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2 Chebyshev's Estimate and Some Consequences

2 Chebyshev's Estimate and Some Consequences

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148



4 The Density of Primes



Lemma 4.2.1 nk represents the number of ways of choosing k objects out of n

without replacement and without order.

Clearly the number of ways of choosing k objects out of n objects also counts the

number of possible subsets of size k in a finite set with n elements.

n

k



Corollary 4.2.1



= the number of subsets of size k in a finite set with n elements.



Lemma 4.2.2 (The Binomial Theorem) For any real numbers a, b, and natural number n we have

n

n k n−k

(a + b)n =

a b

k

k=0

Letting a = b = 1 in the Binomial Theorem we get

n

n

k=0 k



Corollary 4.2.2 (1 + 1)n = 2n =

0 ≤ k ≤ n.



. In particular



n

k



< 2n for all k with



Combining Corollaries 4.2.1 and 4.2.2 we obtain the well-known result that the

number of subsets of a set with n element is 2n . Consider a set with n elements. Then

total number of subsets =

number of subsets of size 0 + · · · + number of subsets of size n

=

Lemma 4.2.3



n

k



+



n

n

n

+

+ ··· +

=

0

1

n

n

k−1



=



n+1

k



n



k=0



n

= 2n

k



.



This last lemma is the basis of Pascal’s Triangle in which each row consists of

the set of binomial coefficients for that numbered row.

1

1

1

1



1

2



3



1

3



1

1 4 6 4 1

1 5 ... 5 1

Each subsequent row is formed by placing a one on the outside and each subsequent number is placed between 2 numbers in the previous row and is their sum. For

example,

1 3 3 1

1 4 6 4 1



4.2 Chebyshev’s Estimate and Some Consequences



149



Since

1 + 3 = 4, 3 + 3 = 6, 3 + 1 = 4.

The final standard idea we will need is that of Stirling’s approximation (see

Section 3.1.6)



n

2πn( )n .

n!

e

For Chebyshev’s estimate we need the following results which are deeper and use

number theory. π(n) in the lemma is the prime number function.

Lemma 4.2.4 (i) n π(2n)−π(n) <

≤ 22n .

(ii) 2n ≤ 2n

n



2n

n



≤ (2n)π(2n) ,



Proof If p is a prime let e p be the highest power such that p e p |n!. Then by an easy

induction we have

tp

n

[ i]

ep =

p

i=1

where [ ] is the greatest integer function and t p is the first integer such that p t p +1 > n.

Clearly such a t p exists for each prime p. Now consider

2n

(2n)!

(2n)(2n − 1) · · · (n + 1)

=

=

=

n

n!n!

n!

Given a prime p, let m p be the highest power such that p m p |

above

kp

2n

n

([ i ] − 2[ i ])

mp =

p

p

i=1



n



(

j=1

2n

n



where here k p is the first integer such that p k p +1 > 2n.

If 1 ≤ i ≤ k p then

[



2n

n

2n

n

] − 2[ i ] < i − 2( i − 1) = 2.

i

p

p

p

p



Since [ 2n

] and 2[ pni ] are integers it follows that

pi

[



2n

n

] − 2[ i ] ≤ 1

pi

p



if 1 ≤ i ≤ k p . This then implies that



n+ j

).

j



. From the observation



150



4 The Density of Primes

kp



2n

n

mp =

([ i ] − 2[ i ]) ≤

p

p

i=1

Therefore



and hence



2n

n



2n



n



1 = k p.

i=1



pk p

p≤2n



(2n) = (2n)π(2n)



pk p ≤

p≤2n



kp



p≤2n



giving one side of the first inequality.

On the other hand if n < p ≤ 2n then p|(2n)! but p does not divide n!. It follows

that

2n

2n

=⇒

p

p≤

.

n

n

n< p≤2n

n< p≤2n

Now



n = n π(2n)−π(n)



p>

n< p≤2n



n< p≤2n



since there are π(2n) − π(n) primes in the range p < n ≤ 2n. Therefore

n π(2n)−π(n) <



2n

n



establishing the other side of the first inequality.

For the second inequality we have

2n

≤ (1 + 1)2n = 22n

n

and from above

2n

=

n



n



(

j=1



Therefore

2n ≤

establishing the second inequality.



n+ j

)≥

j



n



2 = 2n .

j=1



2n

≤ 22n

n



4.2 Chebyshev’s Estimate and Some Consequences



151



We now give the proof of Chebyhsev’s estimate.

Proof (Theorem 4.2.1) We have to show that there exist positive constants A1 and

A2 such that

x

x

< π(x) < A2

A1

ln x

ln x

for all x ≥ 2.

From the previous lemma we have the inequalities

2n

≤ (2n)π(2n)

n



n π(2n)−π(n) <

2n ≤

Hence



2n

≤ 22n .

n



n π(2n)−π(n) < 22n =⇒ (π(2n) − π(n)) ln n ≤ 2n ln 2

=⇒ π(2n) − π(n) ≤



2n ln 2

.

ln n



On the other hand

(2n)π(2n) ≥ 2n =⇒ π(2n) ≥



n ln 2

.

ln(2n)



For a real variable x ≥ 2 let 2n be the greatest even integer not exceeding x, so

that x ≥ 2n, n ≥ 1 and x < 2n + 2. Then

π(x) ≥ π(2n) ≥





n ln 2

n ln 2



ln(2n)

ln x



ln 2 x

(2n + 2) ln 2

>

.

4 ln x

4 ln x



Therefore

π(x) > A1



x

ln x



for all x ≥ 2 with A1 = ln42 .

To find the existence of A2 let 2n = 2t with t ≥ 3. Then

π(2t ) − π(2t−1 ) ≤

Consider the telescoping sum



2t

2t ln 2

=

.

(t − 1) ln 2

t −1



152



4 The Density of Primes

2j



(π(2t ) − π(2t−1 )) = π(22 j ) − π(4).

t=3



Since π(4) ≤ 4 =

sum that



22

2−1



and π(2t ) − π(2t−1 ) ≤

2j



2t

=

t −1



π(22 j ) <

t=2



Now



j



2t

<

t −1



t=2



and



2j



j



2t

t−1



we obtain using the telescoping

2j



2t

2t

+

.

t − 1 t= j+1 t − 1



t=2



j



2t < 2 j+1

t=2



2j



1

2t

2t



≤ 22 j+1 .

t − 1 t= j+1 j

j

t= j+1

It follows that

π(22 j ) < 2 j+1 +

Since j < 2 j we have 2 j+1 <



22 j+1

j



and therefore for j ≥ 2



π(22 j ) < 2(

This implies that



1 2 j+1

2

.

j



22 j+1

).

j



π(22 j )

4

< for all j ≥ 2.

2

j

2

j



Let x ≥ 2 be a real variable. Then there exists an integer j ≥ 1 such that 22 j−2 <

x ≤ 22 j . Hence

π(22 j )

4π(22 j )

π(x)

≤ 2 j−2 =

.

x

2

22 j

Further

2j ≥

Implying the inequality for



π(22 j )

22 j



4

8 ln 2

ln x

=⇒



.

ln 2

j

ln x

gives



π(22 j )

4

π(x)

16

32 ln 2

<

=⇒

<



22 j

j

x

j

ln x



4.2 Chebyshev’s Estimate and Some Consequences



153



=⇒ π(x) < (32 ln 2)

for all x ≥ 2. Therefore

π(x) < A2



x

ln x



x

ln x



for all x ≥ 2 with A2 = 32 ln 2 establishing Chebyshev’s estimates.

We mention again that the proof is somewhat simpler than that originally given by

Chebyshev and arrives at weaker constants. We obtained A1 = ln42 and A2 = 32 ln 2

which were sufficient for the theorem but nowhere near best possible. Chebyshev

showed that A1 = .922 and A2 = 1.105 could be used. His proof actually involved

a careful analysis of a form of Stirling’s approximation. The values in the constants

in Chebyshev’s inequality have been improved upon many times. Sylvester in 1882

improved the values to A1 = .95695 and A2 = 1.04423 for sufficiently large x. It

can now be shown that for all x > 10, A1 = 1 can be used.

This following is an immediate corollary of the estimate, independent of the values

of A1 and A2 .

Corollary 4.2.3



π(x)

x



→ 0 as x → ∞.



Proof From Chebyshev’s estimate we have

0 < π(x) < A2

Since A2 is a constant



A2

ln x



x

π(x)

A2

=⇒ 0 <

<

.

ln x

x

ln x



→ 0 as x → ∞ so clearly



π(x)

x



→ 0 also.



This corollary says that the primes become relatively scarcer as x gets larger. In

probabilistic terms it says that the probability of randomly choosing a prime less than

or equal to x goes to zero as x goes to infinity.

Before continuing and presenting some consequences of Chebyshev’s result we

introduce a convenient notation for describing the order of magnitude of a function.

Definition 4.2.2 Suppose f (x), g(x) are positive real-valued functions. Then

(1) f (x) = O(g(x)) (read f (x) is big O of g(x)) if there exists a constant A independent of x and an x0 such that

f (x) ≤ Ag(x) for all x ≥ x0 .

(2) f (x) = o(g(x)) (read f (x) is little o of g(x)) if

f (x)

→ 0 as x → ∞.

g(x)

In other words g(x) is of a higher order of magnitude than f (x).



154



4 The Density of Primes



(3) If f (x) = O(g(x)) and g(x) = O( f (x)), that is, there exist constants A1 , A2

independent of x and an x0 such that

A1 g(x) ≤ f (x) ≤ A2 g(x) for all x ≥ x0 ,

then we say that f (x) and g(x) are of the same order of magnitude and write

f (x)

(4) If



g(x).



f (x)

→ 1 as x → ∞

g(x)



then we say that f (x) and g(x) are asymptotically equal and we write

f (x) ∼ g(x).

In general, we write O(g) or o(g) to signify an unspecified function f such

that f = O(g) or f = o(g). Hence, for example, writing f = g + o(x) means that

f −g

→ 0 and saying that f is o(1) means that f (x) → 0 as x → ∞.

x

It is clear that being o(g) implies being O(g) but not necessarily the other way

around. Further it is easy to see that

f ∼ g is equivalent to f = g + o(g) = g(1 + o(1)).

In terms of the notation above Chebyshev’s estimate can be expressed as

π(x)



x

.

ln x



Further the prime number theorem can be expressed by

π(x) ∼

or equivalently

π(x) =



x

ln x



x

(1 + o(1)).

ln x



We will use this notation freely as we develop the proof of the prime number theorem.

We now present some consequences of Chebyshev’s estimate. It was mentioned

at the end of the previous section that the prime number theorem is equivalent to

pn ∼ n ln n, where pn denotes the nth prime (Theorem 4.1.2). Chebyshev’s estimate

gives immediately that pn and n ln n are of the same order of magnitude.



4.2 Chebyshev’s Estimate and Some Consequences



155



Theorem 4.2.2 There exist positive constants B1 , B2 such that

B1 n ln n ≤ pn ≤ B2 n ln n.

Equivalently

pn



n ln n.



Proof Let pn be the nth prime. Then clearly π( pn ) = n. From Chebyshev’s estimate

n = π( pn ) ≤ A2



pn

for all n ≥ 2.

ln pn



This implies

1

n ln pn ≤ pn for all n ≥ 2.

A2

However, pn > n so

1

1

n ln n <

n ln pn ≤ pn for all n ≥ 2.

A2

A2

Therefore In general we write

B1 n ln n ≤ pn

for all n ≥ 2 with B1 = A12 .

In the other direction, we have

n = π( pn ) ≥ A1

Since pn > n it follows that

k such that



Hence

n



ln pn



pn



pn

.

ln pn



→ 0 as n → ∞. Therefore, there exists a constant



ln pn

< A1 if n > k.



pn



ln pn

ln pn

≥ A1 > √ if n > k.

pn

pn







pn and so ln pn < 2 ln n if n > k. Let

p3

pk−1

2

p2

,

,...,

}.

B2 = max{ ,

A1 2 ln 2 3 ln 3

(k − 1) ln(k − 1)



It follows that n >



Then

pn ≤ B2 n ln n for all n ≥ 2.



156



4 The Density of Primes



Note that we could have proved Theorem 4.2.2 and then deduced Chebyshev’s

estimate from it. This result also provides a very simple proof of Euler’s Theorem

given in Chapter 3 that the series p 1p diverges.

Corollary 4.2.4 The sum

p,prime



1

p



diverges.

Proof For n ≥ 2 we have p1n ≤ B1 n1ln n from the last theorem. However the series



1

n=1 n ln n diverges by the integral test.

Although there are infinitely many primes and p 1p diverges it still diverges

very slowly. Using the methods applied in the proof of Chebyshev’s estimate we can

actually bound the growth of the series of reciprocals of the primes.

Theorem 4.2.3 There exists a constant k such that



2< p≤x



1

< k ln ln x if x > 3.

p



Proof From Theorem 4.2.2 we have

pn ≥ B1 n ln n.

Therefore



2< p≤x



1

=

p



π(x)

n=2



1

<

pn



π(x)

n=2



1

1

<

B1 n ln n

B1



[x]

n=2



1

.

n ln n



However

1

=

n ln n

since



1

n ln n







1

t ln t



2< p≤x







n

n−1



dt



n ln n



n

n−1



dt

t ln t



on [n − 1, n] if n ≥ 3. Then

1

1

<

p

B1



1

1

+

2B1 ln 2

B1



[x]

n=2

x

2



1

1

1



+

n ln n

2B1 ln 2

B1



[x]



n



n=3



n−1



dt

t ln t



1

1

dt

1

=

+

ln ln x −

ln ln 2

t ln t

2B1 ln 2

B1

B1



4.2 Chebyshev’s Estimate and Some Consequences



=



157



1

ln ln x + C < k ln ln x

B1



taking k large enough.

In a similar vein we get the following result which bounds the product of all the

primes p less than some given x.

Theorem 4.2.4 If x ≥ 2 then



p≤x



p < 4x .



Proof The theorem is clear for 2 ≤ x < 3. Suppose the theorem is true for an odd

integer n with n ≥ 3. Then it is true for n ≤ x < n + 2 since

p=

p≤x



p < 4n < 4 x .

p≤n



Therefore it is sufficient to prove the theorem for odd integers n. We do an induction

on the odd integers. The theorem is true for n = 3 and so we assume that it is true for

or k = n−1

chosen so that

all odd integers less than or equal to n ≥ 5. Let k = n+1

2

2

k is also odd. Then k ≥ 3 and n − k is even. Further n − k = 2k ± 1 − k ≤ k + 1.

If p is a prime with k < p ≤ n then p|n! but p does not divide either k! or (n − k)!.

n!

. It follows that the product of all such primes divides nk

Therefore p| nk = k!(n−k)!

and hence

n

p≤

.

k

k< p≤n

n

Since nk = n−k

and both are in the binomial expansion of (1 + 1)n it follows that

n

< 2n−1 . Therefore using that k < n and the inductive hypothesis

k



p=

p≤n



p < 4k 2n−1 = 2n+2k−1 ≤ 22n = 4n .



p

p≤k



k< p≤n



Finally based on many of these estimates we can provide a proof of Bertrand’s

Theorem (actually proved by Chebyshev) which we introduced in the last chapter.

Recall that this theorem says that given any natural number n there is always a prime

between n and 2n. The proof actually shows that given any real number x > 1 there

exists a prime between x and 2x.

Theorem 4.2.5 (Bertrand’s Theorem) For every natural number n > 1 there is a

prime p such that n < p < 2n.

Proof By direct computation the theorem is easily established for n ≤ 128. Now

suppose that for some n > 128 there is no prime between n and 2n. For a prime p let

and k p the first power such that p k p +1 > 2n

m p be the highest power of p dividing 2n

n



158



4 The Density of Primes



as in the proof of Chebyshev’s Estimate. Then as in the proof of Chebyshev’s estimate,

since we assume no primes in the range n to 2n we have

2n

=

n



pm p =



pm p , m p ≤ k p .

p≤n



p≤2n



Here we use [x] to indicate the greatest integer function, that is, the greatest integer

less than or equal to x.

Now if 2n

< p ≤ n we then have p ≥ 3 and 2 ≤ 2np < 3 and therefore

3

mp = [



2n

n

] − 2[ ] = 2 − 2 = 0.

p

p





then we have p 2 > 2n and hence k p = 1 and so m p ≤ 1.

If 2n < p ≤ 2n

√3

Finally if p ≤ 2n we have p m p ≤ p k p ≤ 2n. Therefore

2n

=

n





p≤ 2n







pm p





p≤ 2n







pm p



(2n)







pm p

2n

3 < p≤n



2n< p≤ 2n

3



p.

2n< p≤ 2n

3



since there are at most x+1

odd

For a real number x ≥ 128 we have π(x) ≤ x+1

2

2

integers less than x so certainly no more than that primes. Further since x ≥ 128 we

− 2 < x2 − 1. It follows

have at least two odd non primes less than x so π(x) ≤ x+1

2



that π( 2n) < n2 − 1 and hence



p≤ 2n



p < (2n)



√n



2 −1



.



Further from Theorem 4.2.4 we have

2n



p<43.

p≤ 2n

3



Therefore



√n

2n

2n

< (2n) 2 −1 4 3 .

n



Now

22n = (1 + 1)2n = 1 +



2n

2n

2n

+ ··· +

+ ··· +

+ 1.

1

n

2n − 1



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