Slide 8- 2 - Edmond Public Schools

```8.1
Sequences
Quick Review
Let f ( x) 
x
. Find the values of f .
x4
1. f (5)
2. f (-1)
Evaluate the expression a   n  1 d for the given values of
a, n, and d .
3. a  -2, n  2, d  3
4. a  1, n  2, d  2
Slide 8- 2
Quick Review
n -1
Evaluate the expression ar for the given values of a, r , and n.
1
5. a  , r  2, n  3
2
6. a  2, r  1.5, n  4
Find the value of the limit.
2x  2
7. lim
4x  x 1
sin  4 x 
8. lim
x
2
x 
2
x 0
Slide 8- 3
Quick Review Solutions
x
. Find the values of f .
x4
5
1. f (5)
9
1
2. f (-1) 3
Let f ( x) 
Evaluate the expression a   n  1 d for the given values of
a, n, and d .
3. a  -2, n  2, d  3 1
4. a  1, n  2, d  2 -5
Slide 8- 4
Quick Review Solutions
n -1
Evaluate the expression ar for the given values of a, r , and n.
1
5. a  , r  2, n  3 2
2
6. a  2, r  1.5, n  4 -6.75
Find the value of the limit.
2x  2
1
7. lim
4x  x 1 2
sin  4 x 
8. lim
4
x
2
x 
2
x 0
Slide 8- 5




Defining a Sequence
Arithmetic and Geometric Sequences
Graphing a Sequence
Limit of a Sequence
…and why
Sequences arise frequently in mathematics and
applied fields.
Slide 8- 6
Defining a Sequence
A sequence a  is a list of numbers written in an explicit order.
n
For example: a   a , a , a ,..., a ,... , where a is the first term
n
1
2
3
n
1
and a is the nth term of the sequence.
n
Let a , a , a ,..., a ,... be a function with domain the set of positive
1
2
3
n
integers and range a , a , a ,..., a ,.... If the domain is finite, then
1
2
3
n
the sequence is a finite sequence. If the domain is infinite, then
the sequence is an infinite sequence.
Slide 8- 7
Example Defining a Sequence Explicitly
Find the first four terms and the 100th term of the sequence a 
n
 1
where a 
.
n 2
Set n equal to 1, 2, 3, 4, and 100.
n
n
2
 1 1


1 2 3
 1 1


2 2 6
1

11
1

18
1

10, 002
1
a
1
2
2
a
2
a
3
a
a
4
100
2
Slide 8- 8
Example Defining a Sequence Recursively
Find the first three terms and the seventh term for the sequence defined
recursively by the conditions: b  4 and b  b  2 for all n  2.
1
n 1
n
Since b  4 and b  b  2, you can find b  2, b  0, and b  8.
1
n
n 1
2
3
7
Slide 8- 9
Arithmetic Sequence
A sequence a  is an arithmetic sequence if it can be written in the
n
form a, a  d , a  2d ,..., a   n -1 d ,... for some constant d .
The number d is the common difference.
Each term in an arithmetic sequence can be obtained recursively
from its preceeding term by adding d :
a  a  d for all n  2.
n
n 1
Slide 8- 10
Example Defining Arithmetic Sequences
Given the arithmetic sequence:  3, 1, 5, 9, ... find
(a) the common difference,
(b) the ninth term,
(c) a recursive rule for the nth term,
(d) an explicit rule for the nth term.
(a) The common difference between terms is 4.
(b) a  3   9 1 (4)  29.
9
(c) a  3 and a  a  4 for all n  2.
1
n
n 1
(d) a  3  (n  1)(4)
n
 4n  7
Slide 8- 11
Geometric Sequence
A sequence a  is a geometric sequence if it can be written in the
n
form a, a  r , a  r ,..., a  r ,... for some nonzero constant r.
2
n 1
The number r is the common ratio.
Each term in a geometric sequence can be obtained recursively
from its preceeding term by multiplying by r:
a  a  r for all n  2.
n
n 1
Slide 8- 12
Example Defining Geometric Sequences
For the geometric sequence 1,  3, 9,  27, ..., find
(a) the common ratio,
(b) the tenth term,
(c) a recursive rule for the nth term,
(d) an explicit rule for the nth term.
(a) The common ratio is  3.
(b) a  (1)  (3)  19683
9
10
(c) The sequence is defined recursively by a  1 and a   3 a
1
n
n 1
for n  2.
(d) The sequence is defined explicitly by a  1 3   3 .
n1
n1
n
Slide 8- 13
Example Constructing a Sequence
The second and fifth term of a geometric sequence are -6 and 48,
respectively. Find the first term, common ratio and an explicit rule
for the nth term.
4
ar
48

ar
6
1
1
r  8
r  2
Then a  r  6 means that a  3.
3
1
1
The sequence is defined explicitly: a  (3)  2   (1)
n 1
n
n 1
 3 2 
n 1
Slide 8- 14
Example Graphing a Sequence Using
Parametric Mode
Draw a graph of the sequence a  with a   1
n
n
n
n 1
, n  1, 2,... .
n
T 1
Let X =T,Y =(  1)
, and graph in dot mode. Set T  1,
T
T  20, and T  1. Choose X  0, X  20, X  2, Y =  2,
T
1T
1T
max
min
step
min
max
scl
min
Y  2, Y  1, and draw the graph.
max
scl
Slide 8- 15
Example Graphing a Sequence Using
Sequence Graphing Mode
Graph the sequence defined recursively by b  4 and
1
b  b  2 for all n  2.
n
n -1
Set the graph in sequence graphing mode and dot mode.
Replace b by u (n). Select nMin  1, u (n)  u (n  1)  2,
n
and u (nMin)  4.
Slide 8- 16
Example Graphing a Sequence Using
Sequence Graphing Mode
Set nMin  1, nMax  10, PlotStart  1, PlotStep  1, and graph in the
0,10 by -5,25 viewing window
Slide 8- 17
Limit
Let L be a real number. The sequence a has limit L as n approaches 
n
if, given any positive number  , there is a positive number M such that for
all n  M we have a - L   .
n
We write lim a  L and say that the sequence converges to L.
n 
n
Sequences that do not have limits diverge.
Slide 8- 18
Properties of Limits
If L and M are real numbers and lim a  L and lim b  M , then
n 
n
n 
n
1. Sum Rule: lim  a  b   L  M
n 
n
n
2. Difference Rule: lim  a  b   L  M
n 
n
n
3. Product Rule: lim  a b   L  M
n 
n
n
4. Constant Multiple Rule: lim  ca   c  L
n 
n
a  L
5. Quotient Rule: lim    , M  0
b  M
n
n 
n
Slide 8- 19
Example Finding the Limit of a Sequence
Determine whether the sequence converges or diverges. If it converges,
find its limit. a 
n
2n  1
n
Analytically, using the Properties of Limits:
lim
n 
2n  1
1

 lim  2  
n
n

n 
1
 lim(2)  lim  
n
 20  2
n 
n 
Slide 8- 20
The Sandwich Theorem for Sequences
If lim a  lim c  L and if there is an integer N for which
n 
n 
n
n
a  b  c for all n  N , then lim b  L.
n
n
n
n 
n