LESSON 18 SEQUENCES
Notation: The symbol   is used to denote the set of positive integers. Thus,
   { 1, 2 , 3, ..... } .
Definition A sequence is a function whose domain is a set of the form
{ n    : n  N } , where N is a positive integer.
Up to this point, we have worked with functions of a real variable. That is, the
domain of the function was either the set of real numbers or was a subset of the set
of real numbers.
Example Consider the sequence f ( n ) 
n  3.
The domain of this sequence is   , and we have that
f ( 1) 
4 2
f ( 2) 
5
f ( 3) 
6
f ( 4) 
7
f ( 5) 
.
.
.
.
8 2 2
If f is a sequence, then for each positive integer n in the domain of the sequence,
there corresponds a real number f ( n ) in the range of the sequence. If the domain

of the sequence is the set { n   : n  N } , where N is a positive integer , then
the numbers in the range of the sequence may be listed by writing
f ( N ) , f ( N  1) , f ( N  2 ) , f ( N  3 ) . . . . . . , f ( N  ( n  1) ) , . . . . . .
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The number f ( N ) is called the first term of the sequence, f ( N  1) is called the
second term of the sequence, f ( N  2 ) is called the third term of the sequence,
f ( N  3 ) is called the fourth term of the sequence, and f ( N  ( n  1) ) is called
the n th term of the sequence. Of course if N  1 , then the numbers in the range of
the sequence would be
f (1) , f ( 2 ) , f ( 3 ) , f ( 4 ) . . . . . . , f ( n ) , . . . . . .
and the number f ( 1 ) is called the first term of the sequence, f ( 2 ) is called the
second term of the sequence, f ( 3 ) is called the third term of the sequence, f ( 4 )
is called the fourth term of the sequence, and f ( n ) is called the n th term of the
sequence.
COMMENT: This is the reason that some people define a sequence to be a
function whose domain is the set of positive integers   . Thus, by their definition,
2n  3
the function f ( n ) 
would not be a sequence because it is not defined at
n 1
n  1 . By our definition, f is a sequence whose domain is the set
{ n    : n  2 }.
NOTATION: When working with a sequence, it is customary to use a subscript
notation. Thus, instead of writing f ( n ) , we write f n . Thus, for our example
above, instead of having f ( n )  n  3 , we would have f n 
would have written the following
f1 
4  2 instead of f (1) 
f2 
5 instead of f ( 2 ) 
5
f3 
6 instead of f ( 3) 
6
f4 
7 instead of f ( 4 ) 
7
f5 
8  2 2 instead of f ( 5 ) 
4 2
8 2 2
.
.
.
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n  3 . Then we
COMMENT: The most common names for functions of a real variable are f, g, h,
and k. The most common names for sequences are a, b, and c.
Examples Find the domain of the following sequences. Then find the first four
terms of the sequence.
1.
an 
4
n 4

The domain of this sequence is the set { n   : n  4 } = { 4 , 5 , 6 , . . . . } .
a4 
4
0  0 , a5 
4
1  1 , a6 

Answer: { n   : n  4 } ; 0, 1,
4
4
2 , and a 7 
2,
4
4
3
3
a 4 is the first term of the sequence
a 5 is the second term of the sequence
NOTE:
a 6 is the third term of the sequence
a 7 is the fourth term of the sequence
an
2.
bn 
 3
is the n th term of the sequence
n  4
3n  2
The domain of this sequence is the set   = { 1, 2 , 3, . . . . } .
b1 
5
6 3
7
8
4
 5 , b 2   , b 3   1 , and b 4 

1
4 2
7
10 5
Answer:   ; 5,
3.
c n  (  1) n
3
4
, 1,
2
5
4n  1
n 6
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
The domain of this sequence is the set { n   : n  7 } = { 7 , 8 , 9 , . . . . } .
c 7  (  1) 7
27
31 31
35
9 35
  27 , c 8  (  1) 8 
 
, c 9  (  1)
, and
1
2
2
3
3
c 10  (  1) 10
39 39

4
4
31
35 39
,  ,
3
4
2

Answer: { n   : n  7 } ;  27 ,
NOTE:
c 7 is the first term of the sequence
c 8 is the second term of the sequence
c 9 is the third term of the sequence
c 10 is the fourth term of the sequence
cn
4.
 6
is the n th term of the sequence
 3 
 n
 10 
NOTE: Sometimes, sequences are given this way.
The domain of this sequence is the set   = { 1, 2 , 3, . . . . } .
The first term of the sequence is
3
 0 .3 .
10
The second term of the sequence is
The third term of the sequence is
The fourth term of the sequence is
3
 0.03 .
100
3
 0.003 .
1000
3
 0.0003 .
10000
Copyrighted by James D. Anderson, The University of Toledo
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Answer:
 ;
3
3
3
3
,
,
,
10 100 1000 10000
or
  ; 0.3 , 0.03 , 0.003 , 0.0003
a n  L , if for every
Definition A sequence { a n } has the limit L, written nlim
 
  0 , there exists a positive integer N such that a n  L   for all n  N .
If a sequence has a limit, then we say that the sequence converges. If a sequence
does not have a limit, then we say that the sequence diverges.
We have the following limit properties for sequences.
Theorem (Properties of Limits for Sequences) If { a n } and { b n } are sequences
and lim a n  L and lim b n  M , and c is any real number, then
n  
1.
2.
3.
4.
5.
n  
lim c a n = c lim a n  c L
n  
n  
lim ( a n  b n ) = lim a n + lim b n  L  M
n  
n  
n  
lim ( a n  b n ) = lim a n  lim b n  L  M
n  
n  
n  
lim a n b n =  lim a n   lim b n   L M
n  n  
n  
lim
n 
an
=
b
n
lim a n
n  
lim b n
n  

L
M  0
M , provided that
Proof Will be provided later.
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Theorem (The Sandwich Theorem for Sequences) Given the sequences { a n } ,
{ b n } , and { c n } . If lim b n  L and lim c n  L and if b n  a n  c n for all
n  
n  
an  L.
n  N , where N is a positive integer, then nlim
 
Proof Will be provided later.
Since sequences are not differentiable, then L’Hopital’s Rule does not hold for
sequences. However, we have the following theorem which will allow us is to
apply L’Hopital’s Rule.

Theorem Let { a n } be a sequence with a domain of { n   : n  N } . Let
f ( n )  a n . Suppose that the function f is defined for all real numbers x  N . If
lim f ( x )  L , then lim a n  L . Also, if lim f ( x )    , then lim a n    .
n  
n  
n  
n  
Proof Will be provided later.

NOTE: Let { a n } and { b n } be sequences with a domain of { n   : n  N } . Let
f ( n )  a n and g ( n )  b n where the functions f and g are defined for all real
numbers x  N . If the functions f and g are differentiable for all real numbers
an
0

lim
x  N and if n  
has
an
indeterminate
form
of
or
, then we can apply
bn
0

L’Hopital’s Rule to xlim
 
f ( x)
.
g( x )
We will need the following theorem in order to determine the limit of certain
sequences.
Theorem If r  1 is a real number, then
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1.
2.
3.
lim r n  0 if r  1
n  
lim r n   if r  1
n  
lim r n does not exist if r   1
n  
Proof Will be provided later.
a n  0 , then lim a n  0 .
Theorem If nlim
 
n  
Proof Will be provided later.
Examples Determine whether the following sequences converge or diverge. If the
sequence converges, then give its limit.
1.
bn 
n  4
3n  2
This sequence is one of our examples given above.
4
n  4
1
n
lim
lim
=
=
n


2
n   3n  2
3
3
n
1
NOTE: If you want to apply L’Hopital’s Rule, then you would have to do
the following.
lim
x  
x  4
1
1
= xlim
=
  3
3x  2
3
Thus, nlim
 
n  4
1
=
3n  2
3
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Answer: Converges;
2.
1
3
n2
a n  4n
e

n2
lim 4 n =
n   e

We will apply L’Hopital’s Rule to xlim
 
x2
.
e 4x
2x
1
1
1
x
1
1
x2
lim
lim 4 x =
lim 4 x =
lim 4 x = lim
=
=
4
x
4
x
x   4e
x   e
2 x   4e
2xe
8xe
1
(0)  0
8
n2
0
Thus, nlim
  e 4n
Answer: Converges; 0
3.
cos 5 n 2
cn 
n
cos 5 n 2
lim
n  
n
2
Since  1  cos 5 n  1 and
n  0 for n  1 , then
1
cos 5 n 2
1



for all n  1 .
n
n
n
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
1 
cos 5 n 2
1


 0 and lim
 0 by the
 0 , then lim
Since nlim

  
n  
n  
n
n
n


Sandwich Theorem.
Answer: Converges; 0
4.
 e n
3
 n



en

lim
=
n   3 n

ex
We will apply L’Hopital’s Rule to xlim
.
  3 x
ex
ex
lim
3 lim x 2 / 3e x   since lim x 2 / 3e x   since
= xlim
=
  1
x   3 x
x  
x  
x  2/3
3
lim x 2 / 3   and lim e x  
x  
x  
en

Thus, nlim
  3 n
Answer: Diverges
a n  e  3 n tan  1
5
5.
7n
n2  9
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1
NOTE: tan
7n
2
is undefined when n  3 since n  9  0 when
n 9
2
n  3 . Thus, the domain of { a n } is the set { n    : n  4 } .
e  3 n tan  1
Now consider nlim
 
5
1
Since 0  tan
0  e  3 n tan  1
5
7n

 3n 5

 0 for n  4 , then
and e
2
n 9 2
5
7n

 e  3 n for all n  4 .
n 9 2
2
0  0 and lim
Since nlim
n  
 
lim e  3 n tan  1
5
n  
7n
.
n 9
2
  3n
e
2
5
=

1
lim 3 n  0 , then
2 ne
5
7n
 0 by the Sandwich Theorem.
n2  9
Answer: Converges; 0
6.
 3 
b n  n 2 sin  2 
n 
 3 
lim n 2 sin  2  =   0
n  
n 
 3 
sin  2 
n 
 3 
 3 
2
2
lim
n
sin
 2 =
Since n sin  2  =
,
then
we
can
write
1
n  
n 
n 
2
n
 3 
sin  2 
n 
0
lim
,
which
has
an
indeterminate
form
of
.
n  
1
0
n2
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 3
sin  2
x
We will apply L’Hopital’s Rule to xlim
 
1
x2



. Thus,
 3 
 3 
 3 
 1 
 3 
cos  2   D x  2 
cos  2   3 D x  2 
sin  2 
x 
x 
x 
x 
x 
lim
lim
lim
=
=
=
x  
x  
x  
1
 1 
 1 
D
D
x
x
2 
2 
x2
x 
x 
3 
 3 

3 lim cos  2  = 3 cos  lim 2  = 3 cos 0  3 (1)  3
x  
x 
x x 
 3 
sin  2 
n  3
 3 
2
lim
lim
n
sin
 2 3
Thus, n  
. Thus, n  
1
n 
n2
Answer: Converges; 3
7.
an 
2
This is a constant sequence since a n 
lim
n  
2 
2
Answer: Converges;
8.
2 for all n  1 .
 2
cn  4    
 3
2
n
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n
2
2
 2
    0.
Since    1 , then by one of the theorems above, nlim
 
3
3
 3
n

 2 
4       4  0  4
Thus, nlim
 
 3  

Answer: Converges; 4
9.
  7  n 
5   
  6  
n
7
7
 1 , then by one of the theorems above, lim     .
Since
n  
6
6
n
7
5   
Thus, nlim
 
6
Answer: Diverges
10.
b n  (  3.14 ) n
(  3.14 ) n =
Since  3.14   1 , then by one of the theorems above, nlim
 
DNE.
Answer: Diverges
11.
3n
an 
n!
Recall: n !  n ( n  1) ( n  2 )     3  2  1
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3n
33333
3 3 3 3 3
3
9 3 3
3

NOTE:
=      =    
n! 1  2  3  4     n
2 4 5
n
1 2 3 4 5
n
93
9 3 3
3
   =  
24
2 4 4
4
3n
93
  
Thus, 0 
n! 2  4 
n  3
for all n  3
n  3
for all n  3
93
0  0 and lim  
Since nlim
n   2
 
4
Sandwich Theorem.
n  3
3n
 0 by the
 0 , then lim
n   n!
Answer: Converges; 0
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