Dirichlet

Thomas Browning February 2020

1 Convergence of Dirichlet Series

Let a1, a2,... be a of complex numbers. Consider the partial sums defined by

n X Sn = ak. k=1

α Suppose that there is an α > 0 such that |Sn| ≤ Cn for all n ≥ n0. Let K be a compact subset of the half-plane Hα = {s ∈ C : Re s > α}. Then

K ⊆ {s ∈ C : Re(s) ≥ α + ε and |s| ≤ M} for some ε > 0 and M > 0. If n0 ≤ n1 ≤ n2 and s ∈ K then

n2 n2 X an X 1 = (S − S ) ns ns n n−1 n=n1+1 n=n1+1

S S n2  1 1  n2 n1 X = s − s + s − s Sn (n2 + 1) (n1 + 1) n (n + 1) n=n1+1 α α n2 Cn1 Cn2 X 1 1 ≤ α+ε + α+ε + s − s |Sn| (n1 + 1) (n2 + 1) n (n + 1) n=n1+1 n2 Z n+1 C C X −s ≤ ε + ε + s+1 dx |Sn| n1 n2 n x n=n1+1 n 2C X2 M ≤ ε + α+ε+1 |Sn| n1 n n=n1+1 n 2C X2 1 ≤ ε + CM 1+ε n1 n n=n1+1 2C Z n2 1 ≤ ε + CM 1+ε dx n1 n1 x 2C CM ≤ ε + ε . n1 εn1 This shows that the Dirichlet series ∞ X an f(s) = ns n=1

converges locally uniformly on Hα.

1 2 The

The result of the previous section shows that the Riemann zeta function ∞ X 1 ζ(s) = ns n=1

converges locally uniformly on H1 and that the ∞ X (−1)n−1 η(s) = ns n=1

converges locally uniformly on H0. On H1, we have the identity

∞ ! ∞ ! ∞ 2 X 1 X 2 X (−1)n−1 (1 − 21−s)ζ(s) = ζ(s) − ζ(s) = − = = η(s). 2s ns (2n)s ns n=1 n=1 n=1 Then the formula 1 ζ(s) = η(s) 1 − 21−s 1−s gives an of ζ(s) to H0 except with isolated singularities at the points where 1−2 = 0. 2πik These are the points s = 1 − log 2 for k ∈ Z. Performing the same trick with 3 instead of 2 gives an analytic 2πij continuation of ζ(s) to H0 except with isolated singularities at the points s = 1 − log 3 for j ∈ Z. However, if 2πik 2πij j k 1 − log 2 = 1 − log 3 then 2 = 3 so j = k = 0. This shows that the only possible non-removable singularity of ζ(s) in H0 is an isolated singularity at s = 1. Now we can expand ∞ X ((1 − s) log 2)n 1 − 21−s = 1 − e(1−s) log 2 = 1 − = (s − 1) log 2 + [higher order terms] n! n=0 so 1 1 = (s − 1)−1 + [higher order terms]. 1 − 21−s log 2 Now η(1) = log 2 so 1 ζ(s) = η(s) = (s − 1)−1 + [higher order terms]. 1 − 21−s This shows that ζ(s) has a simple pole of residue 1 at s = 1.

Theorem 1. Let a1, a2,... be a sequence of complex numbers. Consider the partial sums defined by n X Sn = ak. k=1 α Suppose that there is an α ∈ (0, 1) and a w ∈ C such that |Sn − wn| ≤ Cn for sufficiently large n. Then the Dirichlet series ∞ X an f(s) = ns n=1

has an analytic continuation to Hα except for a simple pole of residue w at s = 1. Proof. We can write ∞ ∞ X an X an − w f(s) = = + wζ(s) ns ns n=1 n=1

where the sum is analytic on Hα and where wζ(s) has an analytic continuation to H0 except for a simple pole of residue w at s = 1.

2 3 Dirichlet L-functions

Lemma 2. If A is a finite abelian group and if χ: A → C× is a group homomorphism then ( X |A| χ = 1, χ(a) = . 0 χ 6= 1. a∈A Proof. If χ = 1 then X X χ(a) = 1 = |A| . a∈A a∈A Now suppose that χ(b) 6= 1 for some b ∈ A. Then X X X χ(a) = χ(ab) = χ(b) χ(a) a∈A a∈A a∈A so X (1 − χ(b)) χ(a) = 0. a∈A P Since χ(b) 6= 1, this shows that a∈A χ(a) = 0. Let N be a positive integer and let χ:(Z/NZ)× → C× be a group homomorphism. The associated Dirichlet series is defined by X χ(n + n ) L(s, χ) = Z . ns n≥1 gcd(n,N)=1

Proposition 3. If χ 6= 1 then L(s, χ) has an analytic continuation to H0. If χ = 1 then L(s, χ) has an ϕ(N) analytic continuation to H0 except for a simple pole of residue N at s = 1. Proof. This follows from Theorem 1 and Lemma 2.

4 Example From Algebraic Number Theory

Let χ:(Z/4Z)× → C× be the nontrivial character. Then we have  1 1 1 1   1 1 1 1  4ζ(s)L(s, χ) = 4 + + + + ··· − + − + ··· 1s 2s 3s 4s 1s 3s 5s 7s 4 4 4 8 4 4 8 8 4 8 4 8 12 = + + + + + + + + + + + + + ··· 1s 2s 4s 5s 8s 9s 10s 13s 16s 17s 18s 20s 25s ∞ X an 2 2 2 = where an = {(x, y) ∈ : x + y = n} . ns Z n=1 Proving this requires some number theory. A more elementary fact is that the partial sums

n X 2 2 2 Sn = ak = {(x, y) ∈ Z : 1 ≤ x + y ≤ n} k=1 √ satisfy |Sn − πn| ≤ C n for sufficiently large n. By Theorem 1, 4ζ(s)L(s, χ) has a simple pole of residue π π at s = 1. Thus, L(1, χ) = 4 which gives the Leibniz formula for π: π 1 1 1 = 1 − + − + ··· . 4 3 5 7

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