Calculus Volume 2

# Chapter 5

### Checkpoint

5.1

$an=(−1)n+13+2nan=(−1)n+13+2n$

5.2

$an=6n−10an=6n−10$

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

$e−1e−1$

5.12

The series diverges.

5.13

The series diverges.

5.14

The series converges.

5.15

$S5≈1.09035,S5≈1.09035,$ $R5<0.00267R5<0.00267$

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.

$an=4n−7an=4n−7$

9.

$an=3.101−n=30.10−nan=3.101−n=30.10−n$

11.

$an=2n−1an=2n−1$

13.

$an=(−1)n−12n−1an=(−1)n−12n−1$

15.

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

17.

$f(n)=n!/2n−2f(n)=n!/2n−2$

19.

Terms oscillate above and below $5/35/3$ and appear to converge to $5/3.5/3.$ 21.

Terms oscillate above and below $y≈1.57...y≈1.57...$ and appear to converge to a limit. 23.

$77$

25.

$00$

27.

$00$

29.

$11$

31.

bounded, decreasing for $n≥1n≥1$

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 $−1n≤an≤1n).−1n≤an≤1n).$

45.

Graph oscillates and suggests no limit. 47.

$n1/n→1n1/n→1$ and $21/n→1,21/n→1,$ so $an→0an→0$

49.

Since $(1+1/n)n→e,(1+1/n)n→e,$ one has $(1−2/n)n≈(1+k)−2k→e−2(1−2/n)n≈(1+k)−2k→e−2$ as $k→∞.k→∞.$

51.

$2n+3n≤2·3n2n+3n≤2·3n$ and $3n/4n→03n/4n→0$ as $n→∞,n→∞,$ so $an→0an→0$ as $n→∞.n→∞.$

53.

$an+1an=n!/(n+1)(n+2)⋯(2n)an+1an=n!/(n+1)(n+2)⋯(2n)$ $=1·2·3⋯n(n+1)(n+2)⋯(2n)<1/2n.=1·2·3⋯n(n+1)(n+2)⋯(2n)<1/2n.$ In particular, $an+1/an≤1/2,an+1/an≤1/2,$ so $an→0an→0$ as $n→∞.n→∞.$

55.

$xn+1=xn−((xn−1)2−2)/2(xn−1);xn+1=xn−((xn−1)2−2)/2(xn−1);$ $x=1+2,x=1+2,$ $x≈2.4142,x≈2.4142,$ $n=5n=5$

57.

$xn+1=xn−xn(ln(xn)−1);xn+1=xn−xn(ln(xn)−1);$ $x=e,x=e,$ $x≈2.7183,x≈2.7183,$ $n=5n=5$

59.

a. Without losses, the population would obey $Pn=1.06Pn−1.Pn=1.06Pn−1.$ The subtraction of $150150$ accounts for fish losses. b. After $1212$ months, we have $P12≈1494.P12≈1494.$

61.

a. The student owes $93839383$ after $1212$ months. b. The loan will be paid in full after $139139$ months or eleven and a half years.

63.

$b1=0,b1=0,$ $x1=2/3,x1=2/3,$ $b2=1,b2=1,$ $x2=4/3−1=1/3,x2=4/3−1=1/3,$ 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.

$∑n=1∞1n∑n=1∞1n$

69.

$∑n=1∞(−1)n−1n∑n=1∞(−1)n−1n$

71.

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

73.

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

75.

$an=Sn−Sn−1=1n−1−1n.an=Sn−Sn−1=1n−1−1n.$ Series converges to $S=1.S=1.$

77.

$an=Sn−Sn−1=n−n−1=1n−1+n.an=Sn−Sn−1=n−n−1=1n−1+n.$ 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.

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

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, $∑n=1001∞1n∑n=1001∞1n$

89.

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

91.

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

93.

$∑n=1∞5·(−1/5)n,∑n=1∞5·(−1/5)n,$ converges to $−5/6−5/6$

95.

$∑n=1∞100·(1/10)n,∑n=1∞100·(1/10)n,$ converges to $100/9100/9$

97.

$x∑n=0∞(−x)n=∑n=1∞(−1)n−1xnx∑n=0∞(−x)n=∑n=1∞(−1)n−1xn$

99.

$∑n=0∞(−1)nsin2n(x)∑n=0∞(−1)nsin2n(x)$

101.

$Sk=2−21/(k+1)→1Sk=2−21/(k+1)→1$ as $k→∞.k→∞.$

103.

$Sk=1−k+1Sk=1−k+1$ diverges

105.

$∑n=1∞lnn−ln(n+1),Sk=−ln(k+1)∑n=1∞lnn−ln(n+1),Sk=−ln(k+1)$

107.

$an=1lnn−1ln(n+1)an=1lnn−1ln(n+1)$ and $Sk=1ln(2)−1ln(k+1)→1ln(2)Sk=1ln(2)−1ln(k+1)→1ln(2)$

109.

$∑n=1∞an=f(1)−f(2)∑n=1∞an=f(1)−f(2)$

111.

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

113.

$2n3−1=1n−1−2n+1n+1,2n3−1=1n−1−2n+1n+1,$ $Sn=(1−1+1/3)+(1/2−2/3+1/4)Sn=(1−1+1/3)+(1/2−2/3+1/4)$ $+(1/3−2/4+1/5)+(1/4−2/5+1/6)+⋯=1/2+(1/3−2/4+1/5)+(1/4−2/5+1/6)+⋯=1/2$

115.

$tktk$ converges to $0.57721…tk0.57721…tk$ 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 $1/x.1/x.$ 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 $TT…THTT…TH$ which is $1/2n.1/2n.$ The probability of coming up heads for the first time on an even flip is $∑n=1∞1/22n∑n=1∞1/22n$ or $1/3.1/3.$

123.

$5/95/9$

125.

$E=∑n=1∞n/2n+1=1,E=∑n=1∞n/2n+1=1,$ 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 $drn−N,drn−N,$ and, in general, the part remaining of the $mthmth$ dose is $drn−mN,drn−mN,$ so

$A(n)=∑l=0mdrn−lN=∑l=0mdrk+(m−l)N=∑q=0mdrk+qN=drk∑q=0mrNq=drk1−r(m+1)N1−rN,n=k+mN.A(n)=∑l=0mdrn−lN=∑l=0mdrk+(m−l)N=∑q=0mdrk+qN=drk∑q=0mrNq=drk1−r(m+1)N1−rN,n=k+mN.$

129.

$SN+1=aN+1+SN≥SNSN+1=aN+1+SN≥SN$

131.

Since $S>1,S>1,$ $a2>0,a2>0,$ and since $k<1,k<1,$ $S2=1+a2<1+(S−1)=S.S2=1+a2<1+(S−1)=S.$ If $Sn>SSn>S$ for some n, then there is a smallest n. For this n, $S>Sn−1,S>Sn−1,$ so $Sn=Sn−1+k(S−Sn−1)Sn=Sn−1+k(S−Sn−1)$ $=kS+(1−k)Sn−1 a contradiction. Thus $Sn 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∗ then $δ=k(S−S∗)>0,δ=k(S−S∗)>0,$ but we can find n such that $S*−Sn<δ/2,S*−Sn<δ/2,$ which implies that $Sn+1=Sn+k(S−Sn)Sn+1=Sn+k(S−Sn)$ $>S*+δ/2,>S*+δ/2,$ contradicting that $SnSn$ is increasing to $S∗.S∗.$ Thus $Sn→S.Sn→S.$

133.

Let $Sk=∑n=1kanSk=∑n=1kan$ and $Sk→L.Sk→L.$ Then $SkSk$ eventually becomes arbitrarily close to $L,L,$ which means that $L−SN=∑n=N+1∞anL−SN=∑n=N+1∞an$ becomes arbitrarily small as $N→∞.N→∞.$

135.

$L=(1+12)∑n=1∞1/2n=32.L=(1+12)∑n=1∞1/2n=32.$

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 $∑n=0N−18N/9N+1=18(1−(8/9)N)/(1−8/9)→1∑n=0N−18N/9N+1=18(1−(8/9)N)/(1−8/9)→1$

as $N→∞.N→∞.$ The total perimeter is $4+4∑n=08N/3N+1→∞.4+4∑n=08N/3N+1→∞.$

### Section 5.3 Exercises

139.

$limn→∞an=0.limn→∞an=0.$ Divergence test does not apply.

141.

$limn→∞an=2.limn→∞an=2.$ Series diverges.

143.

$limn→∞an=∞limn→∞an=∞$ (does not exist). Series diverges.

145.

$limn→∞an=1.limn→∞an=1.$ Series diverges.

147.

$limn→∞anlimn→∞an$ does not exist. Series diverges.

149.

$limn→∞an=1/e2.limn→∞an=1/e2.$ Series diverges.

151.

$limn→∞an=0.limn→∞an=0.$ 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, $p=2e−π>1.p=2e−π>1.$

159.

Series diverges by comparison with $∫1∞dx(x+5)1/3.∫1∞dx(x+5)1/3.$

161.

Series diverges by comparison with $∫1∞x1+x2dx.∫1∞x1+x2dx.$

163.

Series converges by comparison with $∫1∞2x1+x4dx.∫1∞2x1+x4dx.$

165.

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

167.

$2−2lnn=1/n2ln2.2−2lnn=1/n2ln2.$ Since $2ln2−1<1,2ln2−1<1,$ diverges by $pp$-series.

169.

$R1000≤∫1000∞dtt2=−1t|1000∞=0.001R1000≤∫1000∞dtt2=−1t|1000∞=0.001$

171.

$R1000≤∫1000∞dt1+t2=tan−1∞−tan−1(1000)=π/2−tan−1(1000)≈0.000999R1000≤∫1000∞dt1+t2=tan−1∞−tan−1(1000)=π/2−tan−1(1000)≈0.000999$

173.

$RN<∫N∞dxx2=1/N,N>104RN<∫N∞dxx2=1/N,N>104$

175.

$RN<∫N∞dxx1.01=100N−0.01,N>10600RN<∫N∞dxx1.01=100N−0.01,N>10600$

177.

$RN<∫N∞dx1+x2=π/2−tan−1(N),N>tan(π/2−10−3)≈1000RN<∫N∞dx1+x2=π/2−tan−1(N),N>tan(π/2−10−3)≈1000$

179.

$RN<∫N∞dxex=e−N,N>5ln(10),RN<∫N∞dxex=e−N,N>5ln(10),$ okay if $N=12;∑n=112e−n=0.581973....N=12;∑n=112e−n=0.581973....$ Estimate agrees with $1/(e−1)1/(e−1)$ to five decimal places.

181.

$RN<∫N∞dx/x4=4/N3,N>(4.104)1/3,RN<∫N∞dx/x4=4/N3,N>(4.104)1/3,$ 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 $n+n/2+n/3+⋯+n/(n−1).n+n/2+n/3+⋯+n/(n−1).$ 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.

$N=ee100≈e1043N=ee100≈e1043$ terms are needed.

### Section 5.4 Exercises

195.

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

197.

Diverges by comparison with harmonic series, since $2n−1≥n.2n−1≥n.$

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 $n+1−n=1/(n+1+n)≤2/n,n+1−n=1/(n+1+n)≤2/n,$ 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 $4−n.4−n.$

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 $∑n=1∞12n+1,∑n=1∞12n+1,$ which diverges by limit comparison with the harmonic series.

231.

$(a−b)2=a2−2ab+b2(a−b)2=a2−2ab+b2$ or $a2+b2≥2ab,a2+b2≥2ab,$ 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.

$(lnn)−lnn=e−ln(n)lnln(n).(lnn)−lnn=e−ln(n)lnln(n).$ 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 $p−series.p−series.$

235.

$an→0,an→0,$ so $a2n≤|an|a2n≤|an|$ for large $n.n.$ Convergence follows from limit comparison. $∑1/n2∑1/n2$ converges, but $∑1/n∑1/n$ does not, so the fact that $∑n=1∞a2n∑n=1∞a2n$ converges does not imply that $∑n=1∞an∑n=1∞an$ converges.

237.

No. $∑n=1∞1/n∑n=1∞1/n$ diverges. Let $bk=0bk=0$ unless $k=n2k=n2$ for some $n.n.$ Then $∑kbk/k=∑1/k2∑kbk/k=∑1/k2$ converges.

239.

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

241.

By the comparison test, $x=∑n=1∞bn/2n≤∑n=1∞1/2n=1.x=∑n=1∞bn/2n≤∑n=1∞1/2n=1.$

243.

If $b1=0,b1=0,$ then, by comparison, $x≤∑n=2∞1/2n=1/2.x≤∑n=2∞1/2n=1/2.$

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. $10d−10d−1<10d10d−10d−1<10d$ b. $h(d)<9dh(d)<9d$ c. $m(d)=10d−1+1m(d)=10d−1+1$ 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 $1/m(d).1/m(d).$ $∑d=1∞h(d)/m(d)≤∑d=1∞9d/(10)d−1≤90.∑d=1∞h(d)/m(d)≤∑d=1∞9d/(10)d−1≤90.$ 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 $SN=(1+1/N2)(1+1/(N−1)2…(1+1/4)).SN=(1+1/N2)(1+1/(N−1)2…(1+1/4)).$ Then $ln(SN)=ln(1+1/N2)+ln(1+1/(N−1)2)+⋯+ln(1+1/4).ln(SN)=ln(1+1/N2)+ln(1+1/(N−1)2)+⋯+ln(1+1/4).$ Since $ln(1+t)ln(1+t)$ is bounded by a constant times $t,t,$ when $0 one has $ln(SN)≤C∑n=1N1n2,ln(SN)≤C∑n=1N1n2,$ 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.

$limn→∞cos2(1/n)=1.limn→∞cos2(1/n)=1.$ 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, $p=π−ep=π−e$

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 $an=1n−1n+1an=1n−1n+1$ 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.

$ln(N+1)>10,ln(N+1)>10,$ $N+1>e10,N+1>e10,$ $N≥22026;N≥22026;$ $S22026=0.0257…S22026=0.0257…$

283.

$2N+1>1062N+1>106$ or $N+1>6ln(10)/ln(2)=19.93.N+1>6ln(10)/ln(2)=19.93.$ or $N≥19;N≥19;$ $S19=0.333333969…S19=0.333333969…$

285.

$(N+1)2>106(N+1)2>106$ or $N>999;N>999;$ $S1000≈0.822466.S1000≈0.822466.$

287.

True. $bnbn$ need not tend to zero since if $cn=bn−limbn,cn=bn−limbn,$ then $c2n−1−c2n=b2n−1−b2n.c2n−1−c2n=b2n−1−b2n.$

289.

True. $b3n−1−b3n≥0,b3n−1−b3n≥0,$ so convergence of $∑b3n−2∑b3n−2$ 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 $an≥0an≥0$ and $bn=0bn=0$ if $an<0.an<0.$ Then $∑n=1∞anbn=∑n:an≥0an∑n=1∞anbn=∑n:an≥0an$ converges. Similarly, one can show $∑n:an<0an∑n:an<0an$ converges. Since both series converge, the series must converge absolutely.

295.

Not decreasing. Does not converge absolutely.

297.

Not alternating. Can be expressed as $∑n=1∞(13n−2+13n−1−13n),∑n=1∞(13n−2+13n−1−13n),$ which diverges by comparison with $∑13n−2.∑13n−2.$

299.

Let $a+n=ana+n=an$ if $an≥0an≥0$ 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, $∑n=1∞a+n∑n=1∞a+n$ converges.

301.

For $N=5N=5$ one has $|RN|b6=θ10/10!.|RN|b6=θ10/10!.$ When $θ=1,θ=1,$ $R5≤1/10!≈2.75×10−7.R5≤1/10!≈2.75×10−7.$ 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 $θ=π,θ=π,$ $R5≤π10/10!=0.0258.R5≤π10/10!=0.0258.$

303.

Let $bn=1/(2n−2)!.bn=1/(2n−2)!.$ Then $RN≤1/(2N)!<0.00001RN≤1/(2N)!<0.00001$ when $(2N)!>105(2N)!>105$ or $N=5N=5$ and $1−12!+14!−16!+18!=0.540325…,1−12!+14!−16!+18!=0.540325…,$ whereas $cos1=0.5403023…cos1=0.5403023…$

305.

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

$π=3.141592….π=3.141592….$ 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 $π(1/2−x).π(1/2−x).$ 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. 315.

By the alternating series test, $|Sn−S|≤bn+1,|Sn−S|≤bn+1,$ 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 $∑n=1∞1n2n∑n=1∞1n2n$ 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.

$an+1/an→0.an+1/an→0.$ Converges.

319.

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

321.

$an+1an→1/27<1.an+1an→1/27<1.$ Converges.

323.

$an+1an→4/e2<1.an+1an→4/e2<1.$ Converges.

325.

$an+1an→1.an+1an→1.$ Ratio test is inconclusive.

327.

$anan+1→1/e2.anan+1→1/e2.$ Converges.

329.

$(ak)1/k→2>1.(ak)1/k→2>1.$ Diverges.

331.

$(an)1/n→1/2<1.(an)1/n→1/2<1.$ Converges.

333.

$(ak)1/k→1/e<1.(ak)1/k→1/e<1.$ Converges.

335.

$an1/n=1e+1n→1e<1.an1/n=1e+1n→1e<1.$ Converges.

337.

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

339.

$ak+1ak=12k+1→0.ak+1ak=12k+1→0.$ Converges by ratio test.

341.

$(an)1/n→1/e.(an)1/n→1/e.$ Converges by root test.

343.

$ak1/k→ln(3)>1.ak1/k→ln(3)>1.$ Diverges by root test.

345.

$an+1an=an+1an=$ $32n+123n2+3n+1→0.32n+123n2+3n+1→0.$ Converge.

347.

Converges by root test and limit comparison test since $xn→2.xn→2.$

349.

Converges absolutely by limit comparison with $p−series,p−series,$ $p=2.p=2.$

351.

$limn→∞an=1/e2≠0.limn→∞an=1/e2≠0.$ Series diverges.

353.

Terms do not tend to zero: $ak≥1/2,ak≥1/2,$ since $sin2x≤1.sin2x≤1.$

355.

$an=2(n+1)(n+2),an=2(n+1)(n+2),$ which converges by comparison with $p−seriesp−series$ for $p=2.p=2.$

357.

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

359.

$ak≈e−lnk2=1/k2.ak≈e−lnk2=1/k2.$ Series converges by limit comparison with $p−series,p−series,$ $p=2.p=2.$

361.

If $bk=c1−k/(c−1)bk=c1−k/(c−1)$ and $ak=k,ak=k,$ then $bk+1−bk=−c−kbk+1−bk=−c−k$ and $∑n=1∞kck=a1b1+1c−1∑k=1∞c−k=c(c−1)2.∑n=1∞kck=a1b1+1c−1∑k=1∞c−k=c(c−1)2.$

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 Note that the ratio and root tests are inconclusive. Using the hint, there are $2k2k$ terms $rnrn$ for $k2≤n<(k+1)2,k2≤n<(k+1)2,$ and for $r<1r<1$ each term is at least $rk.rk.$ Thus, $∑n=1∞rn=∑k=1∞∑n=k2(k+1)2−1rn∑n=1∞rn=∑k=1∞∑n=k2(k+1)2−1rn$ $≥∑k=1∞2krk,≥∑k=1∞2krk,$ which converges by the ratio test for $r<1.r<1.$ For $r≥1r≥1$ 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.

$a2n/an=12·n+1n+1+xn+2n+2+x⋯2n2n+x.a2n/an=12·n+1n+1+xn+2n+2+x⋯2n2n+x.$ 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 $(1+x/(2n))−n≈e−x/2.(1+x/(2n))−n≈e−x/2.$ Thus for $x>0,x>0,$ $a2nan≤12e−x/2.a2nan≤12e−x/2.$ 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.

$S10≈383,S10≈383,$ $limn→∞Sn=400limn→∞Sn=400$