Number Theory
Abel Summation: Turning Sums into Integrals
Abel summation is the discrete cousin of integration by parts: it lets you swap a hard-to-control sum ∑ aₙ f(n) for a boundary term plus an integral ∑ A(n)Δf(n), trading the wild oscillation of the aₙ for the tame smoothness of f. This one trick is how analytic number theory turns arithmetic sums into estimates — it converts partial sums of the divisor function, the Möbius function, or the von Mangoldt function into integrals you can actually bound.
Precisely: if A(x) = ∑_{n≤x} aₙ is the summatory function of a sequence (aₙ) and f is continuously differentiable on [y, x], then ∑_{y
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What the formula claims
Let (aₙ)_{n≥1} be any sequence of complex numbers and define its summatory function A(x) = ∑_{n≤x} aₙ (a step function, with A(x)=0 for x<1). Let f: [y, x] → ℂ be continuously differentiable. Then
∑_{y The most common special case takes y = 0 (or any point below 1, so A(y)=0) and a real x: ∑_{n≤x} aₙ f(n) = A(x)f(x) − ∫_1^x A(t) f′(t) dt. The point is structural: the arithmetic content (the aₙ, which may oscillate violently — think Möbius μ(n) or χ(n) for a character) is entirely absorbed into A, while f contributes only through its average smoothness via f′. If you have any nontrivial bound on A(t) — cancellation, a main term plus error — the integral inherits it. The identity is exact; there is no approximation. Think of ∑ aₙf(n) as a weighted total where each 'quantum' aₙ arrives at time n and gets scaled by f(n). Abel summation reorganizes the bookkeeping. Instead of paying f(n) to each individual aₙ, it pays the accumulated total A(t) an infinitesimal rate f′(t) dt as f drifts, and settles the balance at the endpoints with the boundary terms A(x)f(x) − A(y)f(y). The mechanism is telescoping. Writing aₙ = A(n) − A(n−1) and rearranging the sum by 'summation by parts' shifts the difference from the erratic aₙ onto the slowly varying f: differences f(n+1) − f(n) ≈ f′(n) replace f itself. Because f is smooth, those differences are small and controllable, whereas the aₙ were not. In the continuous limit the difference operator becomes d/dt and the sum becomes ∫ A(t) f′(t) dt. It is exactly the discrete analogue of integration by parts: you move the derivative off the ugly factor and onto the nice one. The cleanest route is Riemann–Stieltjes integration by parts. By definition A is a right-continuous step function with jumps of size aₙ at each integer n, so dA is the sum of point masses aₙ·δₙ, and ∑_{y ∫_y^x f dA = [f(t)A(t)]_y^x − ∫_y^x A(t) df(t) = A(x)f(x) − A(y)f(y) − ∫_y^x A(t) f′(t) dt, using df = f′(t) dt because f ∈ C¹. Elementarily, one proves it by the finite identity ∑ aₙf(n) = A(N)f(N) − ∑_{n Take aₙ = 1, so A(x) = ⌊x⌋, and f(n) = 1/n, f′(t) = −1/t². For integer x = N, ∑_{n≤N} 1/n = ⌊N⌋·(1/N) − ∫_1^N ⌊t⌋·(−1/t²) dt = 1 + ∫_1^N ⌊t⌋/t² dt. Write ⌊t⌋ = t − {t} where {t} is the fractional part. Then ∫_1^N ⌊t⌋/t² dt = ∫_1^N (1/t) dt − ∫_1^N {t}/t² dt = ln N − ∫_1^N {t}/t² dt. Hence ∑_{n≤N} 1/n = ln N + 1 − ∫_1^N {t}/t² dt. Since {t}/t² is nonnegative and O(1/t²), the integral converges as N → ∞, and 1 − ∫_1^∞ {t}/t² dt = γ ≈ 0.5772156649, the Euler–Mascheroni constant. Abel summation has produced the asymptotic ∑_{n≤N} 1/n = ln N + γ + O(1/N) directly, with an explicit integral for γ — no clever guessing required. f ∈ C¹ is what makes df = f′ dt. If f is merely continuous but not of bounded variation, the Stieltjes integral ∫ A df need not exist and the formula fails; if f has jumps you must use the Riemann–Stieltjes form ∫ f dA = [fA] − ∫ A df and watch for common discontinuities of f and A (they cannot jump at the same point, or the product rule for Stieltjes integrals breaks). You must also use the correct summatory function. A subtle but standard error is the endpoint convention: whether A jumps at n from the left or right dictates whether the term aₙ at n = y or n = x is included. Get the half-open interval (y, x] right. The identity connects to Dirichlet series (setting f(n) = n^{−s} shows a Dirichlet series converges wherever its partial sums are controlled — Abel's theorem on convergence), to Euler–Maclaurin summation (a higher-order refinement using more derivatives), and to Dirichlet's test for convergence, which is Abel summation plus a bounded-partial-sums hypothesis. Abel summation is the workhorse that converts arithmetic into analysis throughout number theory. Setting f(t) = t^{−s} turns any Dirichlet series ∑ aₙ/nˢ into s∫_1^∞ A(t) t^{−s−1} dt (the Mellin transform of A), which is how one meromorphically continues the Riemann zeta function ζ(s) = s/(s−1) − s∫_1^∞ {t} t^{−s−1} dt past s = 1 and locates its pole. It proves that a Dirichlet series converges on a half-plane Re(s) > σ_c determined by the growth of partial sums (Abel's theorem, 1826). It underlies the equivalence of the Prime Number Theorem with ψ(x) ~ x, converting between ∑_{p≤x} 1, ∑ log p, and ∑ Λ(n) by choosing f = log or 1/log. Beyond number theory it gives Dirichlet's and Abel's convergence tests for series, the summation-by-parts step in numerical analysis and in bounding oscillatory sums (van der Corput), and the discrete integration-by-parts used throughout probability (martingale transforms) and signal processing.The picture: reassigning credit to boundaries
Key idea of the proof
Worked example: ∑ 1/n and Euler's constant
Why the hypotheses matter — and connections
Applications and significance
Identity Left side Right side / boundary + interior term Integration by parts ∫_a^b u dv [uv]_a^b − ∫_a^b v du Discrete summation by parts ∑_{n=1}^{N} aₙ bₙ A_N b_N − ∑_{n=1}^{N−1} Aₙ(b_{n+1} − bₙ), with Aₙ = ∑_{k≤n} aₖ Abel's summation formula ∑_{y<n≤x} aₙ f(n) A(x)f(x) − A(y)f(y) − ∫_y^x A(t)f′(t) dt, f ∈ C¹ Riemann–Stieltjes form ∫_y^x f dA [f A]_y^x − ∫_y^x A df (f, A of bounded variation, no common jumps)
Frequently asked questions
Is Abel summation the same as summation by parts?
Essentially yes — they are the same identity in two guises. 'Summation by parts' usually denotes the purely discrete finite identity ∑_{n=1}^{N} aₙbₙ = A_N b_N − ∑_{n=1}^{N−1} Aₙ(b_{n+1}−bₙ). 'Abel summation' or 'Abel's summation formula' names the version where the second index runs continuously and the discrete difference is written as an integral ∫ A(t)f′(t)dt, which requires f ∈ C¹. The integral form is strictly more useful for asymptotics.
Why must f be continuously differentiable?
The term ∫_y^x A(t)f′(t)dt only makes literal sense when f′ exists and is integrable, i.e. f ∈ C¹ (or at least absolutely continuous). More fundamentally, the Riemann–Stieltjes derivation needs df = f′dt; for f merely of bounded variation you keep the Stieltjes form ∫ A df, and for f neither smooth nor of bounded variation the interior integral can fail to exist and the identity breaks.
What is the summatory function A and why the half-open interval?
A(x) = ∑_{n≤x} aₙ accumulates the sequence up to x; it is a right-continuous step function jumping by aₙ at each integer n. The formula sums over (y, x], a half-open interval, precisely because A's jumps are what encode the aₙ. Sloppiness about whether the endpoints y and x are included is the single most common source of off-by-aₙ errors when applying the formula.
How does Abel summation prove ∑ 1/n = ln N + γ + O(1/N)?
Take aₙ = 1 (so A(t) = ⌊t⌋) and f(t) = 1/t. Abel summation gives ∑_{n≤N} 1/n = 1 + ∫_1^N ⌊t⌋/t² dt = ln N + 1 − ∫_1^N {t}/t² dt. Because {t}/t² is O(1/t²), its tail is O(1/N), and 1 − ∫_1^∞ {t}/t² dt equals the Euler–Mascheroni constant γ ≈ 0.5772. The whole asymptotic drops out of one application.
How is it used to analytically continue the Riemann zeta function?
For Re(s) > 1, apply Abel summation to ζ(s) = ∑ n^{−s} with aₙ=1, f(t)=t^{−s}, f′(t) = −s t^{−s−1}. You get ζ(s) = s∫_1^∞ ⌊t⌋ t^{−s−1} dt = s/(s−1) − s∫_1^∞ {t} t^{−s−1} dt. The last integral converges for Re(s) > 0, so this formula continues ζ to Re(s) > 0 and exhibits the simple pole at s=1 with residue 1.
What is the relationship to integration by parts and Euler–Maclaurin?
Abel summation IS integration by parts for the Stieltjes measure dA. Euler–Maclaurin is its higher-order refinement: instead of stopping at one integral you iterate, replacing {t}−½ by successive Bernoulli polynomials to gain extra derivatives f″, f‴, … and thereby smaller error terms. Abel summation is the first-order Euler–Maclaurin step; the full formula extracts an asymptotic expansion when f is smooth enough.