Dirichlet series

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

${\displaystyle \sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}},}$

where s is complex, and ${\displaystyle a_{n}}$ 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:

${\displaystyle {\mathfrak {D}}_{w}^{A}(s)=\sum _{a\in A}{\frac {1}{w(a)^{s}}}=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}}$

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:

${\displaystyle {\mathfrak {D}}_{w}^{A\uplus B}(s)={\mathfrak {D}}_{w}^{A}(s)+{\mathfrak {D}}_{w}^{B}(s).}$

Moreover, if (A, u) and (B, v) are two weighted sets, and we define a weight function w: A × BN by

${\displaystyle w(a,b)=u(a)v(b),}$

for all a in A and b in B, then we have the following decomposition for the Dirichlet series of the Cartesian product:

${\displaystyle {\mathfrak {D}}_{w}^{A\times B}(s)={\mathfrak {D}}_{u}^{A}(s)\cdot {\mathfrak {D}}_{v}^{B}(s).}$

This follows ultimately from the simple fact that ${\displaystyle n^{-s}\cdot m^{-s}=(nm)^{-s}.}$

Examples

The most famous example of a Dirichlet series is

${\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},}$

whose analytic continuation to ${\displaystyle \mathbb {C} }$ (apart from a simple pole at ${\displaystyle s=1}$) 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:

{\displaystyle {\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:

${\displaystyle {\frac {1}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}}$

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

${\displaystyle {\frac {1}{L(\chi ,s)}}=\sum _{n=1}^{\infty }{\frac {\mu (n)\chi (n)}{n^{s}}}}$

where L(χ, s) is a Dirichlet L-function.

Other identities include

${\displaystyle {\frac {\zeta (s-1)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\varphi (n)}{n^{s}}}}$

where ${\displaystyle \varphi (n)}$ is the totient function,

${\displaystyle {\frac {\zeta (s-k)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {J_{k}(n)}{n^{s}}}}$

where Jk is the Jordan function, and

{\displaystyle {\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

{\displaystyle {\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

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

Similarly, we have that

${\displaystyle -\zeta '(s)=\sum _{n=2}^{\infty }{\frac {\log(n)}{n^{s}}},\qquad \Re (s)>1.}$

Here, Λ(n) is the von Mangoldt function. The logarithmic derivative is then

${\displaystyle {\frac {\zeta '(s)}{\zeta (s)}}=-\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}.}$

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

${\displaystyle {\frac {\zeta (2s)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}.}$

Yet another example involves Ramanujan's sum:

${\displaystyle {\frac {\sigma _{1-s}(m)}{\zeta (s)}}=\sum _{n=1}^{\infty }{\frac {c_{n}(m)}{n^{s}}}.}$

Another pair of examples involves the Möbius function and the prime omega function:[1]

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

Analytic properties

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

${\displaystyle f(s)=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}}$

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 ${\displaystyle \{a_{n}\}_{n\in \mathbb {N} }}$ 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

${\displaystyle a_{n}+a_{n+1}+\cdots +a_{n+k}}$

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 ${\displaystyle \sigma }$ is the abscissa of convergence of a Dirichlet series if converges for ${\displaystyle \Re (s)>\sigma }$ and diverges for ${\displaystyle \Re (s)<\sigma .}$ 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

${\displaystyle \sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s_{0}}}}}$

converges for some ${\displaystyle s_{0}\in \mathbb {C} ,\Re (s_{0})>0.}$

Proposition 1. ${\displaystyle A(N):=\sum _{n=1}^{N}a_{n}=o(N^{s_{0}}).}$

Proof. Note that:

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

and define

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

where

${\displaystyle \ell =\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s_{0}}}}.}$

By summation by parts we have

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

Proof. From the definition

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

so that

{\displaystyle {\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 ${\displaystyle N\to \infty }$ whenever ${\displaystyle \Re (s)>\sigma .}$ Hence, for every ${\displaystyle s}$ such that ${\displaystyle \sum _{n=1}^{\infty }a_{n}n^{-s}}$ diverges, we have ${\displaystyle \sigma \geq \Re (s),}$ and this finishes the proof.

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

Proof. Note that

${\displaystyle n^{-s}-(n+1)^{-s}=sn^{-s-1}+O(n^{-s-2})}$

and ${\displaystyle A(N)-f(0)\to 0}$ we have by summation by parts, for ${\displaystyle \Re (s)>0}$

{\displaystyle {\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, ${\displaystyle |A(n)-f(0)|<\varepsilon }$

${\displaystyle s\sum _{n=1}^{\infty }A(n)n^{-s-1}=\underbrace {sf(0)\zeta (s+1)+s\sum _{n=1}^{N}(A(n)-f(0))n^{-s-1}} _{={\mathcal {O}}(1)}+\underbrace {s\sum _{n=N+1}^{\infty }(A(n)-f(0))n^{-s-1}} _{<\varepsilon |s|\int _{N}^{\infty }x^{-\Re (s)-1}\,dx}}$

and hence, for every ${\displaystyle \varepsilon >0}$ there is a ${\displaystyle C}$ such that for ${\displaystyle \sigma >0}$:

${\displaystyle |f(\sigma +it)|[2]

Formal Dirichlet series

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

${\displaystyle D(a,s)=\sum _{n=1}^{\infty }a(n)n^{-s}\ }$

with addition and multiplication defined by

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

where

${\displaystyle (a+b)(n)=a(n)+b(n)\ }$

is the pointwise sum and

${\displaystyle (a*b)(n)=\sum _{k\mid n}a(k)b(n/k)\ }$

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.[3]

Derivatives

Given

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

it is possible to show that

${\displaystyle F'(s)=-\sum _{n=1}^{\infty }{\frac {f(n)\log(n)}{n^{s}}}}$

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

${\displaystyle {\frac {F^{\prime }(s)}{F(s)}}=-\sum _{n=1}^{\infty }{\frac {f(n)\Lambda (n)}{n^{s}}}}$

converges for Re(s) > σ0. Here, Λ(n) is the von Mangoldt function.

Products

Suppose

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

and

${\displaystyle G(s)=\sum _{n=1}^{\infty }g(n)n^{-s}.}$

If both F(s) and G(s) are absolutely convergent for s > a and s > b then we have

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

If a = b and ƒ(n) = g(n) we have

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

Integral and series transformations

The inverse Mellin transform of a Dirichlet series, divided by s, is given by Perron's formula. Additionally, if ${\displaystyle F(z):=\sum _{n\geq 0}f_{n}z^{n}}$ is the (formal) ordinary generating function of the sequence of ${\displaystyle \{f_{n}\}_{n\geq 0}}$, then an integral representation for the Dirichlet series of the generating function sequence, ${\displaystyle \{f_{n}z^{n}\}_{n\geq 0}}$, is given by [4]

${\displaystyle \sum _{n\geq 0}{\frac {f_{n}z^{n}}{(n+1)^{s}}}={\frac {(-1)^{s-1}}{(s-1)!}}\int _{0}^{1}\log ^{s-1}(t)F(tz)dt,\ s\geq 1.}$

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.[5][6]

Relation to power series

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

${\displaystyle \zeta (s)^{m}=\sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}}$

where ζ(s) is the Riemann zeta function, has the ordinary generating function:

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