Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
Sequences
Def.: A sequence is a
1
whose domain is the integers or a subset of the integers.
Note 1: For most of the examples that we will encounter, the domain of the sequence will
be the
integers: { 1 , 2 , 3 , 4 , …}
Note 2: Although a sequence is a function, it is common to represent sequences with
subscript notation,
, rather than with the standard function notation, f(n).
Exercise 1: Let {an} = {3n – 1} .
(i.) Write the sequence in (the generally unused)
function notation.
(ii.) Make a table with the first four terms.
input
n
output
f(1) = a1 = 3(1) – 1 = 2
1
Note 3: {an} = {3n – 1} denotes the
,
whereas an = 3n – 1 denotes only the
of the sequence.
Limit of a Sequence
Def.: Let L be a real number. L is the limit of a sequence {an}, written
if for any  > 0, there exists M > 0 such that
for every
.
If the limit exists, we say that the sequence {an}
If no limit exists, we say that the sequence {an}
f(n) = an
to L.
.
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
2
 1  n 
Exercise 2: Consider the sequence {an} =    .
 2  
(i.) Write out the first four terms of the sequence. What do you suppose the limit, L, is?
(ii.) Suppose, for example, that  = 0.01. How large must n be so that | an – L | <  ?
That is, how large must n be so that | (0.5)n – 0 | < 0.01 ?
By trial and error, we see that for n =
,
| an – L | = | (0.5)n – 0 | =

< 0.01 = 
(iii.) Suppose, for example, that = 0.0001. How large must n be so that | an – L | <  ?
That is, how large must n be so that | (0.5)n – 0 | < 0.0001 ?
Theorem: Let f be a function of x. Suppose that
If f(n) = an for every positive integer n, then
.
.
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
Exercise 3: Let f ( x) 
3
6 x 2  5x  2
.
2x 2
(i.) Evaluate lim f ( x) .
x 
(ii.) Write out the first four terms of {an}, where an =
6(1) 2  5(1)  2
a1 =
=
2(1) 2
,
6n 2  5n  2
.
2n 2
6( ) 2  5( )  2
a2 =
=
2( ) 2
,
a3 =
,
a4 =
(iii.) Evaluate lim a n .
n
Other Properties of Limits of Sequences
Theorem: Suppose that lim a n  L and lim bn  K . Then each of the following is true.
n 
n 
(i.)
(ii.)
(iii.)
(iv.)
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
Exercise 4: Suppose that lim a n  5 and that bn 
n
4
n an  2
. Evaluate lim bn .
n
3n
Exercise 5: Determine whether {an} = {1 + (-1)n } converges or diverges.
a1 = 1 + (-1)1 = 1 – 1

by theT.I. 89
0 ,
a2 = 1 + (-1)2 =
, a3 =
, a4 =
So, {an} =
The terms alternate as 0 and 2. Therefore, the limit
Squeeze Theorem
Theorem: Suppose that lim a n  L and lim bn  L . Suppose also that
n 
n 
Then
Absolute Value Theorem
Theorem: Suppose that lim an  0 . Then
n
Note: If lim an  c  0 , then lim a n may not even exist.
n
Example: {an} =
n
. That is, {an}
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
 1  n 
Exercise 6: Consider the sequence {an} =     . Write out the first four terms of
 2  
the sequence, and determine the limit of the sequence.
Exercise 7: Write an expression for the nth term of the sequence, and then determine the
limit of the sequence.
(i.) 2 , 1 ,
4 5 2
,
,
,…
5 7 3
(ii.) 1 +
1
3
7
15
,1+ ,1+ ,1+
,…
2
4
8
16
5
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
6
Monotonic Sequences
Def.: A sequence {an} is nondecreasing if
Def.: A sequence {an} is strictly increasing if
Def.: A sequence {an} is nonincreasing if
Def.: A sequence {an} is strictly decreasing if
, that is, if
, that is, if
, that is, if
, that is, if
Def.: A sequence {an} is monotonic if it is nondecreasing, strictly increasing,
nonincreasing, or strictly decreasing.
Exercise 8: Determine if the following sequences are monotonic, given the nth term.
(i.) an =
n3
3n  1
(ii.) bn =
3n
1  2n
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
Bounded Sequences
Def.: A sequence {an} is bounded above if
Def.: A sequence {an} is bounded below if
Def.: A sequence {an} is bounded if it is
Exercise 9: Determine if the following sequence is bounded, given the nth term.
bn =
3n
1  2n
7
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 9, Section 1
Theorem: If a sequence is bounded and monotonic, then it
8
.
Exercise 10: Use this theorem to prove that the sequence whose nth term is bn =
3n
1  2n
converges, without determining the limit.
By Exercise 8, we showed that {bn} is
By Exercise 9, we showed that {bn} is
Therefore, by the theorem
Exercise 11: Give an example of a sequence that is monotonic but not bounded.
Exercise 12: Give an example of a sequence that is bounded, not monotonic, and which
does not converge.
Exercise 13: Give an example of a sequence that is bounded, not monotonic, yet which
does converge.