Abel's summation formula

Formula

Let ${\displaystyle (a_n{)}_{n=0}^{\mathrm{\infty }}}$ be a sequence of real or complex numbers. Define the partial sum function ${\displaystyle A}$ by

${\displaystyle A(t)=\sum _{0\le n\le t}{a}_n}$

for any real number ${\displaystyle t}$. Fix real numbers ${\displaystyle x<y}$, and let ${\displaystyle \varphi }$ be a continuously differentiable function on ${\displaystyle [x,y]}$. Then:

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

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

Variations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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