10 1. Elementary Prime Number Theory, I

At this point Perott uses the evaluation ζ(2) =

π2/6.

However, it is simpler

to proceed as follows: Since

t−2

is a decreasing function of t on the positive

real axis,

ζ(2) = 1 +

∞

n=2

1

n2

1 +

∞

n=1

n+1

n

dt

t2

= 1 +

∞

1

dt

t2

= 2.

Referring back to (1.6), we see that A(N)/N is bounded below by a positive

constant. In particular, it must be that A(N) → ∞ as N → ∞.

Remark. As observed by Dressler [Dre75], Perott’s argument also yields

a lower bound on π(N). Note that since every squarefree number ≤ N is a

product of some subset of the π(N) primes up to N, we have

2π(N)

≥ A(N).

The argument above establishes that A(N) ≥ cN for c = 2 − ζ(2) 0, and

so π(N) ≥ log N/ log 2 + O(1).

For the next proof we need the following simple lemma:

Lemma 1.5. Every natural number n can be written in the form

rs2,

where

r and s are natural numbers and r is squarefree.

Proof. Choose the positive integer s so that

s2

is the largest perfect square

dividing n, and put r =

n/s2.

We claim that r is squarefree. Otherwise

p2

| r for some prime p. But then

(ps)2

| n, contrary to the choice of s.

Erd˝ os’s proof of Theorem 1.1. Let N be a positive integer. There are

at most

√

N squares not exceeding N and at most

2π(N)

squarefree integers

below this bound. So Lemma 1.5 implies that

2π(N)

√

N ≥ N.

Dividing by

√

N and taking logarithms yields the lower bound π(N) ≥

log N/ log 4.

A modification of this argument leads to another proof that

∑

1

p

di-

verges:

Erd˝ os’s proof of Theorem 1.4. Suppose that

∑

1/p converges. Then we

can choose an M for which

(1.7)

pM

1

p

1

2

.

Keep this M fixed.

Let N be an arbitrary natural number. The estimate (1.7) implies that

most integers up to N factor completely over the primes not exceeding M.