Sequences
Sequences and Series
Sequences
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The Pinching Theorem:
This tool gives us a means for
comparing a new, perhaps difficult and
mysterious, sequence with other more
familiar sequences.
Theorem:
Suppose that, aj,bj, and cj are
sequences.
Sequences
a  , b 

j
j 1
j

j 1
a j  bj  c j
 
, and c j

j 1
for every j
if
lim j   a j  lim j   c j  l
then
lim j   b j  l
Sequences
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Infinite Sequences of Real Numbers
a1 , a 2 , a 3 
or
a 
j

j 1
Sequences
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Example:
Discuss the sequence 2, 1, 4, 3, 6, 5,….
We see that
a1  2, a2  1, a3  4, a4  3, a5  6, a6  5,....
or
f (1)  2, f (2)  1, f (3)  4, f (4)  5, f (5)  6
f (6)  5,.....
Sequences
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Remark: It would be a mistake to think
that every sequence is given by a rule.
Far from it. But many sequences do
come from rules, and it is always
interesting to determine what that rule
is.
Sequences
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Example:
How does the sequence 1,2,3,4,5,…
Differ from the sequence in previous
example.
This new sequence has the same values as
the sequence in last example. But they occur
in a different order. Since a sequence is by
definition an ordered list.
Sequences
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Occasionally it will prove useful to begin a
sequence with an index different from 1.
An example is
3 j  5 j  4

7, 10, 13, 16,
Sequences
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Example:
Discuss whether the sequence tends to
a limiting value.
j
a j  2 , or
We write
1 1 1
, , ,
2 4 8
2 
j

j 1
Sequences
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Then we see that the terms become
(and remain) arbitrarily close to zero. It
seems plausible to say that the
sequence tends to zero.
Sequences
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Example:
Does the sequence tend to a limit?
a j  ( 1)
j
we write  1, 1,  1, 1,
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The sequence does not seem to tend to any
limit. The sequence does not become and
remain close to a single value. Therefore we
say that it has no limit.
Sequences
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Example:
Does the sequence tend to a limit?
aj  j
3
We write 1,8,27,64,...
Sequences
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Example:
A quantity of radioactive material
decays. At the beginning of each week
there is half as much as there was the
previous week. The initial quantity is 5
grams. Use sequence notation to
express the amount of material at the
beginning of the jth week.
Sequences
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Solution:
The amount of radioactive material at the
first week is 5, second week is 5/2,
The third week is 5/4, and so on, The
amount of radioactive material tends to 0
as time tends to ∞
 1
a j  5 
 2
j 1
5 5 5
or 5, , , ,
2 4 8
Sequences
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Example:
Discuss convergence for the sequence
1, ½, 1/3, ¼,….
We conclude that the sequence
converges.
Sequences
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Example:
Find the limit of the
sequence
a
j

2 j
j 1
or
a
j
 2 
 2
j 1
or
a
j

2
1 1/ j
Sequences
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Example:
Find the limit of the
sequence
10 j  8 j
aj 
10 j
or
 8
a j  1  
 10 
j
Sequences
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Sequences with and without patterns:
Sometimes a sequence will come from
an obvious pattern or rule, and
sometimes not.
Example:
What is the next element of the
sequence
6, 6, 1, 7, 10, 2, 5, 3, 2, 5, 3
Sequences
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Example:
Find the pattern in the sequence and find
the limit:
1 4 3 6 5
, , , ,
2 3 4 5 6
The sequence is rewritten as
1
1
1
1
1
1  ,1  ,1  ,1  ,1  ,
1
2
3
4
5
1
1
1
1 1 1
1    1 . ,1  ( 1) 2  1 . ,1  ( 1) 3 1 . ,1  ( 1) 4  1 . ,
1
2
3
4
the j th term
2,
a j  1  ( 1) j  1 .
Limit j   a j  1
1
j
Sequences
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Example: Evaluate
lim j  
(sin j ) 2
j
(sin j ) 2
1
let b j 
then
0  bj 
j
j
1
a j  0,
cj 
for every j
j
We observe
lim j   a j  lim j   c j  0
lim j  
(sin j ) 2
 lim j   b j  0
j
Sequences
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Some Special Sequences:
Theorem let S be a real number and the
sequence is:
j 
s
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
j1
If s<0 then the sequence converges to 0
If s>0 then the sequence diverges
If s=0 then the sequence is just constant
sequence 1,1,1,…, and converges to 1.
Sequences
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Theorem let t be a real number and the
sequence is:
t 

j




j1
If |t|<1 then the sequence converges to 0
If |t|>1 then the sequence diverges
If t=1 then the sequence is the constant
sequence 1,1,1,1,…which converges to 1
If t=-1 then the sequence diverges.
Sequences
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Examples:
lim j  
lim j  
lim j  
j
2j
2
4j
3/ 2
2
2 j  3j
4j
j.sin(1 / j )
lim j   j 2 (1  cos(1 / j ))