Chapter 4: Sequences and Mathematical
Induction
February 13, 2008
Outline
1 4.1 Sequences
2 4.2 Mathematical Induction I
3 4.3 Mathematical Induction II
4 4.4 Strong Mathematical Induction
Sequences
• Sequence: an ordered set of elements (most often,
numbers)
am , am+1 , am+2 , . . . , an
• The elements of a sequence are called terms of the
sequence. ai is called the i-th term of the sequence.
• Sequences are often infinite:
am , am+1 , am+2 , . . .
Sequences can be given in two ways:
1
analytically, when each term ak is given by some known
formula involving k :
ak = f (k )
where f is a function.
2
recursively, if the first m terms are given explicitly (these
are called initial conditions) and the rest of the terms
am+1 , am+2 , . . . are given by a recursive formula
an = f (an−1 , an−2 , . . . , ai−m ),
n>m
Examples
(a) ak =
(−1)k
2k
1 1 1
− , ,− ,...
2 4 8
(b) a1 = 1 and ak = k · ak −1 .
This is an example of a recursively defined sequence.
a2 = 2 · 1 = 2
a3 = 3 · a2 = 6
a4 = 4 · a3 = 24
..
.
In fact, this sequence is
ak = k !
Series
Definition
A series is the sum of all terms of a sequence. Given a
sequence ai and two positive integers m and n, such that n ≥ m
n
X
ai = am + am+1 + . . . + an
i=m
If the sequence is infinite, we write
∞
X
ai = am + am+1 + am+2 + . . .
i=m
[Of course, with infinite series, this sum may not exist as a real
number; i.e. such a series may not converge.]
Examples
(a)
6
X
i = 2 + 3 + 4 + 5 + 6 = 20
i=2
(b)
3
X
i =3
i=3
(c)
5
X
(−1)i = −1 + 1 − 1 + 1 − 1 = −1
i=1
(d)
4
X
(−1)i
i=1
1 1 1 1
13
=− + − + =−
i +1
2 3 4 5
60
Change of Variable
Let ai be a sequence and consider the two series
n
X
ai = a1 + a2 + . . . + an−1 + an
i=1
n+1
X
ai−1 = a1 + a2 + . . . + an−1 + an
i=2
Therefore, these two are the
P same series.
In general, given a series ni=m ai , we can change the
variable by setting j = i + k (k ∈ Z) to get
n+k
X
j=m+k
aj−k .
Example
Transform the sum
n−1
X
i=1
i
(n − i)2
by making the change of the variable j = i − 1.
Solution: Since j = i − 1, we have i = j + 1, so:
n−2
X
j=0
n−2
X
j +1
j +1
=
2
(n − (j + 1))
(n − j − 1)2
j=0
Product Notation
Given a sequence ai , we define the product of the sequence as
n
Y
ai = am · am+1 · . . . · an
i=m
Example
4
Y
k =1
k
1 2 3 4
1
= · · · =
k +1
2 3 4 5
5
Definition
For each positive number n, n factorial denoted n! is defined as
the product of all integers from 1 to n
n! =
n
Y
i = n · (n − 1) · . . . · 2 · 1
i=1
We also define
0! = 1
For example,
7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040
Properties of Sums and Products
Theorem
If ai and bi are two sequences and n ≥ m:
1
n
X
ak +
k =m
n
X
bk =
k =m
2
c·
n
X
n
Y
k =m
!
ak =
ak
·
n
Y
k =m
(ak + bk )
k =m
k =m
3
n
X
n
X
c · ak
k =m
!
bk
=
n
Y
k =m
(ak · bk )
Mathematical Induction
Suppose we want to prove the following:
For all positive integers n, 1 + 2 + . . . + n =
n(n+1)
2
If we introduce the predicate P(n) as:
1 + 2 + ... + n =
P(n) :
n(n + 1)
2
we want to show that
∀n ∈ Z+ , P(n) is true
Since this is a universal statement, in order to prove it, we have
to check that it is true for every n, which is physically
impossible.
In order to prove such statements about all positive integers,
we need a new method of proof.
Principle of Mathematical Induction
Let P(n) be a property that is defined for integers n, and let a be
a fixed integer. Suppose the following two statements are true:
1
P(a) is true.
2
For all integers k ≥ a, if P(k ) is true then P(k + 1) is true.
Then, the statement
For all integers n ≥ a, P(n)
is true.
One way to visualize the idea of mathematical induction is to
imagine an infinite sequence of domino tiles in a row, so that if
one of them falls backward, it will cause the tile right behind it to
fall backward as well.
Figure: Mathematical Induction and Dominoes
So, if the first domino tile falls backward, it will cause all other
dominoes to fall backward, too.
Method of Proof by Mathematical
Induction
To prove a statement of the form “For all integers n ≥ a, a
property P(n) is true”, proceed as follows:
1
Show that the property holds for n = a (basis step).
2
Suppose that the property holds for some arbitrary n = k
(k ≥ a), i.e. that P(k ) is true [This is called the inductive
hypothesis.]
3
using the inductive hypothesis, show that the property
holds for n = k + 1; i.e. that P(k + 1) is true.
Theorem
For all integers n ≥ 1,
1 + 2 + ... + n =
n(n + 1)
2
Proof. In our case,
P(n) :
1 + 2 + ... + n =
n(n + 1)
2
Show that the property is true for n = 1: This is clear, since
1=
1(1 + 1)
2
Suppose that the property is true for some arbitrary
n = k(≥ 1):
k (k + 1)
1 + 2 + ... + k =
2
Show that the property is true for n = k + 1:
1 + 2 + . . . + (k + 1) = (1 + 2 + . . . + k ) + (k + 1)
k (k + 1)
+ (k + 1)
(by ind. hyp.)
=
2
k (k + 1) + 2(k + 1)
=
2
k 2 + 3k + 2
=
2
(k + 1)(k + 2)
=
2
Since we have proved that, if P(k ) is true then so is P(k + 1),
the theorem has been proved by mathematical induction.
• A geometric sequence is a sequence in which every term
is obtained from the previous one by multiplying it by some
constant factor r ; e.g. if the first term is 1, then the
sequence is
1, r , r 2 , . . . , r n , . . .
• If r 6= 1, the sum of the first n terms is given by the formula
n
X
i=0
ri =
r n+1 − 1
.
r −1
[In Calculus II, the sum of the geometric series is obtained
from this formula by letting n → ∞, assuming |r | < 1.]
• Next, we will prove this formula using mathematical
induction.
Theorem
(Sum of a Geometric Series) For any real number r 6= 1, and
any integer n ≥ 0,
n
X
r n+1 − 1
ri =
r −1
i=0
Proof. Let P(n) be the property
n
X
i=0
ri =
r n+1 − 1
r −1
Also, suppose r is an arbitrary real number, not equal to 1.
Show that the property is true for n = 0:
0
X
i=0
ri = r0 = 1 =
r 0+1 − 1
r −1
Suppose that the property is true for some arbitrary
n = k(≥ 0):
k
X
r k +1 − 1
ri =
r −1
i=0
Show that the property is true for n = k + 1:
kX
+1
r i = r 0 + r 1 + . . . r k +1 = (r 0 + r 1 + . . . + r k ) + r k +1
i=0
r k +1 − 1
+ r k +1 ,
(ind.hyp.)
r −1
r k +1 − 1 + r k +1+1 − r k +1
=
r −1
k
+2
r
−1
=
r −1
=
So, the theorem has been proved by mathematical induction.
Proposition
For all integers n ≥ 0
32n − 1 is divisible by 8
Proof. Show that the property is true for n = 0:
30 − 1 = 0
and 0 is divisible by 8.
Suppose that the property is true for some arbitrary
n = k(≥ 0):
We assume that
32k − 1 is divisible by 8
i.e. that
32k − 1 = 8r ,
for some r ∈ Z
Show that the property is true for n = k + 1:
32(k +1) − 1 = 32k +2 − 1
= 32k · 32 − 1
= 32k · 9 − 1
= 32k · (8 + 1) − 1
= 8 · 32k + (32k − 1)
= 8 · 32k + 8r
(ind.hypothesis)
= 8(32k + r )
So, 32k − 1 is always divisible by 8.
Proposition
For all integers n ≥ 3, 2n + 1 < 2n .
Proof. Let P(n) be the property
2n + 1 < 2n
Show that the property is true for n = 3:
We need to show that
2 · 3 + 1 < 23
which is obviously true.
Suppose that the property is true for some arbitrary
n = k(≥ 3):
2k + 1 < 2k
Show that the property is true for n = k + 1:
2(k + 1) + 1 = 2k + 3
= (2k + 1) + 2
< 2k + 2
< 2k + 2k
=2·2
(by inductive hypothesis)
(since k ≥ 3)
k
= 2k +1
Example
Suppose a sequence d1 , d2 , d3 , . . . is defined by:
d1 = 2
dk −1
dk =
k
Show that, for all n ≥ 1, dn =
2
n! .
Solution: Show that the property is true for n = 1:
d1 = 2 =
2
1!
Suppose that the property is true for some arbitrary
n = k(≥ 1):
2
dk =
k!
Show that the property is true for n = k + 1:
dk +1 =
=
dk
k +1
2
k!
k +1
2
=
k !(k + 1)
2
=
(k + 1)!
( by ind. hyp.)
Strong Induction
• In 4.2 and 4.3 we use the type of mathematical induction
which is called weak induction.
• In weak induction, we prove the case of n = a, and then
attempt to show that, if the assumption is true for n = k , it
will also work for n = k + 1.
• In some proofs, in order to show that P(k + 1) is true, we
may need to assume not only that P(k ) is true, but that the
predicate is valid for all a ≤ n ≤ k .
Principle of Strong Mathematical
Induction
Let P(n) be a property that is defined for all integers n, and let
a and b be some fixed integers with a ≤ b. Suppose the
following two statements are true:
1
P(a), P(a + 1), . . . , P(b) are all true (basis step)
2
For any integer k > b, if P(i) is true for all integers
a ≤ i < k , then P(k ) is true. (inductive step)
Then, the statement
for all integers n ≥ a, P(n)
is true.
Example
Prove that any integer greater than 1 is divisible by a prime
number.
Solution: We can prove this using strong induction. Suppose
P(n) is the predicate
n is divisible by a prime number
Show that the property is true for n = 2 : Obviously true,
since n = 2 is a prime number itself.
For some k, suppose the property is true for all 2 ≤ i < k :
For every i such that 2 ≤ i < k , i is divisible by a prime number.
Show that the property is true for k :
If k itself is prime, than the statement is trivially true. So,
suppose k is a composite positive integer. In other words, there
exist integers m and n, such that
1 < m, n < k
and
k =m·n
Since 2 ≤ m, n < k , the inductive hypothesis applies to both
these numbers, so one of them, say m, has a prime divisor, call
it p.
We have proved the following in Chapter 3:
if p|m and m|k , then p|k
so k has a prime divisor, which is what we wanted to prove.
Therefore, by strong induction, every integer greater than 1 has
a prime divisor.
Example
Let an be the sequence defined by the recursive formula
a0 = 12,
a1 = 29,
an = 5an−1 − 6an−2 ,
for n > 1
Prove that
an = 5 · (3)n + 7 · (2)n ,
for all n ≥ 0
Solution: We will prove this by strong induction on n.
Show that the formula is true for n = 0, 1 :
a0 = 5 · (3)0 + 7 · (2)0 = 12 (true)
a1 = 5 · (3)1 + 7 · (2)1 = 29 (true)
For some k, suppose the formula is true for all 2 ≤ i < k :
ai = 5 · (3)i + 7 · (2)i ,
0≤i <k
Show that the property is true for k :
We know that
ak = 5ak −1 − 6ak −2
so
ak = 5(5 · (3)k −1 + 7 · (2)k −1 ) − 6(5 · (3)k −2 + 7 · (2)k −2 ) (ind.hypoth
= 25 · 3k −1 + 35 · (2)k −1 − 30 · (3)k −2 − 42 · (2)k −2
= 25 · 3k −1 + 35 · (2)k −1 − 10 · (3)k −1 − 21 · (2)k −1
= 15 · (3)k −1 + 14 · (2)k −1
= 5 · (3)k + 7 · (2)k
Theorem
(Binary Representation Theorem) Given any positive integer n,
n has a unique binary representation
n = cr · 2r + cr −1 · 2r −1 + . . . + c2 · 22 + c1 · 2 + c0
where r is some non-negative integer, and for all j, cj = 1 or 0
(cj are called binary digits in the binary representation of n)
Proof. We will prove this theorem using strong induction.
Show that the property is true for n = 1 :
Take r = 0, c0 = 1 and the binary representation is
1 = 1 · 20
Assume that for some k, the property is true for all
1 ≤ i < k : suppose that for all integers 1 ≤ i < k , i can be
written as
i = cr · 2r + cr −1 · 2r −1 + . . . + c2 · 22 + c1 · 2 + c0
for some r and some choice of the binary coefficients cj .
Show that the property is true for k :
We will split the proof into two cases, depending on whether k
is even or odd.
Case 1 (k is even): In this case, k /2 is an integer and
1 ≤ k2 < k , so we can apply the inductive hypothesis to k /2:
k
= cr · 2r + cr −1 · 2r −1 + . . . + c2 · 22 + c1 · 2 + c0
2
Multiply this equality by 2:
k = cr · 2r +1 + cr −1 · 2r + . . . + c2 · 23 + c1 · 22 + c0 · 2
which proves the claim in this case.
Case 1 (k is odd): If k is odd, then (k − 1)/2 is an integer, and
1 ≤ (k − 1)/2 < k and we can apply the inductive hypothesis to
(k − 1)/2:
k −1
= cr · 2r + cr −1 · 2r −1 + . . . + c2 · 22 + c1 · 2 + c0
2
for some integer r and cj being either 0 or 1.
Multiply this equality by 2 and then add 1 to both sides:
k = cr · 2r +1 + cr −1 · 2r + . . . + c2 · 23 + c1 · 22 + c0 · 2 + 1
which proves that the number can be written in the binary form
in this case as well.
Proof of the uniqueness of binary representation:
We will prove this by contradiction. namely, assume that, for
some positive integer n, we can find two different binary
representations.
n = 2r1 + cr1 −1 · 2r1 −1 + . . . + c1 · 2 + c0
n = 2r2 + dr2 −1 · 2r2 −1 + . . . + d1 · 2 + d0
If we equate these two binary representations and cancel all
terms which are common to both of them, we get an equality of
this type
2r +cr −1 ·2r −1 +. . .+c1 ·2+c0 = 2s +ds−1 ·2s−1 +. . .+d1 ·2+d0
r and s are distinct integers, so we may assume that e.g. r < s
(notice that this is equivalent to r + 1 ≤ s), in which case we
have:
2r + cr −1 · 2r −1 + . . . + c1 · 2 + c0 ≤ 2r + 2r −1 + . . . + 2 + 1
= 2r +1 − 1
< 2s
≤ 2s + ds−1 · 2s−1 + . . . + d0
which is a contradiction.