Chapter 4: Sequences and Mathematical Induction · Strong Induction • In 4.2 and 4.3 we use the type of mathematical induction which is called weak induction. • In weak induction,

Chapter 4: Sequences and MathematicalInduction

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 thesequence. 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 termsam+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

−12,14,−1

8, . . .

(b) a1 = 1 and ak = k · ak−1.This is an example of a recursively defined sequence.

a2 = 2 · 1 = 2a3 = 3 · a2 = 6a4 = 4 · a3 = 24

...

In fact, this sequence is

ak = k !

Series

DefinitionA series is the sum of all terms of a sequence. Given asequence ai and two positive integers m and n, such that n ≥ m

n∑i=m

ai = am + am+1 + . . . + an

If the sequence is infinite, we write

∞∑i=m

ai = am + am+1 + am+2 + . . .

[Of course, with infinite series, this sum may not exist as a realnumber; i.e. such a series may not converge.]

Examples

(a)6∑

i=2

i = 2 + 3 + 4 + 5 + 6 = 20

(b)3∑

i=3

i = 3

(c)5∑

i=1

(−1)i = −1 + 1− 1 + 1− 1 = −1

(d)4∑

i=1

(−1)i

i + 1= −1

2+

13− 1

4+

15

= −1360

Change of Variable

Let ai be a sequence and consider the two series

n∑i=1

ai = a1 + a2 + . . . + an−1 + an

n+1∑i=2

ai−1 = a1 + a2 + . . . + an−1 + an

Therefore, these two are the 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∑j=m+k

aj−k .

ExampleTransform the sum

n−1∑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∑j=0

j + 1(n − (j + 1))2 =

n−2∑j=0

j + 1(n − j − 1)2

Product Notation

Given a sequence ai , we define the product of the sequence as

n∏i=m

ai = am · am+1 · . . . · an

Example

4∏k=1

kk + 1

=12· 2

3· 3

4· 4

5=

15

DefinitionFor each positive number n, n factorial denoted n! is defined asthe product of all integers from 1 to n

n! =n∏

i=1

i = n · (n − 1) · . . . · 2 · 1

We also define0! = 1

For example,

7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040

Properties of Sums and Products

TheoremIf ai and bi are two sequences and n ≥ m:

1n∑

k=m

ak +n∑

k=m

bk =n∑

k=m

(ak + bk )

2

c ·n∑

k=m

ak =n∑

k=m

c · ak

3 (n∏

k=m

ak

)·

(n∏

k=m

bk

)=

n∏k=m

(ak · bk )

Mathematical InductionSuppose 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:

P(n) : 1 + 2 + . . . + 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 haveto check that it is true for every n, which is physicallyimpossible.

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 bea 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 toimagine an infinite sequence of domino tiles in a row, so that ifone of them falls backward, it will cause the tile right behind it tofall backward as well.

Figure: Mathematical Induction and Dominoes

So, if the first domino tile falls backward, it will cause all otherdominoes to fall backward, too.

Method of Proof by MathematicalInduction

To prove a statement of the form “For all integers n ≥ a, aproperty 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 inductivehypothesis.]

3 using the inductive hypothesis, show that the propertyholds for n = k + 1; i.e. that P(k + 1) is true.

TheoremFor 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 arbitraryn = k(≥ 1):

1 + 2 + . . . + k =k(k + 1)

2

Show that the property is true for n = k + 1:

1 + 2 + . . . + (k + 1) = (1 + 2 + . . . + k) + (k + 1)

=k(k + 1)

2+ (k + 1) (by ind. hyp.)

=k(k + 1) + 2(k + 1)

2

=k2 + 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 termis obtained from the previous one by multiplying it by someconstant factor r ; e.g. if the first term is 1, then thesequence is

1, r , r2, . . . , rn, . . .

• If r 6= 1, the sum of the first n terms is given by the formula

n∑i=0

r i =rn+1 − 1

r − 1.

[In Calculus II, the sum of the geometric series is obtainedfrom this formula by letting n →∞, assuming |r | < 1.]

• Next, we will prove this formula using mathematicalinduction.

Theorem(Sum of a Geometric Series) For any real number r 6= 1, andany integer n ≥ 0,

n∑i=0

r i =rn+1 − 1

r − 1

Proof. Let P(n) be the property

n∑i=0

r i =rn+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∑i=0

r i = r0 = 1 =r0+1 − 1

r − 1

Suppose that the property is true for some arbitraryn = k(≥ 0):

k∑i=0

r i =r k+1 − 1

r − 1

Show that the property is true for n = k + 1:

k+1∑i=0

r i = r0 + r1 + . . . r k+1 = (r0 + r1 + . . . + r k ) + r k+1

=r k+1 − 1

r − 1+ r k+1, (ind.hyp.)

=r k+1 − 1 + r k+1+1 − r k+1

r − 1

=r k+2 − 1

r − 1

So, the theorem has been proved by mathematical induction.�

PropositionFor 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 arbitraryn = k(≥ 0):We assume that

32k − 1 is divisible by 8

i.e. that32k − 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. �

PropositionFor 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 arbitraryn = 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 (by inductive hypothesis)

< 2k + 2k (since k ≥ 3)

= 2 · 2k

= 2k+1

�

ExampleSuppose a sequence d1, d2, d3, . . . is defined by:

d1 = 2

dk =dk−1

k

Show that, for all n ≥ 1, dn = 2n! .

Solution: Show that the property is true for n = 1:

d1 = 2 =21!

Suppose that the property is true for some arbitraryn = k(≥ 1):

dk =2k !

Show that the property is true for n = k + 1:

dk+1 =dk

k + 1

=2k!

k + 1( by ind. hyp.)

=2

k !(k + 1)

=2

(k + 1)!

Strong Induction

• In 4.2 and 4.3 we use the type of mathematical inductionwhich is called weak induction.

• In weak induction, we prove the case of n = a, and thenattempt to show that, if the assumption is true for n = k , itwill also work for n = k + 1.

• In some proofs, in order to show that P(k + 1) is true, wemay need to assume not only that P(k) is true, but that thepredicate is valid for all a ≤ n ≤ k .

Principle of Strong MathematicalInduction

Let P(n) be a property that is defined for all integers n, and leta and b be some fixed integers with a ≤ b. Suppose thefollowing 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.

ExampleProve that any integer greater than 1 is divisible by a primenumber.Solution: We can prove this using strong induction. SupposeP(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, thereexist integers m and n, such that

1 < m, n < k

andk = m · n

Since 2 ≤ m, n < k , the inductive hypothesis applies to boththese numbers, so one of them, say m, has a prime divisor, callit 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 hasa prime divisor.

ExampleLet 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 all1 ≤ i < k : suppose that for all integers 1 ≤ i < k , i can bewritten 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 kis even or odd.Case 1 (k is even): In this case, k/2 is an integer and1 ≤ k

2 < k , so we can apply the inductive hypothesis to k/2:

k2

= cr · 2r + cr−1 · 2r−1 + . . . + c2 · 22 + c1 · 2 + c0

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, and1 ≤ (k − 1)/2 < k and we can apply the inductive hypothesis to(k − 1)/2:

k − 12

= cr · 2r + cr−1 · 2r−1 + . . . + c2 · 22 + c1 · 2 + c0

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 formin this case as well.

Proof of the uniqueness of binary representation:We will prove this by contradiction. namely, assume that, forsome positive integer n, we can find two different binaryrepresentations.

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 allterms which are common to both of them, we get an equality ofthis 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 wehave:

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.�