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 [math]\displaystyle{ (a_n)_{n=0}^\infty }[/math] be a sequence of real or complex numbers. Define the partial sum function [math]\displaystyle{ A }[/math] by
- [math]\displaystyle{ A(t) = \sum_{0 \le n \le t} a_n }[/math]
for any real number [math]\displaystyle{ t }[/math]. Fix real numbers [math]\displaystyle{ x \lt y }[/math], and let [math]\displaystyle{ \phi }[/math] be a continuously differentiable function on [math]\displaystyle{ [x, y] }[/math]. Then:
- [math]\displaystyle{ \sum_{x \lt n \le y} a_n\phi(n) = A(y)\phi(y) - A(x)\phi(x) - \int_x^y A(u)\phi'(u)\,du. }[/math]
The formula is derived by applying integration by parts for a Riemann–Stieltjes integral to the functions [math]\displaystyle{ A }[/math] and [math]\displaystyle{ \phi }[/math].
Variations
Taking the left endpoint to be [math]\displaystyle{ -1 }[/math] gives the formula
- [math]\displaystyle{ \sum_{0 \le n \le x} a_n\phi(n) = A(x)\phi(x) - \int_0^x A(u)\phi'(u)\,du. }[/math]
If the sequence [math]\displaystyle{ (a_n) }[/math] is indexed starting at [math]\displaystyle{ n = 1 }[/math], then we may formally define [math]\displaystyle{ a_0 = 0 }[/math]. The previous formula becomes
- [math]\displaystyle{ \sum_{1 \le n \le x} a_n\phi(n) = A(x)\phi(x) - \int_1^x A(u)\phi'(u)\,du. }[/math]
A common way to apply Abel's summation formula is to take the limit of one of these formulas as [math]\displaystyle{ x \to \infty }[/math]. The resulting formulas are
- [math]\displaystyle{ \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} }[/math]
These equations hold whenever both limits on the right-hand side exist and are finite.
A particularly useful case is the sequence [math]\displaystyle{ a_n = 1 }[/math] for all [math]\displaystyle{ n \ge 0 }[/math]. In this case, [math]\displaystyle{ A(x) = \lfloor x + 1 \rfloor }[/math]. For this sequence, Abel's summation formula simplifies to
- [math]\displaystyle{ \sum_{0 \le n \le x} \phi(n) = \lfloor x + 1 \rfloor\phi(x) - \int_0^x \lfloor u + 1\rfloor \phi'(u)\,du. }[/math]
Similarly, for the sequence [math]\displaystyle{ a_0 = 0 }[/math] and [math]\displaystyle{ a_n = 1 }[/math] for all [math]\displaystyle{ n \ge 1 }[/math], the formula becomes
- [math]\displaystyle{ \sum_{1 \le n \le x} \phi(n) = \lfloor x \rfloor\phi(x) - \int_1^x \lfloor u \rfloor \phi'(u)\,du. }[/math]
Upon taking the limit as [math]\displaystyle{ x \to \infty }[/math], we find
- [math]\displaystyle{ \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} }[/math]
assuming that both terms on the right-hand side exist and are finite.
Abel's summation formula can be generalized to the case where [math]\displaystyle{ \phi }[/math] is only assumed to be continuous if the integral is interpreted as a Riemann–Stieltjes integral:
- [math]\displaystyle{ \sum_{x \lt n \le y} a_n\phi(n) = A(y)\phi(y) - A(x)\phi(x) - \int_x^y A(u)\,d\phi(u). }[/math]
By taking [math]\displaystyle{ \phi }[/math] to be the partial sum function associated to some sequence, this leads to the summation by parts formula.
Examples
Harmonic numbers
If [math]\displaystyle{ a_n = 1 }[/math] for [math]\displaystyle{ n \ge 1 }[/math] and [math]\displaystyle{ \phi(x) = 1/x, }[/math] then [math]\displaystyle{ A(x) = \lfloor x \rfloor }[/math] and the formula yields
- [math]\displaystyle{ \sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n} = \frac{\lfloor x \rfloor}{x} + \int_1^x \frac{\lfloor u \rfloor}{u^2} \,du. }[/math]
The left-hand side is the harmonic number [math]\displaystyle{ H_{\lfloor x \rfloor} }[/math].
Representation of Riemann's zeta function
Fix a complex number [math]\displaystyle{ s }[/math]. If [math]\displaystyle{ a_n = 1 }[/math] for [math]\displaystyle{ n \ge 1 }[/math] and [math]\displaystyle{ \phi(x) = x^{-s}, }[/math] then [math]\displaystyle{ A(x) = \lfloor x \rfloor }[/math] and the formula becomes
- [math]\displaystyle{ \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. }[/math]
If [math]\displaystyle{ \Re(s) \gt 1 }[/math], then the limit as [math]\displaystyle{ x \to \infty }[/math] exists and yields the formula
- [math]\displaystyle{ \zeta(s) = s\int_1^\infty \frac{\lfloor u\rfloor}{u^{1+s}}\,du. }[/math]
where [math]\displaystyle{ \zeta(s) }[/math] is the Riemann zeta function. This may be used to derive Dirichlet's theorem that [math]\displaystyle{ \zeta(s) }[/math] 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 [math]\displaystyle{ a_n = \mu(n) }[/math] is the Möbius function and [math]\displaystyle{ \phi(x) = x^{-s} }[/math], then [math]\displaystyle{ A(x) = M(x) = \sum_{n \le x} \mu(n) }[/math] is Mertens function and
- [math]\displaystyle{ \frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = s\int_1^\infty \frac{M(u)}{u^{1+s}}\,du. }[/math]
This formula holds for [math]\displaystyle{ \Re(s) \gt 1 }[/math].
See also
References
- Apostol, Tom (1976), Introduction to Analytic Number Theory, Undergraduate Texts in Mathematics, Springer-Verlag.
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