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: