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Lesson 1a
SEQUENCE
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 ๐‘Ž1, ๐‘Ž2, ๐‘Ž3, ๐‘Ž4, . . .
where: ๐‘Ž1 is the first term
๐‘Ž2 is the second term,
๐‘Ž3 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, . . . ,
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 ๐‘›2
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, . . .
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 =
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
.
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, โˆ’
1
3
,
1
9
, โˆ’
1
27
, . . .
c)
1
2
,
3
4
,
5
6
,
7
8
, . . .
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, โˆ’
1
3
,
1
9
, โˆ’
1
27
, . . .
c)
1
2
,
3
4
,
5
6
,
7
8
, . . .
SEQUENCE
When the general term of a sequence
๐‘Ž1, ๐‘Ž2 , ๐‘Ž3 , . . . , ๐‘Ž๐‘› , . . . (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
๐‘Ž๐‘› ๐‘›=1
+โˆž
or ๐‘Ž๐‘› ๐‘›=1
โˆž
A sequence is a function whose domain is a set of integers
LIMIT OF A SEQUENCE
Limit of a Sequence
Since sequence ๐‘Ž๐‘› are functions, it has also limits.
โ€ข A sequence whose terms approach limiting values are said to converge.
โ€ข A sequence that does not converge to some finite limit is said to diverge.
Example 1. Evaluate lim
๐‘›โ†’+โˆž
๐‘›
๐‘›+1
Solution: lim
๐‘›โ†’+โˆž
๐‘›
๐‘›+1
=
โˆž
โˆž
(indeterminate)
transforming the given function by dividing the numerator and
denominator by n, the results are:
lim
๐‘›โ†’+โˆž
๐‘›
๐‘›
๐‘›
๐‘›
+
1
๐‘›
= lim
๐‘›โ†’+โˆž
1
1 +
1
๐‘›
=
1
1 +
1
โˆž
=
1
1 + 0
= 1.
Thus lim
๐‘›โ†’+โˆž
๐‘›
๐‘›+1
= 1 (converges)
EXAMPLE
2. Evaluate: lim
๐‘›โ†’+โˆž
๐‘› + 1)
Solution:
lim
๐‘›โ†’+โˆž
๐‘› + 1) = โˆž + 1 = โˆž (diverges)
3. Determine whether the sequence,
1,
1
2
,
1
22 ,
1
23 , . . .
1
2๐‘› , . . . converge.
Solution:
lim
๐‘›โ†’โˆž
1
2๐‘› =
1
2โˆž =
1
โˆž
= 0 converges)
EXAMPLE
4. Determine whether the sequence
1, 2, 22
, 23
, . . . , 2๐‘›
, . . . converge.
Solution:
lim
๐‘›โ†’+โˆž
2๐‘›
= 2โˆž
= โˆž (diverges)
EXERCISES
Evaluate the limit of the following sequence:
1. lim
๐‘›โ†’+โˆž
๐‘›2
4๐‘› + 1
2. lim
๐‘›โ†’+โˆž
๐‘›
๐‘› + 3
3. lim
๐‘›โ†’+โˆž
3
4. lim
๐‘›โ†’+โˆž
ln
1
๐‘›
5. lim
๐‘›โ†’+โˆž
2๐‘›3
๐‘›3 + 1
EXERCISES
Evaluate the limit of the following sequence:
1. lim
๐‘›โ†’+โˆž
๐‘›2
4๐‘› + 1
2. lim
๐‘›โ†’+โˆž
๐‘›
๐‘› + 3
3. lim
๐‘›โ†’+โˆž
3
4. lim
๐‘›โ†’+โˆž
ln
1
๐‘›
5. lim
๐‘›โ†’+โˆž
2๐‘›3
๐‘›3 + 1
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)
2๐‘›2
๐‘›=1
+โˆž
2. 1 + โˆ’1)๐‘›
๐‘›=1
+โˆž
3.
ln ๐‘›
๐‘› ๐‘›=1
+โˆž
4. ๐‘›2๐‘’โˆ’๐‘›
๐‘›=1
+โˆž
MONOTONE SEQUENCE
Monotone Sequence
๏ฑ A sequence ๐‘Ž๐‘› ๐‘›=1
+โˆž
is called
- strictly increasing if ๐‘Ž1 < ๐‘Ž2 < ๐‘Ž3 <. . . < ๐‘Ž๐‘› โ‰ค . . .
- increasing if ๐‘Ž1 โ‰ค ๐‘Ž2 โ‰ค ๐‘Ž3 โ‰ค. . . โ‰ค ๐‘Ž๐‘› โ‰ค. . .
- strictly decreasing if ๐‘Ž1> ๐‘Ž2 > ๐‘Ž3 >. . . > ๐‘Ž๐‘› > . . .
- decreasing if ๐‘Ž1 โ‰ฅ ๐‘Ž2 โ‰ฅ ๐‘Ž3 โ‰ฅ. . . โ‰ฅ ๐‘Ž๐‘› โ‰ฅ. . .
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.
EXAMPLE
SEQUENCE DESCRIPTION
1.
1
2
,
2
3
,
3
4
,โ€ฆ,
๐‘›
๐‘›+1
, โ€ฆ Strictly increasing
2. 1,
1
2
,
1
3
,โ€ฆ,
1
๐‘›
, โ€ฆ Strictly decreasing
3. 1, 1, 2, 2, 3, 3,โ€ฆ Increasing: not strictly increasing
4. 1, 1,
1
2
,
1
2
,
1
3
,
1
3
,โ€ฆ Decreasing: not strictly decreasing
5. 1, โˆ’
1
2
,
1
3
,โˆ’
1
4
โ€ฆ, โˆ’1)๐‘›+11
๐‘›
, โ€ฆ Neither increasing nor decreasing
EXAMPLE
TESTING FOR MONOTONICITY
EXAMPLE
Examples:
Determine if the following sequence is monotone or strictly monotone.
1.
๐‘›
๐‘› + 1
Solution: Begin by letting ๐‘Ž๐‘› =
๐‘›
๐‘› + 1
.
Then assign n = 1, 2, 3 in the given sequence, ๐‘Ž๐‘›, to get the first
three term and observe the obtained values.
If n = 1, ๐‘Ž1 =
1
2
; If n = 2, ๐‘Ž2 =
2
3
; If n = 3, ๐‘Ž3 =
3
4
Since ๐‘Ž1 < ๐‘Ž2 < ๐‘Ž3 <. . . < ๐‘Ž๐‘› < . . . ๐‘–s strictly increasing. Then the
sequence is strictly monotone.
EXAMPLE
2.
1
๐‘›
Solution: let ๐‘Ž๐‘› =
1
๐‘›
By assigning n = 1, 2, and 3 in the given sequence, ๐‘Ž๐‘›,
the values obtained are: ๐‘› = 1, ๐‘Ž1 = 1; ๐‘› = 2, ๐‘Ž2 =
1
2
;
๐‘› = 3, ๐‘Ž3 =
1
3
Since ๐‘Ž1 > ๐‘Ž2 > ๐‘Ž3 >. . . > ๐‘Ž๐‘› >. . . Is strictly decreasing.
Then the sequence is strictly monotone.
EXAMPLE
3. Use the difference ๐‘Ž๐‘›+1 โˆ’ ๐‘Ž๐‘› to show that the
๐‘› โˆ’ 2๐‘›
๐‘›=1
+โˆž
is strictly increasing or strictly decreasing.
Solution:
๐‘Ž๐‘› = ๐‘› โˆ’ 2๐‘›; ๐‘Ž๐‘›+1 = ๐‘› + 1 โˆ’ 2๐‘›+1
๐‘Ž๐‘›+1โˆ’๐‘Ž๐‘› = ๐‘› + 1 โˆ’ 2๐‘›+1 โˆ’ ๐‘› โˆ’ 2๐‘›)
= 1 + 2๐‘›
โˆ’ 2๐‘›+1
= 1 โˆ’ 2๐‘›
๐‘Ž๐‘›+1โˆ’๐‘Ž๐‘› = 1 โˆ’ 2๐‘› < 0 which proves that the sequence is
strictly decreasing.
EXAMPLE
4. Use the ratio
๐‘Ž๐‘›+1
๐‘Ž๐‘›
to show that the given sequence
๐‘›๐‘›
๐‘›! ๐‘›=1
+โˆž
is
strictly increasing or strictly decreasing.
Solution:
๐‘Ž๐‘› =
๐‘›๐‘›
๐‘›!
, ๐‘Ž๐‘›+1 =
๐‘›+1)๐‘›+1
๐‘›+1)!
Forming the ratio of successive terms we obtain
๐‘Ž๐‘›+1
๐‘Ž๐‘›
=
๐‘›+1)๐‘›+1
๐‘›+1)!
๐‘›๐‘›
๐‘›!
=
๐‘›+1)๐‘›+1
๐‘›+1)!
โˆ™
๐‘›!
๐‘›๐‘›
=
๐‘›+1 ๐‘› ๐‘›+1)
๐‘›+1)๐‘›๐‘› =
๐‘›+1)๐‘›
๐‘›๐‘› =
๐‘›+1
๐‘›
๐‘›
From which we see that
๐‘Ž๐‘›+1
๐‘Ž๐‘›
> 1. This proves that the sequence is
strictly increasing.
EXAMPLE
5. Show that the sequence
10๐‘›
๐‘›! ๐‘›=1
+โˆž
is eventually strictly
decreasing.
Solution:
๐‘Ž๐‘› =
10๐‘›
๐‘›!
and ๐‘Ž๐‘›+1 =
10๐‘›+1
๐‘›+1)!
๐‘Ž๐‘›+1
๐‘Ž๐‘›
=
10๐‘›+1
๐‘›+1)!
10๐‘›
๐‘›!
=
10๐‘›+1๐‘›!
10๐‘› ๐‘›+1)!
= 10
๐‘›!
๐‘›+1 ๐‘›!
=
10
๐‘›+1
๐‘Ž๐‘›+1
๐‘Ž๐‘›
< 1 for all ๐‘› โ‰ฅ 10 , so the sequence is eventually
decreasing as confirmed by the graph.
EXAMPLE
EXERCISES
Determine if the following sequence is monotone or strictly
monotone.
1. 1 โˆ’
1
๐‘›
2.
๐‘›
3๐‘› + 1
3. ๐‘› โˆ’ 2๐‘›
4. ๐‘› โˆ’ ๐‘›2
BOUNDED SEQUENCE
A sequence is bounded above if there is some number N
such that ๐‘Ž๐‘› โ‰ค N for every n, and bounded below if there
is some number N such that ๐‘Ž๐‘› โ‰ฅ N for every n. If a
sequence is bounded above and bounded below it is
bounded. If a sequence ๐‘Ž๐‘› ๐‘›=0
+โˆž
is increasing or non-
decreasing it is bounded below (by ๐‘Ž0), and if it is
decreasing or nonincreasing it is bounded above (by ๐‘Ž0).
BOUNDED SEQUENCE
.
Theorems 9.2.3 and 9.2.4 (p. 611)
Theorems 9.2.3 and 9.2.4 (p. 611)
EXAMPLES
Determine whether or not the sequence is bounded.
1.
1
๐‘›
Solution: let ๐‘Ž๐‘› =
1
๐‘›
Generating some terms in the sequence ๐‘Ž๐‘›
n = 1, ๐‘Ž1 = 1 bounded above since ๐‘Ž๐‘› โ‰ค 1
n = 2, ๐‘Ž2 = ยฝ
n = 3, ๐‘Ž3 = 1/3
lim
๐‘›โ†’โˆž
1
๐‘›
= 0, bounded below since ๐‘Ž๐‘› > 0
Therefore the sequence is bounded.
EXAMPLES
Determine whether or not the sequence is bounded.
2. ๐‘› โˆ’1)๐‘›
Solution: let ๐‘Ž๐‘› = ๐‘› โˆ’1)๐‘›
Generating some terms in the sequence ๐‘Ž๐‘›
n = 1, ๐‘Ž1 = -1 n = 2, ๐‘Ž2 = 2
n = 3, ๐‘Ž3 = -3 n = 4, ๐‘Ž4 = 4
n = 5, ๐‘Ž5 = -5 n = 6, ๐‘Ž6 = 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
EXAMPLES
Determine whether or not the sequence is bounded.
3.
2๐‘›โˆ’3
3๐‘›+4
Solution: let ๐‘Ž๐‘› =
2๐‘›โˆ’3
3๐‘›+4
Generating some terms in the sequence ๐‘Ž๐‘›
n = 1, ๐‘Ž1 =
โˆ’1
7
bounded below since ๐‘Ž๐‘› โ‰ฅ
โˆ’1
7
n = 2, ๐‘Ž2 = 1/10
n = 3, ๐‘Ž3 = 3/13
lim
๐‘›โ†’โˆž
2๐‘›โˆ’3
3๐‘›+4
= 2/3, bounded above since ๐‘Ž๐‘› < 2/3
Therefore, the sequence is bounded
Source: https://www.youtube.com/watch?v=UbNE_beWlhU

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Lesson 1a_Sequence.pptx

  • 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 ๐‘Ž1, ๐‘Ž2, ๐‘Ž3, ๐‘Ž4, . . . where: ๐‘Ž1 is the first term ๐‘Ž2 is the second term, ๐‘Ž3 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 ๐‘›2 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, โˆ’ 1 3 , 1 9 , โˆ’ 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, โˆ’ 1 3 , 1 9 , โˆ’ 1 27 , . . . c) 1 2 , 3 4 , 5 6 , 7 8 , . . .
  • 8. SEQUENCE When the general term of a sequence ๐‘Ž1, ๐‘Ž2 , ๐‘Ž3 , . . . , ๐‘Ž๐‘› , . . . (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 ๐‘Ž๐‘› ๐‘›=1 +โˆž or ๐‘Ž๐‘› ๐‘›=1 โˆž A sequence is a function whose domain is a set of integers
  • 9. LIMIT OF A SEQUENCE Limit of a Sequence Since sequence ๐‘Ž๐‘› are functions, it has also limits. โ€ข A sequence whose terms approach limiting values are said to converge. โ€ข A sequence that does not converge to some finite limit is said to diverge. Example 1. Evaluate lim ๐‘›โ†’+โˆž ๐‘› ๐‘›+1 Solution: lim ๐‘›โ†’+โˆž ๐‘› ๐‘›+1 = โˆž โˆž (indeterminate) transforming the given function by dividing the numerator and denominator by n, the results are: lim ๐‘›โ†’+โˆž ๐‘› ๐‘› ๐‘› ๐‘› + 1 ๐‘› = lim ๐‘›โ†’+โˆž 1 1 + 1 ๐‘› = 1 1 + 1 โˆž = 1 1 + 0 = 1. Thus lim ๐‘›โ†’+โˆž ๐‘› ๐‘›+1 = 1 (converges)
  • 10. EXAMPLE 2. Evaluate: lim ๐‘›โ†’+โˆž ๐‘› + 1) Solution: lim ๐‘›โ†’+โˆž ๐‘› + 1) = โˆž + 1 = โˆž (diverges) 3. Determine whether the sequence, 1, 1 2 , 1 22 , 1 23 , . . . 1 2๐‘› , . . . converge. Solution: lim ๐‘›โ†’โˆž 1 2๐‘› = 1 2โˆž = 1 โˆž = 0 converges)
  • 11. EXAMPLE 4. Determine whether the sequence 1, 2, 22 , 23 , . . . , 2๐‘› , . . . converge. Solution: lim ๐‘›โ†’+โˆž 2๐‘› = 2โˆž = โˆž (diverges)
  • 12. EXERCISES Evaluate the limit of the following sequence: 1. lim ๐‘›โ†’+โˆž ๐‘›2 4๐‘› + 1 2. lim ๐‘›โ†’+โˆž ๐‘› ๐‘› + 3 3. lim ๐‘›โ†’+โˆž 3 4. lim ๐‘›โ†’+โˆž ln 1 ๐‘› 5. lim ๐‘›โ†’+โˆž 2๐‘›3 ๐‘›3 + 1
  • 13. EXERCISES Evaluate the limit of the following sequence: 1. lim ๐‘›โ†’+โˆž ๐‘›2 4๐‘› + 1 2. lim ๐‘›โ†’+โˆž ๐‘› ๐‘› + 3 3. lim ๐‘›โ†’+โˆž 3 4. lim ๐‘›โ†’+โˆž ln 1 ๐‘› 5. lim ๐‘›โ†’+โˆž 2๐‘›3 ๐‘›3 + 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) 2๐‘›2 ๐‘›=1 +โˆž 2. 1 + โˆ’1)๐‘› ๐‘›=1 +โˆž 3. ln ๐‘› ๐‘› ๐‘›=1 +โˆž 4. ๐‘›2๐‘’โˆ’๐‘› ๐‘›=1 +โˆž
  • 15. MONOTONE SEQUENCE Monotone Sequence ๏ฑ A sequence ๐‘Ž๐‘› ๐‘›=1 +โˆž is called - strictly increasing if ๐‘Ž1 < ๐‘Ž2 < ๐‘Ž3 <. . . < ๐‘Ž๐‘› โ‰ค . . . - increasing if ๐‘Ž1 โ‰ค ๐‘Ž2 โ‰ค ๐‘Ž3 โ‰ค. . . โ‰ค ๐‘Ž๐‘› โ‰ค. . . - strictly decreasing if ๐‘Ž1> ๐‘Ž2 > ๐‘Ž3 >. . . > ๐‘Ž๐‘› > . . . - decreasing if ๐‘Ž1 โ‰ฅ ๐‘Ž2 โ‰ฅ ๐‘Ž3 โ‰ฅ. . . โ‰ฅ ๐‘Ž๐‘› โ‰ฅ. . . 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.
  • 16. EXAMPLE SEQUENCE DESCRIPTION 1. 1 2 , 2 3 , 3 4 ,โ€ฆ, ๐‘› ๐‘›+1 , โ€ฆ Strictly increasing 2. 1, 1 2 , 1 3 ,โ€ฆ, 1 ๐‘› , โ€ฆ Strictly decreasing 3. 1, 1, 2, 2, 3, 3,โ€ฆ Increasing: not strictly increasing 4. 1, 1, 1 2 , 1 2 , 1 3 , 1 3 ,โ€ฆ Decreasing: not strictly decreasing 5. 1, โˆ’ 1 2 , 1 3 ,โˆ’ 1 4 โ€ฆ, โˆ’1)๐‘›+11 ๐‘› , โ€ฆ Neither increasing nor decreasing
  • 19. EXAMPLE Examples: Determine if the following sequence is monotone or strictly monotone. 1. ๐‘› ๐‘› + 1 Solution: Begin by letting ๐‘Ž๐‘› = ๐‘› ๐‘› + 1 . Then assign n = 1, 2, 3 in the given sequence, ๐‘Ž๐‘›, to get the first three term and observe the obtained values. If n = 1, ๐‘Ž1 = 1 2 ; If n = 2, ๐‘Ž2 = 2 3 ; If n = 3, ๐‘Ž3 = 3 4 Since ๐‘Ž1 < ๐‘Ž2 < ๐‘Ž3 <. . . < ๐‘Ž๐‘› < . . . ๐‘–s strictly increasing. Then the sequence is strictly monotone.
  • 20. EXAMPLE 2. 1 ๐‘› Solution: let ๐‘Ž๐‘› = 1 ๐‘› By assigning n = 1, 2, and 3 in the given sequence, ๐‘Ž๐‘›, the values obtained are: ๐‘› = 1, ๐‘Ž1 = 1; ๐‘› = 2, ๐‘Ž2 = 1 2 ; ๐‘› = 3, ๐‘Ž3 = 1 3 Since ๐‘Ž1 > ๐‘Ž2 > ๐‘Ž3 >. . . > ๐‘Ž๐‘› >. . . Is strictly decreasing. Then the sequence is strictly monotone.
  • 21. EXAMPLE 3. Use the difference ๐‘Ž๐‘›+1 โˆ’ ๐‘Ž๐‘› to show that the ๐‘› โˆ’ 2๐‘› ๐‘›=1 +โˆž is strictly increasing or strictly decreasing. Solution: ๐‘Ž๐‘› = ๐‘› โˆ’ 2๐‘›; ๐‘Ž๐‘›+1 = ๐‘› + 1 โˆ’ 2๐‘›+1 ๐‘Ž๐‘›+1โˆ’๐‘Ž๐‘› = ๐‘› + 1 โˆ’ 2๐‘›+1 โˆ’ ๐‘› โˆ’ 2๐‘›) = 1 + 2๐‘› โˆ’ 2๐‘›+1 = 1 โˆ’ 2๐‘› ๐‘Ž๐‘›+1โˆ’๐‘Ž๐‘› = 1 โˆ’ 2๐‘› < 0 which proves that the sequence is strictly decreasing.
  • 22. EXAMPLE 4. Use the ratio ๐‘Ž๐‘›+1 ๐‘Ž๐‘› to show that the given sequence ๐‘›๐‘› ๐‘›! ๐‘›=1 +โˆž is strictly increasing or strictly decreasing. Solution: ๐‘Ž๐‘› = ๐‘›๐‘› ๐‘›! , ๐‘Ž๐‘›+1 = ๐‘›+1)๐‘›+1 ๐‘›+1)! Forming the ratio of successive terms we obtain ๐‘Ž๐‘›+1 ๐‘Ž๐‘› = ๐‘›+1)๐‘›+1 ๐‘›+1)! ๐‘›๐‘› ๐‘›! = ๐‘›+1)๐‘›+1 ๐‘›+1)! โˆ™ ๐‘›! ๐‘›๐‘› = ๐‘›+1 ๐‘› ๐‘›+1) ๐‘›+1)๐‘›๐‘› = ๐‘›+1)๐‘› ๐‘›๐‘› = ๐‘›+1 ๐‘› ๐‘› From which we see that ๐‘Ž๐‘›+1 ๐‘Ž๐‘› > 1. This proves that the sequence is strictly increasing.
  • 23. EXAMPLE 5. Show that the sequence 10๐‘› ๐‘›! ๐‘›=1 +โˆž is eventually strictly decreasing. Solution: ๐‘Ž๐‘› = 10๐‘› ๐‘›! and ๐‘Ž๐‘›+1 = 10๐‘›+1 ๐‘›+1)! ๐‘Ž๐‘›+1 ๐‘Ž๐‘› = 10๐‘›+1 ๐‘›+1)! 10๐‘› ๐‘›! = 10๐‘›+1๐‘›! 10๐‘› ๐‘›+1)! = 10 ๐‘›! ๐‘›+1 ๐‘›! = 10 ๐‘›+1 ๐‘Ž๐‘›+1 ๐‘Ž๐‘› < 1 for all ๐‘› โ‰ฅ 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 โˆ’ 1 ๐‘› 2. ๐‘› 3๐‘› + 1 3. ๐‘› โˆ’ 2๐‘› 4. ๐‘› โˆ’ ๐‘›2
  • 26. BOUNDED SEQUENCE A sequence is bounded above if there is some number N such that ๐‘Ž๐‘› โ‰ค N for every n, and bounded below if there is some number N such that ๐‘Ž๐‘› โ‰ฅ N for every n. If a sequence is bounded above and bounded below it is bounded. If a sequence ๐‘Ž๐‘› ๐‘›=0 +โˆž is increasing or non- decreasing it is bounded below (by ๐‘Ž0), and if it is decreasing or nonincreasing it is bounded above (by ๐‘Ž0).
  • 27. BOUNDED SEQUENCE . Theorems 9.2.3 and 9.2.4 (p. 611) Theorems 9.2.3 and 9.2.4 (p. 611)
  • 28. EXAMPLES Determine whether or not the sequence is bounded. 1. 1 ๐‘› Solution: let ๐‘Ž๐‘› = 1 ๐‘› Generating some terms in the sequence ๐‘Ž๐‘› n = 1, ๐‘Ž1 = 1 bounded above since ๐‘Ž๐‘› โ‰ค 1 n = 2, ๐‘Ž2 = ยฝ n = 3, ๐‘Ž3 = 1/3 lim ๐‘›โ†’โˆž 1 ๐‘› = 0, bounded below since ๐‘Ž๐‘› > 0 Therefore the sequence is bounded.
  • 29. EXAMPLES Determine whether or not the sequence is bounded. 2. ๐‘› โˆ’1)๐‘› Solution: let ๐‘Ž๐‘› = ๐‘› โˆ’1)๐‘› Generating some terms in the sequence ๐‘Ž๐‘› n = 1, ๐‘Ž1 = -1 n = 2, ๐‘Ž2 = 2 n = 3, ๐‘Ž3 = -3 n = 4, ๐‘Ž4 = 4 n = 5, ๐‘Ž5 = -5 n = 6, ๐‘Ž6 = 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. 2๐‘›โˆ’3 3๐‘›+4 Solution: let ๐‘Ž๐‘› = 2๐‘›โˆ’3 3๐‘›+4 Generating some terms in the sequence ๐‘Ž๐‘› n = 1, ๐‘Ž1 = โˆ’1 7 bounded below since ๐‘Ž๐‘› โ‰ฅ โˆ’1 7 n = 2, ๐‘Ž2 = 1/10 n = 3, ๐‘Ž3 = 3/13 lim ๐‘›โ†’โˆž 2๐‘›โˆ’3 3๐‘›+4 = 2/3, bounded above since ๐‘Ž๐‘› < 2/3 Therefore, the sequence is bounded Source: https://www.youtube.com/watch?v=UbNE_beWlhU