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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Checkpoint

5.1

an=(−1)n+13+2nan=(−1)n+13+2n

5.2

an=6n10an=6n10

5.3

The sequence converges, and its limit is 0.0.

5.4

The sequence converges, and its limit is 2/3.2/3.

5.5

22

5.6

0.0.

5.7

The series diverges because the kthkth partial sum Sk>k.Sk>k.

5.8

10.10.

5.9

5/75/7

5.10

475/90475/90

5.11

e1e1

5.12

The series diverges.

5.13

The series diverges.

5.14

The series converges.

5.15

S51.09035,S51.09035, R5<0.00267R5<0.00267

5.16

The series converges.

5.17

The series diverges.

5.18

The series converges.

5.19

0.047620.04762

5.20

The series converges absolutely.

5.21

The series converges.

5.22

The series converges.

5.23

The comparison test because 2n/(3n+n)<2n/3n2n/(3n+n)<2n/3n for all positive integers n.n. The limit comparison test could also be used.

Section 5.1 Exercises

1.

an=0an=0 if nn is odd and an=2an=2 if nn is even

3.

{an}={1,3,6,10,15,21,…}{an}={1,3,6,10,15,21,…}

5.

an=n(n+1)2an=n(n+1)2

7.

an=4n7an=4n7

9.

an=3.101n=30.10nan=3.101n=30.10n

11.

an=2n1an=2n1

13.

an=(−1)n12n1an=(−1)n12n1

15.

f(n)=2nf(n)=2n

17.

f(n)=n!/2n2f(n)=n!/2n2

19.

Terms oscillate above and below 5/35/3 and appear to converge to 5/3.5/3.

This is a graph of an oscillating sequence. Terms oscillate above and below 5/3 and seem to converge to 5/3.
21.

Terms oscillate above and below y1.57...y1.57... and appear to converge to a limit.

This is a graph of the oscillating sequence. Terms oscillate above and below y = 1.57 and seem to converse to 1.57.
23.

77

25.

00

27.

00

29.

11

31.

bounded, decreasing for n1n1

33.

bounded, not monotone

35.

bounded, decreasing

37.

not monotone, not bounded

39.

anan is decreasing and bounded below by 2.2. The limit aa must satisfy a=2aa=2a so a=2,a=2, independent of the initial value.

41.

00

43.

0:|sinx||x|0:|sinx||x| and |sinx|1|sinx|1 so 1nan1n).1nan1n).

45.

Graph oscillates and suggests no limit.

This is a graph that oscillates between 1 and -1 from 0 to 25 on the x axis. There appears to be no limit.
47.

n1/n1n1/n1 and 21/n1,21/n1, so an0an0

49.

Since (1+1/n)ne,(1+1/n)ne, one has (12/n)n(1+k)−2ke−2(12/n)n(1+k)−2ke−2 as k.k.

51.

2n+3n2·3n2n+3n2·3n and 3n/4n03n/4n0 as n,n, so an0an0 as n.n.

53.

an+1an=n!/(n+1)(n+2)(2n)an+1an=n!/(n+1)(n+2)(2n) =1·2·3n(n+1)(n+2)(2n)<1/2n.=1·2·3n(n+1)(n+2)(2n)<1/2n. In particular, an+1/an1/2,an+1/an1/2, so an0an0 as n.n.

55.

xn+1=xn((xn1)22)/2(xn1);xn+1=xn((xn1)22)/2(xn1); x=1+2,x=1+2, x2.4142,x2.4142, n=5n=5

57.

xn+1=xnxn(ln(xn)1);xn+1=xnxn(ln(xn)1); x=e,x=e, x2.7183,x2.7183, n=5n=5

59.

a. Without losses, the population would obey Pn=1.06Pn1.Pn=1.06Pn1. The subtraction of 150150 accounts for fish losses. b. After 1212 months, we have P121494.P121494.

61.

a. The student owes $9383$9383 after 1212 months. b. The loan will be paid in full after 139139 months or eleven and a half years.

63.

b1=0,b1=0, x1=2/3,x1=2/3, b2=1,b2=1, x2=4/31=1/3,x2=4/31=1/3, so the pattern repeats, and 1/3=0.010101.1/3=0.010101.

65.

For the starting values a1=1,a1=1, a2=2,…,a2=2,…, a1=10,a1=10, the corresponding bit averages calculated by the method indicated are 0.5220,0.5220, 0.5000,0.5000, 0.4960,0.4960, 0.4870,0.4870, 0.4860,0.4860, 0.4680,0.4680, 0.5130,0.5130, 0.5210,0.5210, 0.5040,0.5040, and 0.4840.0.4840. Here is an example of ten corresponding averages of strings of 10001000 bits generated by a random number generator: 0.4880,0.4880, 0.4870,0.4870, 0.5150,0.5150, 0.5490,0.5490, 0.5130,0.5130, 0.5180,0.5180, 0.4860,0.4860, 0.5030,0.5030, 0.5050,0.5050, 0.4980.0.4980. There is no real pattern in either type of average. The random-number-generated averages range between 0.48600.4860 and 0.5490,0.5490, a range of 0.0630,0.0630, whereas the calculated PRNG bit averages range between 0.46800.4680 and 0.5220,0.5220, a range of 0.0540.0.0540.

Section 5.2 Exercises

67.

n=11nn=11n

69.

n=1(−1)n1nn=1(−1)n1n

71.

1,3,6,101,3,6,10

73.

1,1,0,01,1,0,0

75.

an=SnSn1=1n11n.an=SnSn1=1n11n. Series converges to S=1.S=1.

77.

an=SnSn1=nn1=1n1+n.an=SnSn1=nn1=1n1+n. Series diverges because partial sums are unbounded.

79.

S1=1/3,S1=1/3, S2=1/3+2/4>1/3+1/3=2/3,S2=1/3+2/4>1/3+1/3=2/3, S3=1/3+2/4+3/5>3·(1/3)=1.S3=1/3+2/4+3/5>3·(1/3)=1. In general Sk>k/3.Sk>k/3. Series diverges.

81.

S1=1/(2.3)=1/6=2/31/2,S2=1/(2.3)+1/(3.4)=2/12+1/12=1/4=3/41/2,S3=1/(2.3)+1/(3.4)+1/(4.5)=10/60+5/60+3/60=3/10=4/51/2,S4=1/(2.3)+1/(3.4)+1/(4.5)+1/(5.6)=10/60+5/60+3/60+2/60=1/3=5/61/2.S1=1/(2.3)=1/6=2/31/2,S2=1/(2.3)+1/(3.4)=2/12+1/12=1/4=3/41/2,S3=1/(2.3)+1/(3.4)+1/(4.5)=10/60+5/60+3/60=3/10=4/51/2,S4=1/(2.3)+1/(3.4)+1/(4.5)+1/(5.6)=10/60+5/60+3/60+2/60=1/3=5/61/2.

The pattern is Sk=(k+1)/(k+2)1/2Sk=(k+1)/(k+2)1/2 and the series converges to 1/2.1/2.

83.

00

85.

−3−3

87.

diverges, n=10011nn=10011n

89.

convergent geometric series, r=1/10<1r=1/10<1

91.

convergent geometric series, r=π/e2<1r=π/e2<1

93.

n=15·(−1/5)n,n=15·(−1/5)n, converges to −5/6−5/6

95.

n=1100·(1/10)n,n=1100·(1/10)n, converges to 100/9100/9

97.

xn=0(x)n=n=1(−1)n1xnxn=0(x)n=n=1(−1)n1xn

99.

n=0(−1)nsin2n(x)n=0(−1)nsin2n(x)

101.

Sk=221/(k+1)1Sk=221/(k+1)1 as k.k.

103.

Sk=1k+1Sk=1k+1 diverges

105.

n=1lnnln(n+1),Sk=ln(k+1)n=1lnnln(n+1),Sk=ln(k+1)

107.

an=1lnn1ln(n+1)an=1lnn1ln(n+1) and Sk=1ln(2)1ln(k+1)1ln(2)Sk=1ln(2)1ln(k+1)1ln(2)

109.

n=1an=f(1)f(2)n=1an=f(1)f(2)

111.

c0+c1+c2+c3+c4=0c0+c1+c2+c3+c4=0

113.

2n31=1n12n+1n+1,2n31=1n12n+1n+1, Sn=(11+1/3)+(1/22/3+1/4)Sn=(11+1/3)+(1/22/3+1/4) +(1/32/4+1/5)+(1/42/5+1/6)+=1/2+(1/32/4+1/5)+(1/42/5+1/6)+=1/2

115.

tktk converges to 0.57721tk0.57721tk is a sum of rectangles of height 1/k1/k over the interval [k,k+1][k,k+1] which lie above the graph of 1/x.1/x.

This is a graph of a sequence in quadrant one that begins close to 0 and appears to converge to 0.57721.
117.

N=22,N=22, SN=6.1415SN=6.1415

119.

N=3,N=3, SN=1.559877597243667...SN=1.559877597243667...

121.

a. The probability of any given ordered sequence of outcomes for nn coin flips is 1/2n.1/2n. b. The probability of coming up heads for the first time on the nn th flip is the probability of the sequence TTTHTTTH which is 1/2n.1/2n. The probability of coming up heads for the first time on an even flip is n=11/22nn=11/22n or 1/3.1/3.

123.

5/95/9

125.

E=n=1n/2n+1=1,E=n=1n/2n+1=1, as can be shown using summation by parts

127.

The part of the first dose after nn hours is drn,drn, the part of the second dose is drnN,drnN, and, in general, the part remaining of the mthmth dose is drnmN,drnmN, so

A(n)=l=0mdrnlN=l=0mdrk+(ml)N=q=0mdrk+qN=drkq=0mrNq=drk1r(m+1)N1rN,n=k+mN.A(n)=l=0mdrnlN=l=0mdrk+(ml)N=q=0mdrk+qN=drkq=0mrNq=drk1r(m+1)N1rN,n=k+mN.

129.

SN+1=aN+1+SNSNSN+1=aN+1+SNSN

131.

Since S>1,S>1, a2>0,a2>0, and since k<1,k<1, S2=1+a2<1+(S1)=S.S2=1+a2<1+(S1)=S. If Sn>SSn>S for some n, then there is a smallest n. For this n, S>Sn1,S>Sn1, so Sn=Sn1+k(SSn1)Sn=Sn1+k(SSn1) =kS+(1k)Sn1<S,=kS+(1k)Sn1<S, a contradiction. Thus Sn<SSn<S and an+1>0an+1>0 for all n, so SnSn is increasing and bounded by S.S. Let S=limSn.S=limSn. If S<S,S<S, then δ=k(SS)>0,δ=k(SS)>0, but we can find n such that S*Sn<δ/2,S*Sn<δ/2, which implies that Sn+1=Sn+k(SSn)Sn+1=Sn+k(SSn) >S*+δ/2,>S*+δ/2, contradicting that SnSn is increasing to S.S. Thus SnS.SnS.

133.

Let Sk=n=1kanSk=n=1kan and SkL.SkL. Then SkSk eventually becomes arbitrarily close to L,L, which means that LSN=n=N+1anLSN=n=N+1an becomes arbitrarily small as N.N.

135.

L=(1+12)n=11/2n=32.L=(1+12)n=11/2n=32.

137.

At stage one a square of area 1/91/9 is removed, at stage 22 one removes 88 squares of area 1/92,1/92, at stage three one removes 8282 squares of area 1/93,1/93, and so on. The total removed area after NN stages is n=0N18N/9N+1=18(1(8/9)N)/(18/9)1n=0N18N/9N+1=18(1(8/9)N)/(18/9)1

as N.N. The total perimeter is 4+4n=08N/3N+1.4+4n=08N/3N+1.

Section 5.3 Exercises

139.

limnan=0.limnan=0. Divergence test does not apply.

141.

limnan=2.limnan=2. Series diverges.

143.

limnan=limnan= (does not exist). Series diverges.

145.

limnan=1.limnan=1. Series diverges.

147.

limnanlimnan does not exist. Series diverges.

149.

limnan=1/e2.limnan=1/e2. Series diverges.

151.

limnan=0.limnan=0. Divergence test does not apply.

153.

Series converges, p>1.p>1.

155.

Series converges, p=4/3>1.p=4/3>1.

157.

Series converges, p=2eπ>1.p=2eπ>1.

159.

Series diverges by comparison with 1dx(x+5)1/3.1dx(x+5)1/3.

161.

Series diverges by comparison with 1x1+x2dx.1x1+x2dx.

163.

Series converges by comparison with 12x1+x4dx.12x1+x4dx.

165.

2lnn=1/nln2.2lnn=1/nln2. Since ln2<1,ln2<1, diverges by pp-series.

167.

2−2lnn=1/n2ln2.2−2lnn=1/n2ln2. Since 2ln21<1,2ln21<1, diverges by pp-series.

169.

R10001000dtt2=1t|1000=0.001R10001000dtt2=1t|1000=0.001

171.

R10001000dt1+t2=tan−1tan−1(1000)=π/2tan−1(1000)0.000999R10001000dt1+t2=tan−1tan−1(1000)=π/2tan−1(1000)0.000999

173.

RN<Ndxx2=1/N,N>104RN<Ndxx2=1/N,N>104

175.

RN<Ndxx1.01=100N−0.01,N>10600RN<Ndxx1.01=100N−0.01,N>10600

177.

RN<Ndx1+x2=π/2tan−1(N),N>tan(π/210−3)1000RN<Ndx1+x2=π/2tan−1(N),N>tan(π/210−3)1000

179.

RN<Ndxex=eN,N>5ln(10),RN<Ndxex=eN,N>5ln(10), okay if N=12;n=112en=0.581973....N=12;n=112en=0.581973.... Estimate agrees with 1/(e1)1/(e1) to five decimal places.

181.

RN<Ndx/x4=4/N3,N>(4.104)1/3,RN<Ndx/x4=4/N3,N>(4.104)1/3, okay if N=35;N=35;

n=1351/n4=1.08231.n=1351/n4=1.08231. Estimate agrees with the sum to four decimal places.

183.

ln(2)ln(2)

185.

T=0.5772...T=0.5772...

187.

The expected number of random insertions to get BB to the top is n+n/2+n/3++n/(n1).n+n/2+n/3++n/(n1). Then one more insertion puts BB back in at random. Thus, the expected number of shuffles to randomize the deck is n(1+1/2++1/n).n(1+1/2++1/n).

189.

Set bn=an+Nbn=an+N and g(t)=f(t+N)g(t)=f(t+N) such that ff is decreasing on [t,).[t,).

191.

The series converges for p>1p>1 by integral test using change of variable.

193.

N=ee100e1043N=ee100e1043 terms are needed.

Section 5.4 Exercises

195.

Converges by comparison with 1/n2.1/n2.

197.

Diverges by comparison with harmonic series, since 2n1n.2n1n.

199.

an=1/(n+1)(n+2)<1/n2.an=1/(n+1)(n+2)<1/n2. Converges by comparison with p-series, p=2.p=2.

201.

sin(1/n)1/n,sin(1/n)1/n, so converges by comparison with p-series, p=2.p=2.

203.

sin(1/n)1,sin(1/n)1, so converges by comparison with p-series, p=3/2.p=3/2.

205.

Since n+1n=1/(n+1+n)2/n,n+1n=1/(n+1+n)2/n, series converges by comparison with p-series for p=1.5.p=1.5.

207.

Converges by limit comparison with p-series for p>1.p>1.

209.

Converges by limit comparison with p-series, p=2.p=2.

211.

Converges by limit comparison with 4n.4n.

213.

Converges by limit comparison with 1/e1.1n.1/e1.1n.

215.

Diverges by limit comparison with harmonic series.

217.

Converges by limit comparison with p-series, p=3.p=3.

219.

Converges by limit comparison with p-series, p=3.p=3.

221.

Diverges by limit comparison with 1/n.1/n.

223.

Converges for p>1p>1 by comparison with a pp series for slightly smaller p.p.

225.

Converges for all p>0.p>0.

227.

Converges for all r>1.r>1. If r>1r>1 then rn>4,rn>4, say, once n>ln(2)/ln(r)n>ln(2)/ln(r) and then the series converges by limit comparison with a geometric series with ratio 1/2.1/2.

229.

The numerator is equal to 11 when nn is odd and 00 when nn is even, so the series can be rewritten n=112n+1,n=112n+1, which diverges by limit comparison with the harmonic series.

231.

(ab)2=a22ab+b2(ab)2=a22ab+b2 or a2+b22ab,a2+b22ab, so convergence follows from comparison of 2anbn2anbn with a2n+b2n.a2n+b2n. Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.

233.

(lnn)lnn=eln(n)lnln(n).(lnn)lnn=eln(n)lnln(n). If nn is sufficiently large, then lnlnn>2,lnlnn>2, so (lnn)lnn<1/n2,(lnn)lnn<1/n2, and the series converges by comparison to a pseries.pseries.

235.

an0,an0, so a2n|an|a2n|an| for large n.n. Convergence follows from limit comparison. 1/n21/n2 converges, but 1/n1/n does not, so the fact that n=1a2nn=1a2n converges does not imply that n=1ann=1an converges.

237.

No. n=11/nn=11/n diverges. Let bk=0bk=0 unless k=n2k=n2 for some n.n. Then kbk/k=1/k2kbk/k=1/k2 converges.

239.

|sint||t|,|sint||t|, so the result follows from the comparison test.

241.

By the comparison test, x=n=1bn/2nn=11/2n=1.x=n=1bn/2nn=11/2n=1.

243.

If b1=0,b1=0, then, by comparison, xn=21/2n=1/2.xn=21/2n=1/2.

245.

Yes. Keep adding 1-kg1-kg weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the 1-kg1-kg weights, and add 0.1-kg0.1-kg weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last 0.1-kg0.1-kg weight. Start adding 0.01-kg0.01-kg weights. If it balances, stop. If it tips to the side with the weights, remove the last 0.01-kg0.01-kg weight that was added. Continue in this way for the 0.001-kg0.001-kg weights, and so on. After a finite number of steps, one has a finite series of the form A+n=1Nsn/10nA+n=1Nsn/10n where AA is the number of full kg weights and dndn is the number of 1/10n-kg1/10n-kg weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the NthNth partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most 1/10N.1/10N.

247.

a. 10d10d1<10d10d10d1<10d b. h(d)<9dh(d)<9d c. m(d)=10d1+1m(d)=10d1+1 d. Group the terms in the deleted harmonic series together by number of digits. h(d)h(d) bounds the number of terms, and each term is at most 1/m(d).1/m(d). d=1h(d)/m(d)d=19d/(10)d190.d=1h(d)/m(d)d=19d/(10)d190. One can actually use comparison to estimate the value to smaller than 80.80. The actual value is smaller than 23.23.

249.

Continuing the hint gives SN=(1+1/N2)(1+1/(N1)2(1+1/4)).SN=(1+1/N2)(1+1/(N1)2(1+1/4)). Then ln(SN)=ln(1+1/N2)+ln(1+1/(N1)2)++ln(1+1/4).ln(SN)=ln(1+1/N2)+ln(1+1/(N1)2)++ln(1+1/4). Since ln(1+t)ln(1+t) is bounded by a constant times t,t, when 0<t<10<t<1 one has ln(SN)Cn=1N1n2,ln(SN)Cn=1N1n2, which converges by comparison to the p-series for p=2.p=2.

Section 5.5 Exercises

251.

Does not converge by divergence test. Terms do not tend to zero.

253.

Converges conditionally by alternating series test, since n+3/nn+3/n is decreasing. Does not converge absolutely by comparison with p-series, p=1/2.p=1/2.

255.

Converges absolutely by limit comparison to 3n/4n,3n/4n, for example.

257.

Diverges by divergence test since limn|an|=e.limn|an|=e.

259.

Does not converge. Terms do not tend to zero.

261.

limncos2(1/n)=1.limncos2(1/n)=1. Diverges by divergence test.

263.

Converges by alternating series test.

265.

Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, p=πep=πe

267.

Diverges; terms do not tend to zero.

269.

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

271.

Converges absolutely by limit comparison with p-series, p=3/2,p=3/2, after applying the hint.

273.

Converges by alternating series test since n(tan−1(n+1)tan−1n)n(tan−1(n+1)tan−1n) is decreasing to zero for large n.n. Does not converge absolutely by limit comparison with harmonic series after applying hint.

275.

Converges absolutely, since an=1n1n+1an=1n1n+1 are terms of a telescoping series.

277.

Terms do not tend to zero. Series diverges by divergence test.

279.

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

281.

ln(N+1)>10,ln(N+1)>10, N+1>e10,N+1>e10, N22026;N22026; S22026=0.0257S22026=0.0257

283.

2N+1>1062N+1>106 or N+1>6ln(10)/ln(2)=19.93.N+1>6ln(10)/ln(2)=19.93. or N19;N19; S19=0.333333969S19=0.333333969

285.

(N+1)2>106(N+1)2>106 or N>999;N>999; S10000.822466.S10000.822466.

287.

True. bnbn need not tend to zero since if cn=bnlimbn,cn=bnlimbn, then c2n1c2n=b2n1b2n.c2n1c2n=b2n1b2n.

289.

True. b3n1b3n0,b3n1b3n0, so convergence of b3n2b3n2 follows from the comparison test.

291.

True. If one converges, then so must the other, implying absolute convergence.

293.

Yes. Take bn=1bn=1 if an0an0 and bn=0bn=0 if an<0.an<0. Then n=1anbn=n:an0ann=1anbn=n:an0an converges. Similarly, one can show n:an<0ann:an<0an converges. Since both series converge, the series must converge absolutely.

295.

Not decreasing. Does not converge absolutely.

297.

Not alternating. Can be expressed as n=1(13n2+13n113n),n=1(13n2+13n113n), which diverges by comparison with 13n2.13n2.

299.

Let a+n=ana+n=an if an0an0 and a+n=0a+n=0 if an<0.an<0. Then a+n|an|a+n|an| for all nn so the sequence of partial sums of a+na+n is increasing and bounded above by the sequence of partial sums of |an|,|an|, which converges; hence, n=1a+nn=1a+n converges.

301.

For N=5N=5 one has |RN|b6=θ10/10!.|RN|b6=θ10/10!. When θ=1,θ=1, R51/10!2.75×10−7.R51/10!2.75×10−7. When θ=π/6,θ=π/6, R5(π/6)10/10!4.26×10−10.R5(π/6)10/10!4.26×10−10. When θ=π,θ=π, R5π10/10!=0.0258.R5π10/10!=0.0258.

303.

Let bn=1/(2n2)!.bn=1/(2n2)!. Then RN1/(2N)!<0.00001RN1/(2N)!<0.00001 when (2N)!>105(2N)!>105 or N=5N=5 and 112!+14!16!+18!=0.540325,112!+14!16!+18!=0.540325, whereas cos1=0.5403023cos1=0.5403023

305.

Let T=1n2.T=1n2. Then TS=12T,TS=12T, so S=T/2.S=T/2. 6×n=110001/n2=3.140638;6×n=110001/n2=3.140638; 12×n=11000(−1)n1/n2=3.141591;12×n=11000(−1)n1/n2=3.141591;

π=3.141592.π=3.141592. The alternating series is more accurate for 10001000 terms.

307.

N=6,N=6, SN=0.9068SN=0.9068

309.

ln(2).ln(2). The 3nth3nth partial sum is the same as that for the alternating harmonic series.

311.

The series jumps rapidly near the endpoints. For xx away from the endpoints, the graph looks like π(1/2x).π(1/2x).

This shows a function in quadrants 1 and 4 that begins at (0, 0), sharply increases to just below 1.5 close to the y axis, decreases linearly, crosses the x axis at 0.5, continues to decrease linearly, and sharply increases just before 1 to 0.
313.

Here is a typical result. The top curve consists of partial sums of the harmonic series. The bottom curve plots partial sums of a random harmonic series.

This shows two curves. The top is an increasing concave down curve. The bottom is a jagged, random harmonic series plot that stays close to 0.
315.

By the alternating series test, |SnS|bn+1,|SnS|bn+1, so one needs 104104 terms of the alternating harmonic series to estimate ln(2)ln(2) to within 0.0001.0.0001. The first 1010 partial sums of the series n=11n2nn=11n2n are (up to four decimals) 0.5000,0.6250,0.6667,0.6823,0.6885,0.6911,0.6923,0.6928,0.6930,0.69310.5000,0.6250,0.6667,0.6823,0.6885,0.6911,0.6923,0.6928,0.6930,0.6931 and the tenth partial sum is within 0.00010.0001 of ln(2)=0.6931.ln(2)=0.6931.

Section 5.6 Exercises

317.

an+1/an0.an+1/an0. Converges.

319.

an+1an=12(n+1n)21/2<1.an+1an=12(n+1n)21/2<1. Converges.

321.

an+1an1/27<1.an+1an1/27<1. Converges.

323.

an+1an4/e2<1.an+1an4/e2<1. Converges.

325.

an+1an1.an+1an1. Ratio test is inconclusive.

327.

anan+11/e2.anan+11/e2. Converges.

329.

(ak)1/k2>1.(ak)1/k2>1. Diverges.

331.

(an)1/n1/2<1.(an)1/n1/2<1. Converges.

333.

(ak)1/k1/e<1.(ak)1/k1/e<1. Converges.

335.

an1/n=1e+1n1e<1.an1/n=1e+1n1e<1. Converges.

337.

an1/n=(ln(1+lnn))(lnn)0an1/n=(ln(1+lnn))(lnn)0 by L’Hôpital’s rule. Converges.

339.

ak+1ak=12k+10.ak+1ak=12k+10. Converges by ratio test.

341.

(an)1/n1/e.(an)1/n1/e. Converges by root test.

343.

ak1/kln(3)>1.ak1/kln(3)>1. Diverges by root test.

345.

an+1an=an+1an= 32n+123n2+3n+10.32n+123n2+3n+10. Converge.

347.

Converges by root test and limit comparison test since xn2.xn2.

349.

Converges absolutely by limit comparison with pseries,pseries, p=2.p=2.

351.

limnan=1/e20.limnan=1/e20. Series diverges.

353.

Terms do not tend to zero: ak1/2,ak1/2, since sin2x1.sin2x1.

355.

an=2(n+1)(n+2),an=2(n+1)(n+2), which converges by comparison with pseriespseries for p=2.p=2.

357.

ak=2k1·2k(2k+1)(2k+2)3k(2/3)kak=2k1·2k(2k+1)(2k+2)3k(2/3)k converges by comparison with geometric series.

359.

akelnk2=1/k2.akelnk2=1/k2. Series converges by limit comparison with pseries,pseries, p=2.p=2.

361.

If bk=c1k/(c1)bk=c1k/(c1) and ak=k,ak=k, then bk+1bk=ckbk+1bk=ck and n=1kck=a1b1+1c1k=1ck=c(c1)2.n=1kck=a1b1+1c1k=1ck=c(c1)2.

363.

6+4+1=116+4+1=11

365.

|x|1|x|1

367.

|x|<|x|<

369.

All real numbers pp by the ratio test.

371.

r<1/pr<1/p

373.

0<r<1.0<r<1. Note that the ratio and root tests are inconclusive. Using the hint, there are 2k2k terms rnrn for k2n<(k+1)2,k2n<(k+1)2, and for r<1r<1 each term is at least rk.rk. Thus, n=1rn=k=1n=k2(k+1)21rnn=1rn=k=1n=k2(k+1)21rn k=12krk,k=12krk, which converges by the ratio test for r<1.r<1. For r1r1 the series diverges by the divergence test.

375.

One has a1=1,a1=1, a2=a3=1/2,…a2n=a2n+1=1/2n.a2=a3=1/2,…a2n=a2n+1=1/2n. The ratio test does not apply because an+1/an=1an+1/an=1 if nn is even. However, an+2/an=1/2,an+2/an=1/2, so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.

377.

a2n/an=12·n+1n+1+xn+2n+2+x2n2n+x.a2n/an=12·n+1n+1+xn+2n+2+x2n2n+x. The inverse of the kthkth factor is (n+k+x)/(n+k)>1+x/(2n)(n+k+x)/(n+k)>1+x/(2n) so the product is less than (1+x/(2n))nex/2.(1+x/(2n))nex/2. Thus for x>0,x>0, a2nan12ex/2.a2nan12ex/2. The series converges for x>0.x>0.

Chapter Review Exercises

379.

false

381.

true

383.

unbounded, not monotone, divergent

385.

bounded, monotone, convergent, 00

387.

unbounded, not monotone, divergent

389.

diverges

391.

converges

393.

converges, but not absolutely

395.

converges absolutely

397.

converges absolutely

399.

1212

401.

,, 0,0, x0x0

403.

S10383,S10383, limnSn=400limnSn=400

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