In analytic number theory, they denote a complex variable by ${s = \sigma + it}$. Its an unfortunate choice of notation but has become too banal to be able to change it. In this post, we shall study the important notion of a Dirichlet series and its convergence in the complex plane. One important tool in analytic number theory is that by defining various meromorphic functions, we look at the residue at their poles. This residue gives a lot of number-theoretic information. We define the Dirichlet series to be of the form

$\displaystyle f(s) = \displaystyle\sum_{n\in\mathbb N} \frac{a_n}{n^s}, \qquad a_n \in \mathbb C.$

The `prime’ example of a Dirichlet series is the famous Riemann ${\zeta}$-function defined by

$\displaystyle \zeta(s) = \displaystyle\sum_{n\in\mathbb N} \frac{1}{n^s}.$

Finding its zeros (roots) is a central open problem in number theory, called the Riemann Hypothesis and may be one of the toughest ways to be a millionaire!

There are other generalizations like Dedekind’s ${\zeta}$-function given by

$\displaystyle \zeta_K(s) = \displaystyle\sum \frac{1}{\mathbb N(\mathfrak a)^s},$

where the sum is taken over all proper ideals of the ring of integers of the number field ${K}$ and $\mathbb{N}$ denotes the norm of the ideal. There are also the Dirichlet ${L}$-series and Hecke ${L}$-series which further generalize this but we won’t define them now. We shall see the convergence of all these in the complex plane follow by using a lemma whose proof uses only basic complex analysis. Interested readers can look up for a proof in Janusz’s Algebraic Number Fields, Ch. IV Prop. 2.1.

Lemma 1 Let ${f(s)}$ be the Dirichlet series as defined above and let ${S(x) = \sum_{n \leq x} a_n}$. Suppose there exist positive constants ${a}$ and ${b}$ such that ${|S(x)| \leq a x^b}$ for all ${x >> 0}$. Then the series ${f(s)}$ is uniformly convergent in ${D(b, \delta, \varepsilon)}$ with any positive ${\delta, \varepsilon}$, where

$\displaystyle D(b, \delta, \varepsilon) = \{ s : \sigma \geq b + \delta, |arg(s-b)\leq \pi/2 - \varepsilon \}.$

In particular, ${f(s)}$ is analytic in the half plane ${Re(s) > b}$.

We use this lemma to prove interesting stuff about poles at the Riemann-${\zeta}$ function. The proof involves a smart trick.

Proposition 2

1. ${\zeta(s)}$ is analytic for ${Re(s) > 1}$.
2. ${\zeta(s)}$ extends to a meromorphic function for ${Re(s) > 0}$.
3. ${\zeta(s)}$ has only one pole in ${Re(s) > 0}$; its located at ${s = 1}$ and is a simple pole.

Proof:

1. This follows from the lemma since ${S(x) \leq x}$.

2. Here is a cool trick: define

${\zeta_2(s) = 1 - \displaystyle\frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} \cdots }$

and observe that

$\displaystyle \zeta_2(s) = \displaystyle \sum \frac{1}{n^s} - 2. \frac{1}{2^s} \sum \frac{1}{n^s}$

so that

$\displaystyle \zeta(s) = \zeta_2(s) . \frac{1}{1 - 2^{(1-s)}}. \ \ \ \ \ (1)$

Since ${S(x) = 0, 1}$ for ${\zeta_2}$, it follows by the lemma that ${\zeta_2}$ is analytic on ${Re(s) > 0}$. Since the other fraction in equation (1)above is meromorphic, it follows that ${\zeta}$ too is meromorphic. This settles (2).

3. By (1), the poles of ${\zeta}$ must be poles of the function ${\displaystyle\frac{1}{1-2^{(1-s)}}}$, which are precisely the points ${s = 1 + \displaystyle\frac{2\pi i m}{\ln 2}, \; m \in\mathbb Z}$.

Now similar to ${\zeta_2}$, define

$\displaystyle \zeta_3 (s) = 1 + \displaystyle \frac{1}{2^s} - \frac{2}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} - \frac{2}{6^s} \cdots .$

Conclude that

$\displaystyle \zeta(s) = \zeta_3(s) . \displaystyle\frac{1}{1-3^{(1-s)}},$

so the possible poles of ${\zeta(s)}$ are ${s = 1 + \displaystyle\frac{2 \pi i n }{\ln 3}, \; n \in \mathbb Z}$. Since the only common pole of ${\zeta_2}$ and ${\zeta_3}$ is ${1}$, it must be the only pole of ${\zeta}$! Further, the order of the pole of ${\zeta}$ at ${1}$ must equal the order of ${\displaystyle\frac{1}{1-2^{(1-s)}}}$ which is 1. The pole is hence, simple. $\Box$

As I said earlier, the residue at this pole for various Dirichlet series gives number-theoretic information but I will discuss that after learning further.