Sequences and

Summations
Section Summary

► Sequences.
► Examples: Geometric Progression, Arithmetic Progression
► Recurrence Relations
► Example: Fibonacci Sequence
► Summations
► Special Integer Sequences (optional)
Introduction

► Sequences are ordered lists of elements.

► 1, 2, 3, 5, 8
► 1, 3, 9, 27, 81, …….
► Sequences arise throughout mathematics, computer science, and in many
other disciplines, ranging from botany to music.
► We will introduce the terminology to represent sequences and sums of the
terms in the sequences.
Sequences

Definition: A sequence is a function from a subset of the integers (usually either

the set {0, 1, 2, 3, 4, …..} or {1, 2, 3, 4, ….} ) to a set S.
► The notation an is used to denote the image of the integer n. We can think
of an as the equivalent of f(n) where f is a function from {0,1,2,…..} to S.
We call an a term of the sequence.
Sequences

Example: Consider the sequence where

Geometric Progression

Definition: A geometric progression is a sequence of the form:

where the initial term a and the common ratio r are real numbers.
Examples:
1. Let a = 1 and r = −1. Then:

2. Let a = 2 and r = 5. Then:

3. Let a = 6 and r = 1/3. Then:

Arithmetic Progression

Definition: A arithmetic progression is a sequence of the form:

where the initial term a and the common difference d are real numbers.

Examples:
1. Let a = −1 and d = 4:

2. Let a = 7 and d = −3:

3. Let a = 1 and d = 2:
Strings

Definition: A string is a finite sequence of characters from a finite set (an

alphabet).
► Sequences of characters or bits are important in computer science.
► The empty string is represented by λ.
► The string abcde has length 5.
Recurrence Relations

Definition: A recurrence relation for the sequence {an} is an equation that

expresses an in terms of one or more of the previous terms of the sequence,
namely, a0, a1, …, an-1, for all integers n with n ≥ n0, where n0 is a nonnegative
integer.
► A sequence is called a solution of a recurrence relation if its terms satisfy the
recurrence relation.
► The initial conditions for a sequence specify the terms that precede the first
term where the recurrence relation takes effect.
Questions about Recurrence
Relations
Example 1: Let {an} be a sequence that satisfies the recurrence relation an =
an-1 + 3 for n = 1,2,3,4,…. and suppose that a0 = 2. What are a1 , a2 and a3?
[Here a0 = 2 is the initial condition.]

Solution: We see from the recurrence relation that

a1 = a 0 + 3 = 2 + 3 = 5
a2 = 5 + 3 = 8
a3 = 8 + 3 = 11
Questions about Recurrence
Relations
Example 2: Let {an} be a sequence that satisfies the recurrence relation an =
an-1 – an-2 for n = 2,3,4,…. and suppose that a0 = 3 and a1 = 5. What are a2 and
a3?
[Here the initial conditions are a0 = 3 and a1 = 5. ]

Solution: We see from the recurrence relation that

a2 = a 1 - a 0 = 5 – 3 = 2
a3 = a2 – a1 = 2 – 5 = –3
Fibonacci Sequence

Definition: Define the Fibonacci sequence, f0 ,f1 ,f2,…, by:

► Initial Conditions: f0 = 0, f1 = 1
► Recurrence Relation: fn = fn-1 + fn-2

Example: Find f2 ,f3 ,f4 , f5 and f6 .

Answer:
f2 = f1 + f0 = 1 + 0 = 1 ,
f3 = f2 + f1 = 1 + 1 = 2 ,
f4 = f3 + f2 = 2 + 1 = 3 ,
f5 = f4 + f3 = 3 + 2 = 5 ,
f6 = f5 + f4 = 5 + 3 = 8 .
Solving Recurrence Relations

► Finding a formula for the nth term of the sequence generated by a recurrence
relation is called solving the recurrence relation.
► Such a formula is called a closed formula.
Iterative Solution Example

Method 1: Working upward, forward substitution

Let {an} be a sequence that satisfies the recurrence relation an = an-1 + 3 for n
= 2,3,4,…. and suppose that a1 = 2.
a2 = 2 + 3
a3 = (2 + 3) + 3 = 2 + 3 ∙ 2
a4 = (2 + 2 ∙ 3) + 3 = 2 + 3 ∙ 3
.
.
.
an = an-1 + 3 = (2 + 3 ∙ (n – 2)) + 3 = 2 + 3(n – 1)
Iterative Solution Example

Method 2: Working downward, backward substitution

Let {an} be a sequence that satisfies the recurrence relation an = an-1 + 3 for n = 2,3,4,…. and suppose that a1 = 2.

an = an-1 + 3
= (an-2 + 3) + 3 = an-2 + 3 ∙ 2

= (an-3 + 3 )+ 3 ∙ 2 = an-3 + 3 ∙ 3
.
.
.

= a2 + 3(n – 2) = (a1 + 3) + 3(n – 2) = 2 + 3(n – 1)

Financial Application

Example: Suppose that a person deposits $10,000.00 in a savings account at a

bank yielding 11% per year with interest compounded annually. How much will
be in the account after 30 years?
Let Pn denote the amount in the account after 30 years. Pn satisfies the
following recurrence relation:
Pn = Pn-1 + 0.11Pn-1 = (1.11) Pn-1
with the initial condition P0 = 10,000

Continued on next slide 🡪

Financial Application

Pn = Pn-1 + 0.11Pn-1 = (1.11) Pn-1

with the initial condition P0 = 10,000
Solution: Forward Substitution
P1 = (1.11)P0
P2 = (1.11)P1 = (1.11)2P0
P3 = (1.11)P2 = (1.11)3P0
:
Pn = (1.11)Pn-1 = (1.11)nP0 = (1.11)n 10,000
Pn = (1.11)n 10,000 (Can prove by induction, covered in Chapter 5)
P30 = (1.11)30 10,000 = $228,992.97
Questions on Special Integer Sequences
(opt)
Example 1: Find formulae for the sequences with the following first five terms:
1, ½, ¼, 1/8, 1/16
Solution: Note that the denominators are powers of 2. The sequence with an = 1/2n
is a possible match. This is a geometric progression with a = 1 and r = ½.
Example 2: Consider 1,3,5,7,9
Solution: Note that each term is obtained by adding 2 to the previous term. A
possible formula is an = 2n + 1. This is an arithmetic progression with a =1 and d
= 2.
Example 3: 1, -1, 1, -1,1
Solution: The terms alternate between 1 and -1. A possible sequence is an = (−1)n .
This is a geometric progression with a = 1 and r = −1.
Useful Sequences
Guessing Sequences (optional)

Example: Conjecture a simple formula for an if the first 10 terms of the

sequence {an} are 1, 7, 25, 79, 241, 727, 2185, 6559, 19681, 59047.

Solution: Note the ratio of each term to the previous approximates 3. So now
compare with the sequence 3n . We notice that the nth term is 2 less than
the corresponding power of 3. So a good conjecture is that an = 3n − 2.
Integer Sequences (optional)

► Here are three interesting sequences to try from the OESIS site. To solve each
puzzle, find a rule that determines the terms of the sequence.
► Guess the rules for forming for the following sequences:
► 2, 3, 3, 5, 10, 13, 39, 43, 172, 177, ...
► Hint: Think of adding and multiplying by numbers to generate this sequence.

► 0, 0, 0, 0, 4, 9, 5, 1, 1, 0, 55, ...
► Hint: Think of the English names for the numbers representing the position in the sequence and
the Roman Numerals for the same number.

► 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, ...

► Hint: Think of the English names for numbers, and whether or not they have the letter ‘e.’
Summations

► Sum of the terms

from the sequence
► The notation:

represents

► The variable j is called the index of summation. It runs through all the
integers starting with its lower limit m and ending with its upper limit n.
Summations

► More generally for a set S:

► Examples:
Product Notation (optional)

► Product of the terms

from the sequence

► The notation:

represents
Geometric Series

Sums of terms of geometric progressions

To compute Sn , first multiply both sides of the

Proof Let equality by r and then manipulate the resulting
: sum as follows:

Continued on next slide 🡪

Geometric Series

From previous slide.

Shifting the index of summation with k = j + 1.

Removing k = n + 1 term and

adding k = 0 term.

Substituting S for summation formula

∴
if r ≠1

if r = 1
Some Useful Summation Formulae

Geometric Series:
We just proved this.

Later we
will prove
some of
these by
induction.

Proof in text
(requires calculus)
Cardinality of Sets
Section Summary

► Cardinality
► Countable Sets
► Computability
Cardinality

Definition: The cardinality of a set A is equal to the cardinality of a set B,

denoted
|A| = |B|,
if and only if there is a one-to-one correspondence (i.e., a bijection) from A
to B.
► If there is a one-to-one function (i.e., an injection) from A to B, the
cardinality of A is less than or the same as the cardinality of B and we write
|A| ≤ |B|.
► When |A| ≤ |B| and A and B have different cardinality, we say that the cardinality
of A is less than the cardinality of B and write |A| < |B|.
Cardinality

► Definition: A set that is either finite or has the same cardinality as the set of
positive integers (Z+) is called countable. A set that is not countable is
uncountable.
► The set of real numbers R is an uncountable set.
► When an infinite set is countable (countably infinite) its cardinality is ℵ0
(where ℵ is aleph, the 1st letter of the Hebrew alphabet). We write |S| = ℵ0 and
say that S has cardinality “aleph null.”
Showing that a Set is Countable

► An infinite set is countable if and only if it is possible to list the elements

of the set in a sequence (indexed by the positive integers).
► The reason for this is that a one-to-one correspondence f from the set of
positive integers to a set S can be expressed in terms of a sequence
a1,a2,…, an ,… where a1 = f(1), a2 = f(2),…, an = f(n),…
Hilbert’s Grand Hotel
David Hilbert
The Grand Hotel (example due to David Hilbert) has countably infinite number
of rooms, each occupied by a guest. We can always accommodate a new guest
at this hotel. How is this possible?

Explanation: Because the rooms of Grand

Hotel are countable, we can list them as
Room 1, Room 2, Room 3, and so on. When
a new guest arrives, we move the guest in
Room 1 to Room 2, the guest in Room 2 to
Room 3, and in general the guest in Room n
to Room n + 1, for all positive integers n.
This frees up Room 1, which we assign to
the new guest, and all the current guests The hotel can also accommodate a
still have rooms. countable number of new guests, all the
guests on a countable number of buses
where each bus contains a countable
number of guests (see exercises).
Showing that a Set is Countable

Example 1: Show that the set of positive even integers E is countable set.
Solution: Let f(x) = 2x.
1 2 3 4 5 6 …..

2 4 6 8 10 12 ……
Then f is a bijection from N to E since f is both one-to-one and onto. To show
that it is one-to-one, suppose that f(n) = f(m). Then 2n = 2m, and so n =
m. To see that it is onto, suppose that t is an even positive integer. Then
t = 2k for some positive integer k and f(k) = t.
Showing that a Set is Countable

Example 2: Show that the set of integers Z is countable.

Solution: Can list in a sequence:
0, 1, − 1, 2, − 2, 3, − 3 ,………..
Or can define a bijection from N to Z:
► When n is even: f(n) = n/2
► When n is odd: f(n) = −(n−1)/2
The Positive Rational Numbers are
Countable
► Definition: A rational number can be expressed as the ratio of two integers p
and q such that q ≠ 0.
► ¾ is a rational number
► √2 is not a rational number.
Example 3: Show that the positive rational numbers are countable.
Solution:The positive rational numbers are countable since they can be
arranged in a sequence:
r1 , r2 , r3 ,…
The next slide shows how this is done. →
 The Positive Rational Numbers are
Countable
Constructing the List
First row q = 1.
Second row q = 2.
First list p/q with p + q = 2.
etc.
Next list p/q with p + q = 3

And so on.

1, ½, 2, 3, 1/3,1/4, 2/3, ….
Strings

Example 4: Show that the set of finite strings S over a finite alphabet A is
countably infinite.
Assume an alphabetical ordering of symbols in A
Solution: Show that the strings can be listed in a sequence. First list
1. All the strings of length 0 in alphabetical order.
2. Then all the strings of length 1 in lexicographic (as in a dictionary) order.
3. Then all the strings of length 2 in lexicographic order.
4. And so on.
This implies a bijection from N to S and hence it is a countably infinite set.
The set of all Java programs is
countable.
Example 5: Show that the set of all Java programs is countable.
Solution: Let S be the set of strings constructed from the characters which
can appear in a Java program. Use the ordering from the previous example.
Take each string in turn:
► Feed the string into a Java compiler. (A Java compiler will determine if the input
program is a syntactically correct Java program.)
► If the compiler says YES, this is a syntactically correct Java program, we add the
program to the list.
► We move on to the next string.
In this way we construct an implied bijection from N to the set of Java
programs. Hence, the set of Java programs is countable.
 Georg Cantor
The Real Numbers are Uncountable (1845-1918)

Example: Show that the set of real numbers is uncountable.

Solution: The method is called the Cantor diagnalization argument, and is a proof by contradiction.
1. Suppose R is countable. Then the real numbers between 0 and 1 are also countable (any subset of a countable set is
countable - an exercise in the text).
2. The real numbers between 0 and 1 can be listed in order r1 , r2 , r3 ,… .
3. Let the decimal representation of this listing be

4. Form a new real number with the decimal expansion

where
5. r is not equal to any of the r1 , r2 , r3 ,... Because it differs from ri in its ith position after the decimal point.
Therefore there is a real number between 0 and 1 that is not on the list since every real number has a unique decimal
expansion. Hence, all the real numbers between 0 and 1 cannot be listed, so the set of real numbers between 0 and 1
is uncountable.
6. Since a set with an uncountable subset is uncountable (an exercise), the set of real numbers is uncountable.
Computability (Optional)

► Definition: We say that a function is computable if there is a computer

program in some programming language that finds the values of this function.
If a function is not computable we say it is uncomputable.
► There are uncomputable functions. We have shown that the set of Java
programs is countable. Exercise 38 in the text shows that there are
uncountably many different functions from a particular countably infinite set
(i.e., the positive integers) to itself. Therefore (Exercise 39) there must be
uncomputable functions.