CHAPTER final.pdf

1
CHAPTER 1
INTRODUCTION
Prime numbers have been the focus of mathematicians for centuries. When Euclid’s
Elements were first published circa 300 B.C.E., there were already several important results of
prime numbers that had been discovered. In his ninth book, Euclid proved that there are infinitely
many prime numbers. In the same book, Euclid also included a proof of the fundamental theorem of
arithmetic, which allows us to break a number into a distinct product of its prime factors: an
immensely useful tool when working in the area of number theory. For some time after that, the
main focuses of prime number theory were developing methods of determining whether a number
was prime, and how to find exceedingly larger prime numbers. During this time, the Greek
mathematician Eratosthenes developed one of the most effective methods for determining larger
primes up to a given limit. However, this method is only efficient for relatively small primes, no
higher than 10 million. After this, there was very little development in prime number theory for over
a millennium.
Around the beginning of the seventeenth century, Fermat, Mersenne, and Euler made
some significant contributions to the field of prime number theory, which included Fermat’s Little
Theorem as well as the idea of Mersenne Numbers. [3] A Mersenne number is one of the form
2 − 1, where if is prime, then the number 2 − 1 could potentially be prime. If was not prime, it
was proved that 2 − 1 cannot be prime. Throughout the century, Mersenne numbers that were
determined to be prime were the largest prime numbers that could be found. For the next few
centuries, Cataldi, Euler, and Lucas proved the existence of the Mersenne numbers with =
19,31,127, respectively, with centuries between each discovery. To date, 45 Mersenne primes have
been discovered, the largest of which with = 43112609. The study of primes naturally led to the
study of their distribution, which will be the main focus of this project.
During the late 18th century, both Gauss and Legendre made similar conjectures
regarding ( ), the number of prime numbers less than or equal to . In particular, impressive

2
hatattheageof15or16,Gauss conjectured a similar o[2]in l798 Legendre estimated
that ( )cou1dbeestimated to ,
.Legendre It is also which was that ( ) ≈
actually show that ( ) .This resultwasobtained byTchebychev
inl852,using ∫ 2
∼ ,
. Nowwefocus ourattention tothefirstpersonto completely elementary
methods in number theory. The work done by Gauss, Legendre, and especially Tchebychev laid the
ground work for solving what is now known as the Prime Number Theorem. The Prime Number
Theorem states that as → ∞, then
( ) = , which was proved later by Jacques Hadamard and Charles ‐Jean ‐
Gustave - Nicholas de la Vallée Poussin in 1896 using methods from complex
analysis. [4] As a result of the Prime Number Theorem, it has been proven that given
any > 1, the probability that a randomly selected number is prime, with ≤ , is .

3
CHAPTER-2
1. PRELIMINARIES
For our study of Tchebychev’s theorem, we will require some basic knowledge of number theory.
This chapter will include many important results that will, in effect, enable us to work through the
proof of Tchebychev’s theorem later on. It is interesting to note that we can work through a
relatively complicated and convoluted proof, with reasonably elementary methods. We begin by
formally defining what a prime number is:
DEFINITION . . A number is prime if and only if its only divisors are 1 and itself.
EXAMPLE . . Given the number = 10, we can clearly break it up into it’s prime
factors so it will have the form 10 = 2 5. Because 10 is divisible by 1, 2, 5, and 10, we
say that 10 is not prime.
EXAMPLE . . Given the number = 23, we can see that 23 cannot be factored
Further than what it already is, thus only 1 and 23 divide 23. Because only 1 and 23
divide 23, we say that 23 is prime
DEFINITION . . For ∈ ℤ we say that is the integer part of a number if
≤ < + 1,
and we denote it by = [ ].
EXAMPLE . . We will look at several different cases: [0.2] = 0, [0.9] = 0, [3.8] = 3,
[−3.2] = −4, [−5.9] = −6, [2] = 2, and lastly [−7] = −7.
COROLLARY . . Let ∈ ℝ Then − 1 < [ ] ≤ for all .
PROOF.
By the definition, we know that if ≤ < + 1, then = [ ]. Setting = [ ]
in the inequality, we obtain = [ ] ≤ < + 1. For the other side of the inequality, we
simply subtract 1 from each side of the inequality. This gives us − 1 ≤ − 1 < , and
setting = [ ] again, we find that − 1 < [ ]. Combining our two inequalities, we

4
obtain − 1 < [ ] ≤ , our desired result.
LEMMA . . Given any real number , then [ − ] = [ ] − .
PROOF.
Clearly, [ ] − ≤ − because in general [ ] ≤ .
We also know that − 1 < [ ] which implies that < [ ] + 1.
Using this, we find that − < [ ] + 1 − .
Bringing these facts together, we obtain [ ] − ≤ − < [ ] − + 1.
This is the definition of the integer part of − , thus [ ] − = [ − ], and
therefore we obtain the result.
DEFINITION . .
We define an ‐combination as the number of ways to choose a particular
number of elements from a set of size , with ∈ ℤ, ≥ 0, and 0 ≤ ≤ . In choosing
our elements, the order of the selected elements does not matter. We say choose ,
and denote it by
( , ) = =
!
! ( − )
.
EXAMPLE . .
Suppose we are asked to find the number of possible ways to choose 6
blocks from a group of 10 distinct blocks, regardless of their order. Then the number of
ways to do this is equal to
(10,6) =
10
6
=
10!
6!(10 − 6)
=
10!
6! ⋅ 4
= 210.
Next we will introduce the Binomial Theorem, which helps us begin our proof of
Tchebychev’s theorem. The proof of the Binomial Theorem is irrelevant to this paper,
and is therefore omitted.

5
THEOREM . .
(Binomial Theorem) Let and be variables and let be a nonneg‐
active integer. Then
∑ = ( + )
DEFINITION . .
We define the factorial function as : ℕ → ℤ , denoted by ( ) = !. The value of
( ) = ! is the product of the first positive integers, so ( ) = 1 ⋅ 2 ⋅ 3 ⋅ … ( − 1) ⋅ ,
and (0) = 0! = 1.
EXAMPLE . .
Take 5!. By the definition, 5! = 5 4 3 2 1 = 120. Similarly, 13! = 13 ⋅ 12 ⋅ 11
10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 6,227,020,800.
THEOREM . .
(Fundamental Theorem of Arithmetic) Every positive integer greater than 1 can
be written uniquely as a prime or as the product of two or more primes where the
prime factors are written in order of node creasing size.
EXAMPLE . .
Take for instance = 21. Then by the Fundamental Theorem of Arithmetic,
21 = 3 ⋅ 7. Similarly, for = 420, we see that 420 = 2 ⋅ 3 ⋅ 5 ⋅ 7.

6
CHAPTER 3
Tchebychev’s Theorem
We begin this section with some necessary definitions and results that will prove to be
useful in the proof of Tchebychev’s Theorem. Our first definition plays a central role throughout the
duration of this paper.
DEFINITION . .
Given any number > 1, we define the number of prime numbers less than or equal
to as ( ).
LEMMA . .
Let , ∈ ℤ. Then ( + ) < ( ) + .
PROOF
We know that the number of prime numbers less than or equal to is ( ).
Similarly, the number of prime numbers less than or equal to + is ( + ) . If we take the
difference between the number of prime numbers less than + , and the number of prime
numbers less than , this difference is clearly less than or equal to . We can see this with the
following inequality:
( + ) − ( ) ≤ .
Rearranging, we obtain ( + ) ≤ ( ) + , the required result.
LEMMA . . Let be any real number. Then
(0.1) 2 [ ] =
[2 ] if − [ ] <
[2 ] − 1 if − [ ] ≥
PROOF.
To prove this, we will have to consider two cases, the first of which being if − [ ] < .

7
Rearranging, we obtain < [ ] + . From here, clearly2 < 2[ ] + 1, and since in general[ ] ≤ ,
then 2 [ ] ≤ 2 < 2[ ] + 1. This inequality implies that 2[ ]is the integer part of2 , and hence
2[ ] = [2 ]. For the second case, we have that − [ ] ≥ . Rearranging, we obtain[ ] ≤ − .
From here, we can see that 2 [ ] ≤ 2 − 1, where 2 − 1 is also equal to 2 − 2 + 1. Factoring,
we obtain2( − 1) + 1, and since in general we have that − 1 < [ ], then clearly 2 − 1 =
2( − 1) + 1 < 2[ ] + 1. Simplifying the inequality, we find that 2 [ ] ≤ 2 − 1 < 2[ ] + 1. This
by definition implies that 2[ ]is the integer part of 2 − 1, and hence 2 [ ] = [2 − 1]. Using
Lemma (1.7), we have that 2 [ ] = [2 − 1] = [2 ] − 1, which proves the second case.
LEMMA . . Given any positive integer ,
+
2
+ ⋯ +
2
=
−
2
1 −
1
2
.
Obtain = + ⋯ + .Now,ubtra ting fro , we indthat −
′
= + … −
+ + ⋯ + . Wcan … eta11themidd1eterms
othep . . ′
right hand side cancel, and we are left
with − = − . Factoring out , we have that 1 − = − , which after further
rearrangement gives us = . Now, since = + + ⋯ + , we obtain + + ⋯ +
= , the result we are looking for.
With the above information, we now have the necessary tools to prove Tchebychev’s
Theorem.
We will now formally introduce this very important Theorem, and its proof will immediately
follow.

8
THEOREM . .
(Tchebychev’s Theorem) Let be any positive integer greater than or equal to 1.
Then ( )has the following upper and lower bounds:
0.1
log
< ( ) < 4
log
.
PROOF. Given two numbers and , the binomial theorem states that
∑ = ( + )
Substituting2 for , and setting both and equal to 1, then we have
∑ 2
(1) (1) = 1 +
2
2
+
2
3
+ ⋯ +
2
+ ⋯ +
2
−12
+1 = (1 + 1) = 2 It
is clear that
2
> 0 for any , thus by rearranging, we can see that
2
< 2
We now claim that > 2 for all < . Indeed, multiplying both sides by ( − ) , which is
positive when < , we obtain
( )( )
( )
> 2 ⋅ ( − ) .
Simplifying on the left, and expanding on the right, we obtain 2 − > 2 − 2 . After further
simplification, we reach our result as it is clear that 2 > .
Next, we expand the combination
2
.
2
=
(2 )!
( !)
=
2 ⋅ (2 − 1)(2 − 2)(2 − 3) … 2 − ( − 1) (2 − )!
( !)
=
2 ⋅ (2 − 1)(2 − 2)(2 − 3) … ( + 1)( !)
( !)

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=
2 ⋅ (2 − 1)(2 − 2)(2 − 3) … ( + 1)
!
=
2 2 − 1
− 1
2 − 2
− 2
…
+ 1
1
Times
= 2
From above, we obtain the following expression
4) 2 ≤
2
. Now, from (0.3) and (0.4), we can see that
2 ≤
2
< 2
We will now look at the prime factorization of
2
. First, we will look at the
prime Factorization of !for any positive integer . Letting be aprime number less than
or equal to , we can write = + , where has order , ∈ ℕ, and ℓ < Clearly,
since ℓ < , then = + ℓ < + = ( + 1) . From here, we obtain the
following inequality
=
− ℓ
≤ < + 1,
Therefore, by definition, is the integer part of , thus =
Now, expanding !we obtain the following expression,
! = 1 ⋅ 2 ⋅ … ⋅ ⋅ … ⋅ 2 ⋅ … ⋅ ⋅ … ⋅ 2 ⋅ … ⋅ ⋅ … ⋅ 2 ⋅ … ,
Where is any prime less than or equal to . Therefore we obtain = multiples of
which are less than , = multiples of which are less than , and in general
= Multiples of which are less than . This allows us to count all the multiples
of of order 1, as well as all the multiples of with order 2, and so on, up to all
multiples of of order . For this reason, the power of in the prime factorization of !is

10
= + + ⋯ + , where is the largest integer such that < If we now apply
this knowledge to the expansion of
2
=
( )!
( !)
, we find that it will contain the
following,
⋯
⋯
⋯
=
⋯
=
2
+
2
+ ⋯ +
2
− 2
p
+ + ⋯ +
=
⋯
,
with aprime less than or equal to2 and the largest integer such that ≤ 2 . Next,
using Lemma2.3, by setting = , we obtain
2
2
=
[ ]
[ ] − 1
if − < if − ≥
Now, we will examine the expression [ ] − 2 . From (0.8), if 2 = [ ], then
[ ] − 2 = [ ] − [ ] = 0. On the other hand, if 2 = [ ] − 1 then[ ] − 2 = [ ] −
([ ] − 1) = 1. From this, it is clear that
7) [ ] − 2 ≤ 1 for all .
Returning to the following expression,
( 2
− 2 ) +
2
− 2 + ⋯+
2
− 2 ,

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If we set = from (0.7), then = . This can therefore be used to show that
2
− 2 +
2
− 2 + ⋯ +
2
− 2 ≤ 1 + 1 + ⋯ + 1 =
Now, if we denote by the prime numbers in the prime factorization of
2
, then
We can see that each will have a power less than or equal to , where ≤ 2 . This
is shown below:
Since there are many primes less than or equal to2 , we can set = (2 ) . This
Means that
2
≤ 2 ( )
.
Now, we will take a look back at the following expression:
2
=
2 ⋅ (2 − 1)( + 1)
⋅ ( − 1)1
.
We will take all the primes that divide
2
, and put them into two groups. First of all,
we denote the primes that divide
2
by , , …, , , …, . We will make our first
group of primes all the primes less than or equal to , and they will be the primes
, , …, . Our second group of primes will consist of all the primes greater than but
less than or equal to2 , and they will be the remaining primes; , , …, . Clearly,
the product of all of the primes in our second group divides
2
=
( )!
( !)
since they
appear in the numerator only. Thus:
2
≥ ⋅ > ⋅ ⋅ … ⋅ =
Now since up to consist of many primes, and to consist of many primes,
then it is clear that the amount of primes from to is exactly equal to − . We
may also note that the primes up to are all less than or equal to as defined above,
which by definition means that there are exactly ( )of them. By setting = ( ) , we

12
obtain the following expression:
<
2
≤ 2 ( )
which implies ( ) ( )
<
2
≤ 2 ( )
.
By comparing (0.5) with the above result, we see that 2 ≤
2
≤ 2 ( )
which
implies 2 ≤ 2 ( )
. Taking the logarithm of both sides, we find that
log (2 ) ≤ log 2 ( )
⋅ log (2) ≤ (2 ) ⋅ log (2 )
⋅ log (2)
log (2 )
≤ (2 )
2
log (2 )
log (2)
2
≤ (2 )
( )
(0.1505149 …) ≤ (2 ) .
Therefore for any even number = 2 , we have shown the first half of the desired
inequality. For odd, we can use a similar argument to satisfy the inequality. We first
claim that 2 ≥ (2 + 1) for ≥ 1. Clearly, 6 ≥ 2(2 + 1) = 4 + 2. Collecting like
terms we find that 2 ≥ 2, which is equivalent to ≥ 1. Now, using this fact, we take
any odd = 2 + 1. We know from above that ⋅ log (2) ≤ (2 ) ⋅ log (2 ) . We can use
this to show the following:
⋅ log (2) ≤ (2 ) ⋅ log (2 ) ≤ (2 + 1) ⋅ log (2 + 1)
2
3
(2 + 1)
log (2)
2
≤ 2 ⋅
log (2)
2
≤ (2 + 1) ⋅ log (2 + 1)
(2 + 1)
log (2 + 1)
2
3
log (2)
2
≤ (2 + 1)
( )
( )
(0.1003433 …) ≤ (2 + 1) .
Since for any even we have ⋅ (0.1505149 … ) ≤ ( ) , and for any odd we have
(0.1003433 …) ≤ ( ) , then we can conclude that for all > 1,
(0.1003433 …) ≤ ( ) .

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We now begin the proof of the second half of the desired inequality. From (0.5) and
we find that ( ) ( )
< 2 Taking the logarithm of both sides, we find that
log ( ) ( )
< (2 )
(2 ) − ( ) ⋅ log < 2 ⋅ log 2
(2 ) − ( ) < 2 ⋅ log 2 ⋅
log
= (0.60206)
log
.
We now suppose that > 1, for ∈ ℝ. Setting = and letting = from
our previous result ( we find that either [ ] = 2 or [ ] = 2 + 1. For the first if2 =
[ ] − 1,then with = [
[
, we indthat2 = [ ] − case’′if 2 = [ ],then sincen=
]
], ′wehavethat[ ] = 2 lwhichimplies thatFor thesecond case[ ] = 2 + 1.
We now observe two results. The first result is that ( ) = ([ ]) = (2 + 1) ≤
(2 ) + 1, using Lemma (2.2). This is true because[ ]is either 2 + 1 or2 . The second
result is that = = ( ) , as = . From these two results, we obtain the
following expression:
( ) −
2
≤ (2 ) + 1 − ( ) = (2 ) − ( ) + 1
< 2 ⋅ log 2 ⋅
log
+ 1
< 2 ⋅ log 2 ⋅
log
+
log
= (2 ⋅ log (2) + 1)
log
= (1.60206)
log
.
We now claim that < for ≥ 3 and < . We begin by looking at the

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root of the first few integers:
√2 = 1.41 √3 = 1.44 √4 = 1.41 √5 = 1.37 …
We can see that for ≥ 3, √ . Taking the logarithm of both sides, we find
weobin
(
<
)
. Nowthat ,
⋅ 1 +< ⋅ 10since = then for ≥ 3,
implies that for > , < . Rearranging,
( ) −
2
< (2 ⋅ log (2) + 1)
log
.
We can also show that the above inequality holds for < 3. First, if < 3, this is
equivalent to saying < 6. But we know for any < 10, then log < 1, so clearly
(
) > 1.Using whatwehavejust − . C1ear1y, forall > , shown, we now look at
the same expression as before
( ) −
2
< ( ) < <
log
< (2 ⋅ log (2) + 1)
log
.
We now look at the following inequality:
( ) log ( ) −
2
log
2
= ( ) log ( ) −
2
log ( ) +
2
log ( ) −
2
log
2
= ( ) −
2
⋅ log ( ) +
2
log ( ) − log
2
< (2 ⋅ log (2) + 1)
log ( )
⋅ log ( ) +
2
log ( ) − log ( ) − log (2)
< (2 ⋅ log (2) + 1) ⋅ +
2
[ log (2)]
= 2 ⋅ log (2) + 1 +
log (2)
2
⋅
= (1.75257 …). .
Now, taking to be any arbitrary positive integer, from above we obtain the following
set of inequalities:

15
( ) log ( ) − log < (1.75257 …). ,
2
log
2
−
4
log
4
< (1.75257… )
2
,
4
log
4
−
8
log
8
< (1.75257… )
4
,
2
log
2
−
2
log
2
< (1.75257 …)
2
.
We now choose so that 2 > . Adding all the inequalities from the previous step, we
obtain the following:
( ) log ( ) −
2
log
2
+
2
log
2
−
4
log
4
+ ⋯ +
2
log
2
−
2
log
2
+
2
log
2
−
2
log
2
< (1.75257 …). + (1.75257 …) + ⋯ + (1.75257… )
= (1.75257 …) + + ⋯ +
The above expression can simplify to
( ) log ( ) − log < (1.75257…) + + ⋯ +
Now, using Lemma2.4, we find that
( ) log ( ) − log < (1.75257…)
= (3.50514) −
< (3.50514). < 4 ,
using the fact that 0 < < 1. Setting = 0 due to the fact that < 1, we can
therefore conclude that
( ) log ( ) −
2
log
2
< 4
( ) log ( ) − 0 ⋅ log
2
< 4

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( ) log ( ) < 4
( ) < 4 ⋅
log
.
Comparing the above inequality with (0.9), we find that
(0.1003433…) ≤ ( ) < 4 ⋅ ,
Thus completing the proof.
We can also complete this proof using the natural logarithm, as opposed to the loga‐
rithm of base 10. The proof remains the exact same, except for when computing values
in certain parts of the proof. Instead of computing values using log base 10, we can use
log base , which is equivalent toln. Computing these values, we find that (0.9) can be
changed to
(0.2310406) ≤ ( ) ,
and the other side of the inequality can be changed to
( ) < 6 ⋅
ln
This implies that in terms of the natural logarithm, ( )can be approximated in the
following way:
(0.23) ≤ ( ) < 6 ⋅ .

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CHAPTER 4
Mertens’ First Theorem
In this section, we use what we have proved in the previous chapter to examine an
application of the distribution of prime numbers. We begin by stating some useful tools
which will be helpful in the proof Mertens’ First Theorem.
LEMMA . . Let be an integer and a prime number. Then
̇
+
̇
+
̇
+ ⋯ + =
−
1 −
1
.
PROOF. We begin by setting = + + + … + . Multiplying by we
obtain = + + ⋯ + . Subtracting from we obtain
−
̇
=
̇
+
̇
+
̇
+ ⋯ + −
̇
+
̇
+ ⋯ +
=
̇
+
̇
+
̇
+ ⋯ + −
̇
−
̇
− ⋯ − −
=
̇
−
?
.
Factoring out from the left side, we obtain 1 − = − . Dividing both sides
by 1 − , and substituting + + + ⋯ + for , we obtain + + + ⋯ +
= , the result we are looking for.
LEMMA . . Take ∈ ℕ, and let > 1 for ∈ ℝ. Then
1 +
1
2
+
1
3
+
1
4
+ ⋯ +
1
tends to a limit lying between and .
LEMMA . . Let be any positive integer. Then !satisfies the following inequality:

18
4
5
⋅ ⋅ ⋅ − < ! < ⋅ ⋅ −
LEMMA3.4. Let ∈ ℝ. Then
2 log < √ ,
for > 0.
PROOF. To show this, we rearrange our expression to look like √ − 2 log > 0.
Treating this expression like a function, it is sufficient to show that this function is
always positive for > 0. Taking the derivative of the function ( ) = √ − 2 log ,
we obtain
( ) =
1
2√
−
2 log
=
− 4√ (0.43429)
2 √
=
√ − 1.73716
2
.
Setting our derivative equal to0, we find that
0 =
√ − 1.73716
2
⇒ √ = 1.73716 ⇒ = 3.01772,
therefore we must have a maximum or minimum at a = 3.01772, and an asymptote at
= 0. Substituting values that are in neighborhood of a = 3.01772 into our derivative,
we find that (3)is negative, and (3.1)is positive, meaning that our function
decreases to = 3.01772 and then increases indefinitely. This is illustrated by
substituting values into our original function:
(3) = 0.777808 (3.01772) = 0.777801 (3.1) = 0.77795
Because (3.01772)is our local minimum, and it is positive, we can therefore conclude
that for > 0, ( ) = √ − 2 log is positive. This implies that √ − 2 log > 0 and
therefore 2 log < √ .

19
THEOREM . .
(Mertens’ First Theorem) Let 2, 3, 5, 7, 11, 13,…, be the primes not
exceeding a given integer N. Then there exists a constant , that can be taken equal to
4, such that
|
log 2
2
+
log 3
3
+
log 5
5
+ ⋯ +
log
− log | < .
PROOF.
We begin by decomposing a given integer !into its prime factors. As we
have seen in the proof of Tchebychev’s Theorem, !has the following form:
(0.10)
! = ⋅ … ,
with , , …, the primes less than or equal to , and
= −
̇
̇ ̇
] + ⋯ + −
where ? ≤ . Taking logarithms of both sides of (0.10), we obtain
log ! = log ⋅
= log + log + ⋯ + log
= log + log + ⋯ + log .
Now, we can estimate log !in two ways. By the definition of , we can see that the
right hand side of the above equation has the form
+ + + ⋯ + log
+( − − − … + − log
+ ⋯ + ( − − − … + − log .

20
We now wish to simplify the above expression by changing
−
̇
−
̇
̇
] + ⋯ + − to
̇
+
?
+
̇
+ ⋯ + ,
Where remains less than for 1 ≤ ≤ , and with ≤ . By removing the integer
part notation, we introduce an error of no more than 1 per term in the estimation of
log !. Indeed, by definition
− ≤ − 1,
thus, for each sum of (0.11), (0.12), and (0.13), we have an error of , , …,
respectively, as there are exactly many terms in the sum of
−
̇
−
̇ ̇
] + ⋯ + −
We have now found a total error on our estimation of log !, namely
⋅ log + ⋅ log + ⋅ log + ⋯ + ⋅ log
= log + log + log + ⋯ + log
times
In the above expression, we can see that is equal to the number of primes less than or
equal to , thus by Tchebychev’s Theorem,
= ( ) < ′’
where can be taken equal to 4. We can now see that our estimate of log !is bounded
by
log <
log
⋅ log = .
This allows us to construct an upper and lower bound for log !:
− + + + ⋯ +
+ + + ⋯ + ) log + ⋯ . . . + + + + ⋯ +

21
log ≥ log ! >
+ + + … + log + ⋯ +
− + + + ⋯ +
+ + + ⋯ +
) log − .
We can now modify Lemma3.1in the following way:
̇
+
̇
+ + ⋯ + =
−
1 −
<
N
1 −
=
− 1
′′
and we also note that
̇
≤
̇
+
̇
+
̇
+ ⋯ + .
Using these two facts, we find that
(∗) log + log + ⋯ + log ≥ log !
> log + log + ⋯ + log − .
We wish to simplify the above inequality even further, so as it has the following form:
(∗∗) + + ⋯ + + > !
> + + ⋯+ − ,
where is a positive constant. To do this, we must find the difference between the left
and right hand sides of (∗) and (∗∗) . We can begin with the obvious case, which is the
right side. Clearly, has simply been factored from all the terms on the right hand side
of (∗) , thus the right hand sides of (∗) and (∗∗) are equivalent. For the former case, we
first find how much the left hand side of (∗) differs from
log
+
log
+ ⋯ +
log
This difference will ultimately be our value. After factoring out from (∗) , we find
that the difference for each term is equal to
log
− 1
−
log
=
log − ( − 1) log
( − 1)
=
log
( ? ⋅ −1)
,

22
For = 1,2,3, …, , thus the total difference between the left hand side of (∗) and (0.14)
is
log
( − 1)
+
log
( − 1)
+ ⋯ +
log
( − 1)
We now claim that for any ≥ 2
log
( − 1)
<
1
√
.
It is clear that the above inequality implies log <
√
, which is equivalent to 2
log < 2 √
. Working with the right hand side of the inequality, we find that for
≥ 2,
2
− 1
√
= 2
√ ( − 1)
= 2 √ −
√
≥ 2 √ −
√
2
= √ .
Now, using Lemma3.4, since
2 log < √ and 2
√
≥ √ a,
we can therefore conclude that 2 log < 2
√
.
We can now find an upper bound for the sum inside the brackets of (0.15):
log
( − 1)
+
log
( − 1)
+ ⋯ +
log
( − 1)
<
1
+
1
+ ⋯ +
1
< 1 +
1
2√2
+
1
3√3
+
1
4√4
+ ⋯ +
1
√
.
Using Lemma4.2, taking = , we find that the last sum of our previous inequality is
bounded:
1 +
1
2√2
+
1
3√3
+
1
4√4
+ ⋯+
1
√
<
3
2 − 1
3
2
= 3.
We can therefore conclude that the difference between the left hand side of (∗) and

23
(0.14), namely , can be taken equal to 3.
We now find another estimate for log !using Lemma3.3. Recall that
√ . − < ! < √ . −
Where = and = . Taking logarithms of each part of the inequality, we obtain
log ⋅ ⋅ < ! < ⋅ ⋅ −
log + log √ + log − < ! < + log √ + log −
(§) log + log + ( log − log ) < ! < + log + ( log − log ) .
We proved previously that < for ≥ 3 and < . Rearranging this
inequality, we find that
log
<
log
.
Thus, it is clear that for ≥ 3,
1
2
log
<
1
2
log 3
3
< 0.08
and
log
<
log
3
< 0.15.
Using these two inequalities, we can modify (§) in the following way
log +
1
2
log + ( log − log ) < ! < +
1
2
log + ( log − log )
( log − log ) < ! <
log
+
log
2
+ ( log − log )
( log − log ) < ! <
log
+
1
2
log
+ log − log
( log − log ) < ! < (0.15 + 0.08 + log − log )
( log − log ) < ! < ( log − log + 0.23) .

24
Now, we recall from (∗∗) that
log
+
log
+ ⋯ +
log
+ > !
>
log
+
log
+ ⋯ +
log
−
Comparing this inequality with the previous inequality, we come to two conclusions. First,
log
+
log
+ ⋯ +
log
+ > − log ,
and second,
log
+
log
+ ⋯ +
log
− < − log + 0.23.
Rearranging the two inequalities, we obtain
log
+
log
+ ⋯ +
log
− log > − − log
and
log
+
log
+ ⋯+
log
− log < − log + 0.23.
Using the fact that and can be taken equal to 3 and 4 respectively, we can calculate
the right hand sides of the two expressions above. Thus, − − log = −3 − 0.43429 =
−3.43429 and − log + 0.23 = 4 − 0.43429 + 0.23 = 3.79571. Taking the larger of the
absolute value of these two numbers, namely | − 3.43429| < |3.79571| < 4, we obtain
−4 <
log
+
log
+ ⋯ +
log
− log < 4.
This was the result we were looking for, which therefore completes this proof.

25
CHAPTER 5
Mertens’ Second Theorem
We begin this chapter by introducing an extremely useful formula which, with the
help of Mertens’ First Theorem, will enable us to prove Mertens’ Second Theorem.
THEOREM . . (Abel’s Formula) Consider the sum
= + + +. . . + ,
where , , , …, and , , , …, are any two sequences of numbers, and denote
the sums , + , + + , …, + + + … + by , , , …, respectively.
Then
= ( − ) + ( − ) + ( − ) + ⋯ + ( − ) + .
THEOREM . . (Mertens’ Second Theorem) Given any > 1, for all < with
prime, the expression
1
2
+
1
3
+
1
5
+
1
7
+ ⋯ +
1
− ln ln
is less than some relatively small constant , where is independent of .
PROOF.
To begin this proof, we first examine the series
=
1
+
1
+
1
+
1
+ ⋯ +
1
,
where , , …, are all the primes less than or equal to any integer . We now set
=
1
log
′′ =
1
log
′′ =
1
log
, … , =
1
log
=
log
, =
log
, =
log
, … , =
log
.

26
Using Theorem . , denoting the sums , + , + + , …, + + + ⋯ +
by , , , …, respectively, we obtain
=
log
,
=
log
+
log
,
=
log
+
log
+ ⋯ +
log
.
We can now write our series in terms of , , …, , and , , …, . Clearly
+ + +. . . +
=
1
log
log
+
1
log
log
+ ⋯ +
1
log
log
=
1
+
1
+
1
+
1
+ ⋯ +
1
,
but by Theorem . ,
+ + + ⋯ +
= ( − ) + ( − ) + ( − ) +. . . +( − ) + , thus
1
+
1
+
1
+
1
+ ⋯ +
1
=
1
log
−
1
log
+
1
log
−
1
log
+
1
log
−
1
log
+ ⋯ +
1
log
−
1
log
+
1
log
⋅
(∗) = − + −
+
1
log
−
1
log
+ ⋯ +
1
log
−
1
log
+
1
log
−
1
log
+
1
log
⋅ .

27
Now by Theorem3.5, we can deduce that:
|
log
− log | < ⇒ | − log | < ⇒ < + ,
|
log
+
log
− log | < ⇒ | − log | < ⇒ < + ,
|
log
+
log
+
log
− log | < ⇒ < + ,
|
log
+
log
+ ⋯ +
log
− log | < ⇒ < + ,
|
log
+
log
+ ⋯ +
log
− log | < ⇒ < + ,
where can be taken equal to 4. Thus we can conclude that
< + , < + , < + , …
. . . , < + , and < + .
From above, it follows that + + + + ⋯ +
<
1
log
−
1
log
( log + ) +
1
log
−
1
log
( log + )
+
1
log
−
1
log
( log + ) + ⋯ +
1
log
−
1
log
log +
+ − ( log + ) + ( log + )
=
1
log
−
1
log
log +
log
−
log
+
1
log
−
1
log
log +
log
−
log
+
1
log
−
1
log
log +
log
−
log
+ ⋯
+
1
log
−
1
log
log +
log
−
log
+
1
log
−
1
log
log
+
log
−
log
+
1
log
log +
log
.
The bound on our series = + + + + ⋯ + can simplify to
1 +
− log
log
+
log
+
1
log
−
1
log
log +
1
log
−
1
log
log

28
+ ⋯ +
1
log
−
1
log
log +
1
log
−
1
log
log +
1
log
log ,
which after further rearrangement yields
(∗∗) 1 + [ + − log + − log + ⋯
+
1
log
−
1
log
log +
1
log
−
1
log
log +
1
log
log ] +
log
.
Now we will focus on the expression found inside the square brackets. We can expand
the expression to look like
− log
log
+
log
log
−
log
log
+
log
log
−
log
log
+ ⋯
. . . + − + − + ,
then factor the common denominators, which will yield the following
( log − log )
1
log
+ ( log − log )
1
log
+ ( log − log )
1
log
+ ⋯ + ( log − log ) + ( log − log ) . We can now estimate the above
expression geometrically by constructing a graph to
model the series. We can depict the expression by the sum of − 1 rectangles. The
area of the first rectangle will have a base of length ( log − log ) and a height of
length , as per the first term in our expression. The second rectangle will have a
base of length ( log − log ) , and a height of length . This pattern continues up
to our ( − 1) rectangle, which will have a base of length ( log − log ) and a
height of length . these rectangles are plotted in Figure (4.1). We now observe that
when = log , that = . Similarly, when = log , = , and so on, up to
= log , and = . In general we find that = , which is a hyperbola that
intersects the top right corner of each rectangle. This hyperbola is also shown in

29
Now that we have defined the hyperbola such that the shaded rectangles are all
below its graph, we have thus found an upper bound on all the rectangles, and
therefore an upper bound on the series . We simply need to calculate the area below
the hyperbola = between the points = log and = log . This is done easily
using calculus. Clearly
∫ = ( ln )| = ln ( log ) − ln ( log ) .
We can therefore conclude that the expression in the square brackets from (∗∗) , is
bounded by ln ( log ) − ln ( log ) , and thus
(0.18) + + + + ⋯ + < 1 + ( log ) − ln ( log ) + .
We now focus our attention on finding a lower bound for our series = + + +
+ ⋯ + . Recall that
=
log
+
log
+ ⋯ +
log
,
where = (1,2, … , ), and ≤ . Adding and subtracting , where > , we
find that
>
log
+
log
+ ⋯ +
log
+
log
− .
Here, we can let = , as for ≥ 3, < . Now, as a result of Mertens’ First
Theorem, we find that − < + + + ⋯ + + − log ,
Which implies
log − <
log
+
log
+
log
+ ⋯ +
log
+
log ?
.
Returning to equation (0.19), we can conclude that

30
>
log
+
log
+
log
+ ⋯ +
log
+
log
−
> ( log ? − ) − ,
for 1 ≤ < . Using the same argument, we find that
log
+
log
+
log
+ ⋯ +
log
> − − ,
with the − added in for convenience. From above, we can conclude that
> − − , > − − , …
. . . , > − − , > − − .
Now, referring back to (∗) , we find that
1
+
1
+
1
+
1
+ ⋯ +
1
>
1
log
−
1
log
( log − − ) +
1
log
−
1
log
( log − − )
+
1
log
−
1
log
( log − − ) + ⋯ +
1
log
−
1
log
( log − − )
+ − ( log − − ) + ( log − − )
=
1
log
−
1
log
log −
+
log
+
+
log
+
1
log
−
1
log
log
−
+
log
+
+
log
+
1
log
−
1
log
log −
+
log
+
+
log
+ ⋯
. . . + − log − + + − log
−
+
log
+
+
log
+
1
log
log −
+
log
.
With further rearrangement, we obtain
=
1
log
−
1
log
log −
+
log
+
1
log
−
1
log
log
+
1
log
−
1
log
log + ⋯ +
1
log
−
1
log
log
+
1
log
−
1
log
log +
1
log
log
=
log
log
−
log
log
+
log
log
−
log
log
+
log
log
−
log
log
+ ⋯ +
log
log
−
log
log

31
+
log
log
−
log
log
+
1
log
log −
+
log
= [( log − log )
1
log
+ ( log − log )
1
log
+ ( log − log )
1
log
+ ⋯
. . . +( log − log ) ] + 1 − .
In this last step, we omitted the − term, and replaced it with − . Now, as we
have previously done with the upper bound on , we can see that the sum inside the
square brackets above is simply the sum of differently sized rectangles. This time, the
secondrectang1ewi11have abase of1ength( log − log )andaheight of1en first
rectang1ewi11have abase of1ength ( log − log )andaheight of1ength
,
. e and
so on, up to a rectangle with a base of length ( log − log ) and height of length
. These rectangles are shown in Figure4.2.
Figure4.2
We can see that when = log , = . This pattern holds for all , with ≤ . We
also notice that, as before, these coordinates define the function = , only this
Time, the curve intersects the top left hand corner of the rectangles, as shown in
Figure 4.2. This implies that the area bounded by the curve between log and log is
less than the area of the shaded rectangles. We simply need to calculate this area,
which will in turn give us a lower bound for our series, but we have already calculated
this to be ln ( log ) − ln ( log ) from
Now that we have a lower bound for the sum of the rectangles, we find that
+ + + + ⋯ + > ( log ) − ln ( log ) + 1 − .
Next, combining equations (0.18) and (0.20), we find that for = + + + + ⋯ +

32
(0.21) ln ( log ) − ln ( log ) + 1 − < < 1 + ( log ) − ln ( log ) + .
Now, as opposed to using logarithms of two different bases in our expression, we will
use the fact that
log = ln ,
where = log . This changes (0.21) to
ln ( ln ) − ln ( log ) + 1 −
+
log
< < 1 + ( ln ) − ln ( log ) +
log
ln ln + ln − ln ( log ) + 1 − < < 1 + + ln − ln ( log ) + .
From here we can simply evaluate our expression using the fact that = log =
0.43429 …, = 2, = 4, and = = 0.15904…. Clearly,
ln ln + ln − ln ( log ) + 1 −
+
log
= ln ln + ln (0.43429) − ln ( log 2) + 1 −
4 + 0.15904
log 2
= ln ln + (−0.83404) − (−1.20054) + 1 − (13.81603)
= ln ln − 12.44953 > − 13.
Similarly,
1 + ln ln + ln − ln ( log ) +
log
= ln ln + 1 + (−0.83404) − (−1.20054) +
4
0.30102
= ln ln + 14.65465 < + 15.
Comparing the two inequalities above, we finally obtain
ln ln − <
1
+
1
+
1
+
1
+ ⋯ +
1
< + ,
where = 15, the larger of the upper and lower bounds. This completes the proof.

33
CHAPTER 6
CONCLUSION
This project explored how prime numbers are distributed throughout the set of real
numbers. We proved that there exists an upper and lower bound on ( ), which is dependent on . It is
fascinating to see how such a complicated yet relatively elementary proof can yield such an elegant
result. We used this result to explore further problems in the study of prime numbers, specifically,
Mertens’ First and Second Theorem.
In Mertens’ First Theorem, we used similar arguments as in the proof of Tchebychev’s
Theorem, and simply found the magnitude of errors between different series. The proof of Mertens’
Second Theorem was more geometrically intuitive. We used Mertens’ First Theorem as well as Abel’s
Formula to transform our series into another more convenient series, which could be interpreted
geometrically. This allowed us to find an upper and lower bound using integration, which enabled us to
ultimately complete the proof.
From this point, the next step in research could be exploring the proof the the Prime
Number Theorem. This would involve a much more complicated proof, using methods from Complex
Analysis.

34
REFERENCE
[1] A.M. Yaglom, I.M. Yaglom, Challenging Mathematical Problems with Elementary Solutions. New
York, Dover Publication, (1987).
[2] D. Goldfeld, The Elementary Proof of the Prime Number Theorem: An Historical Perspective.
Columbia University, New York. 1
[3] J.J. O’Connor, E.F. Robertson, Prime Numbers, The MacTutor History of Mathematics Archive.
May 2009. Web. http://www‐history.mcs.st‐and.ac.uk/HistTopics/Prime‐numbers.html 1
[4] K.H. Rosen, Discrete Mathematics and Its Applications, New York, McGraw‐Hill, (2007). 2

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