Number Theory: An Introduction to Mathematics

404 X A Character Study

Proof Any positive integerNcan be written in the formN=qm+r,whereq≥ 0
and 1≤r≤m.Sinceχ(a)=χ(b)ifa≡bmodm,wehave

∑N

n= 1

χ(n)=

(∑m

n= 1

+

∑^2 m

n=m+ 1

+···+

∑qm

n=(q− 1 )m+ 1

)

χ(n)+

qm∑+r

n=qm+ 1

χ(n)

=q

∑m

n= 1

χ(n)+

∑r

n= 1

χ(n).

But

∑m

n= 1 χ(n)=0, sinceχ=χ^1. Hence

∑N

n= 1

χ(n)=

∑r

n= 1

χ(n)=−

∑m

n=r+ 1

χ(n).

Since|χ(n)|=1 or 0 according as(n,m)=1or(n,m)=1, and sinceφ(m)is the
number of positive integersn≤msuch that(n,m)=1, the result follows.

With each Dirichlet characterχ, there is associated aDirichlet L-function

L(s,χ)=

∑∞

n= 1

χ(n)/ns.

Since|χ(n)|≤1foralln, the series is absolutely convergent forσ:=Rs>1. We
are going to show that ifχ=χ 1 , then the series is also convergent forσ>0. (It does
not converge ifσ≤0, since then|χ(n)/ns|≥1 for infinitely manyn.)
Put

H(x)=

∑

n≤x

χ(n).

Then

∑

n≤x

χ(n)n−s=

∫x+

1 −

t−sdH(t)

=H(x)x−s+s

∫x

1

H(t)t−s−^1 dt.

SinceH(x)is bounded, by Lemma 3, on lettingx→∞we obtain

L(s,χ)=s

∫∞

1

H(t)t−s−^1 dt forσ> 0.

Moreover the integral on the right is uniformly convergent in any half-planeσ ≥δ,
whereδ>0, and henceL(s,χ)is a holomorphic function forσ>0.
The following discussion of DirichletL-functions and the prime number theorem
for arithmetic progressions runs parallel to that of the Riemannζ-function and the
ordinary prime number theorem in the previous chapter. Consequently we will be more
brief.