1
Br. Bradley
Quoc Phung
230 A
Sequences
Definition: A sequence sn  converges to L 
  0, n  N  n  N  d  sn , L   
n
converges to 1
n n
Example: Prove Sn =
Solution:
 Choose   0 ,

n
1  
n n
Sn  1 
n
 n
1 
n n
n n

n
n n
n
n
1


n
1
1
n  n 2



Then

If we choose n 
For N 

1
2
1

2
, then
n
n
1
1
1 


n n
n n
n 
, n  N  Sn  1   .
Therefore : Sn =
n
converges to 1
n n
2
Br. Bradley
Quoc Phung
230 A
1. Theorem: Convergent sequences are bounded
Proof:

Suppose Sn  converges to L.
Choose  = 1, then


N
ie:
lim Sn  L
such that
n > N
Let r = max { 1, d ( S1 , L ) , d ( S2 , L )
x 
 d ( Sn , L ) <  = 1
…. , d ( Sn , L ) }
Then take the ball B( L, r+1). This will contain all terms of the sequence.

Sequence
Sn  is bounded.
Note:
1. Intervals of the form (a   , a   )  {x  R : x  a   .}
i.e: The interval an - neighborhood of a.
Denote (a -r, a+ r ) by the open ball B(a, r) centered at a with radius r
 B(a, r) =  x  R n : x  a  r

2.
A subset U of Rn is open set   a  U, there is some r = r(a)> 0 so that U  B(a, r)
A set S  R n is bounded   r > 0 such that S  B( o, r)
3
Br. Bradley
Quoc Phung
230 A
2. Theorem: If Sn  converges to s
Sn 
converges to t
Then s = t
Proof:

Since Sn 


2

converges to t, then  N1 such that n > N2  | Sn - t | <
2
Since Sn  converges to s, then  N1 such that n > N1  | Sn - s | <
Let N = max{ N1, N 2 }
Then n > N  | s-t |
= | s- Sn + Sn – t |
 | s- Sn | + | Sn – t |


=
+
2
2
=

This is true for any choice of   s = t.
Theorem: A real sequence Sn  converges to s,
then every open ball B ( s , r ) contains all ,
but a finite members of terms form the sequence.
Definition Non-decreasing Of Sequence
A real sequence Sn  is non- decreasing  n  N, S n + 1  S n.
A real sequence Sn  is non- increasing  n  N, S n + 1  S n.
A real sequence Sn  is monotone

non- decreasing or non- increasing.
4
Br. Bradley
Quoc Phung
230 A
Theorem: A monotone real value sequence is convergent

It is bounded
Proof:
Convergent  Bounded.
Bounded  Convergent.
Suppose Sn  is a bounded monotone.
Let


Sn  be Non-decreasing.
Since Sn  is bounded, the set S = { S n | n
 S has a suppremum.
Let L = sup S.
 From definition of suppremum,
Given  > 0 ,  element sn*  S such that
Since
Sn 
is Non-decreasing,
  0, n* such that n > n* 
ie:

Sn 
= 1,2,3,4,.. }is bounded above.
L    sn*  L  
if n > n* , then sn

sn*

L    sn  L  L  

L    sn  L  
d ( L, s n ) < 
converges to L.
Note: All we need have is that the sequence is eventually bounded and monotone.