A Dirichlet series is a series of the form
\[
\alpha(s) = \sum_{n = 1}^{\infty} a_n n^{-s}.
\] It is a general theme in analytic number theory to study a sequence ( arithmetic function) by means of its Dirichlet series. By studying analytic properties of Dirichlet series $\alpha(s)$ we can glean information about the sequence $\{a_n\}$. The theory of Dirichlet series parallels the theory of power series in many ways. We highlight two such similarities below:
- The points where a power series converges are separated from the points where it does not by a circle, i.e., if $P(z) = \sum_{n = 1}^{\infty} a_n z^n$ is a power series, then there exists a radius of convergence $r_c \in [0, \infty]$ such that $P(z)$ converges for $|z| < r_c$ and diverges for $|z| > r_c$. As for the Dirichlet series, the points where it converges are separated from the points it does not by a vertical line, i.e., given a Dirichlet series $\alpha(s)$ there exists a $\sigma_c \in [-\infty, \infty]$ which we call the abscissa of convergence such that $\alpha(s)$ converges for $\sigma > \sigma_c$ and diverges for $\sigma < \sigma_c$.
- Just as in the case of power series which is analytic inside its disc of convergence, a Dirichlet series is also analytic to the right of its abscissa of convergence.
We also note two key differences between power series and Dirichlet series.
- A power series converges absolutely inside its disc of convergence but the analogous statement does not hold in the case of Dirichlet series; if $\sigma > \sigma_c$, then the Dirichlet series $\alpha(s)$ need not converge absolutely. There is however a vertical line $\sigma = \sigma_a$ which separates the points where $\alpha(s)$ converges absolutely from those points where it does not. We shall see later that $\sigma_a$ and $\sigma_c$ cannot be too far apart.
- An important property of power series is that they converge until a singularity is encountered. Precisely speaking, if $r_c$ is the radius of convergence of a power series $P(z) = \sum_{n = 0}^{\infty} a_n z^n$, then $P(z)$ cannot be analytically continued to an open set containing the closed disc of radius $r_c$. This fails however for Dirichlet series, i.e., a Dirichlet series $\alpha(s)$ can possibly be analytically continued to the left of the line $\sigma = \sigma_c$ and in some cases to the entire complex plane. We will later see that if however $a_n \geq 0$ for every (sufficiently large) $n$, then $\alpha(s)$ must have a singularity at $\sigma_c$.
It is easy to see that there exists a $\sigma_a \in [-\infty, \infty]$ such that $\alpha(s)$ converges absolutely for $\sigma > \sigma_a$ and does not converge absolutely for $\sigma < \sigma_a$. Note that if $\alpha(s)$ converge absolutely at the point $s = s_0$, then for $\sigma \geq \sigma_0$ we have
\[
\sum_{n = 1}^{\infty} |a_n n^{-s}| = \sum_{n = 1}^{\infty} |a_n| n^{- \sigma} \leq \sum_{n = 1}^{\infty} |a_n|n^{-\sigma_0} = \sum_{n = 1}^{\infty} |a_n n^{-s_0}|.
\] Now $\sigma_a$ is just the infimum of all $\sigma \in \mathbb{R}$ for which $\alpha(\sigma)$ converges absolutely. Note that if $a_n \ll 1$, then $\sigma_a \leq 1$.
We now show that tail of a Dirichlet series decays exponentially with $\sigma$. Observe that if $N \in \mathbb{N}$ and $\sigma_0 > \sigma_a$ is fixed, then for $\sigma \geq \sigma_0$ we have
\[
\left| \sum_{n \geq N} a_n n^{-s} \right| \leq \sum_{n \geq N} |a_n| n^{-\sigma} = \sum_{n \geq N} |a_n| n^{-\sigma_0} n^{\sigma_0 – \sigma} \leq N^{\sigma_0 – \sigma} \sum_{n =1}^{\infty} |a_n| n^{-\sigma_0}.
\]We can write this concisely as
\[
\sum_{n = N}^{\infty} a_n n^{-s} \ll_{N} N^{-\sigma}
\] where $\sigma \geq \sigma_0$. An interesting consequence of this bound is that the first coefficient of a Dirichlet series is uniquely determined. This is because
\[
a_1 = \lim_{\sigma \to \infty} \alpha(s).
\] Using induction it is not difficult to show that all of the coefficients of a Dirichlet series are uniquely determined. But we prove this in a slightly different and more general way as follows:
Theorem. Let $\alpha(s) = \sum_{n = 1}^{\infty} a_n n^{-s}$ and $\beta(s) = \sum_{n = 1}^{\infty} b_n n^{-s}$ be Dirichlet series both of which converge absolutely for $\sigma > \sigma_0$. Let $\{s_k\}_{k = 1}^{\infty}$ be a sequence of complex numbers such that $\sigma_k > \sigma_0$ for every $k$ and $\sigma_k \to \infty$ as $k \to \infty$. If $\alpha(s_k) = \beta(s_k)$ for every $k$, then we have $\{a_n\} = \{b_n\}$.
Proof. Let $c_n = a_n- b_n$. Then $\gamma(s) = \sum_{n = 1}^{\infty} c_n n^{-s}$ converges absolutely for $\sigma > \sigma_0$ and $\gamma(s_k) = 0$ for every $k$. Our goal is to show that $\{c_n\} = 0$ . Suppose for the sake of contradiction that this is not the case. Then take $N$ to be smallest $n$ such that $c_n \neq 0$. Now note that if $\gamma(s) = 0$, then
\[
c_N= -N^s \sum_{n > N} c_n n^{-s}.\] Without loss of generality assume that $\sigma_0 > \sigma_a(\gamma)$ (for if $\sigma_0 = \sigma_a(\gamma)$, then we can take $\sigma_0$ to be slightly bigger). Then it follows from above result about decay of tail of Dirichlet series that if $\gamma(s) = 0$ and $\sigma \geq \sigma_0$, then
\[
|c_N| \leq N^{\sigma} (N +1)^{\sigma_0 – \sigma} \sum_{n = 1}^{\infty} |c_n|n^{-\sigma_0} \ll \left( \frac{N}{N + 1} \right)^{\sigma},
\] where the implicit constant depends on $\sigma_0$ and $N$ but is independent of $s$. Since $\gamma(s_k) = 0$ for every $k$ we have
\[
|c_N| \ll \left( \frac{N}{N + 1} \right)^{\sigma_k}.
\]The quantity to the right tends to $0$ as $k \to \infty$ and so we conclude that $c_N = 0$, a contradiction. Thus we must have $c_n = 0$ for every $n$.
It follows from above result that if $\{a_n\} \neq 0$ and $\sigma_a < \infty$, then there exists a half plane $\sigma > \sigma_0$ in which the Dirichlet series $\alpha(s) = \sum_{n = 1}^{\infty} a_n n^{-s}$ does not vanish. We now show that the sequence $\{a_n\}$ of coefficients of a Dirichlet series $\alpha(s)$ is determined completely by values of $\alpha(s)$ along a vertical line $\sigma = \sigma_0$ to the right of abscissa of absolute convergence ($\sigma_0 > \sigma_a$). The fact that $\{a_n\}$ is determined uniquely by values of $\alpha(s)$ along the vertical line $\sigma = \sigma_a$ follows from the previous result by analytic continuation.
Theorem. Let $\{a_n\}$ be a sequence and let $a_x = 0$ for $x > 0$ and $x$ not an integer. Then we have
\[
a_x = \lim_{T \to \infty} \frac{1}{2 T} \int_{-T}^{T} \alpha(\sigma + it) x^{\sigma + it} \, dt
\] for $\sigma > \sigma_a$.
Proof. Note that
\[
\int_{-T}^{T} \alpha(\sigma + it) x^{\sigma + it} \, dt = x^{\sigma} \int_{-T}^{T} \sum_{n = 1}^{\infty} a_n n^{-\sigma} \left( \frac{x}{n} \right)^{it} \, dt.
\] Since the Dirichlet series converges uniformly on the line $\sigma- i \infty, \sigma + i \infty$, we can interchange the order of summation and integration to obtain
\[
\int_{-T}^{T} \alpha(\sigma + it) x^{\sigma + it} \, dt = x^{\sigma} \sum_{n = 1}^{\infty} a_n n^{-\sigma} \int_{-T}^{T} e^{it \log (x/n)} \, dt.
\] Note that if $n = x$, then the integral is $2T$. However, if $n \neq x$, then we have
\[
\int_{-T}^{T} e^{it \log(x/n)} \, dt = \frac{2 \sin(T \log (x/n))}{\log (x/n)}
\] Observe that we have $\log (x/n) \gg 1$ for $n \neq x$, where the implicit constant depends on $x$. It thus follows that for a fixed $x$ we have
\[
\int_{-T}^{T} e^{it \log (x/n)} \, dt \ll 1
\] for $n \neq x$. This now yields the estimate
\[
\int_{-T}^{T} \alpha(\sigma + it) x^{\sigma + it} \, dt = 2Ta_x + O \left( x^{\sigma} \sum_{n = 1}^{\infty} |a_n| n^{-\sigma} \right),
\] where the implicit constant depends on $x$ only. This immediately leads to the desired formula.
Let $\alpha(s) = \sum_{n =1}^{\infty} a_n n^{-s}$ and $\beta(s) = \sum_{n = 1}^{\infty} b_n n^{-s}$ be two Dirichlet series and let $\gamma(s) = \alpha(s) \beta(s)$. If $\alpha(s)$ and $\beta(s)$ both converge absolutely, then
\[
\gamma(s) = \sum_{n = 1}^{\infty} a_n n^{-s} \sum_{m = 1}^{\infty} b_m m^{-s} = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} a_n b_m (nm)^{-s}.
\] Because of absolutely convergence we can rearrange the terms and obtain
\[
\gamma(s) = \sum_{k = 1}^{\infty} \left(\sum_{nm = k} a_n b_m \right) k^{-s}.
\] Thus $\gamma(s)$ is a Dirichlet series, $\gamma(s) = \sum_{k =1}^{\infty} c_k k^{-s}$, with coefficients $c_k = \sum_{nm = k} a_n b_m$. Moreover, $\gamma(s)$ also converges absolutely.
Let $\{a_n\}$ be such that $a_1 \neq 0$ and let $\{b_n\}$ be the Dirichlet inverse of $\{a_n\}$, i.e., $\sum_{d \, | \, n} a_n b_{n/d} = e(n)$, where $e(1) = 1$ and $e(n) = 0$ for $n > 1$. If the Dirichlet series $\alpha(s) = \sum_{n = 1}^{\infty} a_n n^{-s}$ and $\beta(s) = \sum_{n = 1}^{\infty} b_n n^{-s}$ both converge absolutely for $\sigma > \sigma_0$, then $\alpha(s) \beta(s) = 1$ for $\sigma > \sigma_0$. In particular, $\alpha(s) \neq 0$ for $\sigma > \sigma_0$. If $\{a_n\}$ represents a completely multiplicative arithmetic function, then $\{b_n\} = \{\mu(n) a_n\}$ is the Dirichlet inverse of $\{a_n\}$ and so $\beta(s)$ converges absolutely for $\sigma > \sigma_a$, where $\sigma_a$ is the abscissa of absolute convergence of $\alpha(s)$. Hence, $\alpha(s) \neq 0$ for $\sigma > \sigma_a$ and
\[
\beta(s) = \sum_{n = 1}^{\infty} b_n n^{-s} = \sum_{n = 1}^{\infty} \mu(n) a_n n^{-s} = \alpha(s)^{-1}
\] for $\sigma > \sigma_a$. This shows that the Dirichlet series of a completely multiplicative function does not vanish in its half plane of absolute convergence. As an example, the Riemann zeta function
\[
\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}
\] does not vanish for $\sigma > 1$ and
\[
\sum_{n = 1}^{\infty} \mu(n) n^{-s} = \frac{1}{\zeta(s)}.
\]