Dirichlet Series

Dirichlet series are functions of a complex variable \( s \) that are defined by certain infinite series. They are generalizations of the Riemann zeta function, and are important in number theory due to their deep connections with the distribution of prime numbers. They have interesting connections with multiplicative functions and Dirichlet convolution.

A Dirichlet series is an expression of the form

\[\sum_{n=1}^{\infty} \frac{a_n}{n^s}.\]

Here \( s \) is a complex variable and \( a_n \) is a sequence of complex numbers. If we write the \( a_n \) as the values of an arithmetic function \( f(n) = a_n, \) we say that the above series is the Dirichlet series associated with \( f.\)

When \( a_n = 1 \) for all \( n, \) the associated Dirichlet series is \( \sum \frac1{n^s}, \) the Riemann zeta function \( \zeta(s).\) So the Dirichlet series associated with the function \( \mathbf{1} \) is \( \zeta(s). \)

Given two Dirichlet series \( F(s) \) and \( G(s) \), the Dirichlet series representation of the product \( F(s)G(s) \) turns out to be related to Dirichlet convolution.

Let

\[F(s) = \sum_{n=1}^{\infty} \frac{f(n)}{n^s}, \ \ \ G(s) = \sum_{n=1}^{\infty} \frac{g(n)}{n^s}.\]

Suppose that the series both converge for a given value of \( s \) and at least one of them converges absolutely. Then \( \sum_{n=1}^{\infty} \frac{(f*g)(n)}{n^s} \) converges, where \( f*g \) is the Dirichlet convolution of \( f \) and \( g \), and

\[F(s)G(s) = \sum \frac{(f*g)(n)}{n^s}.\]

Partial proof:
Ignoring questions of convergence, consider the product \( F(s)G(s):\)

\[F(s)G(s) = \left( f(1) + \frac{f(2)}{2^s} + \cdots \right) \left( g(1) + \frac{g(2)}{2^s} + \cdots \right).\]

This expands to give a sum of fractions of the form

\[\frac{f(i)g(j)}{(ij)^s}.\]

The terms with denominator \( n^s \) will be ones where \( j =\frac ni. \) So the coefficient of \( \frac1{n^s}\) will be \( \sum_{i|n} f(i)g\left(\frac ni\right), \) which is just \( f*g \). \(_\square\)

The Riemann zeta function has a well-known product formula, which is derived from unique factorization and the formula for the sum of a geometric series:

\[\begin{align} \sum_{n=1}^{\infty} \frac1{n^s} &= 1 + \frac1{2^s} + \frac1{3^s} + \cdots \\ &= \left(1+\frac1{2^s}+\frac1{4^s}+\cdots\right)\left(1+\frac1{3^s}+\frac1{9^s} +\cdots\right) \left(1+\frac1{5^s}+\frac1{25^s}+\cdots\right)\left(\cdots\right)\\ &= \frac1{1-\frac{1}{2^s}} \cdot \frac1{1-\frac{1}{3^s}} \cdot \frac1{1-\frac{1}{5^s}} \cdots \\ &= \prod_{p \ \text{prime}} \left( 1-\frac1{p^s} \right)^{-1}.
\end{align}\]

We can generalize this to any Dirichlet series associated with a multiplicative function:

Given that \( F(s) = \sum_{n=1}^{\infty} \frac{f(n)}{n^s} \) converges absolutely,

• if \( f \) is a multiplicative function, then \[F(s) = \prod_{p \ \text{prime}} \left( 1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots \right);\]
• if \( f \) is completely multiplicative, then \[F(s) = \prod_{p \ \text{prime}} \left( 1-\frac{f(p)}{p^s} \right)^{-1}.\]

The proof is essentially the same as the above derivation for the Riemann zeta function.

Show that

\[\sum_{n=1}^\infty \frac{\mu(n)}{n^s}= \frac1{\zeta(s)}.\]

---

The product formula gives

\[\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \prod_{p \ \text{prime}} \left( 1 - \frac1{p^s} \right),\]

which is the reciprocal of the product formula for \( \zeta(s) \). Another way to solve the problem is to use the result on products of Dirichlet series: note that \( \mu * \mathbf{1} = e,\) the convolution identity function, so the product of the associated Dirichlet series is \( \sum \frac{e(n)}{n^s}= 1 \). \(_\square \)

Show that

\[\sum_{n=1}^\infty \frac{\phi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)},\]

where \( \phi \) is the Euler phi function.

---

\( \phi = \mu * I \), where \( I(n) = n \) (see Dirichlet convolution for details), so the theorem above gives

\[\begin{align} \sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} &= \left( \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \right) \left( \sum_{n=1}^{\infty} \frac{n}{n^s} \right) \\ &= \frac1{\zeta(s)} \sum_{n=1}^\infty \frac1{n^{s-1}} \\ &= \frac{\zeta(s-1)}{\zeta(s)}. \ _\square \end{align}\]

(Exercise: do this example a different way, by using the product formula as in the previous example.)