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.
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