# 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
    $A(t)=\sum _{0\leq n\leq t}a_{n}$
for any real number $t$. Fix real numbers $x<y$, and let $\phi$ be a continuously differentiable function on $[x,y]$. Then:
    $\sum _{x<n\leq y}a_{n}\phi (n)=A(y)\phi (y)-A(x)\phi (x)-\int _{x}^{y}A(u)\phi '(u)\,du.$
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
    $\sum _{0\leq n\leq x}a_{n}\phi (n)=A(x)\phi (x)-\int _{0}^{x}A(u)\phi '(u)\,du.$
If the sequence $(a_{n})$ is indexed starting at $n=1$, then we may formally define $a_{0}=0$. The previous formula becomes
    $\sum _{1\leq n\leq x}a_{n}\phi (n)=A(x)\phi (x)-\int _{1}^{x}A(u)\phi '(u)\,du.$
A common way to apply Abel's summation formula is to take the limit of one of these formulas as $x\to \infty$. The resulting formulas are
    ${\begin{aligned}\sum _{n=0}^{\infty }a_{n}\phi (n)&=\lim _{x\to \infty }{\biglA(x)\phi (x){\bigr )}-\int _{0}^{\infty }A(u)\phi '(u)\,du,\\\sum _{n=1}^{\infty }a_{n}\phi (n)&=\lim _{x\to \infty }{\biglA(x)\phi (x){\bigr )}-\int _{1}^{\infty }A(u)\phi '(u)\,du.\end{aligned}}$
These equations hold whenever both limits on the right-hand side exist and are finite.
A particularly useful case is the sequence $a_{n}=1$ for all $n\geq 0$. In this case, $A(x)=\lfloor x+1\rfloor$. For this sequence, Abel's summation formula simplifies to
    $\sum _{0\leq n\leq x}\phi (n)=\lfloor x+1\rfloor \phi (x)-\int _{0}^{x}\lfloor u+1\rfloor \phi '(u)\,du.$
Similarly, for the sequence $a_{0}=0$ and $a_{n}=1$ for all $n\geq 1$, the formula becomes
    $\sum _{1\leq n\leq x}\phi (n)=\lfloor x\rfloor \phi (x)-\int _{1}^{x}\lfloor u\rfloor \phi '(u)\,du.$
Upon taking the limit as $x\to \infty$, we find
    ${\begin{aligned}\sum _{n=0}^{\infty }\phi (n)&=\lim _{x\to \infty }{\bigl\lfloor x+1\rfloor \phi (x){\bigr )}-\int _{0}^{\infty }\lfloor u+1\rfloor \phi '(u)\,du,\\\sum _{n=1}^{\infty }\phi (n)&=\lim _{x\to \infty }{\bigl\lfloor x\rfloor \phi (x){\bigr )}-\int _{1}^{\infty }\lfloor u\rfloor \phi '(u)\,du,\end{aligned}}$
assuming that both terms on the right-hand side exist and are finite.
Abel's summation formula can be generalized to the case where $\phi$ is only assumed to be continuous if the integral is interpreted as a Riemann–Stieltjes integral:
    $\sum _{x<n\leq y}a_{n}\phi (n)=A(y)\phi (y)-A(x)\phi (x)-\int _{x}^{y}A(u)\,d\phi (u).$
By taking $\phi$ 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 $n\geq 1$ 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 $n\geq 1$ 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 $A(x)=M(x)=\sum _{n\leq x}\mu (n)$ 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 $\Re (s)>1$.