TheoremComplete

Multiplication of Dirichlet Series

One of the most beautiful aspects of Dirichlet series is how their multiplication corresponds exactly to Dirichlet convolution of coefficient sequences.

TheoremDirichlet Series Multiplication

Let F(s)=βˆ‘n=1∞an/nsF(s) = \sum_{n=1}^{\infty} a_n/n^s and G(s)=βˆ‘n=1∞bn/nsG(s) = \sum_{n=1}^{\infty} b_n/n^s be Dirichlet series converging absolutely for β„œ(s)>Οƒ1\Re(s) > \sigma_1 and β„œ(s)>Οƒ2\Re(s) > \sigma_2 respectively.

Then their product converges absolutely for β„œ(s)>max⁑(Οƒ1,Οƒ2)\Re(s) > \max(\sigma_1, \sigma_2) and:

F(s)β‹…G(s)=βˆ‘n=1∞cnnsF(s) \cdot G(s) = \sum_{n=1}^{\infty} \frac{c_n}{n^s}

where cn=βˆ‘d∣nadbn/d=(aβˆ—b)(n)c_n = \sum_{d|n} a_d b_{n/d} = (a * b)(n) is the Dirichlet convolution.

ExampleZeta Function Identities

The multiplication theorem immediately gives:

  • ΞΆ(s)2=βˆ‘n=1βˆžΟ„(n)ns\zeta(s)^2 = \sum_{n=1}^{\infty} \frac{\tau(n)}{n^s} (since Ο„=1βˆ—1\tau = 1 * 1)
  • ΞΆ(s)β‹…1ΞΆ(s)=1\zeta(s) \cdot \frac{1}{\zeta(s)} = 1, so βˆ‘n=1∞μ(n)/ns=1/ΞΆ(s)\sum_{n=1}^{\infty} \mu(n)/n^s = 1/\zeta(s)
  • ΞΆ(sβˆ’1)ΞΆ(s)=βˆ‘n=1βˆžΟ†(n)ns\frac{\zeta(s-1)}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\varphi(n)}{n^s} (since Ο†βˆ—1=id\varphi * 1 = \text{id})
ExampleGenerating Arithmetic Functions

We can generate new arithmetic functions via convolution:

  • Let fβˆ—g=hf * g = h. Then F(s)β‹…G(s)=H(s)F(s) \cdot G(s) = H(s)
  • If f(n)=1f(n) = 1 for all nn, then F(s)=ΞΆ(s)F(s) = \zeta(s)
  • If g(n)=ΞΌ(n)g(n) = \mu(n), then G(s)=1/ΞΆ(s)G(s) = 1/\zeta(s)
  • So h=1βˆ—ΞΌ=Ξ΄h = 1 * \mu = \delta gives H(s)=1H(s) = 1, confirming ΞΆ(s)β‹…(1/ΞΆ(s))=1\zeta(s) \cdot (1/\zeta(s)) = 1
Remark

This theorem transforms the ring structure of arithmetic functions (under Dirichlet convolution) into the multiplicative structure of Dirichlet series. It's a form of "transform theory" analogous to how Fourier/Laplace transforms convert convolution to multiplication.

ExampleInverting Dirichlet Series

To find the inverse of F(s)F(s), we need G(s)G(s) such that F(s)β‹…G(s)=1F(s) \cdot G(s) = 1. This means:

(aβˆ—b)(n)=Ξ΄(n)=[n=1](a * b)(n) = \delta(n) = [n=1]

This is solvable recursively:

  • b1=1/a1b_1 = 1/a_1
  • For n>1n > 1: bn=βˆ’1a1βˆ‘d∣n,d<nan/dbdb_n = -\frac{1}{a_1} \sum_{d|n, d<n} a_{n/d} b_d

This requires a1β‰ 0a_1 \neq 0.

The multiplication theorem is fundamental for translating arithmetic identities (convolution relations) into analytic identities (products of Dirichlet series), enabling powerful analytical techniques to study arithmetic functions.