Download Chapter 4.1

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
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 3
Example 4.1.1
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 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  > 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 .
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 5
Example 4.1.3
Consider the sequence
 1n   1,
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)

 = 0.3
1
It appears that lim sn = 0.
(3, 1/3) (4, 1/4)

(5, 1/5)

2
3
4

…
5
6
Let’s show that this is true.
If, for example,  = 0.3, how should N be chosen so that n  N implies
|1/n – 0| <  ?
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.
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
2
It appears that lim sn = 0.
(3, 1/3) (4, 1/4)

(5, 1/5)

4
3

…
5
6
Let’s show that this is true.
Given any  > 0, the Archimedean property says there exists N 
0 < 1/N <  .
such that
Thus for any n  N we have
1
1
– 0 =
n
n

1
< .
N
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Section 4.1, Slide 7
*Similar to 4.1.6, but
n2 + 5n
Example 4.1.6*
Show that lim 3
= 0.
with different numbers.
n –7
n2 + 5n
–0 < .
Given any  > 0, we want to make
n3 – 7
By making n  2 we can remove the absolute value signs since n3 – 7 will be positive.
n2 + 5n
So we want to know how large n has to be in order for
< .
n3 – 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
n2 + 5n

n3 – 7
bn2
b
=
cn3
c
1
n
and the latter expression is relatively easy to make small.
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 8
n2 + 5n
Example 4.1.6*
Show that lim 3
= 0.
n –7
n2 + 5n
–0 < .
Given any  > 0, we want to make
n3 – 7
n2 + 5n
bn2
With n  2 we want 3

.
n –7
cn3
Let b = 2, so that n2 + 5n  2n2.
We have n2  5n
Any b > 1 would work.
Why b = 2?
and n  5.
1
1 3
1
, so that n3 – 7 
n . Why c = ?
2
2
2
We have 1 n3  7, n3  14, and n  3.
2
Now let c =
All three conditions will be satisfied if n  5.
Any c with
0 < c < 1 would work.
In that case we have
n2 + 5n
2n2
4
–
0

=
n3 – 7
(1/2)n3
n
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 9
n2 + 5n
Example 4.1.6*
Show that lim 3
= 0.
n –7
n2 + 5n
–0 < .
Given any  > 0, we want to make
n3 – 7
When n  5 we have
To make
n2 + 5n
4
–
0

n3 – 7
n
4
4
< , we need n > .
n

There are two conditions to be satisfied.
We can accomplish both by letting N be a natural number such that
4
N > max {5, 
}.
This is possible because of
the Archimedean property.
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*
n2 + 5n
Show that lim 3
= 0.
n –7
Given any  > 0, take N 
such that N > max
{5, 4 }.
4
Then for n  N we have n > 5 and n >  .
Since n > 5 we have n2 + 5n  2n2 and n3 – 7  (1/2)n3.
Thus for n  N we have
n2 + 5n
n2 + 5n
–0 =
n3 – 7
n3 – 7

2n2
4
=
< . 
(1/2)n3
n
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  . If for some k > 0
and some m  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  > 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 sn = s. 
 
sn  s  k an < k     .
k
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 12
5n2 – 6
5
Example 4.1.9* Show that lim
=
.
8n2 – 3n
8
To apply the theorem we need to find an upper bound for
5n2 – 6
5
–
8n2 – 3n
8
=
15n – 48
8(8n2 – 3n)
when n is sufficiently large.
The numerator is easy, since |15n – 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
5n2 – 6
5
–
8n2 – 3n
8
=
15n – 48
8(8n2 – 3n)
<
15n
15 1
=
8(7n2)
56 n
2–6
5n
5
Since lim (1/n) = 0, Theorem 4.1.8 implies that lim
=
.
8n2 – 3n
8
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
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. From the definition of
convergence with  = 1, we obtain N  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. 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 14
Theorem 4.1.14
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| <
And there exists N2 

2
, for every n  N1.
such that |sn – t | <

2
, for every n  N2.
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. 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
Section 4.1, Slide 15