Английская Википедия:Abel's summation formula
Шаблон:Short description Шаблон:Dablink
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
Шаблон:Wikibooks Let <math>(a_n)_{n=0}^\infty</math> be a sequence of real or complex numbers. Define the partial sum function <math>A</math> by
- <math>A(t) = \sum_{0 \le n \le t} a_n</math>
for any real number <math>t</math>. Fix real numbers <math>x < y</math>, and let <math>\phi</math> be a continuously differentiable function on <math>[x, y]</math>. Then:
- <math>\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.</math>
The formula is derived by applying integration by parts for a Riemann–Stieltjes integral to the functions <math>A</math> and <math>\phi</math>.
Variations
Taking the left endpoint to be <math>-1</math> gives the formula
- <math>\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>(a_n)</math> is indexed starting at <math>n = 1</math>, then we may formally define <math>a_0 = 0</math>. The previous formula becomes
- <math>\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>x \to \infty</math>. The resulting formulas are
- <math>\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>a_n = 1</math> for all <math>n \ge 0</math>. In this case, <math>A(x) = \lfloor x + 1 \rfloor</math>. For this sequence, Abel's summation formula simplifies to
- <math>\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>a_0 = 0</math> and <math>a_n = 1</math> for all <math>n \ge 1</math>, the formula becomes
- <math>\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>x \to \infty</math>, we find
- <math>\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>\phi</math> is only assumed to be continuous if the integral is interpreted as a Riemann–Stieltjes integral:
- <math>\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).</math>
By taking <math>\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>a_n = 1</math> for <math>n \ge 1</math> and <math>\phi(x) = 1/x,</math> then <math>A(x) = \lfloor x \rfloor</math> and the formula yields
- <math>\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>H_{\lfloor x \rfloor}</math>.
Representation of Riemann's zeta function
Fix a complex number <math>s</math>. If <math>a_n = 1</math> for <math>n \ge 1</math> and <math>\phi(x) = x^{-s},</math> then <math>A(x) = \lfloor x \rfloor</math> and the formula becomes
- <math>\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>\Re(s) > 1</math>, then the limit as <math>x \to \infty</math> exists and yields the formula
- <math>\zeta(s) = s\int_1^\infty \frac{\lfloor u\rfloor}{u^{1+s}}\,du.</math>
where <math>\zeta(s)</math> is the Riemann zeta function. This may be used to derive Dirichlet's theorem that <math>\zeta(s) </math> has a simple pole with residue 1 at Шаблон:Math.
Reciprocal of Riemann zeta function
The technique of the previous example may also be applied to other Dirichlet series. If <math>a_n = \mu(n)</math> is the Möbius function and <math>\phi(x) = x^{-s}</math>, then <math>A(x) = M(x) = \sum_{n \le x} \mu(n)</math> is Mertens function and
- <math>\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>\Re(s) > 1</math>.
See also
References
