Student Note - Sequence

8/12/2019 Student Note - Sequence

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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


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3/9


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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

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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

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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.

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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

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converges/diverges.

Solution:Since1< cos n


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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

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Student Note - Sequence