Analysis with an introduction to proof 5th by steven lay ch04

Chapter 4
Sequences

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 1


Section 4.1
Convergence

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 2


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:

( 1n )

=

( 1,

1 , 1 , 1 , ...

2 3 4

)

Sometimes we want to change the domain to include 0 or start at something
other than 1.

∞
( sn )∞
or
(
s
)
n=0
n n=m

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Section 4.1, Slide 3


Example 4.1.1
n
Consider the sequence (sn ) given by sn = 1 + (–1) .
Here are the first few terms:
s1 = 1 + (–1) = 0
2
s2 = 1 + (–1) = 1 + 1 = 2
3
s3 = 1 + (–1) = 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 sequence
will always have infinitely many terms.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 4


Sometimes the values in a sequence get “close” to a fixed number.
Like (1/n) gets close to 0.

Definition 4.1.2
A sequence (sn) is said to converge to the real number s provided that
for every ε > 0 there exists a natural number N such that for all n ∈ ,
n ≥ N implies that |sn – s| < ε.
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 ε.
We don’t have to find one N that works for all ε.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 5


Example 4.1.3

( 1n ) = ( 1,

Consider the sequence

1 , 1 , 1 , 1 , ...
2 3 4 5

).

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

1

(1, 1)
•

(2, 1/2)

ε

•

(3, 1/3)
•

= 0.3

1

2

It appears that lim sn = 0.

3

(4, 1/4)
•

(5, 1/5)

4

•

•

5

6

…

Let’s show that this is true.

If, for example, ε = 0.3, how should N be chosen so that n ≥ N implies
|1/n – 0| < ε ?
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 ε > 0 there exists an N that works.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 6


Example 4.1.3

( 1n ) = ( 1,

Consider the sequence

1 , 1 , 1 , 1 , ...
2 3 4 5

).

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

(1, 1)

1

•

(2, 1/2)
•

1

(3, 1/3)

2

It appears that lim sn = 0.

(4, 1/4)

•

•

4

3

(5, 1/5)
•

•

5

6

…

Let’s show that this is true.

Given any ε > 0, the Archimedean property says there exists N ∈

such that

0 < 1/N < ε .
Thus for any n ≥ N we have
1
n

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

– 0 =

1
n

≤

1
N

< ε.

Section 4.1, Slide 7


Example 4.1.6*

2
n + 5n
= 0.
3
n –7

Show that lim

*Similar to 4.1.6, but
with different numbers.

2
n + 5n

Given any ε > 0, we want to make

–0 < ε.

3
n –7

3
By making n ≥ 2 we can remove the absolute value signs since n – 7 will be positive.
2
n + 5n

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

3
n –7

< ε.

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.
2
2
2
For large values of n, the numerator behaves like n , so we want n + 5n ≤ bn .
3
3
3
And the denominator behaves like n , so we want n – 7 ≥ cn .

Then we will have

2
n + 5n
3
n –7

≤

=

bn
cn

2

3

b

1

c

n

and the latter expression is relatively easy to make small.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 8


Example 4.1.6*

2
n + 5n
= 0.
3
n –7

Show that lim

2
n + 5n

Given any ε > 0, we want to make
2
n + 5n
≤
3
n –7

With n ≥ 2 we want

bn

.

cn

2

3

2
2
Let b = 2, so that n + 5n ≤ 2n .
2
We have n ≥ 5n
1

Now let c =
2

1

3
n .

1 3
2

Any b > 1 would work.

Why b = 2?

and n ≥ 5.

3

, so that n – 7 ≥

We have

–0 < ε.

3
n –7

2

3
n ≥ 14,

n ≥ 7,

2
n + 5n
3
n –7

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≤

2

0 < c < 1 would work.

In that case we have

2

–0

Any c with

and n ≥ 3.
if n ≥ 5.

All three conditions will be satisfied

1

?

Why c =

2n

3
(1/2)n

=

4
n

Section 4.1, Slide 9

Example 4.1.6*

Show that lim

2
n + 5n
= 0.
3
n –7
2
n + 5n

Given any ε > 0, we want to make

2
n + 5n

When n ≥ 5 we have

4
To make
n

< ε, we need n >

–0 < ε.

3
n –7

–0

3
n –7

.

≤

4
n

4

ε

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.

ε
So when n ≥ N, both conditions are satisfied.
Now let’s organize this into a proof.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 10


Example 4.1.6*

2
n + 5n
= 0.
3
n –7

Show that lim

Given any ε > 0, take N ∈

such that

N > max

Then for n ≥ N we have n > 5 and n >

{5, }.4
ε

.

4

ε
3
Since n > 5 we have n + 5n ≤ 2n and n – 7 ≥ (1/2)n .
2

2

3

Thus for n ≥ N we have
2
n + 5n
3
n –7

–0 =

2
n + 5n
3
n –7

≤

2n

2

(1/2)n

3

=

4
n

< ε. ♦

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

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 11


Theorem 4.1.8
Let (sn) and (an) be sequences of real numbers and let s ∈
and some m ∈

.

If for some k > 0

we have
|sn – s| ≤ k | |,
an

for all n ≥ m,

and if lim an = 0, then it follows that lim sn = s.
Proof: Given any ε
implies that | an | < ε /k.

> 0, since lim an = 0 there exists N1 ∈

such that n ≥ N1

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

and n ≥ N1, so that

ε 
sn − s ≤ k an < k  ÷ = ε .
k

Thus lim sn = s. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 12


Example 4.1.9*

Show that lim

2
5n – 6

5
=
.
2
8n – 3n
8

To apply the theorem we need to find an upper bound for
2
5n – 6

5

2
8n – 3n

8

–

=

15n – 48

when n is sufficiently large.

2
8(8n – 3n)

The numerator is easy, since |15n – 48| < 15n for all n ≥ 2.

2
2
2
For the denominator, we have 8n – 3n ≥ 7n when n ≥ 3n and n ≥ 3.
Thus when n ≥ 3, we have
2
5n – 6
2
8n – 3n

5
8

–

=

15n – 48
2
8(8n – 3n)

Since lim (1/n) = 0, Theorem 4.1.8 implies that

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

<

15n
2
8(7n )

=

15 1
56 n

lim

2
5n – 6
2
8n – 3n

=

5
.
8

Section 4.1, Slide 13


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 ∈

.

Theorem 4.1.13
Every convergent sequence is bounded.
Proof:
Let (sn) be a convergent sequence and let lim sn = s.
convergence with ε = 1, we obtain N ∈

From the definition of
such that | sn – s | < 1 whenever n ≥ N.
 

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. ♦
 

 

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Section 4.1, Slide 14


Theorem 4.1.14
If a sequence converges, its limit is unique.
Proof:

Suppose sn → s and sn → t.
there exists N1 ∈

Given any ε > 0,

such that |sn – s| <

And there exists N2 ∈

ε

, for every n ≥ N1.
2

ε

, for every n ≥ N2.

such that |sn – t | <

2

Therefore, if n ≥ max {N1, N2} then from the triangle inequality we have
≤ |s – sn | + |sn – t |

|s – t | = |s – sn + sn – t |
<

ε
2

+

ε
2

= ε.

Since this holds for all ε > 0, we must have s = t. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 4.1, Slide 15