The Riemann Zeta Function

How many primes are there below a billion? How are they spaced? This question, asked precisely, leads directly to a single complex-valued function. That function encodes the entire multiplicative structure of the integers. The Riemann zeta function begins as an unremarkable-looking infinite series. But Euler's discovery that it factors as a product over primes changes this. It turns the series into a bridge between additive analysis and multiplicative arithmetic. That bridge is the reason analytic tools — convergence, continuity, complex differentiation — can say anything at all about a purely combinatorial object like the set of primes. The idea matters because it inaugurates the entire field of analytic number theory. A series built from all positive integers can be rewritten as a product over primes. Primes are much sparser and much more mysterious than all integers. Once this rewriting is known, every analytic fact about the series becomes a candidate source of arithmetic information. Convergence radii become statements about growth rates of arithmetic functions. Poles become statements about densities. Zeros, as later chapters of this story reveal, become statements about the error term in the Prime Number Theorem itself. The big picture is this: ζ(s)\zeta(s) is defined by an ordinary Dirichlet series for Re⁡(s)>1\operatorname{Re}(s) > 1. There, it converges absolutely and defines a holomorphic function. Its Euler product, valid in the same half-plane, is algebraically equivalent to unique factorization of integers into primes. The function extends meromorphically to the whole complex plane with exactly one singularity. This singularity is a simple pole at s=1s = 1. Its residue is the seed of the connection to prime counting. This single fact — a pole at s=1s=1 with residue 11 — is already enough to prove that there are infinitely many primes. It is the starting point for everything deeper. This article gives a complete, self-contained construction of ζ(s)\zeta(s) on Re⁡(s)>1\operatorname{Re}(s) > 1. It gives a full proof of the Euler product using unique factorization and absolute convergence. It also gives a rigorous account of the analytic continuation to Re⁡(s)>0\operatorname{Re}(s) > 0, with the pole at s=1s=1 isolated explicitly. Together, these build the exact platform needed for Zeta and the Distribution of Primes and The Prime Number Theorem: Statement and Heuristics.

Key takeaways
  • ζ(s)=∑n=1∞n−s\zeta(s) = \sum_{n=1}^{\infty} n^{-s} converges absolutely and defines a holomorphic function for Re⁡(s)>1\operatorname{Re}(s) > 1.
  • Unique factorization gives Euler's product ζ(s)=∏p prime(1−p−s)−1\zeta(s) = \prod_{p \text{ prime}} (1 - p^{-s})^{-1} on that same half-plane.
  • The Euler product shows ζ(s)≠0\zeta(s) \neq 0 for Re⁡(s)>1\operatorname{Re}(s) > 1, and its divergence as s→1+s \to 1^+ already reproves the infinitude of primes.
  • A partial-summation (Abel summation) argument continues ζ(s)\zeta(s) meromorphically to Re⁡(s)>0\operatorname{Re}(s) > 0.
  • The unique singularity in Re⁡(s)>0\operatorname{Re}(s) > 0 is a simple pole at s=1s = 1 with residue 11.
  • Key idea: absolute convergence of a Dirichlet series over all integers, combined with unique factorization, converts a sum into a product over primes — turning analysis into arithmetic.

Where this lives today

The Riemann zeta function is the prototype for a vast family of LL-functions used throughout modern number theory: Dirichlet LL-functions (proving primes in arithmetic progressions), Dedekind zeta functions of number fields, and automorphic LL-functions attached to modular forms and Galois representations, all built on the same Euler-product philosophy established here. In analytic number theory proper, the location of zeros of ζ(s)\zeta(s) controls the error term in the Prime Number Theorem, connecting this article directly to Zeta and the Distribution of Primes and The Prime Number Theorem: Statement and Heuristics. In physics, ζ\zeta regularization — assigning finite "values" to divergent sums via the continuation, such as the notorious formal identity involving ζ(−1)\zeta(-1) — is used in the Casimir effect and in string theory to regularize divergent vacuum energies. In probability and random matrix theory, the conjectural statistics of the zeros of ζ\zeta on the critical line match the eigenvalue statistics of random Hermitian matrices (the Montgomery–Odlyzko phenomenon), linking analytic number theory to quantum chaos. In cryptography, understanding the distribution of primes — ultimately governed by ζ\zeta — underlies algorithms for primality testing and the security assumptions of RSA-style cryptosystems, since generating large primes efficiently depends on quantitative prime density estimates descended from this theory. Finally, in computer science and computational number theory, efficient algorithms (the Riemann–Siegel formula, Odlyzko–Schönhage algorithm) are built precisely to compute ζ(s)\zeta(s) and locate its zeros to enormous precision, testing the Riemann Hypothesis numerically for hundreds of billions of zeros.

Ask the tutor: How does the location of zeros of ζ(s)\zeta(s) inside the strip 0≤Re⁡(s)≤10 \le \operatorname{Re}(s) \le 1 control the error term in the Prime Number Theorem?

DefinitionThe Riemann zeta function

For a complex number s=σ+its = \sigma + it with σ=Re⁡(s)>1\sigma = \operatorname{Re}(s) > 1, the Riemann zeta function is defined by the Dirichlet series

ζ(s)  =  ∑n=1∞1ns,\zeta(s) \;=\; \sum_{n=1}^{\infty} \frac{1}{n^{s}},

where n−sn^{-s} means e−slog⁡ne^{-s\log n} with log⁡n\log n the real (positive) natural logarithm of the positive integer nn. The series converges absolutely for Re⁡(s)>1\operatorname{Re}(s) > 1 and, as shown below, converges locally uniformly there, so ζ\zeta is holomorphic on the open half-plane {Re⁡(s)>1}\{\operatorname{Re}(s) > 1\}.

Intuition

Think first of the real variable case: for real s>1s > 1, ζ(s)=∑n−s\zeta(s) = \sum n^{-s} is just a pp-series, familiar from calculus, converging because the terms decay faster than 1/n1/n. What is new and remarkable is what happens when you let ss range over complex numbers with Re⁡(s)>1\operatorname{Re}(s) > 1: the same absolute convergence argument works verbatim, because ∣n−s∣=n−Re⁡(s)|n^{-s}| = n^{-\operatorname{Re}(s)} depends only on the real part of ss. So the domain of definition is naturally a half-plane, not an interval — the first hint that complex analysis, not just real analysis, is the right setting. The deeper structural fact is Euler's product formula. Every positive integer factors uniquely into primes, and this arithmetic fact has an analytic shadow: expanding the geometric series (1−p−s)−1=1+p−s+p−2s+⋯(1-p^{-s})^{-1} = 1 + p^{-s} + p^{-2s} + \cdots for each prime pp and multiplying these together over all primes, distributing the product, produces every term n−sn^{-s} exactly once, by unique factorization. The sum over integers and the product over primes are two descriptions of the same convergent quantity. This is why ζ\zeta is the natural gateway between "additive" analytic tools (series, integrals, contour arguments) and "multiplicative" arithmetic data (primes, factorization). Finally, picture the half-plane Re⁡(s)>1\operatorname{Re}(s) > 1 as the "safe zone" where the naive definitions make sense, with a wall at Re⁡(s)=1\operatorname{Re}(s) = 1 where the series stops converging. The analytic continuation shows the wall is not really a boundary of existence for the function — ζ\zeta can be extended past it — but it is the location of a genuine singularity, a simple pole exactly at s=1s=1. Everything about primes that ζ\zeta will eventually reveal is entangled with the behavior of the function near that single point and, more deeply, with its zeros elsewhere.

Figure
The half-plane of convergence and the pole at s = 1
Re(s) Im(s) absolute convergence Re(s) > 1 continuation 0 < Re(s) < 1 s = 1 (simple pole) 1 0
TheoremEuler product and analytic continuation of zeta

Let ζ(s)=∑n=1∞n−s\zeta(s) = \sum_{n=1}^{\infty} n^{-s} for Re⁡(s)>1\operatorname{Re}(s) > 1. Then: 1. (Holomorphy.) The series converges absolutely, and uniformly on every half-plane {Re⁡(s)≥1+δ}\{\operatorname{Re}(s) \ge 1+\delta\} with δ>0\delta > 0; consequently ζ\zeta is holomorphic on {Re⁡(s)>1}\{\operatorname{Re}(s) > 1\}. 2. (Euler product.) For Re⁡(s)>1\operatorname{Re}(s) > 1,

ζ(s)=∏p prime(1−p−s)−1,\zeta(s) = \prod_{p \text{ prime}} \left(1 - p^{-s}\right)^{-1},

where the product is over all primes pp and converges absolutely; in particular ζ(s)≠0\zeta(s) \neq 0 for Re⁡(s)>1\operatorname{Re}(s) > 1. 3. (Meromorphic continuation.) ζ(s)\zeta(s) extends to a meromorphic function on {Re⁡(s)>0}\{\operatorname{Re}(s) > 0\}, holomorphic there except for a single simple pole at s=1s = 1 with residue 11. Explicitly, for Re⁡(s)>1\operatorname{Re}(s) > 1,

ζ(s)=ss−1−s∫1∞{x} x−s−1 dx,\zeta(s) = \frac{s}{s-1} - s\int_1^\infty \{x\}\, x^{-s-1}\, dx,

where {x}=x−⌊x⌋\{x\} = x - \lfloor x \rfloor is the fractional part, and the integral on the right converges and is holomorphic for Re⁡(s)>0\operatorname{Re}(s) > 0, so this formula defines the continuation.

LemmaAbsolute and locally uniform convergence

For real σ0>1\sigma_0 > 1, the series ∑n=1∞n−s\sum_{n=1}^\infty n^{-s} converges absolutely and uniformly on {s:Re⁡(s)≥σ0}\{s : \operatorname{Re}(s) \ge \sigma_0\}.

Proof

For s=σ+its = \sigma + it with σ≥σ0\sigma \ge \sigma_0,

∣n−s∣=∣e−slog⁡n∣=e−σlog⁡n=n−σ≤n−σ0.|n^{-s}| = |e^{-s \log n}| = e^{-\sigma \log n} = n^{-\sigma} \le n^{-\sigma_0}.

Since σ0>1\sigma_0 > 1, the real series ∑n=1∞n−σ0\sum_{n=1}^\infty n^{-\sigma_0} converges (comparison with the convergent pp-series, or the integral test: ∑n=2Nn−σ0≤∫1Nx−σ0 dx=1−N1−σ0σ0−1<1σ0−1\sum_{n=2}^N n^{-\sigma_0} \le \int_1^N x^{-\sigma_0}\,dx = \frac{1 - N^{1-\sigma_0}}{\sigma_0 - 1} < \frac{1}{\sigma_0-1}, a bound independent of NN, so the partial sums are bounded and increasing, hence convergent). By the Weierstrass MM-test with Mn=n−σ0M_n = n^{-\sigma_0}, the series ∑n−s\sum n^{-s} converges absolutely and uniformly for Re⁡(s)≥σ0\operatorname{Re}(s) \ge \sigma_0. ■\blacksquare

∎
Proof

We prove the theorem in three parts, each building on the previous.

Part 1: Holomorphy. By the Lemma, for every δ>0\delta > 0 the series converges uniformly on {Re⁡(s)≥1+δ}\{\operatorname{Re}(s) \ge 1+\delta\}. Each term n−s=e−slog⁡nn^{-s} = e^{-s\log n} is an entire function of ss. A locally uniform limit of holomorphic functions on an open set is holomorphic on that set (this is a standard consequence of Morera's theorem / Weierstrass's theorem on uniform limits of holomorphic functions: uniform limits preserve the vanishing of contour integrals around triangles, so Morera's theorem applies to the limit). Since {Re⁡(s)>1}=⋃δ>0{Re⁡(s)≥1+δ}\{\operatorname{Re}(s) > 1\} = \bigcup_{\delta > 0} \{\operatorname{Re}(s) \ge 1+\delta\} and holomorphy is a local property, ζ\zeta is holomorphic on the full half-plane {Re⁡(s)>1}\{\operatorname{Re}(s) > 1\}.

Part 2: Euler product. Fix ss with Re⁡(s)=σ>1\operatorname{Re}(s) = \sigma > 1. For each prime pp, since ∣p−s∣=p−σ<1|p^{-s}| = p^{-\sigma} < 1, the geometric series

(1−p−s)−1=∑k=0∞p−ks(1-p^{-s})^{-1} = \sum_{k=0}^\infty p^{-ks}

converges absolutely. Let p1<p2<⋯<prp_1 < p_2 < \cdots < p_r be the first rr primes and set

Pr(s)=∏j=1r(1−pj−s)−1=∏j=1r∑k=0∞pj−ks.P_r(s) = \prod_{j=1}^r \left(1 - p_j^{-s}\right)^{-1} = \prod_{j=1}^r \sum_{k=0}^\infty p_j^{-ks}.

Because each factor converges absolutely, we may multiply out the finite product of absolutely convergent series term by term (rearrangement is justified since a finite product of absolutely convergent series is itself an absolutely convergent series, by Mertens' theorem on products of absolutely convergent series, or directly: absolute convergence of each factor lets us expand and regroup by any bijection without changing the sum). This gives

Pr(s)=∑n∈Arn−s,P_r(s) = \sum_{n \in A_r} n^{-s},

where ArA_r is the set of positive integers all of whose prime factors lie among p1,…,prp_1,\dots,p_r — because expanding the product picks out, for each choice of nonnegative exponents (k1,…,kr)(k_1,\dots,k_r), the term n−sn^{-s} with n=p1k1⋯prkrn = p_1^{k_1}\cdots p_r^{k_r}, and by unique factorization (the Fundamental Theorem of Arithmetic) this correspondence (k1,…,kr)↔n(k_1,\dots,k_r) \leftrightarrow n is a bijection onto ArA_r. Every integer n≤prn \le p_r has all its prime factors ≤n≤pr\le n \le p_r, hence among p1,…,prp_1,\dots,p_r (since p1,…,prp_1,\dots,p_r are all primes up to prp_r). Thus {1,2,…,⌊pr⌋}⊆Ar\{1,2,\dots, \lfloor p_r \rfloor \}\subseteq A_r, and consequently

∣ζ(s)−Pr(s)∣=∣∑n∉Arn−s∣≤∑n>prn−σ  →r→∞0,\left|\zeta(s) - P_r(s)\right| = \left|\sum_{n \notin A_r} n^{-s}\right| \le \sum_{n > p_r} n^{-\sigma} \;\xrightarrow[r\to\infty]{} 0,

because ∑n−σ\sum n^{-\sigma} converges (Lemma) and hence its tail beyond pr→∞p_r \to \infty vanishes. Therefore

ζ(s)=lim⁡r→∞Pr(s)=lim⁡r→∞∏j=1r(1−pj−s)−1,\zeta(s) = \lim_{r\to\infty} P_r(s) = \lim_{r \to \infty} \prod_{j=1}^r \left(1-p_j^{-s}\right)^{-1},

i.e. the infinite product ∏p(1−p−s)−1\prod_p (1-p^{-s})^{-1} converges to ζ(s)\zeta(s). Absolute convergence of the product: an infinite product ∏p(1+ap)\prod_p (1+a_p) with ap=(1−p−s)−1−1a_p = (1-p^{-s})^{-1} - 1 converges absolutely (meaning ∑p∣ap∣<∞\sum_p |a_p| < \infty) since

ap=p−s1−p−s,∣ap∣≤p−σ1−2−σ≤p−σ1−2−1=2p−σa_p = \frac{p^{-s}}{1-p^{-s}}, \qquad |a_p| \le \frac{p^{-\sigma}}{1-2^{-\sigma}} \le \frac{p^{-\sigma}}{1 - 2^{-1}} = 2p^{-\sigma}

for σ>1\sigma > 1 (using p≥2p \ge 2 so p−σ≤2−σ≤1/2p^{-\sigma}\le 2^{-\sigma} \le 1/2), and ∑pp−σ≤∑nn−σ<∞\sum_p p^{-\sigma} \le \sum_n n^{-\sigma} < \infty. Absolute convergence of ∑p∣ap∣\sum_p |a_p| implies the infinite product converges to a nonzero value if and only if no factor is zero; here each factor (1−p−s)−1≠0(1-p^{-s})^{-1} \ne 0, so ζ(s)≠0\zeta(s) \ne 0 for Re⁡(s)>1\operatorname{Re}(s) > 1.

Part 3: Meromorphic continuation to Re⁡(s)>0\operatorname{Re}(s) > 0. We use Abel summation (partial summation), a standard technique from Dirichlet Series and Euler Products. For Re⁡(s)=σ>1\operatorname{Re}(s) = \sigma > 1 and integer N≥1N \ge 1, write ⌊x⌋\lfloor x \rfloor for the greatest integer ≤x\le x. Using the Riemann–Stieltjes representation

∑n=1Nn−s=∫1−Nx−s d⌊x⌋,\sum_{n=1}^{N} n^{-s} = \int_{1^-}^{N} x^{-s}\, d\lfloor x\rfloor,

integrate by parts (justified since ⌊x⌋\lfloor x\rfloor is a step function and x−sx^{-s} is continuously differentiable):

∑n=1Nn−s=N1−s+s∫1N⌊x⌋ x−s−1 dx.\sum_{n=1}^N n^{-s} = N^{1-s} + s\int_1^N \lfloor x \rfloor \, x^{-s-1}\,dx.

(This identity is checked directly: differentiating the right side or, more elementarily, verifying it by comparing jumps at each integer; it is the standard Abel-summation formula for ∑n−s\sum n^{-s} and is proved in the prerequisite treatment of Dirichlet series.) Now write ⌊x⌋=x−{x}\lfloor x \rfloor = x - \{x\}:

∑n=1Nn−s=N1−s+s∫1Nx⋅x−s−1 dx−s∫1N{x} x−s−1 dx=N1−s+s⋅N1−s−11−s−s∫1N{x} x−s−1 dx.\sum_{n=1}^N n^{-s} = N^{1-s} + s\int_1^N x\cdot x^{-s-1}\,dx - s\int_1^N \{x\}\,x^{-s-1}\,dx = N^{1-s} + s\cdot\frac{N^{1-s}-1}{1-s} - s\int_1^N \{x\}\,x^{-s-1}\,dx.

Let N→∞N \to \infty. Since Re⁡(s)=σ>1\operatorname{Re}(s) = \sigma > 1, N1−s=N1−σ⋅N−it→0N^{1-s} = N^{1-\sigma}\cdot N^{-it} \to 0 because ∣N1−s∣=N1−σ→0|N^{1-s}| = N^{1-\sigma}\to 0. The integral ∫1∞{x} x−s−1 dx\int_1^\infty \{x\}\, x^{-s-1}\,dx converges absolutely for Re⁡(s)>0\operatorname{Re}(s) > 0: since 0≤{x}<10 \le \{x\} < 1,

∣∫1∞{x} x−s−1 dx∣≤∫1∞x−σ−1 dx=1σ<∞for σ>0.\left|\int_1^\infty \{x\}\,x^{-s-1}\,dx\right| \le \int_1^\infty x^{-\sigma-1}\,dx = \frac1\sigma < \infty \quad\text{for } \sigma > 0.

Taking N→∞N \to \infty in the displayed identity (valid termwise since all pieces converge) gives, for Re⁡(s)>1\operatorname{Re}(s) > 1,

ζ(s)=ss−1−s∫1∞{x} x−s−1 dx.(∗)\zeta(s) = \frac{s}{s-1} - s\int_1^\infty \{x\}\,x^{-s-1}\,dx. \tag{$\ast$}

Now consider the right-hand side of (∗)(\ast) as a function F(s)F(s) of ss, defined for all Re⁡(s)>0\operatorname{Re}(s) > 0. We claim F(s)F(s) is holomorphic on {Re⁡(s)>0}\{\operatorname{Re}(s) > 0\} except for a simple pole at s=1s=1 with residue 11.

  • The term ss−1=1+1s−1\frac{s}{s-1} = 1 + \frac{1}{s-1} is meromorphic on all of C\mathbb C with a single simple pole at s=1s=1, residue lim⁡s→1(s−1)⋅ss−1=1\lim_{s\to 1}(s-1)\cdot\frac{s}{s-1} = 1.
  • The term I(s):=∫1∞{x} x−s−1 dxI(s) := \int_1^\infty \{x\}\,x^{-s-1}\,dx is holomorphic on {Re⁡(s)>0}\{\operatorname{Re}(s) > 0\}. To see this, fix δ>0\delta>0 and restrict to Re⁡(s)≥δ\operatorname{Re}(s) \ge \delta; the integrand {x}x−s−1\{x\} x^{-s-1} is, for each fixed x≥1x\ge 1, an entire function of ss, and it is dominated by the integrable-in-xx, ss-independent bound x−δ−1x^{-\delta - 1} (integrable near infinity since δ+1>1\delta+1>1, and there is no singularity at x=1x=1). By Morera's theorem combined with Fubini/Morera-type differentiation under the integral sign (a standard fact: an integral of a family of holomorphic functions, locally uniformly bounded by an integrable dominating function, is holomorphic and its derivative may be computed by differentiating under the integral sign), I(s)I(s) is holomorphic on {Re⁡(s)≥δ}\{\operatorname{Re}(s) \ge \delta\} for every δ>0\delta > 0, hence on {Re⁡(s)>0}\{\operatorname{Re}(s) > 0\}.

Therefore F(s)=ss−1−sI(s)F(s) = \frac{s}{s-1} - sI(s) is meromorphic on {Re⁡(s)>0}\{\operatorname{Re}(s)>0\}, holomorphic except for the single simple pole at s=1s=1 inherited from ss−1\frac{s}{s-1} (the factor −sI(s)-sI(s) is holomorphic there, including at s=1s=1, since II is holomorphic on the whole half-plane), and the residue of FF at s=1s=1 equals the residue of ss−1\frac{s}{s-1} at s=1s=1, namely 11 (subtracting a holomorphic function does not change a residue). By (∗)(\ast), F(s)=ζ(s)F(s) = \zeta(s) for Re⁡(s)>1\operatorname{Re}(s) > 1. Since two holomorphic functions agreeing on the open set {Re⁡(s)>1}\{\operatorname{Re}(s)>1\}, which has an accumulation point in the connected domain {Re⁡(s)>0}∖{1}\{\operatorname{Re}(s) > 0\}\setminus\{1\}, must agree on the whole connected domain (the identity theorem for holomorphic functions), FF is the meromorphic continuation of ζ\zeta to {Re⁡(s)>0}\{\operatorname{Re}(s) > 0\}, and we rename FF as ζ\zeta on this larger domain. This continuation is unique: any two meromorphic continuations of ζ\zeta to a connected open set agree on a set with a limit point, hence agree everywhere, by the identity theorem. This proves Part 3 and completes the proof of the theorem. ■\blacksquare

∎
CorollaryInfinitude of primes, via zeta

There are infinitely many primes.

Proof

Suppose, for contradiction, that there were only finitely many primes p1,…,prp_1,\dots,p_r. By the Euler product (Part 2 of the Theorem), for real s>1s>1,

ζ(s)=∏j=1r(1−pj−s)−1.\zeta(s) = \prod_{j=1}^r (1-p_j^{-s})^{-1}.

The right-hand side is a finite product of continuous functions of ss, each factor tending to the finite value (1−pj−1)−1(1-p_j^{-1})^{-1} as s→1+s \to 1^+; hence the right side stays bounded as s→1+s\to 1^+. But by Part 3, ζ(s)=1s−1+(holomorphic near s=1)\zeta(s) = \frac{1}{s-1} + (\text{holomorphic near } s=1), so ζ(s)→+∞\zeta(s) \to +\infty as s→1+s\to 1^+ (through real s>1s>1, where all terms in the original series are positive, so ζ(s)\zeta(s) is manifestly real and increasing to +∞+\infty as s↓1s\downarrow 1, consistent with the pole having positive residue). This contradicts boundedness of the finite product. Hence there are infinitely many primes. ■\blacksquare

∎

Worked examples

Example 1: Bounding ζ(s)\zeta(s) near s=2+3is=2+3i using the Euler product tail. Take s=2+3is = 2+3i, so σ=Re⁡(s)=2\sigma = \operatorname{Re}(s) = 2. We estimate how well the partial Euler product over the first three primes {2,3,5}\{2,3,5\} approximates ζ(s)\zeta(s), and bound the error rigorously. From the proof of Part 2, with ArA_r the set of integers whose prime factors are among p1,…,prp_1,\dots,p_r and prp_r the rr-th prime, we have the tail bound

∣ζ(s)−Pr(s)∣≤∑n>prn−σ.\left|\zeta(s) - P_r(s)\right| \le \sum_{n > p_r} n^{-\sigma}.

With r=3r=3 (primes 2,3,52,3,5, so pr=5p_r = 5) and σ=2\sigma = 2:

∣ζ(2+3i)−P3(2+3i)∣≤∑n=6∞n−2=ζ(2)−(1+14+19+116+125).\left|\zeta(2+3i) - P_3(2+3i)\right| \le \sum_{n=6}^\infty n^{-2} = \zeta(2) - \left(1+\tfrac14+\tfrac19+\tfrac1{16}+\tfrac1{25}\right).

Using ζ(2)=π2/6=1.644934…\zeta(2) = \pi^2/6 = 1.644934\ldots and 1+14+19+116+125=1+0.25+0.111111+0.0625+0.04=1.463611…1+\frac14+\frac19+\frac1{16}+\frac1{25} = 1 + 0.25 + 0.111111 + 0.0625 + 0.04 = 1.463611\ldots, the bound is

∣ζ(2+3i)−P3(2+3i)∣≤1.644934−1.463611=0.181323…\left|\zeta(2+3i) - P_3(2+3i)\right| \le 1.644934 - 1.463611 = 0.181323\ldots

So the three-prime Euler product P3(s)=(1−2−s)−1(1−3−s)−1(1−5−s)−1P_3(s) = (1-2^{-s})^{-1}(1-3^{-s})^{-1}(1-5^{-s})^{-1} already approximates ζ(2+3i)\zeta(2+3i) to within about 0.180.18 in absolute value — and this error bound is completely explicit and rigorous, not asymptotic. This illustrates concretely how the Euler product converges to ζ(s)\zeta(s) term by term as more primes are included, exactly as proved in the theorem, and shows the mechanism (comparison with the tail of ζ(σ)\zeta(\sigma) for real σ\sigma) that makes the convergence proof work quantitatively.

Example 2: Locating and computing the residue at s=1s=1 using the integral formula. We use formula (∗)(\ast),

ζ(s)=ss−1−s∫1∞{x} x−s−1 dx,\zeta(s) = \frac{s}{s-1} - s\int_1^\infty \{x\}\,x^{-s-1}\,dx,

to compute the constant term in the Laurent expansion of ζ(s)\zeta(s) at s=1s=1 beyond just confirming the pole — specifically, to find lim⁡s→1(ζ(s)−1s−1)\lim_{s\to 1}\left(\zeta(s) - \frac{1}{s-1}\right), which the theory of the Riemann zeta function calls the Euler–Mascheroni constant γ\gamma (we verify only the existence and value of the limit here, using the tools already established, not further unproved facts). Write ss−1=1+1s−1\frac{s}{s-1} = 1 + \frac{1}{s-1}. Then

ζ(s)−1s−1=1−s I(s),I(s)=∫1∞{x} x−s−1 dx.\zeta(s) - \frac{1}{s-1} = 1 - s\,I(s), \qquad I(s) = \int_1^\infty \{x\}\,x^{-s-1}\,dx.

Since I(s)I(s) is holomorphic (in particular continuous) at s=1s=1 by Part 3 of the theorem, we may take the limit termwise:

lim⁡s→1(ζ(s)−1s−1)=1−1⋅I(1)=1−∫1∞{x} x−2 dx.\lim_{s\to 1}\left(\zeta(s) - \frac{1}{s-1}\right) = 1 - 1\cdot I(1) = 1 - \int_1^\infty \{x\}\,x^{-2}\,dx.

We now evaluate ∫1∞{x}x−2 dx\int_1^\infty \{x\} x^{-2}\,dx directly by splitting into unit intervals [n,n+1)[n,n+1), on which {x}=x−n\{x\} = x-n:

∫1∞{x}x−2 dx=∑n=1∞∫nn+1(x−n)x−2 dx.\int_1^\infty \{x\}x^{-2}\,dx = \sum_{n=1}^\infty \int_n^{n+1} (x-n)x^{-2}\,dx.

Compute the nn-th integral: ∫nn+1(x−n)x−2dx=∫nn+1x−1dx−n∫nn+1x−2dx=log⁡n+1n−n(1n−1n+1)=log⁡n+1n−1n+1\int_n^{n+1}(x-n)x^{-2}dx = \int_n^{n+1} x^{-1}dx - n\int_n^{n+1}x^{-2}dx = \log\frac{n+1}{n} - n\left(\frac1n - \frac1{n+1}\right) = \log\frac{n+1}{n} - \frac{1}{n+1}. Summing the first several terms numerically to see the partial value (the series telescopes with log⁡\log, matching the classical identity γ=1−∫1∞{x}x−2dx\gamma = 1-\int_1^\infty \{x\}x^{-2}dx):

  • n=1n=1: log⁡2−12=0.693147−0.5=0.193147\log 2 - \tfrac12 = 0.693147 - 0.5 = 0.193147
  • n=2n=2: log⁡(3/2)−13=0.405465−0.333333=0.072132\log(3/2) - \tfrac13 = 0.405465 - 0.333333 = 0.072132
  • n=3n=3: log⁡(4/3)−14=0.287682−0.25=0.037682\log(4/3) - \tfrac14 = 0.287682 - 0.25 = 0.037682
  • n=4n=4: log⁡(5/4)−15=0.223144−0.2=0.023144\log(5/4) - \tfrac15 = 0.223144 - 0.2 = 0.023144

Partial sum after four terms: 0.193147+0.072132+0.037682+0.023144=0.3261050.193147+0.072132+0.037682+0.023144 = 0.326105, with remaining positive terms continuing to accumulate toward the known value ∫1∞{x}x−2dx=1−γ=1−0.5772157…=0.4227843…\int_1^\infty\{x\}x^{-2}dx = 1-\gamma = 1-0.5772157\ldots = 0.4227843\ldots Hence

lim⁡s→1(ζ(s)−1s−1)=1−(1−γ)=γ=0.5772156649…,\lim_{s\to1}\left(\zeta(s)-\frac{1}{s-1}\right) = 1-(1-\gamma) = \gamma = 0.5772156649\ldots,

consistent with (and here re-derived from) the integral representation (∗)(\ast) proved above, showing the pole at s=1s=1 is simple, has residue 11, and the finite part of the Laurent expansion begins with the Euler–Mascheroni constant.

CounterexampleDropping absolute convergence collapses the Euler product argument

The theorem's proof of the Euler product crucially used absolute convergence of ∑n−s\sum n^{-s} at each fixed ss with Re⁡(s)>1\operatorname{Re}(s)>1, to justify freely rearranging and regrouping the product of geometric series. To see why this hypothesis cannot be dropped, consider the related but merely conditionally convergent Dirichlet series

η(s)=∑n=1∞(−1)n−1ns,\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s},

the Dirichlet eta function, which converges (conditionally, by the alternating series test) for Re⁡(s)>0\operatorname{Re}(s) > 0, in particular at real ss with 0<s≤10 < s \le 1 where ∑n−s\sum n^{-s} itself diverges and there is no absolute convergence at all. If one tried to mimic the Euler product proof for η(s)\eta(s) at, say, s=1/2s = 1/2 (where η(1/2)=(1−21−1/2)ζ(1/2)\eta(1/2) = (1-2^{1-1/2})\zeta(1/2), a finite, well-defined alternating sum), the same term-by-term regrouping into a product over primes is not valid: η\eta has no multiplicative structure of this kind because its coefficients (−1)n−1(-1)^{n-1} are not built from a multiplicative arithmetic function in the way that lets factorization distribute correctly, and more fundamentally, at s=1/2s=1/2 the terms n−1/2n^{-1/2} do not form an absolutely convergent series, so rearranging the sum defining η(1/2)\eta(1/2) (for instance grouping all terms with a fixed smallest prime factor) can change its value — conditionally convergent series are not rearrangement-invariant, by the Riemann rearrangement theorem. Concretely, reordering the alternating harmonic-type series ∑(−1)n−1n−1/2\sum (-1)^{n-1} n^{-1/2} into a different order can be made to converge to a different sum or to diverge entirely, so no Euler-product-style regrouping argument can even be attempted rigorously. This is exactly the point in the proof of Part 2 where absolute convergence was invoked to guarantee that reordering the double sum into primes-and-exponents does not change the total — remove that hypothesis, and the entire derivation breaks down at the first step.

Scope, edge cases & common traps

SituationWhat is trueWhat is not true / common error
Re⁡(s)>1\operatorname{Re}(s) > 1Series and Euler product both converge absolutely; ζ(s)≠0\zeta(s)\ne 0 hereAssuming the series ∑n−s\sum n^{-s} converges at s=1s=1 (it diverges: harmonic series)
Re⁡(s)=1\operatorname{Re}(s) = 1, s≠1s \ne 1ζ(s)\zeta(s) is finite (continuation applies) and, famously, ζ(1+it)≠0\zeta(1+it)\ne0 for real t≠0t\ne0 (used in PNT, not proved here)Assuming the Euler product itself converges on this line — it does not, since ∑p−1\sum p^{-1} diverges
0<Re⁡(s)<10 < \operatorname{Re}(s) < 1ζ(s)\zeta(s) is defined by continuation, generally nonzero except at the (unproven-here) "nontrivial zeros"Plugging values into the original series ∑n−s\sum n^{-s} here — it diverges, and the series formula simply does not apply
s=1s = 1Simple pole, residue 11Treating ζ(1)\zeta(1) as a finite number under any definition
Re⁡(s)≤0\operatorname{Re}(s) \le 0Requires the functional equation (beyond this article's scope) to continue furtherApplying formula (∗)(\ast) blindly outside Re⁡(s)>0\operatorname{Re}(s)>0, where the integral representation used here was only proved to converge

The mathematician behind it

Origins · 1 minBernhard Riemann and how this was discovered
BR
Bernhard Riemann
1826–1866 · Göttingen, analytic number theory

Georg Friedrich Bernhard Riemann was a German mathematician whose 1859 paper "Über die Anzahl der Primzahlen unter einer gegebenen Größe" ("On the Number of Primes Less Than a Given Magnitude") — his only paper in number theory — introduced the complex-variable study of the zeta function, proved its functional equation and meromorphic continuation, and stated the Riemann Hypothesis about the location of its nontrivial zeros. Riemann's broader work founded Riemannian geometry and reshaped complex analysis, but this single eight-page memoir created analytic number theory as a field, by showing that Euler's real-variable series becomes vastly more powerful once treated as a function of a complex variable.

History
- 1737: Euler proves the product formula ∑n−s=∏p(1−p−s)−1\sum n^{-s} = \prod_p (1-p^{-s})^{-1} for real s>1s>1 and uses divergence at s=1s=1 to reprove the infinitude of primes.
- 1837: Dirichlet introduces general Dirichlet series and L-functions to prove infinitude of primes in arithmetic progressions, extending Euler's method.
- 1859: Riemann treats ζ(s)\zeta(s) as a function of a complex variable, proves meromorphic continuation and the functional equation, and conjectures the Riemann Hypothesis.
- 1896: Hadamard and de la Vallée Poussin independently prove ζ(1+it)≠0\zeta(1+it)\ne 0 for real t≠0t\ne 0, yielding the Prime Number Theorem.
- 1900s–present: The Riemann Hypothesis remains open, listed among Hilbert's 1900 problems and as one of the Clay Millennium Prize Problems.
Why does the series for ζ(s)\zeta(s) only make sense for Re⁡(s)>1\operatorname{Re}(s) > 1 in the first place?

The terms of the series are n−s=e−slog⁡nn^{-s} = e^{-s\log n}, and ∣n−s∣=n−Re⁡(s)|n^{-s}| = n^{-\operatorname{Re}(s)}. If Re⁡(s)≤1\operatorname{Re}(s) \le 1, the terms decay too slowly (or not at all) for the series ∑n−Re⁡(s)\sum n^{-\operatorname{Re}(s)} to converge; at Re⁡(s)=1\operatorname{Re}(s)=1 it is exactly the divergent harmonic series, and for Re⁡(s)<1\operatorname{Re}(s)<1 the terms do not even tend to zero fast enough. So absolute convergence, which is what the comparison test with the real pp-series controls, genuinely requires Re⁡(s)>1\operatorname{Re}(s)>1. This is why the function must be defined by the series only on that half-plane, and extended elsewhere by other means, such as the integral representation proved in this article.

Does the Euler product remain valid after analytic continuation, say for 0<Re⁡(s)<10 < \operatorname{Re}(s) < 1?

No. The Euler product formula was proved using absolute convergence of ∑pp−s\sum_p p^{-s}, which requires Re⁡(s)>1\operatorname{Re}(s) > 1 exactly as for the original series. Once ζ(s)\zeta(s) is extended to 0<Re⁡(s)≤10 < \operatorname{Re}(s) \le 1 via the integral formula, there is no claim, and in fact no truth, that the infinite product over primes still converges to ζ(s)\zeta(s) there. Indeed, if the Euler product converged and were nonzero throughout 0<Re⁡(s)<10<\operatorname{Re}(s)<1, ζ\zeta could never vanish there — but ζ\zeta does have nontrivial zeros in this strip (a deep fact from later chapters), so the product formula's absolute-convergence region genuinely stops at Re⁡(s)=1\operatorname{Re}(s)=1.

How does the pole at s=1s=1 relate to the infinitude of primes?

The Euler product gives, for real s>1s>1, log⁡ζ(s)=−∑plog⁡(1−p−s)=∑pp−s+O(1)\log \zeta(s) = -\sum_p \log(1-p^{-s}) = \sum_p p^{-s} + O(1) (bounded error), so ζ(s)→∞\zeta(s)\to\infty as s→1+s\to 1^+ forces ∑pp−s→∞\sum_p p^{-s} \to \infty as well. If there were only finitely many primes, this sum would stay bounded as s→1+s\to1^+, contradicting the divergence caused by the pole. This is precisely the corollary proved in this article: the pole's residue being positive (in fact equal to 11) is analytically equivalent to there being infinitely many primes, generalizing Euler's original 1737 argument to the complex-analytic setting.

What is the difference between "convergence" of the series and "existence" of ζ(s)\zeta(s) at a point like s=−1s = -1?

The series ∑n−s\sum n^{-s} diverges completely at s=−1s=-1 (it becomes 1+2+3+⋯1+2+3+\cdots, which diverges to +∞+\infty term by term). However, ζ(−1)\zeta(-1) is a perfectly well-defined finite number, namely −1/12-1/12, once one uses the meromorphic continuation (via the functional equation, beyond this article's scope) rather than the original series. The symbol "ζ(−1)\zeta(-1)" always refers to the continued function's value, never to a literal evaluation of the divergent series; conflating the two is the source of the popular but misleading claim that "1+2+3+⋯=−1/121+2+3+\cdots = -1/12."

Why is it called an "Euler product" if Riemann's name is attached to the function?

Euler discovered and proved the product formula ∑n−s=∏p(1−p−s)−1\sum n^{-s} = \prod_p(1-p^{-s})^{-1} over a century earlier, in 1737, for real s>1s>1, and used it to give a new proof that there are infinitely many primes. Riemann's 1859 contribution was to treat ss as a complex variable, proving the function extends meromorphically, satisfies a functional equation relating ss and 1−s1-s, and has zeros whose location controls the distribution of primes. The function is named after Riemann because his complex-analytic viewpoint — not the series or product themselves — is what created the modern theory; the Euler product remains correctly credited to Euler within that theory.

Does this article prove the Riemann Hypothesis or say anything about the zeros of zeta?

No. This article establishes only the foundational analytic facts: convergence of the defining series, the Euler product on Re⁡(s)>1\operatorname{Re}(s)>1 (which already shows ζ\zeta has no zeros there), and the meromorphic continuation to Re⁡(s)>0\operatorname{Re}(s)>0 with its single simple pole at s=1s=1. Questions about zeros inside the strip 0≤Re⁡(s)≤10\le\operatorname{Re}(s)\le1 — including the location of the "trivial zeros," the "critical line" Re⁡(s)=1/2\operatorname{Re}(s)=1/2, and the Riemann Hypothesis itself — require the functional equation and much deeper analysis, and are addressed in Zeta and the Distribution of Primes and beyond.

What role does absolute convergence play beyond just "making the proof easier"?

Absolute convergence is not a mere technical convenience; it is what allows rearrangement of infinite sums and products without changing their value, which is exactly the operation used to convert the sum over integers into a product over primes. Without it (as the counterexample with the Dirichlet eta function shows), the Riemann rearrangement theorem guarantees that reordering terms can change the sum entirely, so any attempt to regroup a series "by prime factorization" would be logically unjustified. Absolute convergence is precisely the hypothesis that makes the additive-to-multiplicative translation rigorous.

Before this

Where this leads