# Dirichlet series

In mathematics, a Dirichlet series is any series of the form

$\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}},$ where s is complex, and $a_{n}$ is a complex sequence. It is a special case of general Dirichlet series.

Dirichlet series play a variety of important roles in analytic number theory. The most usually seen definition of the Riemann zeta function is a Dirichlet series, as are the Dirichlet L-functions. It is conjectured that the Selberg class of series obeys the generalized Riemann hypothesis. The series is named in honor of Peter Gustav Lejeune Dirichlet.

## Combinatorial importance

Dirichlet series can be used as generating series for counting weighted sets of objects with respect to a weight which is combined multiplicatively when taking Cartesian products.

Suppose that A is a set with a function w: AN assigning a weight to each of the elements of A, and suppose additionally that the fibre over any natural number under that weight is a finite set. (We call such an arrangement (A,w) a weighted set.) Suppose additionally that an is the number of elements of A with weight n. Then we define the formal Dirichlet generating series for A with respect to w as follows:

${\mathfrak {D}}_{w}^{A}(s)=\sum _{a\in A}{\frac {1}{w(a)^{s}}}=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}$ Note that if A and B are disjoint subsets of some weighted set (U, w), then the Dirichlet series for their (disjoint) union is equal to the sum of their Dirichlet series:

${\mathfrak {D}}_{w}^{A\uplus B}(s)={\mathfrak {D}}_{w}^{A}(s)+{\mathfrak {D}}_{w}^{B}(s).$ Moreover, if (A, u) and (B, v) are two weighted sets, and we define a weight function w: A × BN by

$w(a,b)=u(a)v(b),$ for all a in A and b in B, then we have the following decomposition for the Dirichlet series of the Cartesian product:

${\mathfrak {D}}_{w}^{A\times B}(s)={\mathfrak {D}}_{u}^{A}(s)\cdot {\mathfrak {D}}_{v}^{B}(s).$ This follows ultimately from the simple fact that $n^{-s}\cdot m^{-s}=(nm)^{-s}.$ ## Examples

The most famous example of a Dirichlet series is

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},$ whose analytic continuation to $\mathbb {C}$ (apart from a simple pole at $s=1$ ) is the Riemann zeta function.

Treating these as formal Dirichlet series for the time being in order to be able to ignore matters of convergence, note that we have:

{\begin{aligned}\zeta (s)&={\mathfrak {D}}_{\operatorname {id} }^{\mathbb {N} }(s)=\prod _{p{\text{ prime}}}{\mathfrak {D}}_{\operatorname {id} }^{\{p^{n}:n\in \mathbb {N} \}}(s)=\prod _{p{\text{ prime}}}\sum _{n\in \mathbb {N} }{\mathfrak {D}}_{\operatorname {id} }^{\{p^{n}\}}(s)\\&=\prod _{p{\text{ prime}}}\sum _{n\in \mathbb {N} }{\frac {1}{(p^{n})^{s}}}=\prod _{p{\text{ prime}}}\sum _{n\in \mathbb {N} }\left({\frac {1}{p^{s}}}\right)^{n}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}\end{aligned}} as each natural number has a unique multiplicative decomposition into powers of primes. It is this bit of combinatorics which inspires the Euler product formula.

Another is:

${\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}$ where μ(n) is the Möbius function. This and many of the following series may be obtained by applying Möbius inversion and Dirichlet convolution to known series. For example, given a Dirichlet character χ(n) one has

${\frac {1}{L(\chi ,s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)\chi (n)}{n^{s}}}$ where L(χ, s) is a Dirichlet L-function.

Other identities include

${\frac {\zeta (s-1)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\varphi (n)}{n^{s}}}$ where $\varphi (n)$ is the totient function,

${\frac {\zeta (s-k)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {J_{k}(n)}{n^{s}}}$ where Jk is the Jordan function, and

{\begin{aligned}&\zeta (s)\zeta (s-a)=\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)}{n^{s}}}\\[6pt]&{\frac {\zeta (s)\zeta (s-a)\zeta (s-2a)}{\zeta (2s-2a)}}=\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n^{2})}{n^{s}}}\\[6pt]&{\frac {\zeta (s)\zeta (s-a)\zeta (s-b)\zeta (s-a-b)}{\zeta (2s-a-b)}}=\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)\sigma _{b}(n)}{n^{s}}}\end{aligned}} where σa(n) is the divisor function. By specialisation to the divisor function d = σ0 we have

{\begin{aligned}\zeta ^{2}(s)&=\sum _{n=1}^{\infty }{\frac {d(n)}{n^{s}}}\\[6pt]{\frac {\zeta ^{3}(s)}{\zeta (2s)}}&=\sum _{n=1}^{\infty }{\frac {d(n^{2})}{n^{s}}}\\[6pt]{\frac {\zeta ^{4}(s)}{\zeta (2s)}}&=\sum _{n=1}^{\infty }{\frac {d(n)^{2}}{n^{s}}}.\end{aligned}} The logarithm of the zeta function is given by

$\log \zeta (s)=\sum _{n=2}^{\infty }{\frac {\Lambda (n)}{\log(n)}}{\frac {1}{n^{s}}},\qquad \Re (s)>1.$ Similarly, we have that

$-\zeta '(s)=\sum _{n=2}^{\infty }{\frac {\log(n)}{n^{s}}},\qquad \Re (s)>1.$ Here, Λ(n) is the von Mangoldt function. The logarithmic derivative is then

${\frac {\zeta '(s)}{\zeta (s)}}=-\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}.$ These last three are special cases of a more general relationship for derivatives of Dirichlet series, given below.

Given the Liouville function λ(n), one has

${\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}.$ Yet another example involves Ramanujan's sum:

${\frac {\sigma _{1-s}(m)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {c_{n}(m)}{n^{s}}}.$ Another pair of examples involves the Möbius function and the prime omega function:

${\frac {\zeta (s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}\equiv \sum _{n=1}^{\infty }{\frac {\mu ^{2}(n)}{n^{s}}}.$ ${\frac {\zeta ^{2}(s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}.$ ## Analytic properties

Given a sequence $\{a_{n}\}_{n\in \mathbb {N} }$ of complex numbers we try to consider the value of

$f(s)=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}$ as a function of the complex variable s. In order for this to make sense, we need to consider the convergence properties of the above infinite series:

If $\{a_{n}\}_{n\in \mathbb {N} }$ is a bounded sequence of complex numbers, then the corresponding Dirichlet series f converges absolutely on the open half-plane Re(s) > 1. In general, if an = O(nk), the series converges absolutely in the half plane Re(s) > k + 1.

If the set of sums

$a_{n}+a_{n+1}+\cdots +a_{n+k}$ is bounded for n and k ≥ 0, then the above infinite series converges on the open half-plane of s such that Re(s) > 0.

In both cases f is an analytic function on the corresponding open half plane.

In general $\sigma$ is the abscissa of convergence of a Dirichlet series if converges for $\Re (s)>\sigma$ and diverges for $\Re (s)<\sigma .$ This is the analogue for Dirichlet series of the radius of convergence for power series. The Dirichlet series case is more complicated, though: absolute convergence and uniform convergence may occur in distinct half-planes.

In many cases, the analytic function associated with a Dirichlet series has an analytic extension to a larger domain.

### Abscissa of convergence

Suppose

$\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s_{0}}}}$ converges for some $s_{0}\in \mathbb {C} ,\Re (s_{0})>0.$ Proposition 1. $A(N):=\sum _{n=1}^{N}a_{n}=o(N^{s_{0}}).$ Proof. Note that:

$(n+1)^{s}-n^{s}=\int _{n}^{n+1}sx^{s-1}\,dx={\mathcal {O}}(n^{s-1}).$ and define

$B(N)=\sum _{n=1}^{N}{\frac {a_{n}}{n^{s_{0}}}}=\ell +o(1)$ where

$\ell =\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s_{0}}}}.$ By summation by parts we have

{\begin{aligned}A(N)&=\sum _{n=1}^{N}{\frac {a_{n}}{n^{s_{0}}}}n^{s_{0}}\\&=B(N)N^{s_{0}}+\sum _{n=1}^{N-1}B(n)\left(n^{s_{0}}-(n+1)^{s_{0}}\right)\\&=(B(N)-\ell )N^{s_{0}}+\sum _{n=1}^{N-1}(B(n)-\ell )\left(n^{s_{0}}-(n+1)^{s_{0}}\right)\\&=o(N^{s_{0}})+\sum _{n=1}^{N-1}{\mathcal {o}}(n^{s_{0}-1})\\&=o(N^{s_{0}})\end{aligned}} Proposition 2. Define
$L={\begin{cases}\sum _{n=1}^{\infty }a_{n}&{\text{If convergent}}\\0&{\text{otherwise}}\end{cases}}$ Then:
$\sigma =\lim \sup _{N\to \infty }{\frac {\ln |A(N)-L|}{\ln N}}=\inf _{\sigma }\left\{A(N)-L={\mathcal {O}}(N^{\sigma })\right\}$ is the abscissa of convergence of the Dirichlet series.

Proof. From the definition

$\forall \varepsilon >0\qquad A(N)-L={\mathcal {O}}(N^{\sigma +\varepsilon })$ so that

{\begin{aligned}\sum _{n=1}^{N}{\frac {a_{n}}{n^{s}}}&=A(N)N^{-s}+\sum _{n=1}^{N-1}A(n)(n^{-s}-(n+1)^{-s})\\&=(A(N)-L)N^{-s}+\sum _{n=1}^{N-1}(A(n)-L)(n^{-s}-(n+1)^{-s})\\&={\mathcal {O}}(N^{\sigma +\varepsilon -s})+\sum _{n=1}^{N-1}{\mathcal {O}}(n^{\sigma +\varepsilon -s-1})\end{aligned}} which converges as $N\to \infty$ whenever $\Re (s)>\sigma .$ Hence, for every $s$ such that $\sum _{n=1}^{\infty }a_{n}n^{-s}$ diverges, we have $\sigma \geq \Re (s),$ and this finishes the proof.

Proposition 3. If $\sum _{n=1}^{\infty }a_{n}$ converges then $f(\sigma +it)=o\left({\tfrac {1}{\sigma }}\right)$ as $\sigma \to 0^{+}$ and where it is meromorphic $f(s)$ has no poles on $\Re (s)=0.$ Proof. Note that

$n^{-s}-(n+1)^{-s}=sn^{-s-1}+O(n^{-s-2})$ and $A(N)-f(0)\to 0$ we have by summation by parts, for $\Re (s)>0$ {\begin{aligned}f(s)&=\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {a_{n}}{n^{s}}}\\&=\lim _{N\to \infty }A(N)N^{-s}+\sum _{n=1}^{N-1}A(n)(n^{-s}-(n+1)^{-s})\\&=s\sum _{n=1}^{\infty }A(n)n^{-s-1}+\underbrace {{\mathcal {O}}\left(\sum _{n=1}^{\infty }A(n)n^{-s-2}\right)} _{={\mathcal {O}}(1)}\end{aligned}} Now find N such that for n > N, $|A(n)-f(0)|<\varepsilon$ $s\sum _{n=1}^{\infty }A(n)n^{-s-1}=\underbrace {sf(0)\zeta (s+1)+s\sum _{n=1}^{N}(A(n)-f(0))n^{-s-1}} _{={\mathcal {O}}(1)}+\underbrace {s\sum _{n=N+1}^{\infty }(A(n)-f(0))n^{-s-1}} _{<\varepsilon |s|\int _{N}^{\infty }x^{-\Re (s)-1}\,dx}$ and hence, for every $\varepsilon >0$ there is a $C$ such that for $\sigma >0$ :

$|f(\sigma +it)| ## Formal Dirichlet series

A formal Dirichlet series over a ring R is associated to a function a from the positive integers to R

$D(a,s)=\sum _{n=1}^{\infty }a(n)n^{-s}\$ with addition and multiplication defined by

$D(a,s)+D(b,s)=\sum _{n=1}^{\infty }(a+b)(n)n^{-s}\$ $D(a,s)\cdot D(b,s)=\sum _{n=1}^{\infty }(a*b)(n)n^{-s}\$ where

$(a+b)(n)=a(n)+b(n)\$ is the pointwise sum and

$(a*b)(n)=\sum _{k\mid n}a(k)b(n/k)\$ is the Dirichlet convolution of a and b.

The formal Dirichlet series form a ring Ω, indeed an R-algebra, with the zero function as additive zero element and the function δ defined by δ(1) = 1, δ(n) = 0 for n > 1 as multiplicative identity. An element of this ring is invertible if a(1) is invertible in R. If R is commutative, so is Ω; if R is an integral domain, so is Ω. The non-zero multiplicative functions form a subgroup of the group of units of Ω.

The ring of formal Dirichlet series over C is isomorphic to a ring of formal power series in countably many variables.

## Derivatives

Given

$F(s)=\sum _{n=1}^{\infty }{\frac {f(n)}{n^{s}}}$ it is possible to show that

$F'(s)=-\sum _{n=1}^{\infty }{\frac {f(n)\log(n)}{n^{s}}}$ assuming the right hand side converges. For a completely multiplicative function ƒ(n), and assuming the series converges for Re(s) > σ0, then one has that

${\frac {F^{\prime }(s)}{F(s)}}=-\sum _{n=1}^{\infty }{\frac {f(n)\Lambda (n)}{n^{s}}}$ converges for Re(s) > σ0. Here, Λ(n) is the von Mangoldt function.

## Products

Suppose

$F(s)=\sum _{n=1}^{\infty }f(n)n^{-s}$ and

$G(s)=\sum _{n=1}^{\infty }g(n)n^{-s}.$ If both F(s) and G(s) are absolutely convergent for s > a and s > b then we have

${\frac {1}{2T}}\int _{-T}^{T}\,F(a+it)G(b-it)\,dt=\sum _{n=1}^{\infty }f(n)g(n)n^{-a-b}{\text{ as }}T\sim \infty .$ If a = b and ƒ(n) = g(n) we have

${\frac {1}{2T}}\int _{-T}^{T}|F(a+it)|^{2}\,dt=\sum _{n=1}^{\infty }[f(n)]^{2}n^{-2a}{\text{ as }}T\sim \infty .$ ## Integral and series transformations

The inverse Mellin transform of a Dirichlet series, divided by s, is given by Perron's formula. Additionally, if $F(z):=\sum _{n\geq 0}f_{n}z^{n}$ is the (formal) ordinary generating function of the sequence of $\{f_{n}\}_{n\geq 0}$ , then an integral representation for the Dirichlet series of the generating function sequence, $\{f_{n}z^{n}\}_{n\geq 0}$ , is given by 

$\sum _{n\geq 0}{\frac {f_{n}z^{n}}{(n+1)^{s}}}={\frac {(-1)^{s-1}}{(s-1)!}}\int _{0}^{1}\log ^{s-1}(t)F(tz)dt,\ s\geq 1.$ Another class of related derivative and series-based generating function transformations on the ordinary generating function of a sequence which effectively produces the left-hand-side expansion in the previous equation are respectively defined in.

## Relation to power series

The sequence an generated by a Dirichlet series generating function corresponding to:

$\zeta (s)^{m}=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}$ where ζ(s) is the Riemann zeta function, has the ordinary generating function:

$\sum _{n=1}^{\infty }a_{n}x^{n}=x+{m \choose 1}\sum _{a=2}^{\infty }x^{a}+{m \choose 2}\sum _{a=2}^{\infty }\sum _{b=2}^{\infty }x^{ab}+{m \choose 3}\sum _{a=2}^{\infty }\sum _{b=2}^{\infty }\sum _{c=2}^{\infty }x^{abc}+{m \choose 4}\sum _{a=2}^{\infty }\sum _{b=2}^{\infty }\sum _{c=2}^{\infty }\sum _{d=2}^{\infty }x^{abcd}+\cdots$ 