2. SEQUENCE
Definition of a Sequence
An infinite sequence, or more simply a sequence, is an unending
succession of numbers, called terms. It is understood that the terms
have a definite order. It is typically written as ๐1, ๐2, ๐3, ๐4, . . .
where: ๐1 is the first term
๐2 is the second term,
๐3 is the third term, and so on and so forth.
Examples: 1, 2, 3, 4, . . . , 1, ยฝ, 1/3, ยผ, . . . ,
2, 4, 6, 8, . . . , 1, -1, 1, -1, . . . ,
3. EXAMPLE
๏ฑEach of these sequences has a definite pattern known as rule or
formula or general term that make it easy to generate additional
terms.
Examples:
a) 2, 4, 6, 8, . . . is a sequence having the rule or general formula 2n
since each term is twice the term number.
b) 1, 4, 9, 16, 25, . . . is a sequence having the general term ๐2
Example 1: In each part, find the general term of the sequence.
a) ยฝ, ยผ, 1/8, 1/16, . . .
b) ยฝ, 2/3, ยพ, 4/5, . . .
4. EXAMPLE
Finding the General Term of the Sequence
Answer to Example 1.
a) ยฝ, ยผ, 1/8, 1/16, . . . ,
1
2๐, . . .
Solution:
Observe the denominators given in the sequence. It can
be expressed as powers of 2 21 = 2, 22 = 4, 23 = 8, 24 =
5. SEQUENCE
b) ยฝ, 2/3, ยพ, 4/5, . . . ,
๐
๐+1
. . .
Solution:
You may notice that the numerator of the four known
terms is the same as their term numbers ( say 1st term = 1,
2nd term = 2, 3rd term = 3, and 4th term = 4) and their
denominators is one greater than their term numbers.
Thus, if we let n be the numerator and n + 1 be the
denominator, the sequence can be expressed as
๐
๐+1
.
6. EXERCISES
Exercise 1
In each part, find the general term of the sequence, starting
with n =1.
a) 1,
1
5
,
1
25
,
1
125
, . . .
b) 1, โ
1
3
,
1
9
, โ
1
27
, . . .
c)
1
2
,
3
4
,
5
6
,
7
8
, . . .
7. EXERCISES
Exercise 1
In each part, find the general term of the sequence, starting
with n =1.
a) 1,
1
5
,
1
25
,
1
125
, . . .
b) 1, โ
1
3
,
1
9
, โ
1
27
, . . .
c)
1
2
,
3
4
,
5
6
,
7
8
, . . .
8. SEQUENCE
When the general term of a sequence
๐1, ๐2 , ๐3 , . . . , ๐๐ , . . . (1)
is known, there is no need to write out the initial terms and it
is common to write only the general term enclosed in braces.
Thus (1) might be written as
๐๐ ๐=1
+โ
or ๐๐ ๐=1
โ
A sequence is a function whose domain is a set of integers
9. LIMIT OF A SEQUENCE
Limit of a Sequence
Since sequence ๐๐ are functions, it has also limits.
โข A sequence whose terms approach limiting values are said to converge.
โข A sequence that does not converge to some finite limit is said to diverge.
Example 1. Evaluate lim
๐โ+โ
๐
๐+1
Solution: lim
๐โ+โ
๐
๐+1
=
โ
โ
(indeterminate)
transforming the given function by dividing the numerator and
denominator by n, the results are:
lim
๐โ+โ
๐
๐
๐
๐
+
1
๐
= lim
๐โ+โ
1
1 +
1
๐
=
1
1 +
1
โ
=
1
1 + 0
= 1.
Thus lim
๐โ+โ
๐
๐+1
= 1 (converges)
12. EXERCISES
Evaluate the limit of the following sequence:
1. lim
๐โ+โ
๐2
4๐ + 1
2. lim
๐โ+โ
๐
๐ + 3
3. lim
๐โ+โ
3
4. lim
๐โ+โ
ln
1
๐
5. lim
๐โ+โ
2๐3
๐3 + 1
13. EXERCISES
Evaluate the limit of the following sequence:
1. lim
๐โ+โ
๐2
4๐ + 1
2. lim
๐โ+โ
๐
๐ + 3
3. lim
๐โ+โ
3
4. lim
๐โ+โ
ln
1
๐
5. lim
๐โ+โ
2๐3
๐3 + 1
14. EXERCISES
Write out the first five terms of the sequence, determine
whether the sequence converges, and if so find its limit.
1.
๐+1) ๐+2)
2๐2
๐=1
+โ
2. 1 + โ1)๐
๐=1
+โ
3.
ln ๐
๐ ๐=1
+โ
4. ๐2๐โ๐
๐=1
+โ
15. MONOTONE SEQUENCE
Monotone Sequence
๏ฑ A sequence ๐๐ ๐=1
+โ
is called
- strictly increasing if ๐1 < ๐2 < ๐3 <. . . < ๐๐ โค . . .
- increasing if ๐1 โค ๐2 โค ๐3 โค. . . โค ๐๐ โค. . .
- strictly decreasing if ๐1> ๐2 > ๐3 >. . . > ๐๐ > . . .
- decreasing if ๐1 โฅ ๐2 โฅ ๐3 โฅ. . . โฅ ๐๐ โฅ. . .
A sequence that is either increasing or decreasing is said to
be monotone, and a sequence that is either strictly increasing
or strictly decreasing is said to be strictly monotone.
19. EXAMPLE
Examples:
Determine if the following sequence is monotone or strictly monotone.
1.
๐
๐ + 1
Solution: Begin by letting ๐๐ =
๐
๐ + 1
.
Then assign n = 1, 2, 3 in the given sequence, ๐๐, to get the first
three term and observe the obtained values.
If n = 1, ๐1 =
1
2
; If n = 2, ๐2 =
2
3
; If n = 3, ๐3 =
3
4
Since ๐1 < ๐2 < ๐3 <. . . < ๐๐ < . . . ๐s strictly increasing. Then the
sequence is strictly monotone.
20. EXAMPLE
2.
1
๐
Solution: let ๐๐ =
1
๐
By assigning n = 1, 2, and 3 in the given sequence, ๐๐,
the values obtained are: ๐ = 1, ๐1 = 1; ๐ = 2, ๐2 =
1
2
;
๐ = 3, ๐3 =
1
3
Since ๐1 > ๐2 > ๐3 >. . . > ๐๐ >. . . Is strictly decreasing.
Then the sequence is strictly monotone.
21. EXAMPLE
3. Use the difference ๐๐+1 โ ๐๐ to show that the
๐ โ 2๐
๐=1
+โ
is strictly increasing or strictly decreasing.
Solution:
๐๐ = ๐ โ 2๐; ๐๐+1 = ๐ + 1 โ 2๐+1
๐๐+1โ๐๐ = ๐ + 1 โ 2๐+1 โ ๐ โ 2๐)
= 1 + 2๐
โ 2๐+1
= 1 โ 2๐
๐๐+1โ๐๐ = 1 โ 2๐ < 0 which proves that the sequence is
strictly decreasing.
22. EXAMPLE
4. Use the ratio
๐๐+1
๐๐
to show that the given sequence
๐๐
๐! ๐=1
+โ
is
strictly increasing or strictly decreasing.
Solution:
๐๐ =
๐๐
๐!
, ๐๐+1 =
๐+1)๐+1
๐+1)!
Forming the ratio of successive terms we obtain
๐๐+1
๐๐
=
๐+1)๐+1
๐+1)!
๐๐
๐!
=
๐+1)๐+1
๐+1)!
โ
๐!
๐๐
=
๐+1 ๐ ๐+1)
๐+1)๐๐ =
๐+1)๐
๐๐ =
๐+1
๐
๐
From which we see that
๐๐+1
๐๐
> 1. This proves that the sequence is
strictly increasing.
23. EXAMPLE
5. Show that the sequence
10๐
๐! ๐=1
+โ
is eventually strictly
decreasing.
Solution:
๐๐ =
10๐
๐!
and ๐๐+1 =
10๐+1
๐+1)!
๐๐+1
๐๐
=
10๐+1
๐+1)!
10๐
๐!
=
10๐+1๐!
10๐ ๐+1)!
= 10
๐!
๐+1 ๐!
=
10
๐+1
๐๐+1
๐๐
< 1 for all ๐ โฅ 10 , so the sequence is eventually
decreasing as confirmed by the graph.
25. EXERCISES
Determine if the following sequence is monotone or strictly
monotone.
1. 1 โ
1
๐
2.
๐
3๐ + 1
3. ๐ โ 2๐
4. ๐ โ ๐2
26. BOUNDED SEQUENCE
A sequence is bounded above if there is some number N
such that ๐๐ โค N for every n, and bounded below if there
is some number N such that ๐๐ โฅ N for every n. If a
sequence is bounded above and bounded below it is
bounded. If a sequence ๐๐ ๐=0
+โ
is increasing or non-
decreasing it is bounded below (by ๐0), and if it is
decreasing or nonincreasing it is bounded above (by ๐0).
28. EXAMPLES
Determine whether or not the sequence is bounded.
1.
1
๐
Solution: let ๐๐ =
1
๐
Generating some terms in the sequence ๐๐
n = 1, ๐1 = 1 bounded above since ๐๐ โค 1
n = 2, ๐2 = ยฝ
n = 3, ๐3 = 1/3
lim
๐โโ
1
๐
= 0, bounded below since ๐๐ > 0
Therefore the sequence is bounded.
29. EXAMPLES
Determine whether or not the sequence is bounded.
2. ๐ โ1)๐
Solution: let ๐๐ = ๐ โ1)๐
Generating some terms in the sequence ๐๐
n = 1, ๐1 = -1 n = 2, ๐2 = 2
n = 3, ๐3 = -3 n = 4, ๐4 = 4
n = 5, ๐5 = -5 n = 6, ๐6 = 6
The terms in the sequence are alternating in signs, the positive
terms increasing without bound and the negative terms
decreasing without bound,
Therefore the sequence is not bounded.
Source: https://www.youtube.com/watch?v=UbNE_beWlhU
30. EXAMPLES
Determine whether or not the sequence is bounded.
3.
2๐โ3
3๐+4
Solution: let ๐๐ =
2๐โ3
3๐+4
Generating some terms in the sequence ๐๐
n = 1, ๐1 =
โ1
7
bounded below since ๐๐ โฅ
โ1
7
n = 2, ๐2 = 1/10
n = 3, ๐3 = 3/13
lim
๐โโ
2๐โ3
3๐+4
= 2/3, bounded above since ๐๐ < 2/3
Therefore, the sequence is bounded
Source: https://www.youtube.com/watch?v=UbNE_beWlhU