10 1. Elementary Prime Number Theory, I
At this point Perott uses the evaluation ζ(2) =
However, it is simpler
to proceed as follows: Since
is a decreasing function of t on the positive
ζ(2) = 1 +
= 1 +
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
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
r and s are natural numbers and r is squarefree.
Proof. Choose the positive integer s so that
is the largest perfect square
dividing n, and put r =
We claim that r is squarefree. Otherwise
| r for some prime p. But then
| n, contrary to the choice of s.
Erd˝ os’s proof of Theorem 1.1. Let N be a positive integer. There are
N squares not exceeding N and at most
below this bound. So Lemma 1.5 implies that
N ≥ N.
N and taking logarithms yields the lower bound π(N) ≥
log N/ log 4.
A modification of this argument leads to another proof that
Erd˝ os’s proof of Theorem 1.4. Suppose that
1/p converges. Then we
can choose an M for which
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.