Sequences and Series
From Simple Patterns to Elegant and Profound
Mathematics
David W. Stephens
The Bryn Mawr School
Baltimore, Maryland
PCTM – 28 October 2005
1
Contact Information
Email:
stephensd@brynmawrschool.org
The post office mailing address is:
David W. Stephens
109 W. Melrose Avenue
Baltimore, MD 21210
410-323-8800
The PowerPoint slides will be available on my school website:
http://207.239.98.140/UpperSchool/math/stephensd/StephensFirstPa
ge.htm , listed under “PCTM October 2005”
2
Teaching Sequences and Series
We will look at some ideas for teaching sequences
and series as well as some applications in
mathematics classes at THREE different levels:
I.
Early (Algebra 1, Algebra 2, and Geometry)
II.
Intermediate (Advanced Algebra,
Precalculus)
III. Advanced (AP Calculus, esp. BC Calculus)
3
Teaching Sequences and Series
Many of the topics and examples used today will not
be new to you, but I want you to consider thinking
of them …and talking about them with students …
as sequences and series. It can be a good way for
them to think about these diverse topics as bring
linked mathematically. Just as functions link a lot
of what we teach, the patterns of sequences and
series can tie these ideas together for better
comprehension.
4
Early Sequences and Series
(Algebra 1, Algebra 2, and Geometry)
1.
2.
3.
4.
5.
6.
7.
8.
9.
Looking for patterns
Identifying kinds of sequences
Describing patterns in sequences
Using variables
Summation notation
Strategies for summing
Applications with geometry ideas
Graphing patterns
Data analysis & functions as sequences
5
Intermediate Sequences and Series
(Advanced Algebra, Precalculus)
1. More geometric sequences and exponential
functions
2. Infinite series
3. Convergence and divergence
4. Informal limits
5. More advanced data analysis (“straightening
data”)
6. Applications (compound interest, astronomy,
chemistry, biology, economics, periodic motion
or repeating phenomena)
6
Advanced Sequences and Series
(AP Calculus, esp. BC Calculus)
1. Newton’s method for locating roots
2. Riemann sums
3. Trapezoid rule and Simpson’s rule
4. Euler’s method for differential equations
5. Power series (Maclaurin and Taylor series
polynomials)
6. Convergence tests for series
7
Early Topics
(Algebra 1 , Algebra 2, and Geometry)
I sometimes have begun my Algebra 2 classes in September with this
topic because…
a) New students to the school (and the class) do not feel
“new”.
b) I can use algebraic language.
c) I can review linear functions in a new context.
d) I can sneak in some review which does not feel like
review!
Early Sequences and Series
8
Activity 1
Find the next three numbers in these sequences:
A)
B)
C)
D)
E)
6, 9, 13, 18, 24, …
12, 17, 13, 16, 14, 15, 15, …
5, 10, 20, 40, …
7, -21, 63, -189, …
2, 3, 4, 5, 4, 3, 2, 1, 0, -1, 0, 1, 2, 3, …
Early Sequences and Series
9
Activity 2
Students build their own sequences, and they
challenge their classmates to guess the next few
entries. This can be a neat homework assignment.
(It can be extended to later activities where they
have to code their sequence patterns with variables,
too.)
Early Sequences and Series
10
Activity 3
Describe the pattern in words:
A) 7, 5, 3, 1, -1, -3, …
B) 70, 68, 64, 56, 40, 32, 28, 26, 25, …
C) 1 2 3 4 5 7 81
, , , , , , ,...
4 7 8 9 11 13 17
D) 12, 13, 14, 15, 16, 15, 14, 13, 12, 13, 14, 15, ...
E) 1, 2, 3, 4, 4,3, 2, 1, 3, 5, 7, 7 , 5, 3, 1, 12, 23, 34, 23, …
F) 4, 9, 32, 50, 53, 54, 54, 54, 54, 54.1, 54.13, 54.135,
54.1356 , ...
Early Sequences and Series
11
Activity 4
Learn to code the pattern with variables:
A)
9, 13, 17, 21, 25, 29 , …
Let a0 = 9
an = an-1+ 4
or an = a0+ (n-1)d
(Some texts use tn, where t = term, instead of an)
It could also be coded that a6 = 9 and then a7 = 13,
if you decided to start the count at item #6
Early Sequences and Series
12
Activity 4 (continued)
B)
3, 6, 12, 24, 48, …
Let a0 = 3 an = a0r n-1
= (first term)(ratioterm# - 1)
( or let t0 = 3
tn = a0r n-1)
Early Sequences and Series
13
Activity 5
Introduction to Fibonacci sequences
A)
1, 1, 2, 3, 5, 8, 13, 21, 34
B)
2, 5, 7, 12, 19, 31, 50
an = an-1+ an-2 (recursively defined functions)
Early Sequences and Series
14
Activity 6
Series and Summation Notation
1.
a)
sequence
b) series
1 , 3 , 5 , 7 , 9 , 11
1 + 3 + 5 + 7 + 9 + 11
c) series notation
6
5
10
n1
n0
n5
 2n  1 or  2n  1 or  2n  9
Early Sequences and Series
15
Activity 6 (continued)
2. a) sequence
b) series
2, 6, 18, 54
2 + 6 + 18 + 54
4
c) series notation
2
n
n1
3. a) sequence
b) series
4, -2, 1, , …
4–2+1- + -…
c) series notation
Early Sequences and Series
16
Activity 7
Interleaved and other creative sequences
Find the next three terms, and describe the two
sequences that are interleaved.
A)
B)
1, 3, 4, 9, 7, 27, 10, 81, 13, 243 , …
1, 3, 4, 9, 7, 27, 10, 81, 13, 243 , …
5, 1, 7, 4, 9, 7, 11, 10, 13, 13,
5, 1, 7, 4, 9, 7, 11, 10, 13, 13, …
Early Sequences and Series
17
Introduction to Series
How to add up an arithmetic series efficiently:
Example: sn = 6 + 9 + 12 + 15 + 18 + … + 219
• Add the first and last terms, the second and the
second to the last, etc. What do you notice?
• How many pairs are there?
• What if there are an odd number of terms to
add?
Sn = n (s  s )
2
Early Sequences and Series
1
n
18
Introduction to Series
How to add up a geometric series efficiently:
Example: sn = 5 + 10 + 20 + 40 + 80 + 160 + 320
= a0 + a0r + a0r2 + a0r3+ … + a0rn-1
rsn =
a0r + a0r2 + a0r3 + a0r4 + …
+a0rn
Then sn – rsn = a0 – a0rn
This provides us with the usual formula for a
geometric series:
a0 (1  r n )
sn =
1 r
Early Sequences and Series
19
Activity 8
For the series,
sn = 5 + 10 + 20 + 40 + … + 320 + …
calculate s28
Early Sequences and Series
20
Activity 9
Write an arithmetic series and a geometric series
so that the value of the sum for the arithmetic
series is greater than the sum for the geometric
series for the first 10 terms, but
a) the arithmetic series still exceeds the
geometric series for the first 20 terms
b) the geometric series exceeds the arithmetic
series for the first 20 terms
Early Sequences and Series
21
Activity 10
Find specific terms in sequences
a) For 6, 11, 16, 21, … calculate t41
b) For an arithmetic sequence with t1 = 7 and
t4 = 16, calculate t 91
c) For an arithmetic sequence with t4 = -9 and
t6 = 7, calculate t50
d) For a geometric sequence with t1 = 9 and
r = .49, calculate t102
Early Sequences and Series
22
Data Analysis
Building Functions from Data
Treat the list of x and y coordinates as sequences!
Example 1: x 1 2 3 4 5 6 7 8
y 7 10 13 16 19 22 25 28
x is an arithmetic sequence
y is an arithmetic sequence
This is sometimes called an “add-add” property.
Thus, y = f(x) is LINEAR
What is the actual function?
Ans: f(x) = 3x + 4
3 = the common difference
= d in “sequence language”
4 = a0 in the sequence
Early Sequences and Series
23
Data Analysis
Building Functions from Data
Example 2:
x 3 5 7
y 4 8 12
9
16
11 13 15 17
20 24 28 32
x is an arithmetic sequence
y is an arithmetic sequence
Thus f(x) is LINEAR again.
What is the actual function?
Ans: common difference = d = 2 = slope
If a3 = 4 and d = 2, then a0 = -2
So f(x) = 2x - 2
Early Sequences and Series
24
Data Analysis
Building Functions from Data
Example 3:
x
y
1 2 3 4 5 6 7 8
6 9 14 21 30 41 54 69
x is an arithmetic sequence
 y is an arithmetic sequence, and y is
constant
Thus f(x) is QUADRATIC.
What is the actual function?
Ans: f(x) = x2 +5
Early Sequences and Series
25
Geometry
Angles of Polygons
What is the general formula for the sum of the
interior angles of a polygon with n sides?
(n,  measures of interior angles) :
(3, 180) , (4, 360), (5 , 540) , (6 , 720) , …
(n , 180(n-2))
Early Sequences and Series
26
Geometry
A Modeling Application
Handshake Problem:
If n people shakes hands with everyone else
at a meeting, how many handshakes
occur?
1. Visualize this as a geometry problem.
2. Consider a simpler version with just a few
number of people.
3. Generalize the data, and consider the data
as sequence.
Early Sequences and Series
27
Geometry
A Modeling Application
Handshake
Problem:
E
D
F
A
C
A
Early Sequences and Series
B
28
Geometry
A Modeling Application
n = number of people
h(n) = number of handshakes
n
1
h(n) 0
2
1
3
3
4
6
5
10
6
15
7
21
29
Geometry
A Derivation of

Find the perimeter of a sequence of regular polygons
which are inscribed in a unit circle, and emphasize that
the sequence of results is important to watch.
s = length of one side of the polygon
p = perimeter of the polygon
Early Sequences and Series
30
Geometry
m AB = 2.01 cm
4 1.00 cm
CD =
A Derivation of
4

2
2
A
A
s = length of
one
side
-5
C
E
p=
perimeter of
inscribed
polygon
D -5
B
5
E
C
D
B
-2
-2
s=
s= 2
3
p = 3 3 = 5.196
Early Sequences and Series
-4
p=4 2
= 5.657
31
Geometry
4
A Derivation
of
CD = 1.00 cm

4
2
2
A
E
A
-5
E
C
D
-5
5
F
G
I
H
-2
-2
-4
s = 2 sin(36) = 1.176
s=1
p = 5(1.176) = 5.878
p=6
Early Sequences and Series
-4
32
Geometry
A Derivation of

In general, the length of one-half of a side of an
inscribed regular polygon is sin(. 5 * 360 )
n
So a side measures
2 sin(
180
)
n
and the
perimeter of the polygon measures
Since p  2 
, then
2n sin(
180
)
n
 can be calculated.
Early Sequences and Series
33
Geometry
A Derivation of
CD = 2.98 cm

A
1
0.5
-2
G -1
1I
2
-0.5
The central angle
for each side is 360
n
-1
-1.5
Each half-side has length
equal to the sine of one-half
the central angle.
Early Sequences and Series
34
Geometry
A Derivation of

Here are the
perimeters of the
polygons from
the TI-83 as a list
(L2)
Note: Ignore L3.
Early Sequences and Series
35
Intermediate Sequences and Series
(Advanced Algebra, Precalculus)
1. More geometric sequences and exponential
functions
2. Infinite series
3. Convergence and divergence
4. Informal limits
5. More advanced data analysis (“straightening
data”)
6. Applications (compound interest, astronomy,
chemistry, biology, economics, periodic motion
or repeating phenomena)
Intermediate Sequences and Series
36
Data Analysis
Building Functions from Data
Example 4:
x 1 2 3 4 5
6
7
8
y 3 9 27 81 243 729 2187 6561
x is an arithmetic sequence
y is a geometric sequence
This is sometimes called an “add-multiply” property
So y = f(x) is EXPONENTIAL
What is the actual function?
Ans: f(x) = 3x
( where r = 3 in the geometric sequence)
Intermediate Sequences and Series
37
Data Analysis
Building Functions from Data
Example 5:
x 1 2 3 4 5
6
7
8
y 5 11 29 83 245 731 2189 6563
y 3 9 27 81 243 729 2187 6561
x is an arithmetic sequence
y is not exactly a geometric sequence
But if the sequence of y-values is compared with the last set of y’s,
then we see that this sequence is 2 more than a geometric
sequence.
So y = 3x + 2
Intermediate Sequences and Series
38
Data Analysis
Building Functions from Data
Example 6:
x
y
1
6
4
48
7
384
10
3072
13
24,576
x is an arithmetic sequence
y is not exactly a geometric sequence
Since the two sequences have the “add-multiply” property,
then y is a geometric sequence, and it is exponential. Notice
that the x’s do not have to be consecutive.
We have to find the “r” value as if we are calculating
geometric means
Intermediate Sequences and Series
39
Data Analysis
Building Functions from Data
Example 7:
x 1
y 6
4
48
7
384
10
3072
13
24,576
a1 = 6 and a4 = 48, and we need to fill in the sequence so
that we know the y-values for terms 2 and 3. Since the
desired sequence is geometric, we need to know what to
multiply a1 by repeatedly three times to get 48. This
suggests that r*r*r= 48/6.
So r =
38
= 2 , and
y = 3 * 2x
Intermediate Sequences and Series
40
Data Analysis
Building Functions from Data
Example 7:
x
1
2 3
4
5 6
y
6
12 24 48 96 192 384 768 1536 3072
r = 3 8 = 2 , and
7
8
9
10
11
6144
y = 3 * 2x
Intermediate Sequences and Series
41
An Historical Diversion
Let’s take a look at the pairing of an arithmetic
and a geometric sequence.
n
1
2
3
4
5
6
an
2
4
8
16
32
64
Let’s suppose that we wanted intermediate
terms:
n
1
an
2
3/2 2 5/2 3 7/2 4 9/2 5
4
8
Intermediate Sequences and Series
16
32
6
64
42
An Historical Diversion
n
an
1
2
3/2 2 5/2 3 7/2 4 9/2 5
4
8
16
32
Thinking about an as a geometric sequence, we need
a geometric mean to fill in the missing terms.
Our desired multiplier, r, is 2 .
an
2 2 2 4 4 2 8 8 2 16 16 2 32
an
2 2 3/2 4 2 5/2 8 2 7/2 16 2 9/2
32
Intermediate Sequences and Series
43
An Historical Diversion
So when we write an = 2n , then the sequence,
n, becomes the exponents, or the
logarithms, for the geometric sequence.
This is part of the history of Henry Briggs, John
Napier, Jobst Burgi, John Wallis, and Johann
Bernoulli from 1620 to 1749 in the development
of logarithms.
Intermediate Sequences and Series
44
Function Transformations using
Sequences
If functions are considered as lists of data, and one function is a
transformation of another one, then the alterations to the
sequence of function values is the key to decoding the
transformation.
X
0
f(x) 6
1
3
2
8
3
9
4
6
8
5
7
14 10 21 43 8
-3 5
1
5
-4
g(x) 11
13 5
3
8
9
14 10
6
-3
5
1
5
-4
We want to write g(x) as a transformation of f(x), so
9
6
21
g(x) = f(x – 3)
45
Function Transformations using
Sequences
Preliminary questions:
A. When a transformation such as f(x + a) is used, what happens to the y values?
B. When a transformation such as f(x) + a is used, what happens to the y values?
C. When a transformation such as a*f(x) is used, what happens to the y values?
X
0
1
2
3
4
5
6
7
8
9
f(x) 6
3
8
9
14
10
21
43
8
6
6
7
4
9
10
15
-3
g(x) 4
5
8
1
-1
5
0
-3
5
1
5
g(x) = f(x – 5) + 1
46
Function Transformations using
Sequences
x
0
f(x) 6
1
3
-3
g(x) -8
2
8
3
9
5
3
1
12
-6
4
14
5
6
1-
5
10
6
21
7
43
8
8
9
6
-4
16 18
28
20
42
86
2
1-
-8
g(x) = 2f(x – 2)
47
Infinite Sequences, Series and
Convergence
There are some really good opportunities to lead
students to important conclusions, as well as to
challenge their intuition with some sophisticated
ideas … with infinite sequences and series.
We can extend their numerical sense as well as
exploiting their graphical skills to help generate
conclusions.
Intermediate Sequences and Series
48
Infinite Sequences, Series and
Convergence
Suppose an: 1, 3, 5, 7, 9, …
Where does an go as n gets large?
Suppose bn: 1, 1.01, 1.02 , 1.03 , 1.04 , …
Where does bn go as n gets large?
Suppose cn: 1, 2, 4, 8, 16, …
Where does cn go as n gets large?
Intermediate Sequences and Series
49
Infinite Sequences, Series and
Convergence
1 3 5 7 9
Suppose dn:
, , , , ,...
2 4 8 16 32
Where does dn go as n gets large?
Since this is the ratio of two sequences, each
of which approaches infinity, explain your
answer to this question.
Intermediate Sequences and Series
50
Infinite Sequences, Series and
Convergence
Suppose en: 1, 0, -1 , 1 , 0 , -1 , 1 , 0 , -1 , …
Where does en go as n gets large?
Suppose fn: 40 , 32, 25.6, 20.48, 16.384, …
Where does fn go as n gets large?
Type 40 on the calculator and hit ENTER.
Type * .8 and hit ENTER
The screen will read ANS * .8
Repeatedly hit enter to generate the sequence.
Intermediate Sequences and Series
51
Infinite Sequences, Series and
Convergence
4
Suppose gn: 60, 90, 108, 120, 128 , 135, …
7
Where does gn go as n gets large?
What sequence is this?
Suppose hn: 120, 90, 72, 60, 51
3 , 45, …
7
Where does hn go as n gets large?
What sequence is this?
Intermediate Sequences and Series
52
Infinite Sequences, Series and
Convergence
Suppose in:
0 1 2 3
98
8341
, , , ,...,
,...,
,...
2 3 4 5
100
8343
Where does in go as n gets large?
Intermediate Sequences and Series
53
Intermediate Level Applications
Sequence Mode on the Calculator
Suppose we want to generate the sequence as
an iterated function (recursive function).
So: 2, 5, 8, 11, 14, 17, …
could be an = 2 + 3n or an = an-1 + 3
Intermediate Sequences and Series
54
Intermediate Level Applications
Sequences and Series on the Calculator
To generate sequences on the HOME screen, go to
LIST (2nd STAT)/OPS/<Option 5>
which will give seq(
The inputs required for seq( are:
seq(expression, variable, begin, end [increment])
Example: an = 2n+1  1, 3, 5, 7, 9, …
Intermediate Sequences and Series
55
Intermediate Level Applications
Sequence and Series on the Calculator
Example: an = 2n+1  1, 3, 5, 7, 9, …
Notice that the name of the variable does
not matter, as long as it is specified.
Intermediate Sequences and Series
56
Intermediate Level Applications
Sequence and Series on the Calculator
If the series is desired, the sum( function is used.
Example: an = 2n+1  1, 3, 5, 7, 9
sn = sum(an)  1+3+5+7+9 = 25
“Sum(“ is found
in LIST (2nd
STAT)/MATH/<
Option 5>
Intermediate Sequences and Series
57
Intermediate Level Applications
Sequence and Series on the Calculator
Partial sums can also be generated, and this is
helpful if there is an application where the sums
should be considered as making a sequence,
perhaps if their convergence is being considered.
The function cumSum( is found under
LIST (2nd STAT)/OPS/<Option 6>
Example: an = 2n+1  1, 3, 5, 7, 9
cumSum(an)  1, 4, 9, 16, 25
Intermediate Sequences and Series
58
Intermediate Level Applications
Sequence and Series on the Calculator
On the calculator, cumSum( {1, 3, 5, 7, 9})
or …
cumSum(seq(2N+1, N, 0, 4))
If a list is already in the
calculator, perhaps in L1, then
cumSum(L1) or sum(L1) will
give series results.
Intermediate Sequences and Series
59
Intermediate Level Applications
Sequence Mode on the Calculator
First term #
First term
value
Recursive
function
Intermediate Sequences and Series
60
Intermediate Level Applications
Sequence Mode on the Calculator
Using the same recursive function:
an = an-1 + 3
or u(n) = u(n-1) +3,
suppose that we want to build a sequence in a list
on the calculator.
Intermediate Sequences and Series
61
Intermediate Level Applications
Sequence Mode on the Calculator
It is also possible to use the sequence mode to graph
some more complicated ideas.
Suppose that we are trying to convince a student
that the geometric sequence 100, 80, 64, 51.2, …
converges.
Set the Window to
Intermediate Sequences and Series
62
Intermediate Level Applications
Sequence Mode on the Calculator
Go to the 2nd ZOOM [Format] key, and make
sure that TIME is selected at the top.
Hit GRAPH.
u(n)=0.8u(n-1)
This is a
scatterplot of
the (n, an)
Intermediate Sequences and Series
63
Intermediate Level Applications
Sequence Mode on the Calculator
Instead, select 2nd Zoom [Format] and choose
Web.
Set the Window to 0 < x< 105 and 0 < y < 105
Hit Graph and
Trace. Hit the
right arrow to
iterate the web.
Intermediate Sequences and Series
64
Intermediate Level Applications
Sequence Mode on the Calculator
Let’s look at an = 10(-.8)n,
which becomes u(n) = -.8u(n-1)
You have to think about the WINDOW, but it has a
web which looks like:
Intermediate Sequences and Series
65
Intermediate Level Applications
Sequence Mode on the Calculator
An application which is stretching toward the advanced is the idea
of a predator-prey model *. The populations of the two
populations depend on the size of the other population.
Depending on various parameters, the populations will either die
out, grow without bound (!), or move into an equilibrium.
Two sequence functions can be used:
Rn = Rn-1(1+0.05 -.001*Wn-1)
“rabbits”
Wn = Wn-1(1+0.0002Rn-1 – 0.03) “wolves”
(This example is from the TI-83 manual, page 6-13)
* There is a long document on my website about predator-prey models that I cowrote as a NSA sponsored project in June 2004.
Intermediate Sequences and Series
66
Intermediate Level Applications
Sequence Mode on the Calculator
Intermediate Sequences and Series
67
Intermediate Level Applications
Sequence Mode on the Calculator
Using a WINDOW of nMin = 0 and nMax = 400
PlotStart = 1 PlotStep =1
XMin = 0 XMax = 400 Xscl = 100
YMin = 0 YMax = 300 YScl = 100
Under FORMAT, use the TIME
choice.
It makes for great
classroom discussion to
interpret these graphs.
Intermediate Sequences and Series
68
Intermediate Level Applications
Sequence Mode on the Calculator
With the sequence mode, we can do something
quite interesting on the calculator. The first graph
was showing the separate rabbit and wolf
populations as time progressed. But what if we
want to see how the graphs of the two populations
look relative to each other, i.e., (rabbits, wolves).
To do this select the FORMAT key and then find
the uv choice at the top.
Intermediate Sequences and Series
69
Intermediate Level Applications
Sequence Mode on the Calculator
Experimentation with the data suggests that the new WINDOW
be:
XMin = 80 XMax = 250
Xscl = 50
YMin = 0 YMax = 100
YScl = 10
Intermediate Sequences and Series
70
Infinite Sequences, Series and
Convergence
When series go on forever, we call them infinite
series. Let’s look at arithmetic series first.
Sn = 4 + 7+ 10 + 13 + …
= n( a1  an )
2
What is the sum s100?
s1000 ? s 1,000,000 ?
Intermediate Sequences and Series
71
Infinite Sequences, Series and
Convergence
Compare the results for each of these arithmetic
series:
1. Sn = 4 + 7 + 10 + 13 + …
2. Sn = 1 + 1.1 + 1.2 + 1.3 + 1.4 + …
3. Sn = 5+ 5.001 + 5.000001 + …
4. Sn = 4 + 3.5 + 3 + 2.5 + 2 + …
Conclusion…… ?
Intermediate Sequences and Series
72
Infinite Sequences, Series and
Convergence
Moving onto geometric series, consider the
behavior of these sums by taking the
number of terms to be higher and higher.
1. sn = 2 + 4 + 8 + 16 + …
sn =
s10 =
a1 (1  r n )
1 r
s100 =
Intermediate Sequences and Series
s1000 =
73
Infinite Sequences, Series and
Convergence
2. sn = 2 + 2(1.02) + 2(1.02)2 + 2(1.02)3 + …
3. sn=2 + 2(0.98) + 2(.98)2+ 2(.98)3 + …
sn = 1 – 3 + 9 – 27 + 81 + …
1 1 1
5. sn = 1 +
   ...
2 4 8
4.
1 1 1
   ...
6. sn = 1 +
2 4 8
Intermediate Sequences and Series
74
Infinite Sequences, Series and
Convergence
It eventually becomes obvious that there are
geometric series which converge and others which
diverge.
The idea is that convergence depends on the value of
r (the common ratio).
Conclusion: An infinite geometric series converges
when 1 < r < 1 or |r| < 1
Intermediate Sequences and Series
75
Infinite Sequences, Series and
Convergence
Looking at the sequences graphically makes some
strong connections with algebra, and the visual
impact helps with understanding about
convergence and divergence.
Let’s look at some ideas about series first
(because the graphs of sequences vs. the graphs of
series is also an important distinction).
Intermediate Sequences and Series
76
Graphs of Sequences and Series
Examples:
an =
1
2 
2
cn =
n
n
bn = 2   1 


3


2(3 )
8
 an
n 1
8
 bn
n 1
dn =
n
2
n 1
3
8
 cn
n 1
Intermediate Sequences and Series
8
 dn
n 1
77
Graphs of Sequences and Series
Intermediate Sequences and Series
78
Graphs of Sequences and Series
Intermediate Sequences and Series
79
Intermediate Level Applications
Deer Populations
In this application, the various quantities affect each
other. This is part of a discrete mathematics topic.
The sequences involved (and note why they are not
series!) affect each other. Whether or not they
converge is the important point, since this involves
whether the populations remain stable, or whether
they explode or become extinct. There are intuitive
ideas of limits here.
Intermediate Sequences and Series
80
Intermediate Level Applications
Deer Populations
Newborn
Yearling
Adult Male
Adult Female
N
Y
AM
AF
N = 0.20 AF
Y = 0.90 N
AM = 0.90 AM + 0.45 Y
AF = 0.90 AF + 0.48 Y
TOTAL
1
20
16
90
65
191
2
13
18
88
66
185
3
13
11
87
68
179
4
13
11
83
66
173
5
13
11
79
64
167
6
12
11
76
62
161
7
12
10
73
61
156
8
12
10
70
59
151
9
11
10
67
57
145
10
11
9
64
56
140
Intermediate Sequences and Series
81
Intermediate Level Applications
Deer Populations
Newborn
Yearling
Adult Male
Adult Female
N
Y
AM
AF
N = 0.20 AF
Y = 0.90 N
AM = 0.90 AM + 0.45 Y
AF = 0.90 AF + 0.48 Y
TOTAL
11
11
9
61
54
135
12
10
9
58
52
129
13
10
9
56
51
126
14
10
9
54
50
123
15
10
9
52
49
120
16
9
9
50
48
116
17
9
8
49
47
113
18
9
8
47
46
110
19
9
8
45
45
107
20
9
8
44
44
105
21
8
8
43
43
102
22
8
7
42
42
99
Intermediate Sequences and Series
82
Intermediate Level Applications
Suppose that we earn simple interest on a
bank account. Let’s say that the interest
rate is 5% on a principal of $1,000.
Compound
Interest
a0 = 1000
a1 = 1000 + 1000(.05) = 1050
a2= 1050 + 1000(.05) = 1100
a3 = 1100 + 1000(.05) = 1150
a n:
1000, 1050 , 1100 , 1150 , 1200 , …
Intermediate Sequences and Series
83
Intermediate Level Applications
Instead, suppose that we earn 5% interest
on a $1,000 principal, compounded
annually.
Compound
Interest
a0 = 1000
a1 = 1000 + 1000(.05) = 1000(1.05) = 1050
a2 = 1050 + 1050(.05) = 1050(1.05) =
1000(1.05)2
a3 = 1000(1.05)3
 at = 1000(1.05)t
Intermediate Sequences and Series
84
Intermediate Level Applications
Compound Interest
Most banks and financial institutions offer compound interest
which is awarded more frequently than annually, and it is
important for students to realize that there is an advantage to
getting a fraction of the annual interest more frequently so that
more compounding can occur earlier in time.
If yo is the initial principal,
r = the annual percentage rate,
t =the number of years for the money to be invested,
n = the number of times per year that compounding will occur,
yt = yo(1 +
r
nt
n )
Intermediate Sequences and Series
85
Intermediate Level Applications
Compound
Interest
If the number of compoundings is discrete,
then this formula is fine. But what if the
number of compoundings each year becomes
more and more frequent?
1 n
Investigate the sequence of (1 +
) as n
n
increases.
Intermediate Sequences and Series
86
Intermediate Level Applications
Compound Interest
N
1
( 1+ n ) n
100
2.7048
200
2.7115
1000
2.7169
10,000
2.7181
1,000,000
2.7183
Intermediate Sequences and Series
87
Intermediate Level Applications
Compound
Interest
Note that if n > 1012, the calculator will be
subject to some serious roundoff errors. This is
because the memory of the calculator only holds
about 12 digits, and larger numbers than that
overwhelm the capabilities of the machine.
The sequence is (for n = 1, 2, 3, 4, 5…)
2, 2.25, 2.370, 2.441, 2.448, 2.522, 2.545, … ,
2.7048 , … 2.7115 , 2.7169 , … , 2.7181 , …
2.7183

e
Intermediate Sequences and Series
88
Intermediate Level Applications
Compound Interest
There are some wonderful problems for students to
solve with interest, and their interest (bad pun…) is
piqued with some challenges, such as …
Two people each have $10,000. One invests the
money at a 5.1% interest rate, compounded
monthly. The other invests at 5% compounded
daily. Which investment is better after 8 years?
When will they be equal? Which is better after
many years?
Intermediate Sequences and Series
89
Intermediate Level Applications
Linear and Exponential Functions Compared
Consider two scenarios:
1) Invest $5000 with 5% compound interest earned
annually.
2) Invest $5000 and add $500 each year to the account.
No interest is earned.
Which investment is better?
Intermediate Sequences and Series
90
Intermediate Level Applications
Linear and Exponential Functions Compared
The first situation is modeled with an exponential
function, since it is geometric sequence.
The second situation is modeled with a linear
function, since it is an arithmetic sequence.
Eventually…..if both sequences increase, a
geometric sequence will exceed an arithmetic
sequence.
Intermediate Sequences and Series
91
Intermediate Level Applications
Linear and Exponential Functions Compared
Suppose that person has a debt obligation which is subject
to a compound annual interest rate of 18% (such as a credit
card). The amount owed is $50,000. If the minimum
monthly payment is 2.5% of the remaining balance, and the
minimum payment is what is made each month, what
happens to the debt?
Question: Is a (geometric sequence – arithmetic sequence) a
good strategy to pay back a debt? Could it be fine if the
minimum payment is high enough?
Intermediate Sequences and Series
92
Intermediate Level Applications
Astronomy and Sequences
In the middle of the 19th century, data concerning the
distance of the planets in our solar system from the sun
indicated that there was a remarkable sequence … with a
missing number:
Planet
Mercury
Venus
Earth
Mars
Jupiter
Dist sun* 36
67.2
92.9
141.6
483.7
A.U.**
0.3875
0.7234
1.0000
1.5242
5.2067
Planet
Saturn
Uranus
Pluto
Dist sun* 890.6
1777
2654.4
A.U.**
19.1281
39.3369
9.5867
* (in millions of miles) ** astronomical units
Intermediate Sequences and Series
93
Intermediate Level Applications
Astronomy and Sequences
It seemed that there were two “holes” in the location of
the planets, and the location …even the existence (ah,
such a word for a mathematician) … of a possible
planet was discovered by calculation rather than by
observation.
The conclusion was that there was another body
pulling Uranus out of the orbit predicted by Bode’s
Law, so Adams (England) and Leverrier (France)
solved g = m1m2 to calculate the place where another
d2
planet ought to be found.
Intermediate Sequences and Series
94
Bode’s
A
Law
B
SUM
4
Intermediate
Level
Applications
Astronomy and
Sequences
SUM/10
4
0.4
4
3
7
0.7
4
6
10
1.0
4
12
16
1.6
4
24
28
2.8
4
48
52
5.2
4
96
100
10.0
4
192
196
19.6
4
384
388
38.8
4
768
772
77.2
Intermediate Sequences and Series
95
Planet
A.U.
A
Intermediate
Level
Applications
Astronomy and
Sequences
Mercury
0.3875
4
Venus
0.7234
4
Earth
1.0000
Mars
1.5242
B
Bode’s
Law
SUM
SUM/
10
4
0.4
3
7
0.7
4
6
10
1.0
4
12
16
1.6
4
24
28
2.8
Jupiter
5.2067
4
48
52
5.2
Saturn
9.5867
4
96
100
10.0
Uranus
19.1281
4 192
196
19.6
30.1335
4 384
388
38.8
39.3369
4 768
772
77.2
Pluto
Intermediate Sequences and Series
96
Intermediate
Level
Applications
Astronomy and Sequences
On September 23, 1846,
astronomers had their telescopes
trained on the piece of the night sky
where Adams and Leverrier had
predicted that a missing planet
might be located.
A mere half hour after they began
looking, Neptune was observed,
only 52 minutes of arc (less than
one degree) off from Leverrier’s
prediction. It was 2.8 billion miles
from earth.
Viva les mathematiques!
Intermediate Sequences and Series
97
Intermediate Level Applications
Chemistry, Data Analysis and Sequences
Looks for
patterns in
atomic weight,
specific heat or
boiling points
across rows or
down columns.
NB: The TI-84 has
a built in periodic
table, and there are
graphical displays
included!
Intermediate Sequences and Series
98
Intermediate Level Applications
Biological Growth and Sequences
If a virus grows from a population of 200 at 8 AM to a
population of 1000 by noon, how many virus will
there be at 4 PM? 6 PM? midnight?
Answer: y0 = 200 (8 AM)
y4 = 1000 (noon, which is 4 hours later)
The sequence is:
t: 0, 4, 8, 12, 16, 20, …
at:
200, 1000, 5000, 25000, 125000, 625000
Intermediate Sequences and Series
99
Intermediate Level Applications
Biological Growth and Sequences
If we are only interested in the virus counts at whole
number of hours, we need the geometric means, and
the multiplier becomes 4
5  1.4953
So the sequence is
at = 500 (1.4963t-1)
Intermediate Sequences and Series
100
Intermediate Level Applications
Biological Growth and Sequences
No wonder healthy
people at 8 AM are
not feeling well at the
ned of a day!
Intermediate Sequences and Series
101
Intermediate Level Applications
Chemical Half-Life: Radioactivity
The half-life of the chemical element technetium is about 6 hours.
This element is used in medicine when tracing body functions,
especially renal function or failure in patients receiving
chemotherapy. Given the short half-life, what percentage of Tc
injected into the body remains after 2 hours? 3 hours? 4 hours?
This is done just as the biological (population) growth, and the
hurly percentages can be thought of as a sequence which
converges to some value.
Intermediate Sequences and Series
102
Intermediate Level Applications
Antibiotic Medications: Sequences and Series
Suppose that an antibiotic medication dissipates
in the body so that 20% of the amount currently
in the body is gone after 4 hours (or 80% of the
medication remains after 4 hours).
A patient is given a 600 mg bolus (a large initial
dosage) to begin the treatment. Then the dosage
is an additional 100 mg every 4 hours. It is
dangerous for the body to have more than 700
mg at any one time, and at least 500 mg is
needed to fight the illness (e.g., strep throat).
Intermediate Sequences and Series
103
Intermediate Level Applications
Antibiotic Medications: Sequences and Series
This is a good example of a problem which can be considered as both a
sequence and as a series.
Sequence: a t = amount of medication given at each 4 hour interval
a1 = 600
a2 = 100 + 600(.8) = 580 mg
a3 = 100 + 100(.8) + 600(.82) = 564 mg
a4 = 100 + 100(.8) + 100(.82) + 600(.83) = 551.2 mg
a5 = 100 + 100(.8) + 100(.82) + 100(.83) + 600(.84) = 540.96 mg
Intermediate Sequences and Series
104
Intermediate Level Applications
Antibiotic Medications: Sequences and Series
The medication after the bolus forms a geometric sequence which
decreases to zero, and the repeated medications form a geometric
series:
sum = 100 + 100(.8) + 100(.82) + 100(.83) + …
=
a1 = 100
1 r
1  .8
= 500
The combined dosages (which are a series) form a sequence which
needs to stay between the effective and the dangerous drug levels.
(What happens to the original bolus?)
Intermediate Sequences and Series
105
Intermediate Level Applications
Cooling of Liquids
A hot cup of coffee ( of cocoa, tea, …) fresh
from the coffeepot has a temperature of
140o F.
a) How does it cool?
b) This can be simulated with a CBL and
TI-83/84.
c) Use appropriate data analysis and
regressions.
Intermediate Sequences and Series
106
Intermediate Level Applications
Cooling of Liquids
Which
sequence of
temperatures
makes the
most sense?
How are each of
the sequences
calculated?
Time (min) Temp1
Temp2
Temp3
Temp4
0
140
140
140
140
1
135
126
133
130
2
130
113.4
126.7
121
3
125
102.1
121
113
4
120
91.9
115.9
106
5
115
82.67
111.3
100
10
90
48.8
94.4
85
20
65
17
78.5
??
Intermediate Sequences and Series
107
Intermediate Level Applications
Cooling of Liquids
It seems to be good, authentic mathematics and science to
guess which of the sequences is most reasonable, and then
try to fit a function to that sequence.
Following such intuition with a data collection with a CBL
on cooling water will give data to verify or refute the
earlier guess.
Newton’s Law of Cooling:
d (T  Tambient ) = k(T-T
-kt
ambient) or T-Ta = (T0-Ta)e
dt
Intermediate Sequences and Series
108
Intermediate Level Applications
Economics
When a yearbook is printed, suppose it costs $9000 to print one
copy, because of the set-up costs for the press, type-setting,
importing photographs, binding, cover set-up, and artwork. It
costs as additional $8 for each book, since the press is already
set up, and only paper, binding, and some ink are needed for
the second copy.
1.
What is the cost of 5 books? 10 books? 100 books? n books?
2.
What is the average cost of n books?
3.
What is the difference in average costs for printing n to (n+1)
books for various values of n?
Intermediate Sequences and Series
109
Intermediate Level Applications
Economics
As a sequence or series problem,
b1 = 9000
b1 average = 9000/1 = 9000
b2 = 9000 + 8
b2 average = 9008/2 = 4504
b3 = 9000 + 8 + 8
b3 average = 9016 / 3 = 3005.33
bn = 9000 + 8(n-1)
bn average = (8992 + 8n) / n
= 8992/n + 8
If 500 yearbooks are ordered, it costs $12,992 to print
them, and the average cost is $25.98
This can be taught as a sequence problem or as a rational
function problem.
Intermediate Sequences and Series
110
Advanced Level
Calculus Examples, especially AP Calculus
Newton’s Method uses the definition of derivative to provide a method
to locate the roots of a function. (It differs from the algorithm ROOT
FINDER in the TI-83/84 calculators which uses the IVT)
x n+1 = xn –
f ( xn )
f '( xn )
This is an iterative algorithm, where the results (output) of each stage
become the input of the next stage. If we look at each xn and its
subsequent xn+1, then the fraction which is subtracted can be
considered as the “correction factor”, which (hopefully) sends us
closer, via a sequence, to the exact location of the root of a function.
Advanced Sequences and Series
111
Advanced Level
Calculus Examples, especially AP Calculus
An example of Newton’s method:
Suppose we want to approximate
3
This is a root of f(x) = x2 – 3
The sequence of values from Newton’s Method looks
like:
x0 = 1 (our choice for a “guess”)
The sequence
seems to
converge.
Advanced Sequences and Series
112
Advanced Level
Calculus Examples, especially AP Calculus
An example of Newton’s method:
Suppose we want to approximate the roots of
f(x) = x2 + 3
x
The sequence of values from Newton’s Method
looks like:
x0 = 1 (our choice for a “guess”)
This time, there is
no convergence,
and we cannot
locate a root.
Advanced Sequences and Series
113
Advanced Level
Calculus Examples, especially AP Calculus
Riemann sums are the basis for evaluating the area under a function,
as sums of the areas of rectangles are used to approximate the exact
area.
It is probably a good idea to mention the words sequence and series
in the explanation for the strategy. After all, the “C” part of the BC
Calculus concerns the ideas of series and convergence, but the ideas
of the convergence of sequences and series can appear very early in
the “A” part of the differential calculus when limits are discussed and
when early ideas about areas under functions are introduced.
Advanced Sequences and Series
114
Advanced Level
Calculus Examples, especially AP Calculus
There is a
sequence of the
areas of each
rectangle, and
there is a
sequence of the
partial sums of
the rectangles.
Convergence of
each of these is
an important
idea.
10
8
f x = x2+1
6
4
2
-10
-5
Advanced Sequences and Series
5
10
115
Advanced Level
Calculus Examples, especially AP Calculus
…and as the number
of partitions goes
from 4 to 8 to 16 to
…, there is a
sequence of
estimates on the
area, and the idea for
calculus students is
to believe that the
sequence of series
converges…to the
exact area.
10
8
f x = x2+1
6
4
2
-10
-5
Advanced Sequences and Series
5
10
116
Advanced Level
Calculus Examples, especially AP Calculus
We usually consider the Trapezoid Rule and Simpson’s Rule as
series, but if we repeat them with more and more partitions,
then sequence of the series should converge.
b
Trap =

f ( x)dx 
a
1
h( y0  2 y1  2 y2  ...  2 yn 1  yn )
2
Simpson =
b

a
1
f ( x)dx  h( y0  4 y1  2 y2  4 y3  2 y4  ...  4 yn 2  2 yn 1  yn )
3
n = an even number of partitions required for Simpson’s rule
Advanced Sequences and Series
117
Advanced Level
Calculus Examples, especially AP Calculus
4
Evaluate
2
x
 dx
using different algorithms.
0
Upper
Lower
Trapezoids
Simpson
n=4
30
14
22
21.3333
N=8
25.5
17.5
21.5
21.3333
n = 20
22.96
19.76
21.36
21.3333
n = 100
21.9776
20.6976
21.3376
21.3333
n = 50
21.56544
21.0144
21.3344
21.3333
n = 1000
21.365
21.301344
21.33334
21.3333
Advanced Sequences and Series
118
Advanced Level
Calculus Examples, especially AP Calculus
Euler’s method is an iterative algorithm to give
approximate solutions to differential equations. It is
really just a linearization method that is used
repeatedly to give a sequence of points which serve
as a numerical function.
dy
yn 1  yn  * x
dx
Advanced Sequences and Series
119
Advanced Level
Calculus Examples, especially AP Calculus
Example: Solve
dy
 xy 2
dx
Use
with the initial condition (1, 2)
x = 0.1
answer: The first point on the solution is (1,1) because
x0 = 2 and y0 = 1
dy
yn 1  yn  * x
dx
= 2 + [(1)(22)] (0.1)
= 2.4
 the second point is (1.1, 2.4)
Advanced Sequences and Series
120
Advanced Level
Calculus Examples, especially AP Calculus
So x1 = 1.1 and y1 = 2.4
dy
yn 1  yn  * x
dx
= 2.4 + [(1.1)(2.42)] (0.1)
= 3.0336
 the third point is (1.2, 3.0336)
We continue the process, generating a sequence of approximate
solutions to the differential equation. (If
x
is smaller, then the theory says that the sequence should more
closely match the function which is the solution to the
differential equation.)
Advanced Sequences and Series
121
Advanced Level
Calculus Examples, especially AP Calculus
The exact solution, using separable differential equation methods, is
2
y
2  x2
It is not always the case that an exact solution can be found, and those
are the examples for which the approximate solutions algorithms are
important.
There are also some algorithms which provide more accurate
approximations. Two of them are called the Improved Euler method
and the Runge-Kutta Method.
Advanced Sequences and Series
122
Advanced Level
Calculus Examples, especially AP Calculus
A summary of the results of these algorithms is:
X
Y
(exact)
Y
(Euler)
Y
(ImprovedEuler)
Y
(Runge-Kutta)
1
2
2
2
2
1.1
2.5316
2.4
2.5168
2.5316
1.2
3.5714
3.0336
3.4848
3.5706
1.3
6.4516
4.1379
5.80101
6.4304
better
best
perfect!
good
Advanced Sequences and Series
123
Advanced Level
Calculus Examples, especially AP Calculus
Maclaurin and Taylor polynomials are a series of polynomial
(power) terms, and they are typically taught near the end of a BC
Calculus course. A suggestion is to introduce them much earlier in
the course, since students only need to be able to do derivatives to
calculate these series. Then the approximation methods that they
provide with “polynomials simulating other function” can be used,
for example, when an indefinite or definite integral is to be done,
and students have not yet learned the antiderivative of that
function.
We want to convince them that the polynomial (or power) series is
a sequence of series that converges.
Advanced Sequences and Series
124
Advanced Level
Calculus Examples, especially AP Calculus
A couple of ideas to emphasize the ideas of sequences of series with
Maclaurin and Taylor polynomials.
a.
Show simultaneously the graphs of
y = sin x
y=
x3 x5 x 7
x     ...
3! 5! 7!
Put increasingly more terms of the series in the calculator to see
how the original function and its Maclaurin series match.
Advanced Sequences and Series
125
Advanced Level
Calculus Examples, especially AP Calculus
b.
Show, graphically, the limited convergence of
y = ln (x + 1)
y=
x 2 x3 x 4
x     ...
2 3 4
This will provide a good foundation for
understanding the “convergence tests” and
“intervals of convergence” ideas which follow.
Advanced Sequences and Series
126
Advanced Level
Calculus Examples, especially AP Calculus
c.
Evaluate
9.08
using a series for y =
x
centered at x = 9.
Show that a Taylor series is easier to evaluate (“easier” = “uses
only simple arithmetic”) and can almost be done without a
calculator at all.
Advanced Sequences and Series
127
Advanced Level
Calculus Examples, especially AP Calculus
James Gregory’s method for estimating
Since

4

(1671)
1
= tan -1 (1) which equals the value of
1
0 1  x 2 dx
then a Maclaurin series for the integrand can be antidifferentiated
and an approximate value can be done with ordinary arithmetic.
Advanced Sequences and Series
128
Advanced Level
Calculus Examples, especially AP Calculus
The sequence of operations that is useful here is:
1.
1
 1  x  x 2  x 3  x 4  ...
1 x
1
2
4
6
8

1

x

x

x

x
 ...
2
1 x
This last series is accomplished by replacing the x’s by x2’s in the
first series. This is a very helpful (and yet unique) feature of
Maclaurin series!
Advanced Sequences and Series
129
Advanced Level
Calculus Examples, especially AP Calculus
2.
Then
1
1
1 3 1 5 1 7
0 1  x 2 dx  x  3 x  5 x  7 x  ...
1
0
= 0.8349206349 (ok, so I did use a calculator to do some
of this!!)
Then

= 4 (0.8349206349) = 3.33968254
Advanced Sequences and Series
130
Advanced Level
Calculus Examples, especially AP Calculus
3.
We can update what James Gregory did, using technology to see
whether his series converges.

1
n 1
(

1)
*
The antiderivative series can be written as 
(2n  1)
n 1
On a TI-83/84,
put the counters n is L1 (as far as you want to go)
the terms of the sequence in L2 as (-1)^(L1+1) /(2L1 -1)
the accumulated sum in to L3 as cumSum(L2)
The series does not converge very quickly, so it is not useful,
but it is a valuable method to teach.
Advanced Sequences and Series
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Sequences and Series
From Simple Patterns to Elegant and Profound Mathematics
Mathematics is all about expressing patterns,
numerically and graphically.
Patterns can indicate some interesting, usual,
unusual, and sometimes complicated simulations
of real phenomena.
So sequences and series ought to be as much a part
of our mathematical language as functions,
formulas, equations, expressions, and shapes.
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