4. The Euler-Riemann zeta function 7

and let

P0(x) :=

p≤x

(

1 + |f(p)| +

|f(p2)|

+ · · ·

)

=

n:p|n⇒p≤x

|f(n)| ≥

n≤x

|f(n)|.

Since P0(x) ≤ P0 for all x, the partial sums

∑

n≤x

|f(n)| form a bounded

increasing sequence. Thus

∑

|f(n)| converges, proving (i).

We can now present Euler’s first proof of the infinitude of the primes.

Euler’s first proof of Theorem 1.1. Let f be defined by f(n) = 1/n for

every n. Assuming that there are only finitely many primes, condition (ii) of

Theorem 1.3 is trivially satisfied, as the product in question only has finitely

many terms. It follows that

∞

n=1

1

n

=

p

1 +

1

p

+

1

p2

+ · · · ∞,

in contradiction with the well-known divergence of the harmonic series.

As pointed out by Euler, this proof gives a much stronger result than

that asserted in Theorem 1.1.

Theorem 1.4. The series

∑

1

p

diverges, where the sum extends over all

primes p.

Proof. Suppose not and let C =

∑

1/p. As in the last proof, we take

f(n) = 1/n and apply Theorem 1.2. Let us check that condition (ii) of that

theorem holds here. First, notice that

p≤x

1 +

1

p

+

1

p2

+ · · · =

p≤x

1

1 −

1

p

=

p≤x

1 +

1

p − 1

≤

p≤x

1 +

2

p

.

Now recall that

et

≥ 1 + t for every nonnegative t; this is clear from trun-

cating the Taylor expansion

et

= 1 + t +

t2/2!

+ . . .. It follows that

p≤x

1 +

2

p

≤

p≤x

e2/p

= exp

⎛

⎝

p≤x

2/p⎠

⎞

≤ exp(2C).

Consequently, the partial products

p≤x

1 +

1

p

+

1

p2

+ · · ·