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

Template:Wikibooks Let $(a_n{)}_{n=0}^{\mathrm{\infty }}$ 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 $\varphi$ be a continuously differentiable function on $[x,y]$. Then:

$\sum _{x<n\le y}{a}_n\varphi (n)=A(y)\varphi (y)-A(x)\varphi (x)-{\int }_x^{y}A(u){\varphi }^{\prime }(u)du.$

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

Variations

Taking the left endpoint to be $-1$ gives the formula

$\sum _{0\le n\le x}{a}_n\varphi (n)=A(x)\varphi (x)-{\int }_0^{x}A(u){\varphi }^{\prime }(u)du.$

If the sequence $(a_n)$ is indexed starting at $n=1$, then we may formally define $a_0=0$. The previous formula becomes

$\sum _{1\le n\le x}{a}_n\varphi (n)=A(x)\varphi (x)-{\int }_1^{x}A(u){\varphi }^{\prime }(u)du.$

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

$\begin{array}{cc}\sum _{n=0}^{\mathrm{\infty }}{a}_n\varphi (n)& =\underset{x\to \mathrm{\infty }}{lim}(A(x)\varphi (x))-{\int }_0^{\mathrm{\infty }}A(u){\varphi }^{\prime }(u)du,\\ \sum _{n=1}^{\mathrm{\infty }}{a}_n\varphi (n)& =\underset{x\to \mathrm{\infty }}{lim}(A(x)\varphi (x))-{\int }_1^{\mathrm{\infty }}A(u){\varphi }^{\prime }(u)du.\end{array}$

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

$\sum _{0\le n\le x}\varphi (n)=\lfloor x+1\rfloor \varphi (x)-{\int }_0^x\lfloor u+1\rfloor {\varphi }^{\prime }(u)du.$

Similarly, for the sequence $a_0=0$ and $a_n=1$ for all $n\ge 1$, the formula becomes

$\sum _{1\le n\le x}\varphi (n)=\lfloor x\rfloor \varphi (x)-{\int }_1^x\lfloor u\rfloor {\varphi }^{\prime }(u)du.$

Upon taking the limit as $x\to \mathrm{\infty }$, we find

$\begin{array}{cc}\sum _{n=0}^{\mathrm{\infty }}\varphi (n)& =\underset{x\to \mathrm{\infty }}{lim}(\lfloor x+1\rfloor \varphi (x))-{\int }_0^{\mathrm{\infty }}\lfloor u+1\rfloor {\varphi }^{\prime }(u)du,\\ \sum _{n=1}^{\mathrm{\infty }}\varphi (n)& =\underset{x\to \mathrm{\infty }}{lim}(\lfloor x\rfloor \varphi (x))-{\int }_1^{\mathrm{\infty }}\lfloor u\rfloor {\varphi }^{\prime }(u)du,\end{array}$

assuming that both terms on the right-hand side exist and are finite.

Abel's summation formula can be generalized to the case where $\varphi$ is only assumed to be continuous if the integral is interpreted as a Riemann–Stieltjes integral:

$\sum _{x<n\le y}{a}_n\varphi (n)=A(y)\varphi (y)-A(x)\varphi (x)-{\int }_x^{y}A(u)d\varphi (u).$

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

Examples

Harmonic numbers

If $a_n=1$ for $n\ge 1$ and $\varphi (x)=1/x,$ 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 $\varphi (x)=x^{-s},$ 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 $\mathrm{\Re }(s)>1$, then the limit as $x\to \mathrm{\infty }$ exists and yields the formula

$\zeta (s)=s{\int }_1^{\mathrm{\infty }}\frac{\lfloor u\rfloor }{u^{1+s}}du.$

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 Template:Math.

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 $\varphi (x)=x^{-s}$, then $A(x)=M(x)=\sum _{n\le x}\mu (n)$ is Mertens function and

$\frac{1}{\zeta (s)}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mu (n)}{n^s}=s{\int }_1^{\mathrm{\infty }}\frac{M(u)}{u^{1+s}}du.$

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

See also

• Summation by parts
• Integration by parts

References

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