-
8/12/2019 Student Note - Sequence
1/9
Sequences
Objectives: At the end of the module students are able to
understand what a converging sequence is,
understand what a diverging sequence is,
identify whether a given sequence is converging or diverging
identify formula or rule for a given sequence
Definition:A sequence is a list of number in a definite
order:
a1, a2, a3, a4, an, a
where a1 is the first term, a2 is the second term, ..., an is
the n-th term. Or it may bedenoted as{an} or{an}n=1.For
example:
i) 1, 2, 3, 4, ...,n, ...
ii) 3, 6, 9, 12, ..., 3n, ...
iii) 2, 2, 2, 2, , 2(1)n+1
iv) 1, 12
, 14
, 18
, , .. 12n1
Example:List the first four terms of the sequence
an= n+ 1
2n 1, n Z , n 1
Solution:
a1, a2, a3, a4, = 2
1,
2 + 1
2(2) 1,
3 + 1
2(3) 1,
4 + 1
2(4) 1,
= 2 , 1 ,45
, 57
,
Example:List the first four terms of the sequence
b0 = 1 , bn+1= 1
bn+ 1, n 0
1
-
8/12/2019 Student Note - Sequence
2/9
-
8/12/2019 Student Note - Sequence
3/9
-
8/12/2019 Student Note - Sequence
4/9
Solution:
To find the limit,
limn
an = limn
n2 + 1
n2
= limn
1 + 1
n2
1
= 1 + 0
1 = 1
The sequence converges to 1.
Example 2:
Find the limit of the sequence
an= n2 + 2
n 1
Solution:
To find the limit,
limn
an = limn
n2 + 2
n 1
= limn
n+ 2n
1 1n
=
+ 0
1 0 =
The sequence is divergent.
II. The Limit Laws
If{an}and{bn}are convergent sequences and cis a constant,
1. limn
(an+bn) = limn
an+ limn
bn
2. limn
(an bn) = limn
an limn
bn
3. limn
c an= c limn
an
4. limn
c= c
5. limn
(an bn) = limn
an limn
bn
4
-
8/12/2019 Student Note - Sequence
5/9
6. limn
anbn
=
limn
an
limn
bn, if lim
n
bn= 0
7. limn
(an)p =
limn
an
p
, provided an 0
Theorem 1:
Let an be a sequence, let f(n) =an and suppose that f(x) exists
for every real numberx 1.
i) If limx
f(x) =L, then limn
f(n) =L
ii) If limx
f(x) = (or), then limn
f(n) = (or)
Example:Ifan= 1 + 1
n, determine whether {an} converges/diverges.
Solution:Let f(n) = 1 + 1
n, then f(x) = 1 + 1
xfor every real number x 1,
limx
f(x) = limx
1 +
1
x
= 1 + 0 = 1
Thus the sequence {an} converges to 1.
Theorem 2:
If limn
|an|= 0 then limn
an= 0. This theorem is useful when the limit is zero.
Sandwich Theorem (Squeeze Theorem)
Ifan bn cn for all n n0 and limn
an = limn
cn= L, then limn
bn=L.
Example:Determine whether the sequence below
i.) an= 1
4n2 1
ii.) an= n 1
2n+ 1
iii.) an=(1)n
n
5
-
8/12/2019 Student Note - Sequence
6/9
iv.) an= (1)n1
v.) an=ln n
n
converges or diverges. If it converges, find the limit.
Solution:
i.)1
4n2 1
n=1= 3
4, 0 , 5
4, 3 ,
To determine whether the sequence converges or diverges, we
check the term atlimit n ,
limn
14
n2 1
=1
4 limn
n2 limn
(1) =
Limit does not exist and the sequence diverges.
ii.) The sequence
n 12n+ 1
n=1
= 0 ,15
, 27
, 39
,
The limit of the sequence at n is
limn
n 1
2n+ 1 = lim
n
n
n 1
n
2n
n+ 1
n
= limn
1 1n
2 + 1n
=1
2
The sequence converges to 12
.
iii.) The sequence (1)n
n
n=1
=1 ,1
2,
1
3,
1
4,
The limit of the sequence at n
limn
(1)n
n
= limn
1
n= 0
ByTheorem 2,
limn
(1)n
n = 0
The sequence converges to 0.
iv.) The terms (1)n1 oscillates between 1 and 1 as follows
(1)
n1n=1= 1 , 1 , 1 , 1 ,
Thus, limn
(1)n1 does not exist and the sequence diverges.
Note: In this case, it is wrong to use Theorem 2to show that
limn
|(1)n1|= limn
|1|= 1
and say that the sequence
(1)n1
converges.
6
-
8/12/2019 Student Note - Sequence
7/9
v.) The sequenceln n
n
n=1
= 0
1,
0.693
2 ,
1.099
3 ,
1.386
4 ,
1.61
5 ,
= 0 , 0.3465 , 0.365 , 0.3465 , 0.322 , 0.298 ,
The limit of the sequence at n is
limn
ln n
n = lim
n
1
n
1 = 0
where we have used the LHopital rule to evaluate the limit of
the sequence.The sequence converges to 0.
Theorem 3:
The sequence {rn
} is convergent if |r| < 1 and r = 1, and divergent if |r|
> 1 andr= 1, i.e.,
limn
rn =
0 if |r|< 1 , (convergent)
1 if r= 1 , (convergent)
if |r|> 1 , (divergent)
undefined if r= 1 , (divergent)
Example:
The limit
limn
2
3
n
is convergent because r = 23
, and thus |r|= | 23
|= 23
1, then by Theorem 3that the limit lim
n
(1.01)n =.
The sequence diverges.
Example:Show whether the sequence represented by
an=cos2 n
3n
7
-
8/12/2019 Student Note - Sequence
8/9
converges/diverges.
Solution:Since1< cos n
-
8/12/2019 Student Note - Sequence
9/9
Definition:
A sequence {an} is called monotonic increasing ifan < an+1
for all n 1, i.e.,
a1 < a2 < a3 a2 > a3 > > an >
Example:Determine whether the sequence below is monotonic
increasing/decreasing or not mono-tonic.
i) an = 1
5n
ii) an = n +1
n
Solution:
i) The sequence
{an}=1
5,
1
52,
1
53 ,
an > an+1 for all n 1. It is monotonic decreasing.
ii) The sequence{an}= 1 + 1 , 2 +
1
2, 3 +
1
3,
an < an+1 for all n 1. It is monotonic increasing.
Exercise:Determine whether the sequence converges or diverges.
If it converges find the limit.
i) an = ln n
ln 2n
ii) an =
1 +
2
n
1n
iii) an =
1 +
3
n
n
9
