Analytic number theory lives on a single recurring trick: a sum you cannot evaluate becomes an integral you can. The bridge is Abel summation — also called summation by parts or partial summation — the discrete twin of the integration by parts you already know from calculus. Where integration by parts trades \int u\,dv for \int v\,du, Abel summation trades a weighted sum \sum a_n f(n) for the partial sums of the a_n against the derivative of the smooth weight f.

Why does this matter so much? Because the arithmetic lives in the coefficients a_n — they might be 1, or \Lambda(n), or an indicator of the primes — while the analysis lives in the smooth factor f, typically 1/n, n^{-s}, or \log n. Abel summation lets you handle each in its own natural language: control the coefficients through their average A(x)=\sum_{n\le x} a_n, and control the weight through ordinary calculus. It is the single most-used identity in the whole subject, and this page teaches it thoroughly, because you will reach for it on almost every page that follows. If the asymptotic notation O(\cdot) and \sim feels shaky, keep it nearby — we use it constantly to name the error terms Abel summation produces.

The formula

Read it as a swap of roles. On the left, the smooth function f is sampled at the integers and weighted by the arithmetic a_n. On the right, the arithmetic has been collected into the step function A(t), and f now appears only through its derivative under a tame integral. The lone A(x)f(x) is the boundary term — the exact analogue of the [uv] bracket in integration by parts.

Where it comes from

There are two clean derivations; both are worth carrying. The first is a bare-hands rearrangement. Because a_n = A(n)-A(n-1), the finite sum \sum_{n=1}^{N} a_n f(n) is

Shifting the index in the second sum and collecting the A(n) terms gives the discrete summation-by-parts identity

Now the difference f(n+1)-f(n) is exactly \int_n^{n+1} f'(t)\,dt by the fundamental theorem of calculus, and A(t) is constant (equal to A(n)) on each interval [n,n+1). So the sum on the right glues into a single integral, \int_1^N A(t)f'(t)\,dt, and we land on the theorem. The step-function nature of A is not a nuisance — it is precisely what turns a sum into an integral.

Worked example — the harmonic sum and Euler's constant

The signature application: pin down \sum_{n\le x} 1/n to within a vanishing error. Take a_n = 1, so A(t)=\lfloor t\rfloor, and f(t)=1/t, so f'(t)=-1/t^2. Abel summation gives

Now split off the fractional part with \lfloor t\rfloor = t - \{t\}. The main piece \int_1^x t/t^2\,dt = \int_1^x dt/t = \log x produces the logarithm; the fractional piece is where the constant hides:

The integral \int_1^\infty \{t\}\,t^{-2}\,dt converges (the integrand is \le t^{-2}), so we write it as its full value minus a tail. Define

the Euler–Mascheroni constant. Using \lfloor x\rfloor/x = 1 - \{x\}/x = 1 + O(1/x) and bounding the tail \int_x^\infty \{t\}t^{-2}\,dt \le \int_x^\infty t^{-2}\,dt = 1/x, everything collapses to the clean asymptotic

That is Abel summation earning its keep in four lines: a sum with no closed form becomes \log x plus a specific constant plus an error that dies like 1/x. The constant \gamma is not an accident of the method — it is defined by the leftover integral the method exposes.

Seeing it: the sum shadows \log x + \gamma

The chart plots the staircase H(x)=\sum_{n\le x} 1/n against the smooth curve \log x + \gamma. The staircase jumps by 1/n at each integer and levels off between jumps; the smooth curve threads through it. What Abel summation proved is that the vertical gap between the two is O(1/x) — already by x=10 they are within a couple of hundredths, and the gap keeps shrinking.

The picture also explains the constant's meaning geometrically: \gamma is the limiting total area of the little slivers trapped between the staircase and the curve 1/t — the accumulated discrepancy between summing and integrating, exactly the quantity the partial-summation integral measures.

Worked example — extending \zeta(s) past the pole

The Dirichlet series \zeta(s)=\sum_{n\ge1} n^{-s} converges only for \Re(s)>1. Abel summation continues it into the strip \Re(s)>0 for free. Take a_n=1 again, so A(t)=\lfloor t\rfloor, and f(t)=t^{-s}, so f'(t)=-s\,t^{-s-1}:

Substitute \lfloor t\rfloor = t-\{t\} and integrate the main term with \int_1^x t^{-s}\,dt = \dfrac{x^{1-s}-1}{1-s}. For \Re(s)>1 let x\to\infty: the boundary term and x^{1-s} vanish, leaving

Here is the magic. The right-hand side makes sense for all \Re(s)>0 (except s=1), because |\{t\}|\le1 makes the integral converge whenever \Re(s)>0. So this formula is the analytic continuation of \zeta into the critical strip, and it lays the pole bare: the only singularity is the s/(s-1) term, a simple pole at s=1 with residue 1. One application of Abel summation has taken a series that diverges at s=1 and handed us \zeta, its pole, and its residue all at once. As a bonus, letting s\to1 in the finite-x version reproduces the harmonic asymptotic above, with \gamma = 1 - \int_1^\infty \{t\}t^{-2}\,dt reappearing as the constant term in the Laurent expansion \zeta(s)=\frac1{s-1}+\gamma+\cdots.

Worked example — a Mertens-type sum

Abel summation is bidirectional: it also lets you feed a known asymptotic for a summatory function into a weighted sum. Suppose we know the harmonic estimate H(x)=\sum_{n\le x}1/n=\log x+\gamma+O(1/x) and we want \sum_{n\le x}\frac{\log n}{n}. Take a_n=1/n so that A(t)=H(t), and f(t)=\log t with f'(t)=1/t:

Substituting H(t)=\log t+\gamma+O(1/t) and integrating term by term (using \int_1^x \frac{\log t}{t}\,dt=\tfrac12\log^2 x and \int_1^x \frac{dt}{t}=\log x) the leading pieces are (\log x+\gamma)\log x - \big(\tfrac12\log^2x+\gamma\log x\big), and the \gamma\log x terms cancel to leave

This is the shape of the estimates Mertens used. The same machine, fed the prime summatory functions instead of the harmonic one, produces Mertens' theorems on \sum_{p\le x}\frac{\log p}{p} and \sum_{p\le x}\frac1p — the point being that partial summation is how you convert a statement about one prime-weighted sum into a statement about another.

Worked example — trading \pi(x), \theta(x), and prime sums

The reason partial summation is the tool of prime-counting is that it converts freely between the different ways of weighting the primes. Take the coefficients a_n = (\log p)\cdot\mathbf 1[n=p\text{ prime}], whose summatory function is Chebyshev's \theta(x)=\sum_{p\le x}\log p. To reach the prime sum \sum_{p\le x}\frac{\log p}{p} we weight by f(t)=1/t:

Feeding in \theta(t)\sim t (the PNT in Chebyshev form) makes the integral \sim\int^x dt/t=\log x, recovering \sum_{p\le x}\frac{\log p}{p}\sim\log x. Going the other direction, to pass from \theta(x)=\sum_{p\le x}\log p to the raw count \pi(x)=\sum_{p\le x}1, weight by f(t)=1/\log t (which strips off the \log):

With \theta(x)\sim x this yields \pi(x)\sim x/\log x, and carrying the integral honestly upgrades the estimate to \pi(x)\sim\operatorname{Li}(x). This is exactly how one shows the several forms of the Prime Number Theorem — \theta(x)\sim x, \psi(x)\sim x, \pi(x)\sim x/\log x — are all equivalent: each is a partial summation away from the next.