6 1. Elementary Prime Number Theory, I

Remark. Without imposing a condition such as (i) or (ii), it is possible for

either the series or the product in (1.3) to converge while the other diverges,

or for both to converge without being equal. See [Win43, §15] for explicit

examples.

If f is not merely multiplicative but completely multiplicative, then the

factors in (1.3) form a geometric series whose convergence is implied by

either of the above conditions. Thus we have the following consequence:

Corollary 1.3. Let f be a completely multiplicative function. Then

∞

n=1

f(n) =

p

1

1 − f(p)

subject to either of the two convergence criteria of Theorem 1.2.

The factorization (1.2) of the zeta function is immediate from this corol-

lary: One takes f(n) =

1/ns

and observes that for s 1, condition (i) holds

(for example) by the integral test.

Proof of Theorem 1.2. Suppose that condition (i) holds and set S0 :=

∑∞

n=1

|f(n)|. For each prime p, the series

∑∞

k=0

f(pk)

converges absolutely,

since

∑

∞

k=0

|f(pk)|

≤ S0. Therefore

P (x) =

p≤x

(

1 + f(p) +

f(p2)

+ · · ·

)

is a finite product of absolutely convergent series. It follows that

P (x) =

n:p|n⇒p≤x

f(n).

If we now set S =

∑∞

n=1

f(n) (which converges absolutely), we have

S − P (x) =

n:p|n for some px

f(n),

which shows

|S − P (x)| ≤

nx

|f(n)| → 0

as x → ∞. Thus P (x) → S as x → ∞, which is the assertion of (1.3).

Now suppose that (ii) holds. We shall show that (i) holds as well, so

that the theorem follows from what we have just done. To see this, let

P0 =

p

(

1 + |f(p)| +

|f(p2)|

+ · · ·

)

,