advanced26 min studyLesson 583 in Mathematics

Dirichlet series and Euler products

Translate divisor convolution into analytic products inside a justified half-plane.

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What you’ll understand

  • Locate the absolute-convergence half-plane of a Dirichlet series.
  • Justify products and coefficient rearrangements by absolute convergence.
  • Distinguish a prime cutoff from an integer cutoff.

A Dirichlet series weights the coefficient at an integer by a complex power of that integer. Because multiplying integers adds their logarithms, its natural coefficient algebra is divisor convolution. Multiplicative coefficients then separate into factors indexed by primes. These identities are powerful only when their analytic domain is specified: a formal factorization and a convergent infinite product are related statements with different obligations.

A half-plane of safe rearrangements

Write s=σ+its=\sigma+it and interpret n−s=exp⁡(−slog⁡n)n^{-s}=\exp(-s\log n) using the real logarithm. Its absolute value is n−σn^{-\sigma}, so increasing the real part improves convergence independently of the imaginary part. The abscissa of absolute convergence is an extended real number separating an open half-plane of absolute convergence from failure to its left. Nothing universal puts that boundary at one for arbitrary coefficients. Coefficients growing like a fixed power shift the elementary comparison boundary accordingly.

Inside this open half-plane, convergence is locally uniform, and the series defines a holomorphic function. If two Dirichlet series converge absolutely at the same real part, their product can be rearranged by the product of the two integer indices. The coefficient at a given integer is exactly the Dirichlet convolution. Without absolute convergence, a convenient rearrangement may change a value or be unjustified. Analytic continuation later supplies new values of a function; it does not retroactively make the original series absolutely convergent there.

F(s)=∑n≥1a(n)ns,F(s)G(s)=∑n≥1(a∗b)(n)nsin a common absolute-convergence half-plane.F(s)=\sum_{n\geq1}\frac{a(n)}{n^s},\qquad F(s)G(s)=\sum_{n\geq1}\frac{(a*b)(n)}{n^s}\quad\text{in a common absolute-convergence half-plane}.
The domain condition permits grouping all pairs of integer indices with the same product.↔ Scroll wide equations horizontally.

Prime-power factors

Suppose the coefficients are multiplicative and their value at one is one. Unique prime factorization expresses every coefficient as the product of its prime-power coefficients. Expanding a finite product of local series therefore sums over integers whose prime factors belong to that finite set. Absolute convergence allows the finite set of primes to expand to all primes. For completely multiplicative coefficients, each local series is geometric; ordinary multiplicativity alone does not replace an arbitrary local series by a geometric factor.

The constant-one coefficients give the Riemann zeta function, absolutely convergent for σ>1\sigma>1. Its local factors are reciprocal geometric denominators. On that same half-plane the product is nonzero: the logarithms of the factors form an absolutely convergent sum, since the sum of p−σp^{-\sigma} over primes is bounded by a convergent integer sum. The Möbius series is its reciprocal, and the divisor-count series is its square. These statements include their domains, rather than asserting Euler products at every continued value.

ζ(s)=∏p(1−p−s)−1,∑n≥1μ(n)ns=1ζ(s),∑n≥1τ(n)ns=ζ(s)2(σ>1).\zeta(s)=\prod_p(1-p^{-s})^{-1},\qquad \sum_{n\geq1}\frac{\mu(n)}{n^s}=\frac1{\zeta(s)},\qquad \sum_{n\geq1}\frac{\tau(n)}{n^s}=\zeta(s)^2\quad(\sigma>1).
All three series and the corresponding products converge absolutely in the stated half-plane.↔ Scroll wide equations horizontally.

What a truncation actually includes

Stopping an Euler product at a prime threshold differs from stopping a Dirichlet series at an integer threshold. Even a single zeta local factor includes arbitrarily high powers of its prime. A product over two primes therefore represents infinitely many integers, while a finite integer sum contains only finitely many. Conversely, the Möbius local factors are finite, so a finite prime product becomes a finite sum over squarefree products of the selected primes. Those products can exceed the prime threshold substantially.

Logarithmic differentiation of the zeta product, justified on compact subsets of σ>1\sigma>1, yields −ζ′(s)/ζ(s)=∑nΛ(n)n−s-\zeta'(s)/\zeta(s)=\sum_n\Lambda(n)n^{-s}. Each local geometric expansion contributes every prime power with the same logarithmic weight. This is the analytic counterpart of the earlier convolution identity between the constant function, von Mangoldt weights, and the logarithm. It explains why weighted prime counts are natural for contour methods. Merely truncating the logarithmic derivative numerically supplies no theorem about the uncomputed tail.

A finite Möbius product

Expand the reciprocal-zeta factors for just the primes two and three at the real argument two, and identify their exact coefficient support.

  1. The factors are 1−2−2=3/41-2^{-2}=3/4 and 1−3−2=8/91-3^{-2}=8/9. Their product is 2/32/3. These local factors have only two terms each because Möbius vanishes on prime squares.

  2. Expanding gives 1−1/4−1/9+1/361-1/4-1/9+1/36. The supporting integers are 1,2,3,61,2,3,6, with their correct Möbius coefficients. The positive term at six comes from selecting the negative term in both factors.

  3. This finite product is neither the integer sum through three nor the full reciprocal 1/ζ(2)=6/π21/\zeta(2)=6/\pi^2. The omitted primes contribute additional factors, so using a familiar zeta value does not identify the truncated product with the full one.

The result

The two-prime Möbius product equals two thirds and corresponds to squarefree integers supported on precisely those primes.

Explore the model

Interactive lab

Interactive lab · spectral

Absolute coefficient bounds and finite cutoffs

Pause/Reset and set mode count 1 and decay 2. At time zero the value at spatial position pi/2 is 1. For a fresh comparison Pause/Reset, set both mode count 3 and decay 2; predict value 8/9 there. The third coefficient subtracts 1/9 while the second contributes zero.

Absolute coefficient bounds and finite cutoffsA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA bound on the sum of absolute omitted coefficients controls an oscillating tail. This finite sine instrument illustrates that reasoning; its modes are not prime factors or a Dirichlet series.

Check your understanding

At a point with real part greater than one, what does the zeta Euler product restricted to primes two and three represent?

Try it yourself

Evaluate the two-prime zeta product at argument two, then find the coefficient of the integer twelve in the square of the full zeta series.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Invert the two local Möbius factors for the first calculation. For the second, count ordered factor pairs with product twelve.

Compare with the solution

The finite-prime zeta value is (1−1/4)−1(1−1/9)−1=(4/3)(9/8)=3/2(1-1/4)^{-1}(1-1/9)^{-1}=(4/3)(9/8)=3/2. It includes all integers supported on two and three, not just integers through six. In the square of the full series, the coefficient at twelve counts the pairs (1,12),(2,6),(3,4),(4,3),(6,2),(12,1)(1,12),(2,6),(3,4),(4,3),(6,2),(12,1), giving six. These are ordered pairs, so replacing the list by unordered factorizations would incorrectly halve the coefficient.

Take it with you

  • State an absolute-convergence half-plane before rearranging infinite arithmetic sums.
  • Multiplicative coefficients produce prime-power factors; complete multiplicativity makes them geometric.
  • Prime truncations and integer truncations describe different sets of coefficients.

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