Chapter 4 Sequences Section - PowerPoint Presentation

Slide1

Chapter 4

Sequences

Section

4.1

Convergence

A

sequence S is a function whose domain is the set of natural numbers.

We usually write

sn instead of S(n).

We may describe the sequence S as (sn) or by listing the elements (s1, s2, s3, …).

Sometimes we just give a formula for the typical nth term:

Sometimes we want to change the domain to include 0 or start at something

other than 1.

Consider the sequence (sn ) given by sn = 1 + (–1)n.

Here are the first few terms:

s1 = 1 + (–1) = 0

s2 = 1 + (–1)2 = 1 + 1 = 2

s3 = 1 + (–1)3 = 1 – 1 = 0

s4 = 1 + (–1)4 = 1 + 1 = 2, and so on.

So, (sn) = (0, 2, 0, 2, 0, 2, …)

Note that the terms of a sequence do not have to be distinct.

We consider s2 and s4 to be different terms, even though their values are both 2.

The range of this sequence is the set of values obtained: {0, 2}.

So the range of a sequence may be finite, even though the sequencewill always have infinitely many terms.

Example 4.1.1

Sometimes the values in a sequence get “close” to a fixed number.

Like (1/n) gets close to 0.

A sequence (sn) is said to converge to the real number s provided that

for every

 > 0 there exists a natural number N such that for all n  ,n  N implies that |sn – s| < .

If (

sn) converges to s, then s is called the limit of the sequence (sn).

We write limn  sn = s,

lim sn = s,

or sn  s.

If a sequence does not converge to a real number, it is said to diverge.

Note the order of the quantifiers in the definition.

The number N may depend on .

We don’t have to find one N that works for all .

Definition 4.1.2

Consider the sequence

If we were to graph this sequence as a function it would look like this:



(1,

1)



(2,

1/2)



(3,

 1/3)



(4,

 1/4)



(5,

 1/5)

1

1

2

3

4

5

6



…



=

0.3

It appears that lim

s

n

= 0.

If, for example,



= 0.3, how should

N

be chosen so that

n  N implies |1/n – 0| <  ?

Let’s show that this is true.

We have |1/n – 0| = 1/n,

and 1/n is less than 0.3 when n  4.

So choose any natural number N such that N  4.

This is an illustration, but not a proof.

We must show that for every  > 0 there exists an N that works.

Example 4.1.3

If we were to graph this sequence as a function it would look like this:



(1,

 1)



(2,

 1/2)



(3,

 1/3)



(4,

 1/4)



(5,

 1/5)

1

1

2

3

4

5

6



…

It appears that lim

s

n

= 0.

Let’s show that this is true.

Thus for any

n



N

we have

Given

any



> 0, the Archimedean property says there exists N  such that 0 < 1/N <  .

<

 .

1

n



1

N

– 0 =

1

n

Consider the sequence

Example 4.1.3

Show that lim = 0.

*Similar to 4.1.6, but

with different numbers.

n

2

+ 5nn3 – 7

Given any



> 0, we want to make

n

2 + 5nn3 – 7

– 0

<  .

By making

n

 2 we can remove the absolute value signs since n3 – 7 will be positive.

So we want to know how large n has to be in order for

n

2

+ 5nn3 – 7

<

 .

This is hard to solve for n, so we seek an estimate of how large the left side will be.

That is, we seek an upper bound for the numerator and a lower bound for the denominator.

For large values of n, the numerator behaves like n2, so we want n2 + 5n  bn2.

And the denominator behaves like n3, so we want n3 – 7  cn3.

Then we will have  =

n

2 + 5nn3 – 7

bn

2

cn3

b

c

1

n

and the latter expression is relatively easy to make small.

Example 4.1.6*

With

n

 2 we want  .

Let b = 2, so that n2 + 5n  2n2.

We have n2  5n

We have n3  7,

n

2 + 5nn3 – 7

bn

2

cn3

4

n

All three conditions will be satisfied

Why

b

= 2?

Any b > 1 would work.

and n  5.

Now let c = , so that n3 – 7  n3.

1

2

Why

c = ?

Any c with 0 < c < 1 would work.

1

2

1

2

1

2

n

3

 14,

and n  3.

if n  5.

In that case we have

2

n2(1/2)n3

n

2

+ 5

n

n3 – 7

– 0 

=

Show that lim = 0.

n

2

+ 5

n

n3 – 7

Given any



> 0, we want to make

n

2 + 5nn3 – 7

– 0 <

 .

Example 4.1.6*

4

n

To make <



, we need

n > .

When n  5 we have

n

2

+ 5

nn3 – 7

– 0 

4

n

4



There are two conditions to be satisfied.

We can accomplish both by letting

N

be a natural number such that

N

> max

{

5, }.

4



This is possible because of

the Archimedean property.

So when n  N, both conditions are satisfied.

Now let’s organize this into a proof.

Show that lim = 0.

n

2 + 5nn3 – 7

Given any



> 0, we want to make

n

2 + 5nn3 – 7

– 0 <

 .

Example 4.1.6*

Then for n  N we have n > 5 and n > .

Since n > 5 we have n2 + 5n  2n2 and n3 – 7  (1/2)n3.

N > max {5, }.

4



Given any



> 0, take N  such that

4



Thus for

n

 N we have

4

n

2

n

2(1/2)n3

n

2

+ 5

n

n3 – 7

– 0 =

=

<

. 

n

2

+ 5

nn3 – 7



This is a lot of work, but it can be reduced somewhat by the following theorem.

Show that lim = 0.

n

2

+ 5nn3 – 7

Example 4.1.6*

If for some

k > 0 and some m  we have

Let (

s

n) and (an) be sequences of real numbers and let s  .

|

sn – s|  k | an |, for all n  m,

and if lim an = 0, then it follows that lim sn = s.

Theorem 4.1.8

Proof:

Given any  > 0, since lim an = 0 there exists N1  such that n ≥ N1 implies that | an | <  /k.

Now let N = max {m, N1}. Then for n ≥ N we have n ≥ m and n ≥ N1, so that

Thus lim

s

n

=

s

.



lim = .

5

n2 – 6 58n2 – 3n 8

To apply the theorem we need to find an upper bound for

5

n

2

– 6 5

8n2 – 3n 8

– =

15

n

– 48

8(8

n

2 – 3n)

when

n

is sufficiently large.

The numerator is easy, since |15

n – 48| < 15n for all n  2.

For the denominator, we have 8n2 – 3n  7n2 when n2  3n and n  3.

Thus when n  3, we have

5

n

2 – 6 58n2 – 3n 8

– =

15

n

– 48

8(8

n

2 – 3n)

<

15

n

8(7

n

2)

15 1

56

n

=

Since lim (1/

n

) = 0, Theorem 4.1.8 implies that

Show that lim = .

5

n

2

– 6 5

8n2 – 3n 8

Example 4.1.9*

From the definition of

convergence with  = 1, we obtain N  such that | sn – s | < 1 whenever n ≥ N.

Note:

A sequence (sn) is bounded if the range {sn : n  } is a bounded set.

That is, if there exists an

M

> 0 such that | sn |  M for all n  .

Every convergent sequence is bounded.

Proof:

Theorem 4.1.13

Let (sn) be a convergent sequence and let lim sn = s.

Thus for n ≥ N we have | sn | < | s | + 1 by Exercise 3.2.6(b).

If we let

M = max {| s1 |, | s2 |, …, | sN |, | s | +1},

then we have |

 sn |  M for all n  , so (sn) is bounded. 

If a sequence converges, its limit is unique.

Proof: Suppose sn  s and sn  t.

Given any  > 0,

there exists

N1  such that |sn – s| < , for every n  N1.



2

And there exists

N

2

 such that |sn – t | < , for every n  N2.



2

Therefore, if

n

 max {N1, N2} then from the triangle inequality we have

|s – t | = |s – sn + sn – t |

 |s – sn | + |sn – t |

< +



2



2

=

 .

Since this holds for all

 > 0, we must have s = t. 

Theorem 4.1.14