Transcript 9 1 9 2 Sequences Series

9-1/9-2: Sequences and Series
Objectives:
Assignment:
1. To determine whether a
sequence converges or
diverges
• P. 602-604: 1-3, 32-36
even, 37-41 odd, 49-77
eoo, 80, 117-120
2. To apply the definition of
a convergent or
divergent series
• P. 612-615: 1, 5, 7-15
odd, 23-27 odd, 39-49
odd, 57-67 odd, 79, 85,
89, 117-120
• Homework Supplement
Warm Up 1
Identify the 𝑛th term of the sequence
1, 2, 6, 24, 120, …
Warm Up 2
Evaluate −1 𝑛 and −1 𝑛+1 for
𝑛 = 1, 2, 3, 4, … What role do you think these
expressions play in the 𝑛th term of
sequence?
Objective 1
You will be able to determine whether
a sequence converges or diverges
Sequences
A sequence refers to an ordered list of numbers.
More formally, a sequence is a function whose
domain is the set of positive integers, ℤ+ .
𝟏
𝑎1
𝟐
𝑎2
𝟑
𝑎3
Terms
𝟒
𝑎4
⋯
⋯
𝒏
𝑎𝑛
𝑛th Term
⋯
⋯
𝑎𝑛
represents
the entire
sequence.
Sequences
A sequence refers to an ordered list of numbers.
More formally, a sequence is a function whose
domain is the set of positive integers, ℤ+ .
𝟎
𝑎0
𝟏
𝑎1
𝟐
𝑎2
Terms
𝟑
𝑎3
⋯
⋯
𝒏
𝑎𝑛
𝑛th Term
⋯
⋯
𝑎𝑛
represents
the entire
sequence.
Exercise 1a
List the first 4 terms of each sequence.
𝑎𝑛 = 3 + −1 𝑛
3 + −1
𝑛
= 3 − 1, 3 + 1, 3 − 1, 3 + 1, …
= 2, 4, 2, 4, …
Exercise 1b
List the first 4 terms of each sequence.
𝑛
𝑏𝑛 =
1 − 2𝑛
𝑛
4
1
2
3
=
,…
,
,
,
1 − 2𝑛
1−2 1−4 1−6 1−8
1 2 3 4
= − ,− ,− ,− ,…
1 3 5 7
Exercise 1c
List the first 4 terms of each sequence.
𝑛2
𝑐𝑛 = 𝑛
2 −1
𝑛2
4
1
9
16
=
,
,
,
,…
2𝑛 − 1
2 − 1 4 − 1 8 − 1 16 − 1
1 4 9 16
= , , , ,…
1 3 7 15
Exercise 1d
List the first 13 terms of each sequence.
𝑑𝑛 , where 𝑑1 = 1, 𝑑2 = 1, and 𝑑𝑛+2 = 𝑑𝑛 + 𝑑𝑛+1
𝑑𝑛 = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …
This strange sequence is defined
recursively; that is, in order to find the
𝑛th term, you must know the terms
that come before it.
Exercise 1e
List the first 13 terms of each sequence.
𝑑𝑛+1
𝑑𝑛
, where 𝑑1 = 1, 𝑑2 = 1, and 𝑑𝑛+2 = 𝑑𝑛 + 𝑑𝑛+1
𝑑𝑛+1
1 2 3 5 8 13 21 34 55 89 144 233
= , , , , , , , , , ,
,
,…
𝑑𝑛
1 1 2 3 5 8 13 21 34 55 89 144
Convergence/Divergence
Besides finding the 𝑛th term of a sequence, an
important consideration is whether or not the terms
of the sequence converge or diverge.
Diverges
Converges
Limit of a Sequence
Let 𝐿 be a real
number. The limit
of a sequence 𝑎𝑛
is 𝐿 if
The terms increase or
decrease without bound.
lim 𝑎𝑛 = 𝐿
𝑛→∞
If this limit exists,
then the sequence
converges to L.
If this limit does not exist,
then the sequence diverges.
Limit of a Sequence
Let 𝐿 be a real
number. The limit
of a sequence 𝑎𝑛
is 𝐿 if
lim 𝑎𝑛 = ±∞
𝑛→∞
lim 𝑎𝑛 = 𝐿
𝑛→∞
If this limit exists,
then the sequence
converges to L.
If this limit does not exist,
then the sequence diverges.
Limit of a Sequence
Usually, we find a function 𝑓 that matches 𝑎𝑛
for every positive integer in its domain.
Let 𝐿 be a real
number. Let 𝑓 be a
function of a real
variable such that
lim 𝑓 𝑥 = 𝐿.
𝑥→∞
If 𝑎𝑛 is a
sequence such that
𝑓 𝑛 = 𝑎𝑛 for every
positive integer 𝑛,
then lim 𝑎𝑛 = 𝐿.
𝑛→∞
Exercise 2
Find the limit of the sequence 𝑎𝑛 =
1+
1 𝑛
𝑛
.
Properties of Limits of Sequences
Let lim 𝑎𝑛 = 𝐿 and lim 𝑏𝑛 = 𝐾.
𝑛→∞
lim 𝑎𝑛 ± 𝑏𝑛 = 𝐿 ± 𝐾
𝑛→∞
lim 𝑎𝑛 𝑏𝑛 = 𝐿𝐾
𝑛→∞
𝑛→∞
lim 𝑐𝑎𝑛 = 𝑐𝐿
𝑛→∞
𝑎𝑛
lim
𝑛→∞ 𝑏𝑛
=
𝐿
,
𝐾
𝑏𝑛 ≠ 0, 𝐾 ≠ 0
Exercise 3
Determine whether the following sequences
converge or diverge.
1.
𝑎𝑛 = 3 + −1
𝑛
2.
𝑏𝑛 =
𝑛
1−2𝑛
3.
𝑐𝑛 =
𝑛2
2𝑛 −1
Exercise 4
Compare the relative growth rates of 2𝑛 and 𝑛!.
𝒏
2𝑛
𝟏
2
𝟐
4
𝟑
8
4
16
5
32
6
64
7
128
𝑛!
1
2
6
24
120
720 5040
8
256
9
512
40320 362880
Exercise 4
Compare the relative growth rates of 2𝑛 and 𝑛!.
𝑛!
lim 𝑛 = ∞
𝑛→∞ 2
𝑛! grows
much faster
than 2𝑛
2𝑛
lim
=0
𝑛→∞ 𝑛!
Exercise 4
Compare the relative growth rates of 2𝑛 and 𝑛!.
𝑛!
lim 𝑛 = ∞
𝑛→∞ 𝑏
𝑛! grows
much faster
than 𝑏 𝑛
𝑏𝑛
lim
=0
𝑛→∞ 𝑛!
Exercise 5
Determine whether 𝑎𝑛 =
diverges.
2𝑛
𝑛!
converges or
Alternating Sequences
Absolute
Value
Theorem
Sometimes sequences contain both positive and
negative terms. Using the Squeeze Theorem, it
can be shown that if 𝑎𝑛 converges to zero, then
𝑎𝑛 converges to zero.
For the sequence 𝑎𝑛 , if
lim 𝑎𝑛 = 0, then lim 𝑎𝑛 = 0.
𝑛→∞
𝑛→∞
Exercise 6
Show that 𝑐𝑛 =
𝑛 1
−1
𝑛!
converges.
Pattern Recognition
If a sequence is not given explicitly as a function of
𝑛, we first have to find the 𝑛th term 𝑎𝑛 before
determining whether the sequence converges or
diverges.
1 1 1 1
1
1 1 1 1
2 4 8
6
𝑛+1 𝑛2 −𝑛+6
𝑎𝑛 = 2, 4, 8, 16, … , 2𝑛
𝑏𝑛 = , , , 15, … ,
1 1 1
𝑐𝑛 = 2, 4, 8,
1 1 1
7
,
62
…,
𝑛2 −3𝑛+3
9𝑛2 −25𝑛+18
𝑑𝑛 = 2, 4, 8, 0, … , −𝑛
𝑛+1 𝑛−4
6 𝑛2 +3𝑛−2
Converges to 0
Converges to 0
1
Converges to 9
Diverges
Exercise 7
Determine whether 𝑎𝑛 converges or diverges.
3𝑛 − 1
1 less than a power of 3
𝑎𝑛 =
2 8
− , ,
1 2
26 80
− , ,
6 24
Alternating signs
242
− ,
120
𝑛!
…
−1 𝑛 3𝑛 −1
𝑛!
Pattern Recognition
Here are some common patterns and their
𝑛th term.
Factorials: 1, 2, 6, 24, 120, … , 𝑛!
Alternating terms: −1
𝑛
or −1
𝑛+1
Terms with even numbers: 2𝑛
Terms with odd numbers: 2𝑛 + 1 or 2𝑛 − 1
Objective 2
You will be able to apply the
definition of a convergent or
divergent series
Sequences vs. Series
Sequence
Series
A sequence is an
ordered list of terms.
An infinite series is the
sum of those terms.
𝑎𝑛 = 𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 , …
∞
𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 + ⋯
𝑛=1
𝑛th Partial Sum
To find the sum of a series, first consider the
sequence of partial sums:
𝑆1 = 𝑎1
𝑆2 = 𝑎1 + 𝑎2
𝑆3 = 𝑎1 + 𝑎2 + 𝑎3
⋮
𝑆𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛
𝑛th partial sum
If this sequence of partial
sums converges, then the
series is said to converge.
Basically, the partial sums
form a sequence, and if the
this sequence converges, the
series converges.
Convergent/Divergent Series
For the infinite series ∞
𝑛=1 𝑎𝑛 , the 𝒏th partial
sum is given by 𝑆𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 .
If the sequence of partial sums 𝑆𝑛 converges
to 𝑆, then the series ∞
𝑛=1 𝑎𝑛 converges.
The limit 𝑆 is called the sum of the series.
𝑆 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 + ⋯
Convergent/Divergent Series
For the infinite series ∞
𝑛=1 𝑎𝑛 , the 𝒏th partial
sum is given by 𝑆𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 .
If 𝑆𝑛 diverges, then the series diverges.
Exercise 8
Zeno’s Paradox can be represented by
1
1
1
1
1
∞
𝑛=1 2𝑛 = 2 + 4 + 8 + 16 + ⋯. Find the first 4 partial
sums and the 𝑛th partial sum. Resolve Zeno’s
Paradox by determining whether the series
converges or diverges.
Start
Finish?
Exercise 8
1 less than
a power of 2
1
𝑆1 =
2
𝑆2 =
1 1 3
+ =
2 4 4
1 1 1 7
𝑆3 = + + =
2 4 8 8
1 1 1 1
15
𝑆4 = + + +
=
2 4 8 16 16
⋮
1 1 1 1
1
𝑆𝑛 = + + +
+ ⋯+ 𝑛
2 4 8 16
2
𝑆𝑛 =
1 3 7 15
, , , ,
2 4 8 16
Powers of 2
𝑆𝑛 =
2𝑛 − 1
…
2𝑛
2𝑛 −1
2𝑛
ln 2 2𝑛
2𝑛 − 1
= lim
=1
lim
𝑛→∞ ln 2 2𝑛
𝑛→∞ 2𝑛
This resolves
Zeno’s
Paradox.
Therefore the
series
converges and
its sum is 1.
Exercise 8
1 less than
a power of 2
1
𝑆1 =
2
𝑆2 =
1 1 3
+ =
2 4 4
1 1 1 7
𝑆3 = + + =
2 4 8 8
1 1 1 1
15
𝑆4 = + + +
=
2 4 8 16 16
⋮
1 1 1 1
1
𝑆𝑛 = + + +
+ ⋯+ 𝑛
2 4 8 16
2
𝑆𝑛 =
1 3 7 15
, , , ,
2 4 8 16
Powers of 2
𝑆𝑛 =
2𝑛 − 1
…
2𝑛
2𝑛 −1
2𝑛
ln 2 2𝑛
2𝑛 − 1
= lim
=1
lim
𝑛→∞ ln 2 2𝑛
𝑛→∞ 2𝑛
Exercise 8
1 less than
a power of 2
1
𝑆1 =
2
𝑆2 =
1 1 3
+ =
2 4 4
1 1 1 7
𝑆3 = + + =
2 4 8 8
1 1 1 1
15
𝑆4 = + + +
=
2 4 8 16 16
⋮
1 1 1 1
1
𝑆𝑛 = + + +
+ ⋯+ 𝑛
2 4 8 16
2
𝑆𝑛 =
1 3 7 15
, , , ,
2 4 8 16
Powers of 2
𝑆𝑛 =
2𝑛 − 1
…
2𝑛
2𝑛 −1
2𝑛
ln 2 2𝑛
2𝑛 − 1
= lim
=1
lim
𝑛→∞ ln 2 2𝑛
𝑛→∞ 2𝑛
Geometric Series
The previous example demonstrated a
geometric series.
∞
First term
𝑎 ∙ 𝑟 𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛 + ⋯
𝑛=0
Common ratio of consecutive terms
Exercise 9
Find the sum of the geometric series
∞
𝑛
𝑎𝑟
.
𝑛=0
𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛−1
Multiply by 𝑟
− 𝑟 ∙ 𝑆𝑛 = 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + 𝑎𝑟 4 + ⋯ + 𝑎𝑟 𝑛
Subtract 𝑆𝑛 − 𝑟𝑆𝑛
𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎 − 𝑎𝑟 𝑛
Factor
𝑆𝑛 1 − 𝑟 = 𝑎 1 − 𝑟 𝑛
𝑎
𝑆𝑛 =
1 − 𝑟𝑛
1−𝑟
𝑎
lim 𝑆𝑛 = lim
1 − 𝑟𝑛
𝑛→∞
𝑛→∞ 1 − 𝑟
Divide by 1 − 𝑟
Take lim 𝑆𝑛
𝑛→∞
=
𝑎
lim 1 − 𝑟 𝑛
1 − 𝑟 𝑛→∞
Exercise 9
Find the sum of the geometric series
∞
𝑛
𝑎𝑟
.
𝑛=0
𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛−1
− 𝑟 ∙ 𝑆𝑛 = 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + 𝑎𝑟 4 + ⋯ + 𝑎𝑟 𝑛
𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎 − 𝑎𝑟 𝑛
𝑆𝑛 1 − 𝑟 = 𝑎 1 − 𝑟 𝑛
𝑎
𝑆𝑛 =
1 − 𝑟𝑛
1−𝑟
𝑎
lim 𝑆𝑛 = lim
1 − 𝑟𝑛
𝑛→∞
𝑛→∞ 1 − 𝑟
If 𝑟 ≥ 1,
then this
limit
diverges.
=
𝑎
lim 1 − 𝑟 𝑛
1 − 𝑟 𝑛→∞
Exercise 9
Find the sum of the geometric series
∞
𝑛
𝑎𝑟
.
𝑛=0
𝑆𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + ⋯ + 𝑎𝑟 𝑛−1
− 𝑟 ∙ 𝑆𝑛 = 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + 𝑎𝑟 4 + ⋯ + 𝑎𝑟 𝑛
𝑆𝑛 − 𝑟𝑆𝑛 = 𝑎 − 𝑎𝑟 𝑛
𝑆𝑛 1 − 𝑟 = 𝑎 1 −
𝑎
𝑆𝑛 =
1−𝑟
lim 𝑆𝑛 = lim
𝑛→∞
𝑛→∞
If 0 < 𝑟 < 1,
then this limit
converges to 0.
𝑟𝑛
1 − 𝑟𝑛
𝑎
1−𝑟
1 − 𝑟𝑛
=
𝑎
lim 1 − 𝑟 𝑛
1 − 𝑟 𝑛→∞
Convergence of Geometric Series
A geometric series with ratio 𝑟
diverges if 𝑟 ≥ 1. If 0 < 𝑟 < 1,
then the series converges to the sum
∞
𝑎𝑟 𝑛
𝑛=0
Always on
the AP Exam
𝑎
=
1−𝑟
Exercise 10
Find the sum of each geometric series.
1.
3
∞
𝑛=0 2𝑛
2.
3 𝑛
∞
𝑛=0 2
Exercise 11
Determine whether the series below converges or
diverges. If it converges, find its sum.
28 112 448 1792
7+
+
+
+
+⋯
5
25 125 625
Properties of Infinite Series
If 𝑎𝑛 = 𝐴, 𝑏𝑛 = 𝐵, and 𝑐 is a real number, then
the following series converge to the indicated sum.
𝑎𝑛 − 𝑏𝑛
𝑐𝑎𝑛 = 𝑐𝐴
=𝐴−𝐵
𝑎𝑛 + 𝑏𝑛
=𝐴+𝐵
Notice the
complete
lack of
products and
quotients.
Exercise 12
Write the contrapositive of the following
theorem:
If
𝑎𝑛 converges, then lim 𝑎𝑛 = 0.
𝑛→∞
Not totally helpful…
𝑛th Term Divergence Test
If lim 𝑎𝑛 ≠ 0, then
𝑛→∞
If lim 𝑎𝑛 = 0,
𝑛→∞
then you can’t
use this test.
𝑎𝑛 diverges.
In other words, if the terms of a sequence
are not approaching zero, then a series
made from those terms will not converge.
Exercise 13
Determine whether each series converges or
diverges.
1.
2𝑛
2.
𝑛!
2𝑛!+1
3.
1
𝑛
9-1/9-2: Sequences and Series
Objectives:
Assignment:
1. To determine whether a
sequence converges or
diverges
• P. 602-604: 1-3, 32-36
even, 37-41 odd, 49-77
eoo, 80, 117-120
2. To apply the definition of
a convergent or
divergent series
• P. 612-615: 1, 5, 7-15
odd, 23-27 odd, 39-49
odd, 57-67 odd, 79, 85,
89, 117-120
• Homework Supplement