Dirichlet series generating functions

A couple days ago I wrote about Dirichlet convolution, and in that post I said

Define the function ζ to be the constant function 1.

This was following the notation from the book I quoted in the post.

Someone might question the use of ζ because it is commonly used to denote the Riemann ζ function. And they’d be onto something: Calling the sequence of all 1’s ζ is sort of a pun on the ζ function.

Dirichlet series generating functions

Given a sequence a1, a2, a3, … its Dirichlet series generating function (dsgf) is defined to be

A(s) = \sum_{n=1}^\infty \frac{a_n}{n^s}

The Dirichlet series generating function for the sequence of all 1’s is the Riemann zeta function ζ(s).

\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}

The definition of ζ as a sequence was alluding to its associated Dirichlet series.

Dirichlet convolution

Recall from the earlier post that the Dirichlet convolution of two sequences is defined by

(f*g)(n) = \sum_{d|n} f(d) \,\,g\left(\frac{n}{d} \right)

A basic theorem connecting Dirichlet series and Dirichlet convolution says that if A(s) is the dsgf of

a1, a2, a3, …

and B(s) is the dsgf of

b1, b2, b3, …

then the dsgf of a*b, the Dirichlet convolution of the two series, is A(s) B(s).

In short, the dsgf of a convolution is the product of the dsgfs. This is an example of the general pattern that transforms turn convolutions into products. Different kinds of transformations have different kinds of convolution. See examples here.

Dirichlet series of an inverse

In the earlier post we defined the inverse of a sequence (with respect to Dirichlet convolution) to be the sequence you convolve it with to get the sequence δ(n) given by

1, 0, 0, 0, …

The dsgf of δ(n) is simply the constant 1. Therefore if a*b = δ, then AB = 1. Said another way, if a is the inverse sequence of b, then the dsgf of a is the reciprocal of the dsgf of b.

The earlier post defined the Möbius function μ(n) to be the inverse of the sequence of 1’s, and so its dsgf is the reciprocal of the zeta function.

\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}

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