Calculus Volume 2

# Chapter 5

### Checkpoint

5.1

$a n = ( −1 ) n + 1 3 + 2 n a n = ( −1 ) n + 1 3 + 2 n$

5.2

$a n = 6 n − 10 a n = 6 n − 10$

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

$2 2$

5.6

$0 . 0 .$

5.7

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

5.8

$10 . 10 .$

5.9

$5 / 7 5 / 7$

5.10

$475 / 90 475 / 90$

5.11

$e − 1 e − 1$

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

$0.04762 0.04762$

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 .

${ a n } = { 1 , 3 , 6 , 10 , 15 , 21 ,… } { a n } = { 1 , 3 , 6 , 10 , 15 , 21 ,… }$

5 .

$a n = n ( n + 1 ) 2 a n = n ( n + 1 ) 2$

7 .

$a n = 4 n − 7 a n = 4 n − 7$

9 .

$a n = 3.10 1 − n = 30.10 − n a n = 3.10 1 − n = 30.10 − n$

11 .

$a n = 2 n − 1 − 1 a n = 2 n − 1 − 1$

13 .

$a n = ( −1 ) n − 1 2 n − 1 a n = ( −1 ) n − 1 2 n − 1$

15 .

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

17 .

$f ( n ) = n ! / 2 n − 2 f ( n ) = n ! / 2 n − 2$

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 .

$7 7$

25 .

$0 0$

27 .

$0 0$

29 .

$1 1$

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 .

$0 0$

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 .

$∑ n = 1 ∞ 1 n ∑ n = 1 ∞ 1 n$

69 .

$∑ n = 1 ∞ ( −1 ) n − 1 n ∑ n = 1 ∞ ( −1 ) n − 1 n$

71 .

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

73 .

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

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 .

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

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 .

$0 0$

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 .

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

99 .

$∑ n = 0 ∞ ( −1 ) n sin 2 n ( x ) ∑ n = 0 ∞ ( −1 ) n sin 2 n ( x )$

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 .

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

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 .

$∑ n = 1 ∞ a n = f ( 1 ) − f ( 2 ) ∑ n = 1 ∞ a n = f ( 1 ) − f ( 2 )$

111 .

$c 0 + c 1 + c 2 + c 3 + c 4 = 0 c 0 + c 1 + c 2 + c 3 + c 4 = 0$

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 .

$5 / 9 5 / 9$

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

$A ( n ) = ∑ l = 0 m d r n − l N = ∑ l = 0 m d r k + ( m − l ) N = ∑ q = 0 m d r k + q N = d r k ∑ q = 0 m r N q = d r k 1 − r ( m + 1 ) N 1 − r N , n = k + m N . A ( n ) = ∑ l = 0 m d r n − l N = ∑ l = 0 m d r k + ( m − l ) N = ∑ q = 0 m d r k + q N = d r k ∑ q = 0 m r N q = d r k 1 − r ( m + 1 ) N 1 − r N , n = k + m N .$

129 .

$S N + 1 = a N + 1 + S N ≥ S N S N + 1 = a N + 1 + S N ≥ S N$

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 .

$L = ( 1 + 1 2 ) ∑ n = 1 ∞ 1 / 2 n = 3 2 . L = ( 1 + 1 2 ) ∑ n = 1 ∞ 1 / 2 n = 3 2 .$

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 .

$R 1000 ≤ ∫ 1000 ∞ d t t 2 = − 1 t | 1000 ∞ = 0.001 R 1000 ≤ ∫ 1000 ∞ d t t 2 = − 1 t | 1000 ∞ = 0.001$

171 .

$R 1000 ≤ ∫ 1000 ∞ d t 1 + t 2 = tan −1 ∞ − tan −1 ( 1000 ) = π / 2 − tan −1 ( 1000 ) ≈ 0.000999 R 1000 ≤ ∫ 1000 ∞ d t 1 + t 2 = tan −1 ∞ − tan −1 ( 1000 ) = π / 2 − tan −1 ( 1000 ) ≈ 0.000999$

173 .

$R N < ∫ N ∞ d x x 2 = 1 / N , N > 10 4 R N < ∫ N ∞ d x x 2 = 1 / N , N > 10 4$

175 .

$R N < ∫ N ∞ d x x 1.01 = 100 N −0.01 , N > 10 600 R N < ∫ N ∞ d x x 1.01 = 100 N −0.01 , N > 10 600$

177 .

$R N < ∫ N ∞ d x 1 + x 2 = π / 2 − tan −1 ( N ) , N > tan ( π / 2 − 10 −3 ) ≈ 1000 R N < ∫ N ∞ d x 1 + x 2 = π / 2 − tan −1 ( N ) , N > tan ( π / 2 − 10 −3 ) ≈ 1000$

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 .

$6 + 4 + 1 = 11 6 + 4 + 1 = 11$

365 .

$| x | ≤ 1 | x | ≤ 1$

367 .

$| x | < ∞ | x | < ∞$

369 .

All real numbers $pp$ by the ratio test.

371 .

$r < 1 / p r < 1 / p$

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 $x>0.x>0.$

### 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 .

$1 2 1 2$

401 .

$∞,∞,$ $0,0,$ $x0x0$

403 .

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

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