Variations

Taking the left endpoint to be ${\displaystyle -1}$ gives the formula

${\displaystyle \sum _{0\leq n\leq x}a_{n}\phi (n)=A(x)\phi (x)-\int _{0}^{x}A(u)\phi '(u)\,du.}$

If the sequence ${\displaystyle (a_{n})}$ is indexed starting at ${\displaystyle n=1}$, then we may formally define ${\displaystyle a_{0}=0}$. The previous formula becomes

${\displaystyle \sum _{1\leq n\leq x}a_{n}\phi (n)=A(x)\phi (x)-\int _{1}^{x}A(u)\phi '(u)\,du.}$

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

${\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }a_{n}\phi (n)&=\lim _{x\to \infty }{\bigl (}A(x)\phi (x){\bigr )}-\int _{0}^{\infty }A(u)\phi '(u)\,du,\\\sum _{n=1}^{\infty }a_{n}\phi (n)&=\lim _{x\to \infty }{\bigl (}A(x)\phi (x){\bigr )}-\int _{1}^{\infty }A(u)\phi '(u)\,du.\end{aligned}}}$

These equations hold whenever both limits on the right-hand side exist and are finite.

A particularly useful case is the sequence ${\displaystyle a_{n}=1}$ for all ${\displaystyle n\geq 0}$. In this case, ${\displaystyle A(x)=\lfloor x+1\rfloor }$. For this sequence, Abel's summation formula simplifies to

${\displaystyle \sum _{0\leq n\leq x}\phi (n)=\lfloor x+1\rfloor \phi (x)-\int _{0}^{x}\lfloor u+1\rfloor \phi '(u)\,du.}$

Similarly, for the sequence ${\displaystyle a_{0}=0}$ and ${\displaystyle a_{n}=1}$ for all ${\displaystyle n\geq 1}$, the formula becomes

${\displaystyle \sum _{1\leq n\leq x}\phi (n)=\lfloor x\rfloor \phi (x)-\int _{1}^{x}\lfloor u\rfloor \phi '(u)\,du.}$

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

${\displaystyle {\begin{aligned}\sum _{n=0}^{\infty }\phi (n)&=\lim _{x\to \infty }{\bigl (}\lfloor x+1\rfloor \phi (x){\bigr )}-\int _{0}^{\infty }\lfloor u+1\rfloor \phi '(u)\,du,\\\sum _{n=1}^{\infty }\phi (n)&=\lim _{x\to \infty }{\bigl (}\lfloor x\rfloor \phi (x){\bigr )}-\int _{1}^{\infty }\lfloor u\rfloor \phi '(u)\,du,\end{aligned}}}$

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

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

${\displaystyle \sum _{x<n\leq y}a_{n}\phi (n)=A(y)\phi (y)-A(x)\phi (x)-\int _{x}^{y}A(u)\,d\phi (u).}$

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

Harmonic numbers

If ${\displaystyle a_{n}=1}$ for ${\displaystyle n\geq 1}$ and ${\displaystyle \phi (x)=1/x,}$ then ${\displaystyle A(x)=\lfloor x\rfloor }$ and the formula yields

${\displaystyle \sum _{n=1}^{\lfloor x\rfloor }{\frac {1}{n}}={\frac {\lfloor x\rfloor }{x}}+\int _{1}^{x}{\frac {\lfloor u\rfloor }{u^{2}}}\,du.}$

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

Representation of Riemann's zeta function

Fix a complex number ${\displaystyle s}$. If ${\displaystyle a_{n}=1}$ for ${\displaystyle n\geq 1}$ and ${\displaystyle \phi (x)=x^{-s},}$ then ${\displaystyle A(x)=\lfloor x\rfloor }$ and the formula becomes

${\displaystyle \sum _{n=1}^{\lfloor x\rfloor }{\frac {1}{n^{s}}}={\frac {\lfloor x\rfloor }{x^{s}}}+s\int _{1}^{x}{\frac {\lfloor u\rfloor }{u^{1+s}}}\,du.}$

If ${\displaystyle \Re (s)>1}$, then the limit as ${\displaystyle x\to \infty }$ exists and yields the formula

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

where ${\displaystyle \zeta (s)}$ is the Riemann zeta function. This may be used to derive Dirichlet's theorem that ${\displaystyle \zeta (s)}$ 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 ${\displaystyle a_{n}=\mu (n)}$ is the Möbius function and ${\displaystyle \phi (x)=x^{-s}}$, then ${\displaystyle A(x)=M(x)=\sum _{n\leq x}\mu (n)}$ is Mertens function and

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

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