Determine whether each of the following geometric series converges or diverges.. 572 CHAPTER 18 Geometric Sums, Geometric Series18.3 A MORE GENERAL DISCUSSION OF INFINITE SERIES In the p
Trang 1P R O B L E M S F O R S E C T I O N 1 8 2
For Problems 1 through 11, determine whether the series converges or diverges If it converges, find its sum.
1 1 − 10 + 100 − · · · + (−10)n
+ · · ·
2 0.3 + 0.03 + 0.003 + 0.0003 + 0.00003 + · · ·
3 23+29+272 + · · · +32n+ · · ·
4 23+ 2 + 6 + · · · + 2(3)n+ · · ·
5 1 −12+14−18+161 + · · ·
6 14−18+161 −321 + · · ·
7 23+ 1 + 32+94+ · · ·
8 32−34+38− · · · +(−1)2n+1n 3+ · · ·
9 e + 1 + e−1+ e−2+ e−3+ · · ·
10 2e + 2e2+ 2e3+ · · · + 2en+ · · ·
11 (2e)−2+ (2e)−3+ (2e)−4+ · · · + (2e)−n+ · · ·
12 Find the sum of the following (If there is no finite sum, say so.)
(a) 3 + 9 + 27 + · · · + 320
(b) 23+232+233+ · · · +23n+ · · ·
(c) (0.2)(10) + (0.2)(100) + (0.2)(1000) + · · ·
(d) 3 + 3(0.8) + 3(0.8)2+ 3(0.8)3+ · · ·
(e) (0.2) + (0.2)(1.3) + (0.2)(1.3)2+ (0.2)(1.3)3+ · · ·
(f ) 1 + x2+ x4+ x6+ · · · for −1 < x < 1
13 Determine whether each of the following geometric series converges or diverges If the series converges, determine to what it converges
(a) −43−12−163 −1289 + · · ·
(b) −1001 +(100)1.12 −(100)1.213 +(100)1.3314 − · · ·
(c) −100007 +110007 −121007 +133107 − · · ·
(d) 1 − x + x2− x3+ · · · for |x| < 1
14 Write each of the following series first as a repeating decimal and then as a fraction (a) 2 +102 +1002 +10002 + · · ·
(b) 3 +10122 +10124 +10126 + · · ·
Trang 2572 CHAPTER 18 Geometric Sums, Geometric Series
18.3 A MORE GENERAL DISCUSSION OF INFINITE SERIES
In the previous four sections we focused on geometric sums and geometric series In this section we broaden our discussion to investigate other infinite series Our focus in this chapter is geometric series, but you will have a better appreciation of geometric series if you have some familiarity with series that are not geometric
Given an infinite series
a1+ a2+ a3+ · · · + an+ · · · the most basic question to consider is whether the series converges or diverges
Suppose all the terms of the infinite series are positive Then Sn= S(n), the sum of the first n terms, is an increasing function We know from our study of functions that an increasing function may increase without bound, or it may increase but be bounded, in which case it will be asymptotic to a horizontal line In the latter case, the function must be increasing at a decreasing rate; in fact, if the function has a horizontal asymptote, its rate of increase must be approaching zero Similarly, if limn→∞Sn= L for some finite constant
L, then the rate at which S(n) is increasing must be approaching zero This translates to the observation that if an infinite series is to have any chance at converging, then its terms must
be approaching zero, that is, limn→∞anmust be 0
partial sums
n
The partial sums are increasing at
a decreasing rate;
The partial sums are bounded.
partial sums
n
The partial sums are increasing at
a decreasing rate;
The partial sums are unbounded.
Figure 18.1
If all of the terms of the series are negative, an analogous argument can be made If some
of the terms of an infinite series are positive and others are negative, it is still true that in order for the series to have any shot at converging the terms must be approaching zero If limn→∞an
bound if k > 0 and eventually decreasing without bound if k < 0.7 Suppose the terms of a series are approaching zero; is this enough to guarantee convergence? In the case of geometric series the answer is “yes”, but in the general case
the answer is NO! The situation in general is much more subtle;8 the next example will convince you of that fact
7 If lim n→∞ a n does not exist, then the partial sums will be bouncing around and will not converge.
8 We know that a function can be increasing at a decreasing rate and have a horizontal asymptote, but it can also be increasing
at a rate tending toward zero and yet be unbounded Consider, for example, f (x) = ln x Its rate of increase is given by1 limn→∞1
= 0, so the rate of increase of ln x tends toward zero as x increases without bound Nevertheless, as x → ∞ we know that ln x → ∞.
Trang 3EXAMPLE 18.7 Does the infinite series 12+13+14+15+ · · · +1n+ · · · converge or diverge? This infinite
series is called the harmonic series.
SOLUTION The terms of this series are going toward zero: limn→∞1n= 0 The harmonic series is not
a geometric series Can we put Sninto closed form? Let’s look at some partial sums
1 2 1
2+1
3 =5 6 1
2+1
3 +1
4=13 12 1
2+1
3 +1
4+1
5 =77 60 1
2+1
3 +1
4+ · · · +n1 =??
Closed form allowed us to get a firm grip on something rather slippery In this example our luck has run out We cannot express the general partial sum in closed form The expression limn→∞21+13+14+ · · · +n1gives us no insight into the convergence or divergence of the series We need to take a different perspective We will compare this series to a familiar series
1
2 =1 2 1
4 +1
4=1
1
3+1
4 >
1 2 1
8 +18+18 +18=12 so 1
5+16 +17+18>1
2
8 · 116
9+101 +111 + · · · +161 >1
2
16 · 1 32
=1
1
17+ 1
18+ 1
19 + · · · + 1
32>
1 2 and so on Think about this in terms of slices of pies How many pies must we bake in order to give out slices as dictated by the harmonic series? If the series converges we need only bake some finite number of pies
1
2 +13 +14+15 +16 +17+18 +19+ · · · +161 + · · · equals
1 2
+ 13 +14
+ 15 +16+17 +18
+ 19 + · · · +161
+ · · ·
half a pie
+more than half a pie
+more than half a pie
+more than half a pie
+ · · · , which is greater than or equal to 12+12+12+12 + · · · But this latter series diverges; consequently, the same must be true of the harmonic series By comparing the harmonic
Trang 4574 CHAPTER 18 Geometric Sums, Geometric Series
series with the divergent series12+12+12+12+ · · · , we see that the harmonic series must diverge (No matter how many pies we bake, we will eventually run out and need more.)
We had previously observed that if an infinite series is to have any chance at converg-ing, then the terms must be going toward zero Although this is a necessary condition for convergence, a look at the harmonic series shows that it is not enough to guarantee conver-gence; the condition limn→∞an= 0 is necessary for convergence but not sufficient.9It is important to realize that if the terms of an infinite series are going to zero, then the series
may converge (as is true for all geometric series), yet on the other hand, the series may
diverge (as in the example of the harmonic series)
A Summary of the Main Principles
The Nth Term Test for Divergence:
If lim
n→∞an 1+ a2+ a3+ · · · + an+ · · · diverges.10
Warning: This is a test for divergence only!
Increasing and Bounded Partial Sums Test:Suppose the terms of a series are all positive Then Snincreases with n If the partial sums are bounded, that is, there exists a constant
Msuch that Sn≤ M for all n, then it can be shown that limn→∞Snexists and is finite Therefore, the series converges
In many cases the question of convergence or divergence of an infinite series is a very subtle one Often, for instance, one can determine that a certain series converges without being able to say exactly what it converges to We will return to infinite series in Chapter
30 Questions of convergence become much simpler if we focus on the special case of the geometric series, and this is our main focus in this chapter In the case of geometric series, convergence and divergence are straightforward to establish
a + ar + ar2+ · · · + arn+ · · ·
a 1−r for |r| < 1 diverges for |r| ≥ 1
In Section 18.4 we introduce some convenient notation for working with series, and in Section 18.5 we apply geometric series to real-world situations
P R O B L E M S F O R S E C T I O N 1 8 3
For Problems 1 through 9, determine whether the series converges or diverges Explain your reasoning.
1 1 − 2 + 3 − · · · + (−1)(n+1)n+ · · ·
2 10001 +10002 +10003 +10004 + · · ·
3 23+24+25+ · · · +2n+ · · ·
9 “Necessary” versus “sufficient”: In order for a polygon to be called a square it is necessary that it have four sides, but this alone is not sufficient to classify it as a square In order to win a race it is necessary to finish it, but this alone is not sufficient.
10 For an infinite series to diverge it is sufficient that limn→∞a n
Trang 54 1
2·2 2 +3·213 +4·214 +5·215 + · · ·
(Hint: Compare this term-by-term to a geometric series you know Choose a convergent
geometric series whose terms are larger than the terms of this series.)
5 (sin 1)3 2 +(sin 2)32 2 +(sin 3)33 2 + · · · +(sin n)3n 2 + · · ·
(Hint: Compare this term-by-term to a geometric series you know Choose a convergent
geometric series whose terms are larger than the terms of this series.)
6 333 +444 +555 + · · ·
7 −12 −13−14−15− · · · − 1n− · · ·
8 12−12+12−12+ · · ·
9 (1 + 1)1+
1 +122+
1 + 133+ · · · +
1 + 1n
n
+ · · ·
10 The sum 1 + 2 + 3 + 4 + 5 + · · · + n = n
k=1kis not geometric, but we can express
it in an easy-to-compute form
Let
S = 1 + 2 + 3 + 4 + 5 + · · · + n (18.1) Writing the terms from largest to smallest gives
S = n + (n − 1) + (n − 2) + · · · + 2 + 1 (18.2) Add equations (18.1) and (18.2) and divide by two to show that
1 + 2 + 3 + 4 + 5 + · · · + n =n(n + 1)2
11 Challenge: Use the same line of reasoning as outlined in Problem 10 to show that
1 + 3 + 5 + 7 + · · · + (2n − 1) = n2
12 Give an example of each of the following
(a) An infinite series that converges and whose partial sums are always increasing (b) An infinite series that converges and whose partial sums oscillate around the sum
of the series
(c) An infinite series that diverges although its terms approach zero
(d) An infinite series that diverges but whose partial sums do not grow without bound
18.4 SUMMATION NOTATION
Sums whose terms follow a consistent pattern can often be written in a more compact way by using summation notation Summation notation is only notation; it is a compact shorthand for writing out a sum We will introduce it via examples
a1+ a2+ a3+ · · · + a28
Trang 6576 CHAPTER 18 Geometric Sums, Geometric Series
can be written in shorthand as
28
i=1
ai, where denotes the summation process.11
The terms of the sum are all of the form ai, where the “i”s form a sequence of consecutive integers starting with the integer indicated at the bottom of and ending with the integer
at the top In this example we successively substitute i = 1, 2, 3, , 28 in place of i in the expression aiand add up these 28 terms i is called the index We can choose any letter we like for the index
28
i=1
ai,
28
j =1
aj, and
28
k=1
ak
all are shorthand for a1+ a2+ a3+ · · · + a28 In fact, 27
q=0aq+1is also equivalent to
a1+ a2+ a3+ · · · + a28; there are infinitely many different ways of putting a sum into summation notation
k=02k is shorthand for 20+ 21+ 22+ · · · + 214 ii) 14
k=0k2 is shorthand for 02+ 12+ 22+ · · · + 142 iii) 49
k=3(−1)kk is shorthand for (−1)33 + (−1)44 + (−1)55 + · · · + (−1)4949
or −3 + 4 − 5 + 6 + · · · − 49
iv) a + ar + ar2+ ar3+ · · · + arn+ · · · can be written as ∞
n=0arn v) 12+212 +213+ · · · +21n + · · · can be written as ∞
k=1 21k, or, if we want this to look more explicitly like the general geometric series in part (iv), it can be written ∞
k=012
1 2
k
vi) 12−212 +213+ · · · +2133 can be written 32
k=012
−1 2
k
vii) f1n1n+ f
2 n
1
3 n
1
n+ · · · + fnn n1 can be written n
i=1f
i n
1
n
NOTEIf a sum is geometric, we can begin by identifying “a” (the first term) and “r” (the ratio of any term to the previous one) Write w
k=0arkand figure out what w ought to be In part (vi) when we use this approach the upper index is 32 not 33 Writing 33
k=1−1−12k would also be correct.12
EXERCISE 18.3 Try these on your own and then check your answers with those given below
(a) Express23+232+233+ · · · +2340in summation notation Do this in two ways; with the index starting at 1 and with the index starting at 0
(b) Express232−235+238−2311+ · · · in summation notation
Answers
(a)23+
2 3
2
+
2 3
3
+ · · · +
2 3
40
can be written as 40
n=1
2 3
n
or as 39 n=0 23
2 3
n
This is a geometric sum with a =23and r =23
11 is the uppercase Greek letter sigma; you can think of it as the letter S for sum It denotes a process, much in the same way that d
dx denotes a process or operation.
12
Trang 7(b) This is a geometric series First identify r by finding the ratio of the second term to the first, r = −233 Once we know r and a, we’re well on our way The series can be written as ∞
k=0
2 3
2
−23
3k
, or, alternatively, ∞
k=2
−23
3k+2
EXERCISE 18.4 Determine whether each of the following statements is always true or not always true.
(a) n
i=1(ai+ bi) = n
i=1ai+ n
i=1bi (c) n
i=1bi
(e) n i=1aibi= n
i=1ai ni=1bi (f) If ai> bifor i = 1, 2, 3, , n, then n
i=1ai> n
i=1bi
(Four of the six statements are always true.)
EXAMPLE 18.9 What rational number has the decimal expansion 5.123232323 ?
SOLUTION
5.1232323 = 5.1 + 0.0232323
= 5.1 + 23
103+ 23
105+ 23
107+ · · · + 23
102n+1+ · · ·13 Following 5.1 is a geometric series with a =10233and r =1012 |r| < 1, so the series converges
We have 5.1 +
∞
n=0
23
103
1 100
n
=51
10+
23 1000
1 − 1001 =
51
10+
23
1000·100 99
=5110+99023 =5072990 =2536495
P R O B L E M S F O R S E C T I O N 1 8 4
For Problems 1 through 10, write the sum using summation notation.
1 23+ 33+ 43+ · · · + 193
2 2 − 3 + 4 − 5 + 6 − · · · + 100
3 23+ 34+ 45+ · · · + 100101
4 (a) 4x2+ 4x3+ 4x4+ 4x5+ · · · + (b) 2x + 3x2+ 4x3+ 5x4+ 6x5+ · · · +
5 1 − 10 + 100 − · · · + (−10)n
+ · · ·
6 0.3 + 0.03 + 0.003 + 0.0003 + 0.00003 + · · ·
13 In order to write a general term for this series we need to write10(odd number)23 An even number can be written as 2n for n a positive integer An odd number can be expressed as 2n + 1 or 2n − 1; hence the general term is given as 23 .
Trang 8578 CHAPTER 18 Geometric Sums, Geometric Series
7 (a) 23+29+272 + · · · +32n+ · · · (b) 23+ 2 + 6 + · · · + 2(3)n+ · · ·
8 (a) 1 −12+14−18+161 + · · · (b) 14−18+161 −321 + · · ·
9 (a) 23+ 1 + 32+94+ · · · (b) 32−34+38− · · · + (−1)2nn3
10 (a) e + 1 + e−1+ e−2+ e−3+ · · · (b) 2e + 2e2+ 2e3+ · · · + 2en+ · · · (c) (2e)−2+ (2e)−3+ (2e)−4+ · · · + (2e)−n+ · · ·
For Problems 11 through 16, do the following.
i Write out the first two terms of the series.
ii Determine whether or not the series converges.
iii If the series converges, determine its sum.
11 (a) ∞
k=1 (−1) n
3 n (b) ∞
k=2(−1) n
3 n
12 ∞ n=3 (−1)
n 3
2 n
13 ∞ n=2 3 n
4 n−1
14 ∞ n=1 (−1) n
3 n
15 ∞ n=10010n
16 Does the series ∞
k=1 ln(k+2) 3k converge or diverge? Explain
17 Consider the sum
q5− q7+ q9− q11+ · · · + q41
(a) Put the sum into closed form
(b) Put the sum into summation notation
(c) Now put −q5+ q7− q9+ q11− · · · − q41into summation notation
18 For each of the following geometric sums, first write the sum using summation notation and then write the sum in closed form
(a) 322 +324 +326 + · · ·3218 (b) 1 − 2 + 22− 23+ 24− · · · + 246 (c) −1001 +1001.1 −1.21100 +1.331100 − · · · −1.1100100 (d) 322 +2323 +2334 + · · · +231617
19 Write the following without using summation notation and answer the following ques-tions
Trang 9(a) Is the series geometric?
(b) Does the series converge? If so, indicate to what it converges
i)
∞
n=0
3 2
5 2
n
ii)
∞
n=0
15
102
1 10
3n
iii)
∞
n=1
3 2
−2 3
n
iv)
∞
n=1
ln n
20 Write the following sums in summation notation In each case, determine the sum; i.e., sum the first and determine what the second converges to
(a) 500 + 500e.1+ 500e.2+ 500e.3+ · · · + 500e2 (b) 53−56+125 −245 + · · ·
18.5 APPLICATIONS OF GEOMETRIC SUMS AND SERIES
EXAMPLE 18.10 Paulina is self-employed On the first day of every month she puts $400 into an account she
has set up for retirement The account pays 0.5% per month, i.e., 6% per year compounded monthly How much money will be in Paulina’s retirement account five years after she sets
up the account immediately, before her 61st payment? (Assume that interest is paid on the last day of every month.)
SOLUTION Making Estimates. If we ignored interest completely, there would be $400 · 60 = $24,000
in the retirement account Because there is interest, we expect the answer to be more than
$24,000 Each payment is in the bank for a different amount of time Money in the account grows according to
M(t ) = M0(1.005)12t or M(t ) = M0(1.005)m, where t is time in years and m is the number of months the money has been in the bank
If all the money were in the bank for five years she would have $24,000(1.005)12·5=
$24,000(1.005)60= $32, 372.40 Thus, we expect the actual balance to be greater than
$24,000 and less than $32,372
Strategy:We will look at each of the 60 payments individually and determine how much each will grow to by the end of the five-year period In other words, we will look at the future value of each of the 60 payments Then we’ll sum these future values
We can represent this diagramatically We’ll push each of the payments to the same point in the future
$400 $400 $400 $400
60th payment 1 month after
60th payment
• • •
The first payment is in the bank for 60 months, the second for 59 months, the third for 58 months, and so on The 60th payment is in the bank for 1 month
Trang 10580 CHAPTER 18 Geometric Sums, Geometric Series
future value of the 1st payment = $400(1.005)60 future value of the 2nd payment = $400(1.005)59 future value of the 3rd payment = $400(1.005)58
future value of the 4th payment = $400(1.005)57
future value of the 59th payment = $400(1.005)2 future value of the 60th payment = $400(1.005)1 The total amount of money in the account after five years is given by14
$400(1.005)1+ $400(1.005)2+ $400(1.005)3+ · · · + $400(1.005)59+ $400(1.005)60 Denote this sum by S “r” = (1.005) Put this sum in closed form You will get
400(1.005) − 400(1.005)61
1 − 1.005 or $28, 047.55.
Look back at our original expectations This answer falls in the interval we expected
EXAMPLE 18.11 Marietta puts $200 into an account every year for four years in order to finance a long
vacation in Greece The bank pays 5% interest per year compounded annually Her first payment is January 1, 2000 She estimates that her vacation will cost $900 Will she be able
to go on the trip immediately after her fourth payment, on January 1, 2003? Will she be able to go on the trip one year after making her fourth payment?
SOLUTION Making Estimates. Certainly if Marietta makes five payments she’ll have over $1000 and
the trip won’t be a problem After making only four payments she will have put a total of
$800 into the vacation account If that total were in the bank for 3 years it would grow to
$800(1.05)3= $926.10, but the total is not in the bank for three years
Strategy:We’ll find the amount of money in the account on January 1st, 2003 We can represent this diagramatically, pushing all the payments to this date
$200 $200 $200 $200
January 1, 2003
On January 1, 2003 the first payment has grown to $200(1.05)3, the second payment has grown to $200(1.05)2, the third payment has grown to $200(1.05)1, the fourth payment remains at $200
14 The order in which the addition is done does not matter Had you chosen to write the sum as
$400(1.005)60+ $400(1.005)59+ $400(1.005)58+ · · · + $400(1.005)2+ $400(1.005)1 you would have r = (1.005) −1
...$24,000(1.005)60< /sup>= $32, 372.40 Thus, we expect the actual balance to be greater than
$24,000 and less than $32,372
Strategy:We will look at each of the 60 payments individually and. .. term) and “r” (the ratio of any term to the previous one) Write w
k=0arkand figure out what w ought to be In part (vi) when we use this approach. .. in the bank for 60 months, the second for 59 months, the third for 58 months, and so on The 60th payment is in the bank for month
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