8 1. Elementary Prime Number Theory, I
form a bounded, increasing sequence, which shows that we have condition
(ii). We conclude that

n=1
1
n
=
p
1
1
1
p
exp(2C),
a contradiction.
Tweaking this argument, it is possible to derive an explicit lower bound
on the partial sums

p≤x
1/p: Note that for x 2,
(1.4)
p≤x
1
1
1
p
=
n:p|n⇒p≤x
1
n

n≤x
1
n
log x.
From the upper bound (1
1/p)−1
= (1 + 1/(p 1)) exp((p
1)−1),
we
deduce (taking the logarithm of (1.4)) that

p≤x
(p
1)−1
log log x. To
derive a lower bound for

p≤x
1/p from this, note that
p≤x
1
p
=
p≤x
1
p 1

p≤x
1
p 1

1
p

p≤x
1
p 1

n≥2
1
n 1

1
n
=


p≤x
1
p 1


1 log log x 1.
(1.5)
The next two proofs also make use of the zeta function and its Euler
factorization, but in a decidedly different manner.
Proof of J. Hacks. We need the well-known result, also due to Euler, that
ζ(2) =
π2/6;
a proof is sketched in Exercise 5 (for alternative arguments see
[AZ04, Chapter 7], [Cha02]). Plugging s = 2 into the Euler factorization
(1.2) we obtain
π2
6
= ζ(2) =
p
1
1
1
p2
.
If there are only finitely many primes, then the product appearing here is
a finite product of rational numbers, so that
π2/6
must also be a rational
number. But this is impossible, since π is well known to be a transcen-
dental number, i.e., not the root of any nonzero polynomial with rational
coefficients. A weaker result, which suffices for the current argument, is the
subject of Exercise 6 (cf. [AZ04, Chapter 6, Theorem 2]).
One can give a similar argument avoiding irrationality considerations:
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