Section 11.2 Series

(1)

Section 11.2 Series

51. Let x = 0.99999 . . .

(a) Do you think that x < 1 or x = 1?

(b) Sum a geometric series to find the value of x.

(c) How many decimal representations does the number 1 have?

(d) Which numbers have more than one decimal representation?

Solution: SECTION 11.2 SERIES ¤ 119

49. (a) Many people would guess that   1, but note that  consists of an infinite number of 9s.

(b)  = 099999    = 9 10 + 9

100+ 9

1000+ 9

10,000+ · · · = ^∞

=1

9

10^, which is a geometric series with 1= 09and

 = 01. Its sum is 09

1 − 01 = 09

09 = 1, that is,  = 1.

(c) The number 1 has two decimal representations, 100000    and 099999    .

(d) Except for 0, all rational numbers that have a terminating decimal representation can be written in more than one way. For example, 05 can be written as 049999    as well as 050000    .

50. 1 = 1, = (5 − )^−1 ⇒ ²= (5 − 2)¹= 3(1) = 3, 3= (5 − 3)²= 2(3) = 6, 4= (5 − 4)³= 1(6) = 6,

5 = (5 − 5)⁴= 0, and all succeeding terms equal 0. Thus, ^∞

=1

=

4

=1

= 1 + 3 + 6 + 6 = 16.

51. 08 = 8 10 + 8

10² + · · · is a geometric series with  = 8

10 and  = 1

10. It converges to 

1 −  = 810 1 − 110 = 8

9.

52. 046 = 46 100+ 46

100² + · · · is a geometric series with  = 46

100and  = 1

100. It converges to 

1 −  = 46100 1 − 1100 = 46

99.

53. 15342 = 153 + 42 10⁴ + 42

10⁶ + · · · . Now 42 10⁴ + 42

10⁶ + · · · is a geometric series with  = 42

10⁴ and  = 1 10². It converges to 

1 −  = 4210⁴

1 − 110² = 4210⁴ 9910² = 42

9900. Thus, 15342 = 153 + 42

9900 = 153 100+ 42

9900 = 15,147 9900 + 42

9900 = 15,189

9900 or 5063 3300. 54. 712345 = 7 +12,345

10⁵ +12,345

10¹⁰ + · · · . Now 12,345

10⁵ +12,345

10¹⁰ + · · · is a geometric series with  = 12,345

10⁵ and  = 1 10⁵. It converges to 

1 −  = 12,34510⁵

1 − 110⁵ = 12,34510⁵

99,99910⁵ = 12,345 99,999. Thus, 712345 = 7 + 12,345

99,999 = 699,993

99,999 +12,345

99,999= 712,338

99,999 or 237,446 33,333.

55. 1234567 = 1234 +567 10⁶ +567

10⁹ + · · · . Now 567 10⁶ + 567

10⁹ + · · · is a geometric series with  = 567 10⁶ and

 = 1

10³. It converges to 

1 −  = 56710⁶

1 − 110³ = 56710⁶

99910³ = 567

999,000 = 21

37,000. Thus, 1234567 = 1234 + 21

37,000= 1234

1000 + 21

37,000 = 45,658

37,000+ 21

37,000= 45,679 37,000.

56. 571358 = 5 +71,358

10⁵ + 71,358

10¹⁰ + · · · . Now71,358

10⁵ +71,358

10¹⁰ + · · · is a geometric series with  = 71,358 10⁵ and

 = 1

10⁵. It converges to 

1 −  = 71,35810⁵

1 − 110⁵ = 71,35810⁵

99,99910⁵ = 71,358

99,999 = 23,786 33,333. Thus, 571358 = 5 + 23,786

33,333 = 166,665

33,333 +23,786

33,333= 190,451 33,333 .

69. If the nth partial sum of a series

∞

P

n=1

an is

sn= n − 1 n + 1 find an and

∞

P

n=1

an. Solution:

SECTION 11.2 SERIES ¤ 121

65. After defining , We use convert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and Simplify in Derive to find that the general term is 3²+ 3 + 1

(²+ )³ = 1

³ − 1

( + 1)³. So the nth partial sum is

=



=1

 1

³ − 1 ( + 1)³



=

 1 − 1

2³

 +

1 2³ − 1

3³



+ · · · +

 1

³ − 1 ( + 1)³



= 1 − 1

( + 1)³ The series converges to lim

→∞= 1. This can be confirmed by directly computing the sum using

sum(f,n=1..infinity); (in Maple), Sum[f,{n,1,Infinity}] (in Mathematica), or Calculus Sum (from 1 to ∞) and Simplify (in Derive).

66. See Exercise 65 for specific CAS commands.

1

⁵− 5³+ 4 = 1

24( − 2)+ 1

24( + 2)− 1

6( − 1)− 1

6( + 1)+ 1

4. So the th partial sum is

= 1 24



=3

 1

 − 2− 4

 − 1+ 6

− 4

 + 1+ 1

 + 2



= 1 24

1 1−4

2+6 3−4

4+ 1 5



+ · · · +

 1

 − 2 − 4

 − 1 + 6

 − 4

 + 1 + 1

 + 2



The terms with denominator 5 or greater cancel, except for a few terms with  in the denominator. So as  → ∞,

→ 1 24

1 1−3

2+ 3 3−1

4



= 1 24

1 4



= 1 96. 67. For  = 1, 1= 0since 1= 0. For   1,

= − ^−1=  − 1

 + 1 −( − 1) − 1

( − 1) + 1 = ( − 1) − ( + 1)( − 2)

( + 1) = 2

( + 1) Also, ^∞

=1

= lim

→∞= lim

→∞

1 − 1

1 + 1 = 1.

68. 1= 1= 3 −¹2 = ⁵₂. For  6= 1,

= − ^−1=

3 − 2^−

−

3 − ( − 1)2^−(−1)

= − 

2^ + − 1 2^⁻¹ · 2

2 = 2( − 1) 2^ − 

2^ =  − 2 2^ Also, ^∞

=1

= lim

→∞= lim

→∞

3 −  2^

= 3because lim

→∞

 2^

= limH

→∞

1 2^ln 2 = 0.

69. (a) The quantity of the drug in the body after the first tablet is 100 mg. After the second tablet, there is 100 mg plus 20% of the first 100-mg tablet; that is, 100 + 020(100) = 120 mg. After the third tablet, the quantity is 100 + 020(120) or, equivalently, 100 + 100(020) + 100(020)². Either expression gives us 124 mg.

(b) From part (a), we see that +1= 100 + 020 .

(c) = 100 + 100(020)¹+ 100(020)²+ · · · + 100(020)^⁻¹

=



=1

100(020)^−1 [geometric with  = 100 and  = 020].

The quantity of the antibiotic that remains in the body in the long run is lim

→∞= 100

1 − 020 = 100

45 = 125 mg.

75. When money is spent on goods and services, those who receive the money also spend some of it. The people receiving some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local government begins the process by spending D dollars. Suppose that each recipient of spent money spends 100c% and saves 100s% of the money that he or she receives. The values c and s are called the marginal propensity to consume and the marginal propensity to save and, of course, c + s = 1.

(a) Let Sn be the total spending that has been generated after n transactions. Find an equation for Sn. (b) Show that lim

n→∞Sn = kD, where k = 1/s. The number k is called the multiplier. What is the multiplier if the marginal propensity to consume is 80%?

Note: The federal government uses this principle to justify deficit spending. Banks use this principle to justify lending a large percentage of the money that they receive in deposits.

1

(2)

Solution:

122 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

70. (a) The concentration of the drug after the first injection is 15 mgL. “Reduced by 90%” is the same as 10% remains, so the concentration after the second injection is 15 + 010(15) = 165 mgL. The concentration after the third injection is 15 + 010(165), or, equivalently, 15 + 15(010) + 15(010)². Either expression gives us 1665 mgL.

(b) = 15 + 15(010)¹+ 15(010)²+ · · · + 15(010)^⁻¹

=



=1

15(010)^⁻¹ [geometric with  = 15 and  = 010].

By (3), = 15[1 − (010)^] 1 − 010 = 15

09[1 − (010)^] = 5

3[1 − (010)^] mgL.

(c) The limiting value of the concentration is lim

→∞= lim

→∞

5

3[1 − (010)^] = ⁵₃(1 − 0) = ⁵3 mgL.

71. (a) The quantity of the drug in the body after the first tablet is 150 mg. After the second tablet, there is 150 mg plus 5%

of the first 150- mg tablet, that is, [150 + 150(005)] mg. After the third tablet, the quantity is [150 + 150(005) + 150(005)²] = 157875 mg. After  tablets, the quantity (in mg) is 150 + 150(005) + · · · + 150(005)^⁻¹. We can use Formula 3 to write this as 150(1 − 005^)

1 − 005 = 3000

19 (1 − 005^).

(b) The number of milligrams remaining in the body in the long run is lim

→∞

3000

19 (1 − 005^)

= ³⁰⁰⁰₁₉ (1 − 0) ≈ 157895, only 002 mg more than the amount after 3 tablets.

72. (a) The residual concentration just before the second injection is ^−; before the third, ^− + ^−2; before the

( + 1)st, ^− + ^−2 + · · · + ^−. This sum is equal to^−

1 − ^−

1 − ^− [Formula 3].

(b) The limiting pre-injection concentration is lim

→∞

^−

1 − ^−

1 − ^− = ^−(1 − 0) 1 − ^− ·^

^ = 

^− 1.

(c) 

^− 1 ≥  ⇒  ≥ 

^ − 1, so the minimal dosage is  = 

^ − 1.

73. (a) The first step in the chain occurs when the local government spends  dollars. The people who receive it spend a fraction  of those  dollars, that is,  dollars. Those who receive the  dollars spend a fraction  of it, that is,

²dollars. Continuing in this way, we see that the total spending after  transactions is

=  +  + ²+ · · · + ^–1= (1 − ^) 1 −  by (3).

(b) lim

→∞ = lim

→∞

(1 − ^)

1 −  = 

1 −  lim

→∞(1 − ^) =  1 − 

since 0    1 ⇒ lim

→∞^= 0

= 

 [since  +  = 1] =  [since  = 1]

If  = 08, then  = 1 −  = 02 and the multiplier is  = 1 = 5.

83. The figure shows two circles C and D of radius 1 that touch at P . The line T is a common tangent line; C1 is the circle that touches C, D, and T ; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C₂. This procedure can be continued indefinitely and produces an infinite sequence of circles {C_n}. Find an expression for the diameter of C_n and thus provide another geometric demonstration of Example 2.

718 chapter 11 Infinite Sequences and Series

84. If o an is divergent and c ± 0, show that o can is divergent.

85. If o an is convergent and o bn is divergent, show that the series o san1bnd is divergent. [Hint: Argue by contradiction.]

86. If o an and o bn are both divergent, is o san1bnd necessarily divergent?

87. Suppose that a series o an has positive terms and its partial sums sn satisfy the inequality sn<1000 for all n. Explain why o an must be convergent.

88. The Fibonacci sequence was defined in Section 11.1 by the equations

f1− 1, f2− 1, fn− fn211fn22 n > 3 Show that each of the following statements is true.

(a) 1

fn21 fn11 − 1

fn21 fn 2 1 fn fn11

(b) _n−2

o

^` _f_n21¹_f_n11 ^{− 1 (c)}_n−2

o

^` _f_n21^f_fⁿ_n11 ^{− 2}

89. The Cantor set, named after the German mathematician Georg Cantor (1845–1918), is constructed as follows. We start with the closed interval [0, 1] and remove the open interval

(

¹₃, ²₃

)

. That leaves the two intervals

f

^0,¹3

g

^and

f

²3, 1

g

^{and we}

remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them.

We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in [0, 1] after all those intervals have been removed.

(a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantor set contains infi- nitely many numbers. Give examples of some numbers in the Cantor set.

(b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0.

90. (a) A sequence hanj is defined recursively by the equation an−¹₂san211an22d for n > 3, where a1 and a2 can be any real numbers. Experiment with various values of a1

and a2 and use your calculator to guess the limit of the sequence.

79. The figure shows two circles C and D of radius 1 that touch at P. The line T is a common tangent line; C1 is the circle that touches C, D, and T; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C2. This procedure can be continued indefinitely and produces an infinite sequence of circles hCnj. Find an expression for the diameter of Cn and thus provide another geometric demonstration of Example 8.

1 1

P C™C£

C¡ D

T C

80. A right triangle ABC is given with /A − and

|

^AC

|

^{− b.}

CD is drawn perpendicular to AB, DE is drawn perpendicular to BC, EF AB, and this process is continued indefi nitely, as shown in the figure. Find the total length of all the perpendiculars

|

^CD

|

¹

|

^DE

|

¹

|

^EF

|

¹

|

^FG

|

¹^{∙ ∙ ∙}

in terms of b and .

A

C E G B

H F

D ¨

b

81. What is wrong with the following calculation?

0 − 0 1 0 1 0 1 ∙ ∙ ∙

−s1 2 1d 1 s1 2 1d 1 s1 2 1d 1 ∙ ∙ ∙ − 1 2 1 1 1 2 1 1 1 2 1 1 ∙ ∙ ∙

− 1 1s21 1 1d 1 s21 1 1d 1 s21 1 1d 1 ∙ ∙ ∙ − 1 1 0 1 0 1 0 1 ∙ ∙ ∙ − 1

(Guido Ubaldus thought that this proved the existence of God because “something has been created out of nothing.”) 82. Suppose that o^`n−1 an san± 0d is known to be a convergent

series. Prove that o^`n−1 1yan is a divergent series.

83. Prove part (i) of Theorem 8.

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Solution:

124 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 75. ^∞

=2

(1 + )^−is a geometric series with  = (1 + )⁻²and  = (1 + )⁻¹, so the series converges when

(1 + )⁻¹

  1 ⇔ |1 + |  1 ⇔ 1 +   1 or 1 +   −1 ⇔   0 or   −2. We calculate the sum of the

series and set it equal to 2: (1 + )⁻²

1 − (1 + )⁻¹ = 2 ⇔

 1

1 + 

2

= 2 − 2

 1

1 + 



⇔ 1 = 2(1 + )²− 2(1 + ) ⇔

2²+ 2 − 1 = 0 ⇔  = ^{−2 ±}4^√¹² = ^±^√³₂^{− 1}. However, the negative root is inadmissible because −2  ⁻^√³₂^{− 1}  0.

So  = ^√³₂^{− 1}.

76. ^∞

=0

^= ^∞

=0

(^)^is a geometric series with  = (^)⁰= 1and  = ^. If ^ 1, it has sum 1

1 − ^, so 1

1 − ^ = 10 ⇒

1

10 = 1 − ^ ⇒ ^= ₁₀⁹ ⇒  = ln10⁹.

77. ^^= ¹⁺¹²⁺¹³⁺^···+^¹ = ¹^12^13· · · ^1 (1 + 1)

1 + ¹₂  1 +¹₃

· · · 1 +_¹

[^ 1 + ]

= 2 1 3 2 4

3· · · + 1

 =  + 1 Thus, ^^   + 1and lim

→∞^^ = ∞. Since {^} is increasing, lim_

→∞= ∞, implying that the harmonic series is divergent.

78. The area between  = ^⁻¹and  = ^for 0 ≤  ≤ 1 is

 1 0

(^⁻¹− ^)  =

^

 − ^+1

 + 1

¹

0

= 1

 − 1

 + 1

= ( + 1) − 

( + 1) = 1

( + 1)

We can see from the diagram that as  → ∞, the sum of the areas between the successive curves approaches the area of the unit square, that is, 1. So ^∞

=1

1

 ( + 1) = 1.

79. Let be the diameter of . We draw lines from the centers of the to the center of  (or ), and using the Pythagorean Theorem, we can write 1²+

1 − ¹212

=

1 + ¹₂12

⇔ 1 =

1 +¹₂12

−

1 −¹212

= 21 [difference of squares] ⇒ ¹ = ¹₂. Similarly,

1 =

1 +¹₂22

−

1 − ¹−¹222

= 22+ 21− ²¹− ¹2

= (2 − ¹)(1+ 2) ⇔

2

(3)

SECTION 11.2 SERIES ¤ 125

2= 1

2 − ¹ − ¹= (1 − ¹)² 2 − ¹ , 1 =

1 +¹₂32

−

1 − ¹− ²− ¹232

⇔ ³= [1 − (¹+ 2)]²

2 − (¹+ 2) , and in general,

+1=

1 −

=1

2

2 −

=1

. If we actually calculate 2and 3from the formulas above, we find that they are 1 6= 1

2 · 3and 1

12 = 1

3 · 4 respectively, so we suspect that in general, = 1

( + 1). To prove this, we use induction: Assume that for all

 ≤ , ^= 1

( + 1) = 1

− 1

 + 1. Then^

=1

= 1 − 1

 + 1 = 

 + 1 [telescoping sum]. Substituting this into our

formula for +1, we get +1=



1 − 

 + 1

2

2 −

 

 + 1

 = 1 ( + 1)²

 + 2

 + 1

= 1

( + 1)( + 2), and the induction is complete.

Now, we observe that the partial sums

=1of the diameters of the circles approach 1 as  → ∞; that is,

∞

=1

= ^∞

=1

1

( + 1) = 1, which is what we wanted to prove.

80. || =  sin , || = || sin  =  sin², | | = || sin  =  sin³,    . Therefore,

|| + || + | | + | | + · · · =  ^∞

=1

sin^ = 

 sin  1 − sin 



since this is a geometric series with  = sin 

and |sin |  1 because 0    ^₂.

81. The series 1 − 1 + 1 − 1 + 1 − 1 + · · · diverges (geometric series with  = −1) so we cannot say that 0 = 1 − 1 + 1 − 1 + 1 − 1 + · · · .

82. If ^∞

=1

is convergent, then lim

→∞= 0by Theorem 6, so lim

→∞

1

 6= 0, and so ^∞

=1

1



is divergent by the Test for Divergence.

83. _∞

=1= lim

→∞



=1= lim

→∞

=1=  lim

→∞



=1= _∞

=1, which exists by hypothesis.

84. If

were convergent, then

(1)() =

would be also, by Theorem 8(i). But this is not the case, so



must diverge.

85. Suppose on the contrary that

(+ )converges. Then

(+ )and

are convergent series. So by Theorem 8(iii),

[(+ ) − ^]would also be convergent. But

[(+ ) − ^] =

, a contradiction, since

is given to be divergent.

86. No. For example, take

=

and

=

(−), which both diverge, yet

(+ ) =0, which converges with sum 0.

95. Consider the series

∞

P

n=1 n (n+1)!.

(a) Find the partial sums s₁, s₂, s₃, and s₄. Do you recognize the denominators? Use the pattern to guess a formula for s_n.

(b) Use mathematical induction to prove your guess.

(c) Show that the given infinite series is convergent, and find its sum.

Solution:

SECTION 11.2 SERIES ¤ 127 90. (a)

1 1 2 4 1 1 1000

2 2 3 1 4 1000 1

3 15 25 25 25 5005 5005

4 175 275 175 325 75025 25075

5 1625 2625 2125 2875 625375 375625

6 16875 26875 19375 30625 687813 313188

7 165625 265625 203125 296875 656594 344406

8 167188 267188 198438 301563 672203 328797

9 166406 266406 200781 299219 664398 336602

10 166797 266797 199609 300391 668301 332699

11 166602 266602 200195 299805 666350 334650

12 166699 266699 199902 300098 667325 333675

The limits seem to be ⁵₃,⁸₃, 2, 3, 667, and 334. Note that the limits appear to be “weighted” more toward 2. In general, we guess that the limit is 1+ 22

3 .

(b) +1− ^= ¹₂(+ −1) − ^= −¹2(− ^−1) = −¹2

₁

2(−1+ −2) − ^−1

= −¹2

−¹2(−1− ^−2)

= · · · =

−¹2

^−1

(2− ¹)

Note that we have used the formula = ¹₂(−1+ −2)a total of  − 1 times in this calculation, once for each  between 3 and  + 1. Now we can write

= 1+ (2− ¹) + (3− ²) + · · · + (^−1− ^−2) + (− ^−1)

= 1+

−1

=1

(+1− ^) = 1+

−1

=1

−¹2

−1

(2− ¹) and so

lim→∞= 1+ (2− ¹) ^∞

=1

−¹2

−1

= 1+ (2− ¹)

 1

1 − (−12)



= 1+²₃(2− ¹) = 1+ 22

3 .

91. (a) For ^∞

=1



( + 1)!, 1 = 1 1 · 2 = 1

2, 2= 1

2+ 2

1 · 2 · 3 = 5

6, 3 = 5

6+ 3

1 · 2 · 3 · 4 = 23 24,

4= 23

24+ 4

1 · 2 · 3 · 4 · 5 = 119

120. The denominators are ( + 1)!, so a guess would be = ( + 1)! − 1 ( + 1)! . (b) For  = 1, 1= 1

2 = 2! − 1

2! , so the formula holds for  = 1. Assume = ( + 1)! − 1 ( + 1)! . Then

+1= ( + 1)! − 1

( + 1)! +  + 1

( + 2)! = ( + 1)! − 1

( + 1)! +  + 1

( + 1)!( + 2) = ( + 2)! − ( + 2) +  + 1 ( + 2)!

= ( + 2)! − 1 ( + 2)!

Thus, the formula is true for  =  + 1. So by induction, the guess is correct.

(c) lim

→∞= lim

→∞

( + 1)! − 1 ( + 1)! = lim

→∞



1 − 1

( + 1)!



= 1and so ^∞

=1



( + 1)! = 1.

3