-thirds ofits previous height. What is the total distance traveled by the ball if it bounces innitely many times?49. Let {bn} be a sequence and let an = bnbn1. Show that

n=1an converges if and only if limnbn exists.solution Let an = bnbn1. The general term in the sequence of partial sums for the series

n=1an is thenSN = (b1b0) +(b2b1) +(b3b2) + +(bN bN1) = bN b0.Now, if limNbN exists, then so does limNSN and

n=1an converges. On the other hand, if

n=1an converges, thenlimNSN exists, which implies that limNbN also exists. Thus,

n=1an converges if and only if limnbn exists.Assumptions Matter Show, by giving counterexamples, that the assertions of Theorem 1 are not valid if theseries

n=0an and

n=0bn are not convergent.Further Insights and ChallengesExercises 5153 use the formula1 +r +r2+ +rN1= 1 rN1 r751. Professor GeorgeAndrews of Pennsylvania State University observed that we canuse Eq. (7) to calculate the derivativeof f (x) = xN(for N 0). Assume that a = 0 and let x = ra. Show thatf

(a) = limxaxNaNx a= aN1limr1rN1r 1and evaluate the limit.solution According to the denition of derivative of f (x) at x = af

(a) = limxaxNaNx a.Now, let x = ra. Then x a if and only if r 1, andf

(a) = limxaxNaNx a= limr1(ra)NaNra a= limr1aN_rN1_a (r 1)= aN1limr1rN1r 1 .By Eq. (7) for a geometric sum,1 rN1 r= rN1r 1 = 1 +r +r2+ +rN1,solimr1rN1r 1 = limr1_1 +r +r2+ +rN1_= 1 +1 +12+ +1N1= N.Therefore, f

(a) = aN1 N = NaN1Pierre de Fermat used geometric series to compute the area under the graph of f (x) = xNover [0, A]. For0 < r < 1, let F(r) be the sum of the areas of the innitely many right-endpoint rectangles with endpoints Arn, asin Figure 6. As r tends to 1, the rectangles become narrower and F(r) tends to the area under the graph.(a) Show that F(r) = AN+1 1 r1 rN+1.(b) Use Eq. (7) to evaluate_ A0xNdx = limr1F(r).53. Verify the GregoryLeibniz formula as follows.(a) Set r = x2in Eq. (7) and rearrange to show that11 +x2 = 1 x2+x4 +(1)N1x2N2+ (1)Nx2N1 +x2May 18, 201122 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)(b) Show, by integrating over [0, 1], that4 = 1 13 + 15 17 + + (1)N12N 1 +(1)N_ 10x2Ndx1 +x2(c) Use the Comparison Theorem for integrals to prove that0 _ 10x2Ndx1 +x2 12N +1Hint: Observe that the integrand is x2N.(d) Prove that4 = 1 13 + 15 17 + 19 Hint: Use (b) and (c) to show that the partial sums SN of satisfy SN 4 12N+1, and thereby conclude thatlimNSN = 4 .solution(a) Start with Eq. (7), and substitute x2for r:1 +r +r2+ +rN1= 1 rN1 r1 x2+x4+ +(1)N1x2N2= 1 (1)Nx2N1 (x2)1 x2+x4+ +(1)N1x2N2= 11 +x2 (1)Nx2N1 +x211 +x2 = 1 x2+x4+ +(1)N1x2N2+ (1)Nx2N1 +x2(b) The integrals of both sides must be equal. Now,_ 1011 +x2 dx = tan1x10= tan11 tan10 = 4while_ 10_1 x2+x4+ +(1)N1x2N2+ (1)Nx2N1 +x2_ dx=_x 13x3+ 15x5+ +(1)N1 12N 1x2N1_+(1)N_ 10x2Ndx1 +x2= 1 13 + 15 + +(1)N1 12N 1 +(1)N_ 10x2Ndx1 +x2(c) Note that for x [0, 1], we have 1 +x2 1, so that0 x2N1 +x2 x2NBy the Comparison Theorem for integrals, we then see that0 _ 10x2Ndx1 +x2 _ 10x2Ndx = 12N +1x2N+110= 12N +1(d) Writean = (1)n 12n 1, n 1and let SN be the partial sums. ThenSN 4 =(1)N_ 10x2Ndx1 +x2=_ 10x2Ndx1 +x2 12N +1May 18, 2011S E C T I ON 10.3 Convergence of Series with Positive Terms (LT SECTION 11.3) 23Thus limNSN = 4 so that4 = 1 13 + 15 17 + 19 . . .Cantors Disappearing Table (following Larry Knop of Hamilton College) Take a table of length L (Figure7). At stage 1, remove the section of length L/4 centered at the midpoint. Two sections remain, each with length lessthan L/2. At stage 2, remove sections of length L/42from each of these two sections (this stage removes L/8 of thetable). Now four sections remain, each of length less than L/4. At stage 3, remove the four central sections of lengthL/43, etc.(a) Show that at the Nth stage, each remaining section has length less than L/2Nand that the total amount of tableremoved isL_14 + 18 + 116 + + 12N+1_(b) Show that in the limit as N , precisely one-half of the table remains.This result is curious, because there are no nonzero intervals of table left (at each stage, the remaining sections havea length less than L/2N). So the table has disappeared. However, we can place any object longer than L/4 on thetable. It will not fall through because it will not t through any of the removed sections.55. The Koch snowake (described in 1904 by Swedish mathematician Helge von Koch) is an innitely jagged fractalcurve obtained as a limit of polygonal curves (it is continuous but has no tangent line at any point). Begin with anequilateral triangle (stage 0) and produce stage 1 by replacing each edge with four edges of one-third the length, arrangedas in Figure 8. Continue the process: At the nth stage, replace each edge with four edges of one-third the length.(a) Show that the perimeter Pn of the polygon at the nth stage satises Pn = 43Pn1. Prove that limnPn = . Thesnowake has innite length.(b) Let A0 be the area of the original equilateral triangle. Show that (3)4n1new triangles are added at the nth stage,each with area A0/9n(for n 1). Show that the total area of the Koch snowake is 85A0.Stage 1 Stage 3 Stage 2FIGURE 8solution(a) Each edge of the polygon at the (n 1)st stage is replaced by four edges of one-third the length; hence the perimeterof the polygon at the nth stage is 43 times the perimeter of the polygon at the (n 1)th stage. That is, Pn = 43Pn1. Thus,P1 = 43P0; P2 = 43P1 =_43_2P0, P3 = 43P2 =_43_3P0,and, in general, Pn =_43_nP0. As n , it follows thatlimnPn = P0 limn_43_n= .(b) When each edge is replaced by four edges of one-third the length, one new triangle is created. At the (n 1)st stage,there are 3 4n1edges in the snowake, so 3 4n1new triangles are generated at the nth stage. Because the area of anequilateral triangle is proportional to the square of its side length and the side length for each new triangle is one-thirdthe side length of triangles from the previous stage, it follows that the area of the triangles added at each stage is reducedby a factor of 19 from the area of the triangles added at the previous stage. Thus, each triangle added at the nth stage hasan area of A0/9n. This means that the nth stage contributes3 4n1 A09n = 34A0_49_nto the area of the snowake. The total area is thereforeA = A0+ 34A0

n=1_49_n= A0+ 34A0491 49= A0+ 34A0 45 = 85A0.10.3 Convergence of Series with Positive Terms (LT Section 11.3)Preliminary Questions1. Let S =

n=1an. If the partial sums SN are increasing, then (choose the correct conclusion):(a) {an} is an increasing sequence.(b) {an} is a positive sequence.solution The correct response is (b). Recall that SN = a1+a2+a3+ +aN; thus, SN SN1 = aN. If SN isincreasing, then SN SN1 0. It then follows that aN 0; that is, {an} is a positive sequence.2. What are the hypotheses of the Integral Test?solution The hypotheses for the Integral Test are: Afunction f (x) such that an = f (n) must be positive, decreasing,and continuous for x 1.May 18, 201124 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)3. Which test would you use to determine whether

n=1n3.2converges?solution Because n3.2= 1n3.2, we see that the indicated series is a p-series with p = 3.2 > 1. Therefore, the seriesconverges.4. Which test would you use to determine whether

n=112n+nconverges?solution Because12n+n 1, f

(x) is negative for x > 1; hence, f is decreasing for x 2. To compute the improper integral,we make the substitution u = ln x, du = 1xdx. We obtain:_ 21x(ln x)2 dx = limR_ R21x(ln x)2 dx = limR_ ln Rln 2duu2= limR_ 1ln R 1ln 2_= 1ln 2.The integral converges; hence, the series

n=21n(ln n)2 also converges.May 18, 201126 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=1ln nn213.

n=112ln nsolution Note that2ln n= (eln 2)ln n= (eln n)ln 2= nln 2.Thus,

n=112ln n =

n=11nln 2.Now, let f (x) = 1xln 2. This function is positive, continuous and decreasing on the interval x 1; therefore, the IntegralTest applies. Moreover,_ 1dxxln 2 = limR_ R1dxxln 2 = 11 ln 2 limR(R1ln 21) = ,because 1 ln 2 > 0. The integral diverges; hence, the series

n=112ln n also diverges.

n=113ln n15. Show that

n=11n3+8nconverges by using the Comparison Test with

n=1n3.solution We compare the series with the p-series

n=1n3. For n 1,1n3+8n 1n3.Since

n=11n3 converges (it is a p-series with p = 3 > 1), the series

n=11n3+8nalso converges by the Comparison Test.Show that

n=21_n23diverges by comparing with

n=2n1.17. Let S =

n=11n +n. Verify that for n 1,1n +n 1n,1n +n 1nCan either inequality be used to show that S diverges? Show that 1n +n 12nand conclude that S diverges.solution For n 1, n +n n and n +n n. Taking the reciprocal of each of these inequalities yields1n +n 1nand 1n +n 1n.These inequalities indicate that the series

n=11n +nis smaller than both

n=11nand

n=11n; however,

n=11nand

n=11nboth diverge so neither inequality allows us to show that S diverges.On the other hand, for n 1, n n, so 2n n +n and1n +n 12n.The series

n=112n= 2

n=11ndiverges, since the harmonic series diverges. The Comparison Test then lets us concludethat the larger series

n=11n +nalso diverges.Which of the following inequalities can be used to study the convergence of

n=21n2+n? Explain.1n2+n 1n,1n2+n 1n2May 18, 2011S E C T I ON 10.3 Convergence of Series with Positive Terms (LT SECTION 11.3) 27In Exercises 1930, use the Comparison Test to determine whether the innite series is convergent.19.

n=11n2nsolution We compare with the geometric series

n=1_12_n. For n 1,1n2n 12n =_12_n.Since

n=1_12_nconverges (it is a geometric series with r = 12), we conclude by the Comparison Test that

n=11n2n alsoconverges.

n=1n3n5+4n +121.

n=11n1/3+2nsolution For n 1,1n1/3+2n 12nThe series

n=112n is a geometric series with r = 12, so it converges. By the Comparison test, so does

n=11n1/3+2n.

n=11_n3+2n 123.

m=14m! +4msolution For m 1,4m! +4m 44m =_14_m1.The series

m=1_14_m1is a geometric series with r = 14, so it converges. By the Comparison Test we can thereforeconclude that the series

m=14m! +4m also converges.

n=4nn 325.

k=1sin2kk2solution For k 1, 0 sin2k 1, so0 sin2kk2 1k2.The series

k=11k2 is a p-series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude thatthe series

k=1sin2kk2 also converges.

k=2k1/3k5/4k27.

n=123n+3nsolution Since 3n> 0 for all n,23n+3n 23n = 2_13_n.The series

n=12_13_nis a geometric series with r = 13, so it converges. By the Comparison Theorem we can thereforeconclude that the series

n=123n+3n also converges.May 18, 201128 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

k=12k2 29.

n=11(n +1)!solution Note that for n 2,(n +1)! = 1 2 3 n (n +1). ,, .n factors 2nso that

n=11(n +1)! = 1 +

n=21(n +1)! 1 +

n=212nBut

n=212n is a geometric series with ratio r = 12, so it converges. By the comparison test,

n=11(n +1)! converges aswell.

n=1n!n3Exercise 3136: For all a > 0 and b > 1, the inequalitiesln n na, na< bnare true for n sufciently large (this can be proved using LHopitals Rule). Use this, together with the ComparisonTheorem, to determine whether the series converges or diverges.31.

n=1ln nn3solution For n sufciently large (say n = k, although in this case n = 1 sufces), we have ln n n, so that

n=kln nn3

n=knn3 =

n=k1n2This is a p-series with p = 2 > 1, so it converges. Thus

n=kln nn3 also converges; adding back in the nite number ofterms for 1 n k does not affect this result.

m=21ln m33.

n=1(ln n)100n1.1solution Choose N so that ln n n0.0005for n N. Then also for n > N, (ln n)100 (n0.0005)100= n0.05. Then

n=N(ln n)100n1.1

n=Nn0.05n1.1 =

n=N1n1.05But

n=N1n1.05 is a p-series with p = 1.05 > 1, so is convergent. It follows that n=N(ln n)100n1.1 is also convergent;adding back in the nite number of terms for n = 1, 2, . . . , N 1 shows that

n=1(ln n)100n1.1 converges as well.

n=11(ln n)1035.

n=1n3nsolution Choose N such that n 2nfor n N. Then

n=Nn3n

n=N_23_nThe latter sum is a geometric series with r = 23 < 1, so it converges. Thus the series on the left converges as well. Addingback in the nite number of terms for n < N shows that

n=1n3n converges.

n=1n52n37. Show that

n=1sin 1n2 converges. Hint: Use the inequality sin x x for x 0.solution For n 1,0 1n2 1 < ;May 18, 2011S E C T I ON 10.3 Convergence of Series with Positive Terms (LT SECTION 11.3) 29therefore, sin 1n2 > 0 for n 1. Moreover, for n 1,sin 1n2 1n2.The series

n=11n2 is a p-series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude thatthe series

n=1sin 1n2 also converges.Does

n=2sin(1/n)ln nconverge?In Exercises 3948, use the Limit Comparison Test to prove convergence or divergence of the innite series.39.

n=2n2n41solution Let an = n2n41. For large n, n2n41 n2n4 = 1n2, so we apply the Limit Comparison Test with bn = 1n2.We ndL = limnanbn= limnn2n411n2= limnn4n41 = 1.The series

n=11n2 is a p-series with p = 2 > 1, so it converges; hence,

n=21n2 also converges. Because L exists, by theLimit Comparison Test we can conclude that the series

n=2n2n41 converges.

n=21n2n41.

n=2n_n3+1solution Let an = n_n3+1. For large n, n_n3+1 nn3 = 1n, so we apply the Limit Comparison test withbn = 1n. We ndL = limnanbn= limnnn3+11n= limnn3_n3+1= 1.The series

n=11nis a p-series with p = 12 < 1, so it diverges; hence,

n=21nalso diverges. Because L > 0, by theLimit Comparison Test we can conclude that the series

n=2n_n3+1diverges.

n=2n3_n7+2n2+143.

n=33n +5n(n 1)(n 2)solution Let an = 3n +5n(n 1)(n 2). For large n, 3n +5n(n 1)(n 2) 3nn3 = 3n2, so we apply the Limit ComparisonTest with bn = 1n2. We ndL = limnanbn= limn3n+5n(n+1)(n+2)1n2= limn3n3+5n2n(n +1)(n +2)= 3.The series

n=11n2 is a p-series with p = 2 > 1, so it converges; hence, the series

n=31n2 also converges. Because Lexists, by the Limit Comparison Test we can conclude that the series

n=33n +5n(n 1)(n 2)converges.May 18, 201130 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=1en+ne2nn245.

n=11n +ln nsolution Letan = 1n +ln nFor large n,n +ln n n, so apply the Comparison Test with bn = 1n. We ndL = limnanbn= limn1n +ln nn1 = limn11 + ln nn= 1The series

n=11nis a p-series with p = 12 < 1, so it diverges. Because L exists, the Limit Comparison Test tells us thethe original series also diverges.

n=1ln(n +4)n5/247.

n=1_1 cos 1n_Hint: Compare with

n=1n2.solution Let an = 1 cos 1n, and apply the Limit Comparison Test with bn = 1n2. We ndL = limnanbn= limn1 cos 1n1n2= limx1 cos 1x1x2= limx 1x2 sin 1x 2x3= 12 limxsin 1x1x.As x , u = 1x 0, soL = 12 limxsin 1x1x= 12 limu0sin uu= 12.The series

n=11n2 is a p-series with p = 2 > 1, so it converges. Because L exists, by the Limit Comparison Test we canconclude that the series

n=1_1 cos 1n_also converges.

n=1(1 21/n) Hint: Compare with the harmonic series.In Exercises 4974, determine convergence or divergence using any method covered so far.49.

n=41n29solution Apply the Limit Comparison Test with an = 1n29 and bn = 1n2:L = limnanbn= limn1n291n2= limnn2n29 = 1.Since the p-series

n=11n2 converges, the series

n=41n2 also converges. Because L exists, by the Limit Comparison Testwe can conclude that the series

n=41n29 converges.

n=1cos2nn251.

n=1n4n +9solution Apply the Limit Comparison Test with an =n4n +9 and bn = 1n:L = limnanbn= limnn4n+91n= limnn4n +9 = 14.The series

n=11nis a divergent p-series. Because L > 0, by the Limit Comparison Test we can conclude that the series

n=1n4n +9 also diverges.May 18, 2011S E C T I ON 10.3 Convergence of Series with Positive Terms (LT SECTION 11.3) 31

n=1n cos nn353.

n=1n2nn5+nsolution First rewrite an = n2nn5+n= n (n 1)n_n4+1_ = n 1n4+1 and observen 1n4+1 1. Thus the series diverges.

n=113n257.

n=21n3/2ln nsolution For n 3, ln n > 1, so n3/2ln n > n3/2and1n3/2ln n

n=N1n1/2which is a divergent p-series. Thus the series on the left diverges as well, and adding back in the nite number of termsfor n < N does not affect the result. Thus

n=21(ln n)4 diverges.May 18, 201132 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=12n3nn63.

n=11n ln n nsolution For n 2, n ln n n n ln n; therefore,1n ln n n 1n ln n.Now, let f (x) = 1x ln x. For x 2, this function is continuous, positive and decreasing, so the Integral Test applies. Usingthe substitution u = ln x, du = 1x dx, we nd_ 2dxx ln x= limR_ R2dxx ln x= limR_ ln Rln 2duu= limR(ln(ln R) ln(ln 2)) = .The integral diverges; hence, the series

n=21n ln nalso diverges. By the Comparison Test we can therefore conclude thatthe series

n=21n ln n ndiverges.

n=11n(ln n)2n65.

n=11nnsolution For n 2, nn 2n; therefore,1nn 12n =_12_n.The series

n=1_12_nis a convergent geometric series, so

n=2_12_nalso converges. By the Comparison Test we cantherefore conclude that the series

n=21nn converges. Hence, the series

n=11nn converges.

n=1n24n3/2n367.

n=11 +(1)nnsolution Letan = 1 +(1)nnThenan =_0 n odd22k = 1k n = 2k evenTherefore, {an} consists of 0s in the odd places and the harmonic series in the even places, so

i=1 an is just the sum ofthe harmonic series, which diverges. Thus

i=1 an diverges as well.

n=12 +(1)nn3/269.

n=1sin 1nsolution Apply the Limit Comparison Test with an = sin 1nand bn = 1n:L = limnsin 1n1n= limu0sin uu= 1,where u = 1n. The harmonic series diverges. Because L > 0, by the Limit Comparison Test we can conclude that theseries

n=1sin 1nalso diverges.

n=1sin(1/n)n71.

n=12n +14nsolution For n 3, 2n +1 < 2n, so2n +14n

n=N1n3/4which is a divergent p-series. Thus the original series diverges as well - as usual, adding back in the nite number ofterms for n < N does not affect convergence.

n=11n3/2ln4n77.

n=14n2+15n3n45n217solution Apply the Limit Comparison Test withan = 4n2+15n3n45n217, bn = 4n23n4 = 43n2We haveL = limnanbn= limn4n2+15n3n45n217 3n24 = limn12n4+45n312n420n268 = limn12 +45/n12 20/n268/n4 = 1Now,

n=1 bn is a p-series with p = 2 > 1, so converges. Since L = 1, we see that

n=14n2+15n3n45n217 converges aswell.

n=1n4n+5n79. For which a does

n=21n(ln n)a converge?solution First consider the case a > 0 but a = 1. Let f (x) = 1x(ln x)a . This function is continuous, positive anddecreasing for x 2, so the Integral Test applies. Now,_ 2dxx(ln x)a = limR_ R2dxx(ln x)a = limR_ ln Rln 2duua = 11 alimR_ 1(ln R)a1 1(ln 2)a1_.BecauselimR1(ln R)a1 =_, 0 < a < 10, a > 1May 18, 201134 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)we conclude the integral diverges when 0 < a < 1 and converges when a > 1. Therefore

n=21n(ln n)a converges for a > 1 and diverges for 0 < a < 1.Next, consider the case a = 1. The series becomes

n=21n ln n. Let f (x) = 1x ln x. For x 2, this function is continuous,positive and decreasing, so the Integral Test applies. Using the substitution u = ln x, du = 1x dx, we nd_ 2dxx ln x= limR_ R2dxx ln x= limR_ ln Rln 2duu= limR(ln(ln R) ln(ln 2)) = .The integral diverges; hence, the series also diverges.Finally, consider the case a < 0. Let b = a > 0 so the series becomes

n=2(ln n)bn. Since ln n > 1 for all n 3, itfollows that(ln n)b> 1 so (ln n)bn>1n.The series

n=31ndiverges, so by the Comparison Test we can conclude that

n=3(ln n)bnalso diverges. Consequently,

n=2(ln n)bndiverges. Thus,

n=21n(ln n)a diverges for a < 0.To summarize:

n=21n(ln n)a converges if a > 1 and diverges if a 1.For which a does

n=21naln nconverge?Approximating Innite Sums In Exercises 8183, let an = f (n), where f (x) is a continuous, decreasing function suchthat f (x) 0 and_1 f (x) dx converges.81. Show that_ 1f (x) dx

n=1an a1+_ 1f (x) dx 3solution From the proof of the Integral Test, we know thata2+a3+a4+ +aN _ N1f (x) dx _ 1f (x) dx;that is,SN a1 _ 1f (x) dx or SN a1+_ 1f (x) dx.Also from the proof of the Integral test, we know that_ N1f (x) dx a1+a2+a3+ +aN1 = SN aN SN.Thus,_ N1f (x) dx SN a1+_ 1f (x) dx.Taking the limit as N yields Eq. (3), as desired.Using Eq. (3), show that5

n=11n1.2 6Thi i l l U l b if h S 5 f N 43 128 d SMay 18, 2011S E C T I ON 10.3 Convergence of Series with Positive Terms (LT SECTION 11.3) 3583. Let S =

n=1an. Arguing as in Exercise 81, show thatM

n=1an+_ M+1f (x) dx S M+1

n=1an+_ M+1f (x) dx 4Conclude that0 S M

n=1an+_ M+1f (x) dx aM+1 5This provides a method for approximating S with an error of at most aM+1.solution Following the proof of the Integral Test and the argument in Exercise 81, but starting with n = M +1 ratherthan n = 1, we obtain_ M+1f (x) dx

n=M+1an aM+1+_ M+1f (x) dx.AddingM

n=1an to each part of this inequality yieldsM

n=1an+_ M+1f (x) dx

n=1an = S M+1

n=1an+_ M+1f (x) dx.SubtractingM

n=1an+_ M+1f (x) dx from each part of this last inequality then gives us0 S M

n=1an+_ M+1f (x) dx aM+1.Use Eq. (4) with M = 43,129 to prove that5.5915810

n=11n1.2 5.591583985. Apply Eq. (4) with M = 40,000 to show that1.644934066

n=11n2 1.644934068Is this consistent with Eulers result, according to which this innite series has sum 2/6?solution Using Eq. (4) with f (x) = 1x2, an = 1n2 and M = 40,000, we ndS40,000+_ 40,001dxx2

n=11n2 S40,001+_ 40,001dxx2 .Now,S40,000 = 1.6449090672;S40,001 = S40,000+ 140,001 = 1.6449090678;and_ 40,001dxx2 = limR_ R40,001dxx2 = limR_1R 140,001_= 140,001 = 0.0000249994.Thus,1.6449090672 +0.0000249994

n=11n2 1.6449090678 +0.0000249994,May 18, 201136 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)or1.6449340665

n=11n2 1.6449340672.Since 26 1.6449340668, our approximation is consistent with Eulers result.Using a CAS and Eq. (5), determine the value of

n=1n6to within an error less than 104. Check that yourresult is consistent with that of Euler, who proved that the sum is equal to 6/945.87. Using a CAS and Eq. (5), determine the value of

n=1n5to within an error less than 104.solution Using Eq. (5) with f (x) = x5and an = n5, we have0

n=1n5M+1

n=1n5+_ M+1x5dx (M +1)5.To guarantee an error less than 104, we need (M + 1)5 104. This yields M 104/5 1 5.3, so we chooseM = 6. Now,7

n=1n5= 1.0368498887,and_ 7x5dx = limR_ R7x5dx = 14 limR_R474_= 14 74 = 0.0001041233.Thus,

n=1n57

n=1n5+_ 7x5dx = 1.0368498887 +0.0001041233 = 1.0369540120.How far can a stack of identical books (of mass m and unit length) extend without tipping over? The stack willnot tip over if the (n +1)st book is placed at the bottom of the stack with its right edge located at the center of massof the rst n books (Figure 5). Let cn be the center of mass of the rst n books, measured along the x-axis, where wetake the positive x-axis to the left of the origin as in Figure 6. Recall that if an object of mass m1 has center of massat x1 and a second object of m2 has center of mass x2, then the center of mass of the system has x-coordinatem1x1+m2x2m1+m2(a) Show that if the (n +1)st book is placed with its right edge at cn, then its center of mass is located at cn+ 12.(b) Consider the rst nbooks as a single object of mass nmwith center of mass at cn and the (n +1)st book as a secondobject of mass m. Show that if the (n +1)st book is placed with its right edge at cn, then cn+1 = cn+ 12(n +1).(c) Prove that limncn = . Thus, by using enough books, the stack can be extended as far as desired withouttipping over.89. The following argument proves the divergence of the harmonic series S =

n=11/n without using the Integral Test.LetS1 = 1 + 13 + 15 + , S2 = 12 + 14 + 16 + Show that if S converges, then(a) S1 and S2 also converge and S = S1+S2.(b) S1 > S2 and S2 = 12S.Observe that (b) contradicts (a), and conclude that S diverges.solution Assume throughout that S converges; we will derive a contradiction. Writean = 1n, bn = 12n 1, cn = 12nfor the nthterms in the series S, S1, and S2. Since 2n 1 n for n 1, we have bn < an. Since S =

an converges,so does S1 = bn by the Comparison Test. Also, cn = 12an, so again by the Comparison Test, the convergence of Simplies the convergence of S2 =

cn. Now, dene two sequencesb

n =_b(n+1)/2 n odd0 n evenc

n =_0 n oddcn/2 n evenThat is, b

n and c

n look like bn and cn, but have zeros inserted in the missing places compared to an. Then an = b

n+c

n;also S1 =

bn =

b

n and S2 =

cn =

c

n. Finally, since S, S1, and S2 all converge, we haveS =

n=1an =

n=1(b

n+c

n) =

n=1b

n+

n=1c

n =

n=1bn+

n=1cn = S1+S2May 18, 2011S E C T I ON 10.3 Convergence of Series with Positive Terms (LT SECTION 11.3) 37Now, bn > cn for every n, so that S1 > S2. Also, we showed above that cn = 12an, so that 2S2 = S. Putting all thistogether givesS = S1+S2 > S2+S2 = 2S2 = Sso that S > S, a contradiction. Thus S must diverge.Further Insights and ChallengesLet S =

n=2an, where an = (ln(ln n))ln n.(a) Show, by taking logarithms, that an = nln(ln(ln n)).(b) Show that ln(ln(ln n)) 2 if n > C, where C = eee2.(c) Show that S converges.91. Kummers Acceleration Method Suppose we wish to approximate S =

n=11/n2. There is a similar telescopingseries whose value can be computed exactly (Example 1 in Section 10.2):

n=11n(n +1)= 1(a) Verify thatS =

n=11n(n +1)+

n=1_ 1n2 1n(n +1)_Thus for M large,S 1 +M

n=11n2(n +1)6(b) Explain what has been gained. Why is Eq. (6) a better approximation to S than isM

n=11/n2?(c) Compute1000

n=11n2, 1 +100

n=11n2(n +1)Which is a better approximation to S, whose exact value is 2/6?solution(a) Because the series

n=11n2 and

n=11n(n +1)both converge,

n=11n(n +1)+

n=1_ 1n2 1n(n +1)_=

n=11n(n +1)+

n=11n2

n=11n(n +1)=

n=11n2 = S.Now,1n2 1n(n +1)= n +1n2(n +1) nn2(n +1)= 1n2(n +1),so, for M large,S 1 +M

n=11n2(n +1).(b) The series

n=1 1n2(n+1) converges more rapidly than

n=11n2 since the degree of n in the denominator is larger.(c) Using a computer algebra system, we nd1000

n=11n2 = 1.6439345667 and 1 +100

n=11n2(n +1)= 1.6448848903.The second sum is more accurate because it is closer to the exact solution 26 1.6449340668.The series S =

k=1k3has been computed to more than 100 million digits. The rst 30 digits areS = 1.202056903159594285399738161511May 18, 201138 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)10.4 Absolute and Conditional Convergence (LT Section 11.4)Preliminary Questions1. Give an example of a series such that

an converges but

|an| diverges.solution The series

(1)n3nconverges by the Leibniz Test, but the positive series

13nis a divergent p-series.2. Which of the following statements is equivalent to Theorem 1?(a) If

n=0|an| diverges, then

n=0an also diverges.(b) If

n=0an diverges, then

n=0|an| also diverges.(c) If

n=0an converges, then

n=0|an| also converges.solution The correct answer is (b): If

n=0an diverges, then

n=0|an| also diverges. Take an = (1)n 1n to see thatstatements (a) and (c) are not true in general.3. Lathika argues that

n=1(1)nn is an alternating series and therefore converges. Is Lathika right?solution No. Although

n=1(1)nn is an alternating series, the terms an =n do not form a decreasing sequencethat tends to zero. In fact, an =n is an increasing sequence that tends to , so

n=1(1)nn diverges by the DivergenceTest.4. Suppose that an is positive, decreasing, and tends to 0, and let S =

n=1(1)n1an. What can we say about |S S100|if a101 = 103? Is S larger or smaller than S100?solution Fromthe text, we knowthat |S S100| < a101 = 103. Also, the Leibniz test tells us that S2N < S < S2N+1for any N 1, so that S100 < S.Exercises1. Show that

n=0(1)n2nconverges absolutely.solution The positive series

n=012n is a geometric series with r = 12. Thus, the positive series converges, and thegiven series converges absolutely.Show that the following series converges conditionally:

n=1(1)n1 1n2/3 = 112/3 122/3 + 132/3 142/3 + In Exercises 310, determine whether the series converges absolutely, conditionally, or not at all.3.

n=1(1)n1n1/3solution The sequence an = 1n1/3 is positive, decreasing, and tends to zero; hence, the series

n=1(1)n1n1/3 convergesby the Leibniz Test. However, the positive series

n=11n1/3 is a divergent p-series, so the original series convergesconditionally.

n=1(1)nn4n3+1May 18, 2011S E C T I ON 10.4 Absolute and Conditional Convergence (LT SECTION 11.4) 395.

n=0(1)n1(1.1)nsolution The positive series

n=0_ 11.1_nis a convergent geometric series; thus, the original series converges abso-lutely.

n=1sin(n4 )n27.

n=2(1)nn ln nsolution Let an = 1n ln n. Then an forms a decreasing sequence (note that n and ln n are both increasing functions ofn) that tends to zero; hence, the series

n=2(1)nn ln nconverges by the Leibniz Test. However, the positive series

n=21n ln ndiverges, so the original series converges conditionally.

n=1(1)n1 + 1n9.

n=2cos n(ln n)2solution Since cos n alternates between +1 and 1,

n=2cos n(lnn)2 =

n=2(1)n(lnn)2This is an alternating series whose general term decreases to zero, so it converges. The associated positive series,

n=21(ln n)2is a divergent series, so the original series converges conditionally.

n=1cos n2n11. Let S =

n=1(1)n+1 1n3.(a) Calculate Sn for 1 n 10.(b) Use Eq. (2) to show that 0.9 S 0.902.solution(a)S1 = 1 S6 = S5 163 = 0.899782407S2 = 1 123 = 78 = 0.875 S7 = S6+ 173 = 0.902697859S3 = S2+ 133 = 0.912037037 S8 = S7 183 = 0.900744734S4 = S3 143 = 0.896412037 S9 = S8+ 193 = 0.902116476S5 = S4+ 153 = 0.904412037 S10 = S9 1103 = 0.901116476(b) By Eq. (2),|S10S| a11 = 1113,soS10 1113 S S10+ 1113,or0.900365161 S 0.901867791.May 18, 201140 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)Use Eq. (2) to approximate

n=1(1)n+1n!to four decimal places.13. Approximate

n=1(1)n+1n4 to three decimal places.solution Let S =

n=1(1)n+1n4 , so that an = 1n4. By Eq. (2),|SN S| aN+1 = 1(N +1)4.To guarantee accuracy to three decimal places, we must choose N so that1(N +1)4 < 5 104or N > 42000 1 5.7.The smallest value that satises the required inequality is then N = 6. Thus,S S6 = 1 124 + 134 144 + 154 164 = 0.946767824.LetS =

n=1(1)n1 nn2+1Use a computer algebra system to calculate and plot the partial sums Sn for 1 n 100. Observe that the partialsums zigzag above and below the limit.In Exercises 15 and 16, nd a value of N such that SN approximates the series with an error of at most 105. If you havea CAS, compute this value of SN.15.

n=1(1)n+1n(n +2)(n +3)solution Let S =

n=1(1)n+1n (n +2) (n +3), so that an = 1n (n +2) (n +3). By Eq. (2),|SN S| aN+1 = 1(N +1)(N +3)(N +4).We must choose N so that1(N +1)(N +3)(N +4) 105or (N +1)(N +3)(N +4) 105.For N = 43, the product on the left hand side is 95,128, while for N = 44 the product is 101,520; hence, the smallestvalue of N which satises the required inequality is N = 44. Thus,S S44 =44

n=1(1)n+1n(n +2)(n +3)= 0.0656746.

n=1(1)n+1ln nn!In Exercises 1732, determine convergence or divergence by any method.17.

n=07nsolution This is a (positive) geometric series with r = 17 < 1, so it converges.

n=11n7.519.

n=115n3nsolution Use the Limit Comparison Test with 15n:L = limn1/(5n3n)1/5n = limn5n5n3n = limn11 (3/5)n = 1But

n=115n is a convergent geometric series. Since L = 1, the Limit Comparison Test tells us that the original seriesconverges as well.

n=2nn2nMay 18, 2011S E C T I ON 10.4 Absolute and Conditional Convergence (LT SECTION 11.4) 4121.

n=113n4+12nsolution Use the Limit Comparison Test with 13n4:L = limn(1/(3n4+12n)1/3n4 = limn3n43n4+12n= limn11 +4n3 = 1But

n=113n4 = 13

n=1 1n4 is a convergent p-series. Since L = 1, the Limit Comparison Test tells us that the originalseries converges as well.

n=1(1)n_n2+123.

n=11_n2+1solution Apply the Limit Comparison Test and compare the series with the divergent harmonic series:L = limn1n2+11n= limnn_n2+1= 1.Because L > 0, we conclude that the series

n=11_n2+1diverges.

n=0(1)nn_n2+125.

n=13n+(2)n5nsolution The series

n=13n5n =

n=1_35_nis a convergent geometric series, as is the series

n=1(1)n2n5n =

n=1_25_n.Hence,

n=13n+(1)n2n5n =

n=1_35_n+

n=1_25_nalso converges.

n=1(1)n+1(2n +1)!27.

n=1(1)nn2en3/3solution Consider the associated positive series

n=1n2en3/3. This series can be seen to converge by the IntegralTest:_ 1x2ex3/3dx = limR_ R1x2ex3/3dx = limRex3/3R1 = e1/3+ limReR3/3= e1/3.The integral converges, so the original series converges absolutely.

n=1nen3/329.

n=2(1)nn1/2(ln n)2solution This is an alternating series with an = 1n1/2(ln n)2. Because an is a decreasing sequence which convergesto zero, the series

n=2(1)nn1/2(ln n)2 converges by the Leibniz Test. (Note that the series converges only conditionally, notabsolutely; the associated positive series is eventually greater than 1n3/4, which is a divergent p-series).May 18, 201142 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=21n(ln n)1/431.

n=1ln nn1.05solution Choose N so that for n N we have ln n n0.01. Then

n=Nln nn1.05

n=Nn0.01n1.05 =

n=N1n1.04This is a convergent p-series, so by the Comparison Test, the original series converges as well.

n=21(ln n)233. Show thatS = 12 12 + 13 13 + 14 14 + converges by computing the partial sums. Does it converge absolutely?solution The sequence of partial sums isS1 = 12S2 = S1 12 = 0S3 = S2+ 13 = 13S4 = S3 13 = 0and, in general,SN =1N, for odd N0, for even NThus, limNSN = 0, and the series converges to 0. The positive series is12 + 12 + 13 + 13 + 14 + 14 + = 2

n=21n;which diverges. Therefore, the original series converges conditionally, not absolutely.The Leibniz Test cannot be applied to12 13 + 122 132 + 123 133 + Why not? Show that it converges by another method.35. Assumptions Matter Show by counterexample that the Leibniz Test does not remain true if the sequencean tends to zero but is not assumed nonincreasing. Hint: ConsiderR = 12 14 + 13 18 + 14 116 + +_1n 12n_+ solution LetR = 12 14 + 13 18 + 14 116 + +_ 1n +1 12n+1_+ This is an alternating series withan =1k +1, n = 2k 112k+1, n = 2kNote that an 0 as n , but the sequence {an} is not decreasing. We will now establish that R diverges.For sake of contradiction, suppose that R converges. The geometric series

n=112n+1May 18, 2011S E C T I ON 10.4 Absolute and Conditional Convergence (LT SECTION 11.4) 43converges, so the sum of R and this geometric series must also converge; however,R +

n=112n+1 =

n=21n,which diverges because the harmonic series diverges. Thus, the series R must diverge.Determine whether the following series converges conditionally:1 13 + 12 15 + 13 17 + 14 19 + 15 111 + 37. Prove that if

an converges absolutely, then

a2n also converges. Then give an example where

an is onlyconditionally convergent and

a2n diverges.solution Suppose the series

an converges absolutely. Because

|an| converges, we know thatlimn|an| = 0.Therefore, there exists a positive integer N such that |an| < 1 for all n N. It then follows that for n N,0 a2n = |an|2= |an| |an| < |an| 1 = |an|.By the Comparison Test we can then conclude that

a2n also converges.Consider the series

n=1(1)nn. This series converges by the Leibniz Test, but the corresponding positive series is adivergent p-series; that is,

n=1(1)nnis conditionally convergent. Now,

n=1a2n is the divergent harmonic series

n=11n.Thus,

a2n need not converge if

an is only conditionally convergent.Further Insights and ChallengesProve the following variant of the Leibniz Test: If {an} is a positive, decreasing sequence with limnan = 0, thenthe seriesa1+a22a3+a4+a52a6+ converges. Hint: Show that S3N is increasing and bounded by a1+ a2, and continue as in the proof of the LeibnizTest.39. Use Exercise 38 to show that the following series converges:S = 1ln 2 + 1ln 3 2ln 4 + 1ln 5 + 1ln 6 2ln 7 + solution The given series has the structure of the generic series from Exercise 38 with an = 1ln(n+1). Because an isa positive, decreasing sequence with limnan = 0, we can conclude from Exercise 38 that the given series converges.Prove the conditional convergence ofR = 1 + 12 + 13 34 + 15 + 16 + 17 38 + 41. Show that the following series diverges:S = 1 + 12 + 13 24 + 15 + 16 + 17 28 + Hint: Use the result of Exercise 40 to write S as the sum of a convergent series and a divergent series.solution LetR = 1 + 12 + 13 34 + 15 + 16 + 17 38 + andS = 1 + 12 + 13 24 + 15 + 16 + 17 28 + For sake of contradiction, suppose the series S converges. From Exercise 40, we know that the series R converges. Thus,the series S R must converge; however,S R = 14 + 18 + 112 + = 14

k=11k,which diverges because the harmonic series diverges. Thus, the series S must diverge.Prove that

n=1(1)n+1(ln n)anconverges for all exponents a. Hint: Show that f (x) = (ln x)a/x is decreasing for x sufciently large.43. We say that {bn} is a rearrangement of {an} if {bn} has the same terms as {an} but occurring in a different order. Showthat if {bn} is a rearrangement of {an} and S =

n=1an converges absolutely, then T =

n=1bn also converges absolutely.(This result does not hold if S is only conditionally convergent.) Hint: Prove that the partial sumsN

n=1|bn| are bounded.It can be shown further that S = T .May 18, 201144 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)solution Suppose the series S =

n=1an converges absolutely and denote the corresponding positive series byS+ =

n=1|an|.Further, let TN =N

n=1|bn| denote the Nth partial sum of the series

n=1|bn|. Because {bn} is a rearrangement of {an}, weknow that0 TN

n=1|an| = S+;that is, the sequence {TN} is bounded. Moreover,TN+1 =N+1

n=1|bn| = TN +|bN+1| TN;that is, {TN} is increasing. It follows that {TN} converges, so the series

n=1|bn| converges, which means the series

n=1bnconverges absolutely.Assumptions Matter In 1829, Lejeune Dirichlet pointed out that the great French mathematician AugustinLouis Cauchy made a mistake in a published paper by improperly assuming the Limit Comparison Test to be validfor nonpositive series. Here are Dirichlets two series:

n=1(1)nn,

n=1(1)nn_1 + (1)nn_Explain how they provide a counterexample to the Limit Comparison Test when the series are not assumed to bepositive.10.5 The Ratio and Root Tests (LT Section 11.5)Preliminary Questions1. In the Ratio Test, is equal to limnan+1anor limnanan+1?solution In the Ratio Test is the limit limnan+1an.2. Is the Ratio Test conclusive for

n=112n? Is it conclusive for

n=11n?solution The general term of

n=112n is an = 12n; thus,an+1an= 12n+1 2n1 = 12,and = limnan+1an= 12 < 1.Consequently, the Ratio Test guarantees that the series

n=112n converges.The general term of

n=11nis an = 1n; thus,an+1an= 1n +1 n1 = nn +1,and = limnan+1an= limnnn +1 = 1.The Ratio Test is therefore inconclusive for the series

n=11n.3. Can the Ratio Test be used to show convergence if the series is only conditionally convergent?solution No. The Ratio Test can only establish absolute convergence and divergence, not conditional convergence.May 18, 2011S E C T I ON 10.5 The Ratio and Root Tests (LT SECTION 11.5) 45ExercisesIn Exercises 120, apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive.1.

n=115nsolution With an = 15n,an+1an= 15n+1 5n1 = 15 and = limnan+1an= 15 < 1.Therefore, the series

n=115n converges by the Ratio Test.

n=1(1)n1n5n3.

n=11nnsolution With an = 1nn,an+1an= 1(n +1)n+1 nn1 = 1n +1_ nn +1_n= 1n +1_1 + 1n_n,and = limnan+1an= 0 1e= 0 < 1.Therefore, the series

n=11nn converges by the Ratio Test.

n=03n +25n3+15.

n=1nn2+1solution With an = nn2+1,an+1an= n +1(n +1)2+1 n2+1n= n +1n n2+1n2+2n +2,and = limnan+1an= 1 1 = 1.Therefore, for the series

n=1nn2+1, the Ratio Test is inconclusive.We can show that this series diverges by using the Limit Comparison Test and comparing with the divergent harmonicseries.

n=12nn7.

n=12nn100solution With an = 2nn100,an+1an= 2n+1(n +1)100 n1002n = 2_ nn +1_100and = limnan+1an= 2 1100= 2 > 1.Therefore, the series

n=12nn100 diverges by the Ratio Test.

n=1n33n29.

n=110n2n2solution With an = 10n2n2 ,an+1an= 10n+12(n+1)2 2n210n = 10 122n+1 and = limnan+1an= 10 0 = 0 < 1.Therefore, the series

n=110n2n2 converges by the Ratio Test.May 18, 201146 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=1enn!11.

n=1ennnsolution With an = ennn,an+1an= en+1(n +1)n+1 nnen = en +1_ nn +1_n= en +1_1 + 1n_n,and = limnan+1an= 0 1e= 0 < 1.Therefore, the series

n=1ennn converges by the Ratio Test.

n=1n40n!13.

n=0n!6nsolution With an = n!6n,an+1an= (n +1)!6n+1 6nn! = n +16 and = limnan+1an= > 1.Therefore, the series

n=0n!6n diverges by the Ratio Test.

n=1n!n915.

n=21n ln nsolution With an = 1n ln n,an+1an= 1(n +1) ln(n +1) n ln n1 = nn +1ln nln(n +1),and = limnan+1an= 1 limnln nln(n +1).Now,limnln nln(n +1)= limxln xln(x +1)= limx1/(x +1)1/x= limxxx +1 = 1.Thus, = 1, and the Ratio Test is inconclusive for the series

n=21n ln n.Using the Integral Test, we can show that the series

n=21n ln ndiverges.

n=11(2n)!17.

n=1n2(2n +1)!solution With an = n2(2n+1)!,an+1an= (n +1)2(2n +3)! (2n +1)!n2 =_n +1n_21(2n +3)(2n +2),and = limnan+1an= 12 0 = 0 < 1.Therefore, the series

n=1n2(2n +1)! converges by the Ratio Test.

n=1(n!)3(3n)!May 18, 2011S E C T I ON 10.5 The Ratio and Root Tests (LT SECTION 11.5) 4719.

n=212n+1solution With an = 12n+1,an+1an= 12n+1+1 2n+11 = 1 +2n2 +2nand = limnan+1an= 12 < 1Therefore, the series

n=212n+1 converges by the Ratio Test.

n=21ln n21. Show that

n=1nk3nconverges for all exponents k.solution With an = nk3n,an+1an= (n +1)k3(n+1)nk3n = 13_1 + 1n_k,and, for all k, = limnan+1an= 13 1 = 13 < 1.Therefore, the series

n=1nk3nconverges for all exponents k by the Ratio Test.Show that

n=1n2xnconverges if |x| < 1.23. Show that

n=12nxnconverges if |x| < 12.solution With an = 2nxn,an+1an= 2n+1|x|n+12n|x|n = 2|x| and = limnan+1an= 2|x|.Therefore, < 1 and the series

n=12nxnconverges by the Ratio Test provided |x| < 12.Show that

n=1rnn! converges for all r.25. Show that

n=1rnnconverges if |r| < 1.solution With an = rnn ,an+1an= |r|n+1n +1 n|r|n = |r| nn +1 and = limnan+1an= 1 |r| = |r|.Therefore, by the Ratio Test, the series

n=1rnnconverges provided |r| < 1.Is there any value of k such that

n=12nnk converges?27. Show that

n=1n!nn converges. Hint: Use limn_1 + 1n_n= e.solution With an = n!nn,an+1an= (n +1)!(n +1)n+1 nnn! =_ nn +1_n=_1 + 1n_n,and = limnan+1an= 1e< 1.May 18, 201148 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)Therefore, the series

n=1n!nn converges by the Ratio Test.In Exercises 2833, assume that |an+1/an| converges to = 13. What can you say about the convergence of the givenseries?

n=1nan29.

n=1n3ansolution Let bn = n3an. Then = limnbn+1bn= limn_n +1n_3an+1an= 13 13 = 13 < 1.Therefore, the series

n=1n3an converges by the Ratio Test.

n=12nan31.

n=13nansolution Let bn = 3nan. Then = limnbn+1bn= limn3n+13nan+1an= 3 13 = 1.Therefore, the Ratio Test is inconclusive for the series

n=13nan.

n=14nan33.

n=1a2nsolution Let bn = a2n. Then = limnbn+1bn= limnan+1an2=_13_2= 19 < 1.Therefore, the series

n=1a2n converges by the Ratio Test.Assume thatan+1/anconverges to = 4. Does

n=1 a1n converge (assume that an = 0 for all n)? 35. Is the Ratio Test conclusive for the p-series

n=11np ?solution With an = 1np ,an+1an= 1(n +1)p np1 =_ nn +1_pand = limnan+1an= 1p= 1.Therefore, the Ratio Test is inconclusive for the p-series

n=11np .In Exercises 3641, use the Root Test to determine convergence or divergence (or state that the test is inconclusive).

n=0110n37.

n=11nnsolution With an = 1nn,nan = n_ 1nn = 1nand limnnan = 0 < 1.Therefore, the series

n=11nn converges by the Root Test.

k=0_ kk +10_kMay 18, 2011S E C T I ON 10.5 The Ratio and Root Tests (LT SECTION 11.5) 4939.

k=0_ k3k +1_ksolution With ak =_ k3k+1_k,kak = k__ k3k +1_k= k3k +1 and limkkak = 13 < 1.Therefore, the series

k=0_ k3k +1_kconverges by the Root Test.

n=1_1 + 1n_n41.

n=4_1 + 1n_n2solution With ak =_1 + 1n_n2,nan = n__1 + 1n_n2=_1 + 1n_nand limnnan = e1< 1.Therefore, the series

n=4_1 + 1n_n2converges by the Root Test.Prove that

n=12n2n! diverges. Hint: Use 2n2= (2n)nand n! nn.In Exercises 4356, determine convergence or divergence using any method covered in the text so far.43.

n=12n+4n7nsolution Because the series

n=12n7n =

n=1_27_nand

n=14n7n =

n=1_47_nare both convergent geometric series, it follows that

n=12n+4n7n =

n=1_27_n+

n=1_47_nalso converges.

n=1n3n!45.

n=1n35nsolution The presence of the exponential term suggests applying the Ratio Test. With an = n35n,an+1an= (n +1)35n+1 5nn3 = 15_1 + 1n_3and = limnan+1an= 15 13= 15 < 1.Therefore, the series

n=1n35n converges by the Ratio Test.

n=21n(ln n)347.

n=21_n3n2solution This series is similar to a p-series; because1_n3n2 1n3 = 1n3/2for large n, we will apply the Limit Comparison Test comparing with the p-series with p = 32. Now,L = limn1n3n21n3/2= limn_ n3n3n2 = 1.The p-series with p = 32 converges and L exists; therefore, the series

n=21_n3n2 also converges.May 18, 201150 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)

n=1n2+4n3n4+949.

n=1n0.8solution

n=1n0.8=

n=11n0.8so that this is a divergent p-series.

n=1(0.8)nn0.851.

n=142n+1solution Observe

n=142n+1=

n=14 (42)n=

n=14_ 116_nis a geometric series with r = 116; therefore, this series converges.

n=1(1)n1n53.

n=1sin 1n2solution Here, we will apply the Limit Comparison Test, comparing with the p-series with p = 2. Now,L = limnsin 1n21n2= limu0sin uu= 1,where u = 1n2. The p-series with p = 2 converges and L exists; therefore, the series

n=1sin 1n2 also converges.

n=1(1)ncos 1n55.

n=1(2)nnsolution Becauselimn2nn= limx2xx= limx2xln 212x= limx2x+1x ln 2 = = 0,the general term in the series

n=1(2)nndoes not tend toward zero; therefore, the series diverges by the Divergence Test.

n=1_ nn +12_nFurther Insights and Challenges57. Proof of the Root Test Let S =

n=0an be a positive series, and assume that L = limnnan exists.(a) Show that S converges if L < 1. Hint: Choose R with L < R < 1 and show that an Rnfor n sufciently large.Then compare with the geometric series

Rn.(b) Show that S diverges if L > 1.solution Suppose limnnan = L exists.(a) If L < 1, let = 1 L2 . By the denition of a limit, there is a positive integer N such that nanL for n N. From this, we conclude that0 nan L +for n N. Now, let R = L +. ThenR = L + 1 L2 = L +12 1, let = L 12 . By the denition of a limit, there is a positive integer N such that nanL for n N. From this, we conclude thatL nanfor n N. Now, let R = L . ThenR = L L 12 = L +12 >1 +12 = 1,andR nan or Rn anfor n N. Because R > 1, the series

n=NRnis a divergent geometric series, so the series

n=Nan diverges by theComparison Test. Therefore, the series

n=0an also diverges.Show that the Ratio Test does not apply, but verify convergence using the Comparison Test for the series12 + 132 + 123 + 134 + 125 + 59. Let S =

n=1cnn!nn , where c is a constant.(a) Prove that S converges absolutely if |c| < e and diverges if |c| > e.(b) It is known that limnenn!nn+1/2 =2. Verify this numerically.(c) Use the Limit Comparison Test to prove that S diverges for c = e.solution(a) With an = cnn!nn ,an+1an= |c|n+1(n +1)!(n +1)n+1 nn|c|nn! = |c|_ nn +1_n= |c|_1 + 1n_n,and = limnan+1an= |c|e1.Thus, by the Ratio Test, the series

n=1cnn!nn converges when |c|e1< 1, or when |c| < e. The series diverges when|c| > e.(b) The table belowlists the value of enn!nn+1/2 for several increasing values of n. Since2 = 2.506628275, the numericalevidence veries thatlimnenn!nn+1/2 =2.n 100 1000 10000 100000enn!nn+1/2 2.508717995 2.506837169 2.506649163 2.506630363(c) With c = e, the series S becomes

n=1enn!nn . Using the result from part (b),L = limnenn!nnn= limnenn!nn+1/2 =2.Because the series

n=1n diverges by the Divergence Test and L > 0, we conclude that

n=1enn!nn diverges by the LimitComparison Test.May 18, 201152 C HA P T E R 10 INFINITE SERIES (LT CHAPTER 11)10.6 Power Series (LT Section 11.6)Preliminary Questions1. Suppose that

anxnconverges for x = 5. Must it also converge for x = 4? What about x = 3?solution The power series

anxnis centered at x = 0. Because the series converges for x = 5, the radius ofconvergence must be at least 5 and the series converges absolutely at least for the interval |x| < 5. Both x = 4 andx = 3 are inside this interval, so the series converges for x = 4 and for x = 3.2. Suppose that

an(x 6)nconverges for x = 10. At which of the points (a)(d) must it also converge?(a) x = 8 (b) x = 11 (c) x = 3 (d) x = 0solution The given power series is centered at x = 6. Because the series converges for x = 10, the radius ofconvergence must be at least |10 6| = 4 and the series converges absolutely at least for the interval |x 6| < 4, or2 < x < 10.(a) x = 8 is inside the interval 2 < x < 10, so the series converges for x = 8.(b) x = 11 is not inside the interval 2 < x < 10, so the series may or may not converge for x = 11.(c) x = 3 is inside the interval 2 < x < 10, so the series converges for x = 2.(d) x = 0 is not inside the interval 2 < x < 10, so the series may or may not converge for x = 0.3. What is the radius of convergence of F(3x) if F(x) is a power series with radius of convergence R = 12?solution If the power series F(x) has radius of convergence R = 12, then the power series F(3x) has radius ofconvergence R = 123 = 4.4. The power series F(x) =

n=1nxnhas radius of convergence R = 1. What is the power series expansion of F

(x)and what is its radius of convergence?solution We obtain the power series expansion for F

(x) by differentiating the power series expansion for F(x)term-by-term. Thus,F

(x) =

n=1n2xn1.The radius of convergence for this series is R = 1, the same as the radius of convergence for the series expansion forF(x).Exercises1. Use the Ratio Test to determine the radius of convergence R of

n=0xn2n . Does it converge at the endpoints x = R?solution With an = xn2n ,an+1an= |x|n+12n+1 2n|x|n = |x|2 and = limnan+1an= |x|2 .By the Ratio Test, the series converges when = |x|2 < 1, or |x| < 2, and diverges when = |x|2 > 1, or |x| > 2.The radius of convergence is therefore R = 2. For x = 2, the left endpoint, the series becomes

n=0(1)n, which isdivergent. For x = 2, the right endpoint, the series becomes

n=0 1, which is also divergent. Thus the series diverges atboth endpoints.Use the Ratio Test to showthat

n=1xnn2n has radius of convergence R = 2. Then determine whether it convergesat the endpoints R = 2.3. Showthat the power series (a)(c) have the same radius of convergence. Then showthat (a) diverges at both endpoints,(b) converges at one endpoint but diverges at the other, and (c) converges at both endpoints.(a)

n=1xn3n (b)

n=1xnn3n (c)

n=1xnn23nsolution(a) With an = xn3n , = limnan+1an= limnxn+13n+1 3nxn= limnx3 =x3May 18, 2011S E C T I ON 10.6 Power Series (LT SECTION 11.6) 53Then < 1 if |x| < 3, so that the radius of convergence is R = 3. F