A Dirichlet series is any series of the form:

$$\sum_{n=1}^{\infty} \frac{a_n}{n^s}$$

where $a_n$ is a sequence of coefficients and $s$ is a complex variable. The Riemann zeta function is the simplest case ($a_n = 1$ for all $n$). But different choices of $a_n$ encode different arithmetic information.

The zeta function as template

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

This is the Dirichlet series with $a_n = 1$. It has an Euler product, analytic continuation, functional equation, and its zeros control the distribution of primes. Every other Dirichlet series is measured against this template.

Dirichlet series from arithmetic functions

Different coefficient sequences $a_n$ produce Dirichlet series that encode different number-theoretic information:

Divisor function $\sigma(n)$ (sum of divisors of $n$):

$$\sum_{n=1}^{\infty} \frac{\sigma(n)}{n^s} = \zeta(s) \cdot \zeta(s-1)$$

The Dirichlet series of $\sigma$ factors as a product of zeta functions. This is not a coincidence — $\sigma$ is a multiplicative function ($\sigma(mn) = \sigma(m)\sigma(n)$ when $\gcd(m,n) = 1$), and multiplicative functions always have Euler products.

Möbius function $\mu(n)$:

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

The Möbius function — which is $(-1)^k$ if $n$ is a product of $k$ distinct primes, and $0$ if $n$ has a squared factor — gives the reciprocal of zeta. This is the heart of Möbius inversion, one of the most useful tools in analytic number theory.

Liouville function $\lambda(n) = (-1)^{\Omega(n)}$:

$$\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}$$

Why Dirichlet series (not power series)?

A power series $\sum a_n x^n$ indexes terms by position ($n$-th term). A Dirichlet series $\sum a_n n^{-s}$ indexes terms by the arithmetic structure of $n$.

The key difference: multiplication of Dirichlet series corresponds to Dirichlet convolution, not ordinary convolution. If $f(s) = \sum a_n n^{-s}$ and $g(s) = \sum b_n n^{-s}$, then:

$$f(s) \cdot g(s) = \sum_{n=1}^{\infty} \frac{c_n}{n^s} \qquad \text{where} \quad c_n = \sum_{d|n} a_d \, b_{n/d}$$

The sum runs over divisors of $n$ — this is multiplicative structure, not additive. The Euler product of the zeta function is a consequence of this: $1 = \mu * 1$ under Dirichlet convolution, so $1/\zeta = \sum \mu(n)/n^s$.

Euler products

A Dirichlet series has an Euler product if it can be written as:

$$\sum_{n=1}^{\infty} \frac{a_n}{n^s} = \prod_{p \text{ prime}} \left(\sum_{k=0}^{\infty} \frac{a_{p^k}}{p^{ks}}\right)$$

This happens precisely when $a_n$ is a multiplicative function. Each prime contributes independently to the product, and each factor is a geometric-like series in $p^{-s}$.

For the zeta function, $a_n = 1$ (multiplicative), so each factor is $\sum p^{-ks} = 1/(1-p^{-s})$ — the geometric series.

Convergence

Every Dirichlet series has an abscissa of convergence $\sigma_c$: the series converges for $\text{Re}(s) > \sigma_c$ and diverges for $\text{Re}(s) < \sigma_c$. For the zeta function, $\sigma_c = 1$. For $\sum \mu(n)/n^s$, $\sigma_c = 1$ as well. The abscissa depends on how fast $a_n$ grows.

There’s also an abscissa of absolute convergence $\sigma_a \geq \sigma_c$. Between the two, the series may converge conditionally — its value depends on the order of summation.