Transcript Infinite Series - El Camino College

College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
9
Sequences
and Series
9.3
Geometric Sequences
Introduction
Geometric sequences occur frequently
in applications to fields such as:
• Finance
• Population growth
Geometric Sequences
Arithmetic Sequence vs. Geometric Sequence
An arithmetic sequence is generated
when:
• We repeatedly add a number d to an initial term a.
A geometric sequence is generated
when:
• We start with a number a and repeatedly
multiply by a fixed nonzero constant r.
Geometric Sequence—Definition
A geometric sequence is a sequence
of the form
a, ar, ar2, ar3, ar4, . . .
• The number a is the first term.
• r is the common ratio of the sequence.
• The nth term of a geometric sequence
is given by:
an = arn–1
Common Ratio
The number r is called the common
ratio because:
• The ratio of any two consecutive terms
of the sequence is r.
E.g. 1—Geometric Sequences
Example (a)
If a = 3 and r = 2, then we have the
geometric sequence
3, 3 · 2, 3 · 22, 3 · 23, 3 · 24, . . .
or
3, 6, 12, 24, 48, . . .
• Notice that the ratio of any
two consecutive terms is r = 2.
• The nth term is: an = 3(2)n–1
E.g. 1—Geometric Sequences
Example (b)
The sequence
2, –10, 50, –250, 1250, . . .
is a geometric sequence
with a = 2 and r = –5
• When r is negative, the terms
of the sequence alternate in sign.
• The nth term is: an = 2(–5)n–1.
E.g. 1—Geometric Sequences
Example (c)
The sequence
1 1 1 1
1, , ,
, ,...
3 9 27 81
is a geometric sequence
with a = 1 and r = ⅓
• The nth term is: an = (⅓)n–1
E.g. 1—Geometric Sequences
Example (d)
Here’s the graph of the geometric
sequence an = (1/5) · 2n – 1
• Notice that the points
in the graph lie on
the graph of the
exponential function
y = (1/5) · 2x–1
E.g. 1—Geometric Sequences
Example (d)
If 0 < r < 1, then the terms of the
geometric sequence arn–1 decrease.
However, if r > 1, then the terms
increase.
• What happens if r = 1?
Geometric Sequences in Nature
Geometric sequences occur naturally.
Here is a simple example.
• Suppose a ball has elasticity such that,
when it is dropped, it bounces up one-third
of the distance it has fallen.
Geometric Sequences in Nature
If the ball is dropped from a height of 2 m,
it bounces up to a height of 2(⅓) = ⅔ m.
• On its second bounce,
it returns to a height
of (⅔)(⅓) = (2/9)m,
and so on.
Geometric Sequences in Nature
Thus, the height hn that the ball reaches
on its nth bounce is given by the geometric
sequence
hn = ⅔(⅓)n–1 = 2(⅓)n
• We can find the nth term of a geometric
sequence if we know any two terms—as
the following examples show.
E.g. 2—Finding Terms of a Geometric Sequence
Find the eighth term of the geometric
sequence
5, 15, 45, . . . .
• To find a formula for the nth term of
this sequence, we need to find a and r.
• Clearly, a = 5.
E.g. 2—Finding Terms of a Geometric Sequence
To find r, we find the ratio of any two
consecutive terms.
• For instance,
r = (45/15) = 3
• Thus,
an = 5(3)n–1
• The eighth term is:
a8 = 5(3)8–1 = 5(3)7 = 10,935
E.g. 3—Finding Terms of a Geometric Sequence
The third term of a geometric sequence
is 63/4, and the sixth term is 1701/32.
Find the fifth term.
• Since this sequence is geometric,
its nth term is given by the formula an = arn–1.
• Thus,
a3 = ar3–1 = ar2
a6 = ar6–1 = ar5
E.g. 3—Finding Terms of a Geometric Sequence
From the values we are given for those two
terms, we get this system of equations:
• We solve this by dividing:
ar 5

2
ar
1701
32
63
4
r3 
27
8
r 
3
2
 634  ar 2
 1701
5
 32  ar
E.g. 3—Finding Terms of a Geometric Sequence
Substituting for r in the first equation,
63/4 = ar2, gives:
2
63
3
4
 a 2
a7
• It follows that the nth term
of this sequence is:
an = 7(3/2)n–1
• Thus, the fifth term is:
a5  7 
3
2

51
 7
3
2

4
 567
16
Partial Sums of
Geometric Sequences
Partial Sums of Geometric Sequences
For the geometric sequence
a, ar, ar2, ar3, ar4, . . . , arn–1, . . . ,
the nth partial sum is:
n
Sn   ar
k 1
k 1
 a  ar  ar  ar  ar      ar
2
3
4
n 1
Partial Sums of Geometric Sequences
To find a formula for Sn, we multiply Sn by r
and subtract from Sn:
Sn  a  ar  ar  ar  ar      ar
2
rSn 
3
4
n 1
ar  ar 2  ar 3  ar 4      ar n 1  ar n
Sn  rSn  a  ar
n
Partial Sums of Geometric Sequences
So,
 
a 1  r 

 r  1
1 r
Sn 1  r   a 1  r
n
n
Sn
• We summarize this result as follows.
Partial Sums of a Geometric Sequence
For the geometric sequence an = arn–1,
the nth partial sum
Sn = a + ar + ar2 + ar3 + ar4 + . . . + arn–1
(r ≠ 1)
is given by:
n
1 r
Sn  a
1 r
E.g. 4—Finding a Partial Sum of a Geometric Sequence
Find the sum of the first five terms
of the geometric sequence
1, 0.7, 0.49, 0.343, . . .
• The required sum is the sum of the first
five terms of a geometric sequence with
a = 1 and r = 0.7
E.g. 4—Finding a Partial Sum of a Geometric Sequence
Using the formula for Sn with n = 5,
we get:
S5  1
1   0.7 
1  0.7
5
 2.7731
• The sum of the first five terms
of the sequence is 2.7731.
E.g. 5—Finding a Partial Sum of a Geometric Sequence
Find the sum
5
 7 
k 1
2
3
k
• The sum is the fifth partial sum of a geometric
sequence with first term a = 7(–⅔) = –14/3
and common ratio r = –⅔.
• Thus, by the formula for Sn, we have:
32
14 1    
14 1  243
770
S5   
  5 
2
3 1   3 
3
243
3
2
3
5
What Is an Infinite Series?
Infinite Series
An expression of the form
a1 + a2 + a3 + a4 + . . .
is called an infinite series.
• The dots mean that we are to continue
the addition indefinitely.
Infinite Series
What meaning can we attach to the sum
of infinitely many numbers?
• It seems at first that it is not possible to add
infinitely many numbers and arrive at a finite
number.
• However, consider the following problem.
Infinite Series
You have a cake and you want to eat it
by:
• First eating half the cake.
• Then eating half of what remains.
• Then again eating half of what remains.
Infinite Series
This process can continue indefinitely
because, at each stage, some of the cake
remains.
• Does this mean that it’s impossible
to eat all of the cake?
• Of course not.
Infinite Series
Let’s write down what you have eaten
from this cake:
1 1 1 1
1
  
   n  
2 4 8 16
2
• This is an infinite series.
• We note two things about it.
Infinite Series
1. It’s clear that, no matter how many terms
of this series we add, the total will never
exceed 1.
2. The more terms of this series we add,
the closer the sum is to 1.
Infinite Series
This suggests that the number 1 can be
written as the sum of infinitely many
smaller numbers:
1 1 1 1
1
1   
   n  
2 4 8 16
2
Infinite Series
To make this more precise, let’s look at
the partial sums of this series:
1
S1 
2
1 1
S2  
2 4
1 1 1
S3   
2 4 8
1 1 1 1
S4    
2 4 8 16
1

2
3

4
7

8
15

16
Infinite Series
In general (see Example 5 of
Section 9.1),
1
Sn  1  n
2
• As n gets larger and larger, we are adding more
and more of the terms of this series.
• Intuitively, as n gets larger, Sn gets closer
to the sum of the series.
Infinite Series
Now, notice that, as n gets large,
(1/2n) gets closer and closer to 0.
• Thus, Sn gets close to 1 – 0 = 1.
• Using the notation of Section 4.6,
we can write:
Sn → 1 as n → ∞
Sum of Infinite Series
In general, if Sn gets close to a finite
number S as n gets large, we say that:
• S is the sum of the infinite series.
Infinite Geometric Series
Infinite Geometric Series
An infinite geometric series is a series
of the form
a + ar + ar2 + ar3 + ar4 + . . . + arn–1 + . . .
• We can apply the reasoning used earlier to
find the sum of an infinite geometric series.
Infinite Geometric Series
The nth partial sum of such a series
is given by:
1 r
Sn  a
1 r
n
 r  1
• It can be shown that, if | r | < 1, rn gets close
to 0 as n gets large.
• You can easily convince yourself of this
using a calculator.
Infinite Geometric Series
It follows that Sn gets close to a/(1 – r)
as n gets large, or
a
Sn 
1 r
as
• Thus, the sum of this infinite
geometric series is:
a/(1 – r)
n
Sum of an Infinite Geometric Series
If | r | < 1, the infinite geometric series
a + ar + ar2 + ar3 + ar4 + . . . + arn–1 + . . .
has the sum
a
S
1 r
E.g. 6—Finding the Sum of an Infinite Geometric Series
Find the sum of the infinite geometric series
2 2
2
2
2 

   n  
5 25 125
5
• We use the formula.
• Here, a = 2 and r = (1/5).
• So, the sum of this infinite series is:
2
5
S

1
1 5 2
E.g. 7—Writing a Repeated Decimal as a Fraction
Find the fraction that represents
the rational number 2.351.
• This repeating decimal can be written as
a series:
23
51
51
51
51




 
10 1000 100,000 10,000,000 1,000,000,000
E.g. 7—Writing a Repeated Decimal as a Fraction
After the first term, the terms of
the series form an infinite geometric
series with:
51
a
and
1000
1
r 
100
E.g. 7—Writing a Repeated Decimal as a Fraction
So, the sum of this part of the series is:
S
51
1000
1
100
1

51
1000
99
100
51 100
51



1000 99 990
• Thus,
23 51 2328 388
2.351 



10 990 990 165