Dirichlet Series and the Abscissa of Convergence

You already know the headline example — the Dirichlet series \sum a(n)/n^{s} that turns a sequence of integers into a function. This page is the graduate upgrade: we stop asking merely "what does the sum equal?" and start pinning down the analytic geometry of the object — exactly where it converges, where it is holomorphic, and why multiplying two of them convolves their coefficients. These are the load-bearing facts behind every generating-function argument in multiplicative number theory.

The single organising picture to hold in your head: a Dirichlet series lives on a vertical half-plane. To its right it converges and defines an analytic function; there is a critical vertical line, the abscissa of convergence, past which everything can fall apart. Power series live on discs; Dirichlet series live on half-planes. Almost every difference between the two theories flows from that one change of shape.

The object and its two critical lines

Fix an arithmetic function a\colon \mathbb{N}\to\mathbb{C} and write s = \sigma + it (Riemann's own notation — \sigma the real part, t the imaginary part). The Dirichlet series of a is

F(s) = \sum_{n=1}^{\infty} \frac{a(n)}{n^{s}}, \qquad \text{where } \frac{1}{n^{s}} = \frac{1}{n^{\sigma}}\,e^{-it\log n}.

Because |n^{-s}| = n^{-\sigma} depends only on the real part \sigma, convergence is governed by \sigma alone — which is precisely why the natural regions are half-planes \sigma > c and not discs. Two thresholds matter, and they are different thresholds:

The values \pm\infty are allowed: \sigma_c = -\infty means the series is entire (e.g. a finite sum), and \sigma_c = +\infty means it converges nowhere (take a(n) = n!).

Why the gap is at most one

The left inequality \sigma_a \ge \sigma_c is free — absolute convergence implies convergence, so the absolute region can only be smaller. The right inequality \sigma_a \le \sigma_c + 1 is the content, and its proof is a clean, worth- memorising estimate.

Suppose F(s) converges at some point s_0 = \sigma_0 + it_0. A convergent series has terms tending to zero, hence bounded: there is M with |a(n)\,n^{-s_0}| \le M, i.e. |a(n)| \le M\,n^{\sigma_0}. Now for any s with \sigma = \Re(s),

\sum_{n=1}^{\infty} \frac{|a(n)|}{n^{\sigma}} \;\le\; M \sum_{n=1}^{\infty} \frac{n^{\sigma_0}}{n^{\sigma}} \;=\; M \sum_{n=1}^{\infty} \frac{1}{n^{\sigma - \sigma_0}},

and the right-hand side is a convergent p-series as soon as \sigma - \sigma_0 > 1. So convergence at \sigma_0 forces absolute convergence for \sigma > \sigma_0 + 1. Letting \sigma_0 \downarrow \sigma_c gives \sigma_a \le \sigma_c + 1. Both endpoints are achieved: for \zeta the two abscissae coincide (\sigma_a=\sigma_c=1), while the alternating series below realises the full gap of exactly 1.

The extreme case: a full unit of separation

The cleanest example where \sigma_a - \sigma_c = 1 is the alternating series with a(n) = (-1)^{n-1}, the Dirichlet eta function:

\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = 1 - \frac{1}{2^{s}} + \frac{1}{3^{s}} - \cdots = \bigl(1 - 2^{1-s}\bigr)\,\zeta(s).

Absolutely, \sum |a(n)|/n^{\sigma} = \sum 1/n^{\sigma} needs \sigma > 1, so \sigma_a = 1. But the sign alternation buys genuine cancellation: by Dirichlet's test the series converges for every \sigma > 0 (partial sums of (-1)^{n-1} are bounded, and n^{-\sigma} decreases to 0), so \sigma_c = 0. The gap is the maximal \sigma_a - \sigma_c = 1. The plot below shows partial sums stabilising well to the left of \sigma = 1, exactly where the absolutely-convergent \zeta series has already given up.

Notice what the picture is not saying: \eta(s) is finite and well-behaved even at \sigma \le 0 because (1-2^{1-s})\zeta(s) continues it there — but the series no longer represents it once \sigma < 0. Convergence of the sum and existence of the function are two different questions; more on that in the Watch-out.

Holomorphy in the half-plane

The engine is a partial-summation (Abel) argument: convergence at a single point s_0 upgrades to uniform convergence throughout any angular sector |\arg(s - s_0)| \le \tfrac{\pi}{2} - \delta — a Stieltjes/Abel wedge opening to the right. Uniform limits of holomorphic functions are holomorphic (Weierstrass), so F is holomorphic on the whole open half-plane. This is the payoff that makes the whole subject analytic number theory: the moment your arithmetic sequence has a convergent Dirichlet series, complex analysis — Cauchy's theorem, contour shifting, residues — is licensed to operate on it.

The algebra: multiplication is Dirichlet convolution

If A(s) = \sum a(n)n^{-s} and B(s) = \sum b(n)n^{-s} both converge absolutely at s, then their product is the Dirichlet series of the Dirichlet convolution a * b:

A(s)\,B(s) = \sum_{n=1}^{\infty} \frac{(a * b)(n)}{n^{s}}, \qquad (a * b)(n) = \sum_{d \mid n} a(d)\,b\!\left(\tfrac{n}{d}\right).

The proof is a one-line bookkeeping miracle that only Dirichlet series can perform. Multiply the two absolutely convergent sums and collect terms by the product of denominators:

A(s)B(s) = \sum_{d=1}^{\infty}\sum_{e=1}^{\infty} \frac{a(d)\,b(e)}{(de)^{s}} = \sum_{n=1}^{\infty} \frac{1}{n^{s}} \underbrace{\sum_{de = n} a(d)\,b(e)}_{=\,(a*b)(n)}.

The exponents add — d^{-s}e^{-s} = (de)^{-s} — so grouping by n = de collects exactly the pairs of divisors. Absolute convergence is what licenses the rearrangement (the double sum may be summed in any order). Thus the ring of Dirichlet series under ordinary multiplication is a faithful copy of the ring of arithmetic functions under *; the multiplicative unit is the series of \varepsilon(n) = [n=1], namely the constant function 1.

Worked example — squaring zeta. With a = b = \mathbf{1} (all ones, A=B=\zeta), the convolution (\mathbf 1 * \mathbf 1)(n) = \sum_{d\mid n} 1 = d(n) counts divisors, so instantly

\zeta(s)^{2} = \sum_{n=1}^{\infty} \frac{d(n)}{n^{s}} \qquad (\Re(s) > 1).

A gallery built from zeta

Four cornerstone identities, each a convolution fact wearing analytic clothing. All hold on the half-plane where both sides converge absolutely — note how the abscissa moves with the size of the coefficients.

IdentityCoefficients a(n)Convolution behind it\sigma_a
\zeta(s) = \sum n^{-s}1—1
\dfrac{1}{\zeta(s)} = \sum \mu(n)\,n^{-s}\mu(n)\mu * \mathbf 1 = \varepsilon1
\zeta(s)^{2} = \sum d(n)\,n^{-s}d(n)\mathbf 1 * \mathbf 1 = d1
\dfrac{\zeta(s-1)}{\zeta(s)} = \sum \varphi(n)\,n^{-s}\varphi(n)\varphi * \mathbf 1 = \mathrm{id}2

The Möbius reciprocal. Because Möbius satisfies \mu * \mathbf 1 = \varepsilon, taking Dirichlet series gives \bigl(\sum \mu(n)n^{-s}\bigr)\zeta(s) = 1, i.e. \sum \mu(n)/n^{s} = 1/\zeta(s). This is the analytic face of Möbius inversion.

The totient quotient. Euler's totient obeys \varphi * \mathbf 1 = \mathrm{id} (i.e. \sum_{d\mid n}\varphi(d) = n). The identity function \mathrm{id}(n) = n has series \sum n/n^{s} = \sum n^{-(s-1)} = \zeta(s-1), so dividing by \zeta(s) yields

\sum_{n=1}^{\infty}\frac{\varphi(n)}{n^{s}} = \frac{\zeta(s-1)}{\zeta(s)} \qquad (\Re(s) > 2).

The abscissa jumped to 2: since \varphi(n) is of size \asymp n, absolute convergence needs \sum n/n^{\sigma} = \sum n^{-(\sigma-1)} to converge, forcing \sigma > 2 — exactly the shift you read off from the \zeta(s-1) in the numerator.

Euler products: multiplicativity factorises the series

If a is multiplicative (a(mn) = a(m)a(n) for \gcd(m,n)=1, a(1)=1) and \sum |a(n)|n^{-\sigma} converges, then the Dirichlet series factors over the primes:

\sum_{n=1}^{\infty}\frac{a(n)}{n^{s}} = \prod_{p \text{ prime}} \left( 1 + \frac{a(p)}{p^{s}} + \frac{a(p^{2})}{p^{2s}} + \cdots \right).

If moreover a is completely multiplicative, each local factor is a geometric series and collapses to \bigl(1 - a(p)p^{-s}\bigr)^{-1}.

The mechanism is the unique factorisation of the integers: expanding the product picks exactly one term a(p^{k})p^{-ks} from each prime's bracket, and multiplying them reconstructs a(p_1^{k_1}\cdots p_r^{k_r})/n^{s} = a(n)/n^{s} for one and only one n. The all-ones sequence is completely multiplicative, giving the prototype \zeta(s) = \prod_p (1 - p^{-s})^{-1} — the analytic embodiment of "every integer factors uniquely into primes." Euler products are the reason Dirichlet series see primes and not just integers.

Uniqueness: coefficients are recoverable

Here is why. Suppose F(s) = \sum c(n)n^{-s} \equiv 0 with not all c(n)=0; let N be the smallest index with c(N)\neq 0. Multiply by N^{s} and isolate the first term:

c(N) = -\sum_{n > N} c(n)\left(\frac{N}{n}\right)^{s}.

Every ratio N/n < 1, so each (N/n)^{s} \to 0 as s \to +\infty along the reals — the tail is dominated by (N/(N+1))^{\sigma}\to 0. The right side vanishes in the limit, forcing c(N) = 0, a contradiction. So the coefficients are an invariant of the analytic function: the map "sequence \mapsto Dirichlet series" is injective. This is what makes reading identities off a series — as we did all through the gallery — rigorous rather than merely suggestive.

Three lines in the plane routinely get conflated. Keep them apart:

1. \sigma_c vs \sigma_a (conditional vs absolute). Between them — on \sigma_c < \sigma < \sigma_a — the series converges but not absolutely. There you may not rearrange terms, multiply two series termwise via the convolution formula, or swap sum and integral without extra justification; those manipulations need \sigma > \sigma_a. The eta series lives entirely in this danger zone for 0 < \sigma < 1.

2. The half-plane of convergence is NOT the domain of the function. The series \sum \mu(n)/n^{s} only converges for \sigma > 1, yet 1/\zeta(s) is defined and holomorphic far to the left (everywhere \zeta is nonzero). Analytic continuation is a separate operation — it extends the function, but the series genuinely diverges outside its half-plane. Writing "\sum \mu(n)/n^{s} = 1/\zeta(s) at s = \tfrac12" is nonsense: the left side does not converge there. Unlike a power series, whose boundary always carries a singularity, a Dirichlet series can be perfectly regular on and past its abscissa of convergence — the line \sigma = \sigma_c is a convergence boundary, not necessarily a singularity boundary.

3. Termwise size does not decide convergence. The Möbius coefficients are bounded by 1 just like the all-ones coefficients, yet people are tempted to think small coefficients "should converge further left." They don't, by themselves — what buys extra ground is cancellation (sign changes), as in the eta series. Absolute convergence ignores signs and so can never beat \sigma_a.

For a power series, 1/R = \limsup |a_n|^{1/n} (Cauchy–Hadamard). Dirichlet series have their own partial-sum version. Writing A(x) = \sum_{n \le x} a(n) for the summatory function, when the series diverges at s=0 the abscissa of convergence is

\sigma_c = \limsup_{x \to \infty} \frac{\log |A(x)|}{\log x},

and the absolute abscissa is the same formula with |a(n)| summed instead. So the abscissa measures the polynomial growth rate of the partial sums of the coefficients — arithmetic size directly controls analytic reach. For \mu, the deep fact that its Mertens sum satisfies M(x) = o(x) (equivalent to the Prime Number Theorem) is a statement about exactly this abscissa on the boundary line \sigma = 1.

The takeaway

A Dirichlet series is a dictionary entry translating between two worlds. On the arithmetic side sit sequences, Dirichlet convolution, and multiplicativity; on the analytic side sit holomorphic functions on half-planes, ordinary multiplication, and Euler products. The abscissae \sigma_c \le \sigma_a \le \sigma_c + 1 tell you where the dictionary is literally valid, uniqueness guarantees the translation is faithful, and holomorphy is the passport that lets complex analysis cross over and attack problems about the integers — starting, in the next pages, with the analytic continuation and functional equation of \zeta itself.