Abel's summation formula

In mathematics, Abel's summation formula, introduced by Niels Henrik Abel, is intensively used in analytic number theory and the study of special functions to compute series.

Formula
Let $$(a_n)_{n=0}^\infty$$ be a sequence of real or complex numbers. Define the partial sum function $$A$$ by
 * $$A(t) = \sum_{0 \le n \le t} a_n$$

for any real number $$t$$. Fix real numbers $$x < y$$, and let $$\phi$$ be a continuously differentiable function on $$[x, y]$$. Then:
 * $$\sum_{x < n \le y} a_n\phi(n) = A(y)\phi(y) - A(x)\phi(x) - \int_x^y A(u)\phi'(u)\,du.$$

The formula is derived by applying integration by parts for a Riemann–Stieltjes integral to the functions $$A$$ and $$\phi$$.

Variations
Taking the left endpoint to be $$-1$$ gives the formula
 * $$\sum_{0 \le n \le x} a_n\phi(n) = A(x)\phi(x) - \int_0^x A(u)\phi'(u)\,du.$$

If the sequence $$(a_n)$$ is indexed starting at $$n = 1$$, then we may formally define $$a_0 = 0$$. The previous formula becomes
 * $$\sum_{1 \le n \le x} a_n\phi(n) = A(x)\phi(x) - \int_1^x A(u)\phi'(u)\,du.$$

A common way to apply Abel's summation formula is to take the limit of one of these formulas as $$x \to \infty$$. The resulting formulas are
 * $$\begin{align}

\sum_{n=0}^\infty a_n\phi(n) &= \lim_{x \to \infty}\bigl(A(x)\phi(x)\bigr) - \int_0^\infty A(u)\phi'(u)\,du, \\ \sum_{n=1}^\infty a_n\phi(n) &= \lim_{x \to \infty}\bigl(A(x)\phi(x)\bigr) - \int_1^\infty A(u)\phi'(u)\,du. \end{align}$$ These equations hold whenever both limits on the right-hand side exist and are finite.

A particularly useful case is the sequence $$a_n = 1$$ for all $$n \ge 0$$. In this case, $$A(x) = \lfloor x + 1 \rfloor$$. For this sequence, Abel's summation formula simplifies to
 * $$\sum_{0 \le n \le x} \phi(n) = \lfloor x + 1 \rfloor\phi(x) - \int_0^x \lfloor u + 1\rfloor \phi'(u)\,du.$$

Similarly, for the sequence $$a_0 = 0$$ and $$a_n = 1$$ for all $$n \ge 1$$, the formula becomes
 * $$\sum_{1 \le n \le x} \phi(n) = \lfloor x \rfloor\phi(x) - \int_1^x \lfloor u \rfloor \phi'(u)\,du.$$

Upon taking the limit as $$x \to \infty$$, we find
 * $$\begin{align}

\sum_{n=0}^\infty \phi(n) &= \lim_{x \to \infty}\bigl(\lfloor x + 1 \rfloor\phi(x)\bigr) - \int_0^\infty \lfloor u + 1\rfloor \phi'(u)\,du, \\ \sum_{n=1}^\infty \phi(n) &= \lim_{x \to \infty}\bigl(\lfloor x \rfloor\phi(x)\bigr) - \int_1^\infty \lfloor u\rfloor \phi'(u)\,du, \end{align}$$ assuming that both terms on the right-hand side exist and are finite.

Abel's summation formula can be generalized to the case where $$\phi$$ is only assumed to be continuous if the integral is interpreted as a Riemann–Stieltjes integral:
 * $$\sum_{x < n \le y} a_n\phi(n) = A(y)\phi(y) - A(x)\phi(x) - \int_x^y A(u)\,d\phi(u).$$

By taking $$\phi$$ to be the partial sum function associated to some sequence, this leads to the summation by parts formula.

Harmonic numbers
If $$a_n = 1$$ for $$n \ge 1$$ and $$\phi(x) = 1/x,$$ then $$A(x) = \lfloor x \rfloor$$ and the formula yields
 * $$\sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n} = \frac{\lfloor x \rfloor}{x} + \int_1^x \frac{\lfloor u \rfloor}{u^2} \,du.$$

The left-hand side is the harmonic number $$H_{\lfloor x \rfloor}$$.

Representation of Riemann's zeta function
Fix a complex number $$s$$. If $$a_n = 1$$ for $$n \ge 1$$ and $$\phi(x) = x^{-s},$$ then $$A(x) = \lfloor x \rfloor$$ and the formula becomes
 * $$\sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n^s} = \frac{\lfloor x \rfloor}{x^s} + s\int_1^x \frac{\lfloor u\rfloor}{u^{1+s}}\,du.$$

If $$\Re(s) > 1$$, then the limit as $$x \to \infty$$ exists and yields the formula
 * $$\zeta(s) = s\int_1^\infty \frac{\lfloor u\rfloor}{u^{1+s}}\,du.$$

where $$\zeta(s)$$ is the Riemann zeta function. This may be used to derive Dirichlet's theorem that $$\zeta(s) $$ has a simple pole with residue 1 at $s = 1$.

Reciprocal of Riemann zeta function
The technique of the previous example may also be applied to other Dirichlet series. If $$a_n = \mu(n)$$ is the Möbius function and $$\phi(x) = x^{-s}$$, then $$A(x) = M(x) = \sum_{n \le x} \mu(n)$$ is Mertens function and
 * $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = s\int_1^\infty \frac{M(u)}{u^{1+s}}\,du.$$

This formula holds for $$\Re(s) > 1$$.