Let
(
a
n
)
n
=
0
∞
{\displaystyle (a_{n})_{n=0}^{\infty }}
sequence of real or complex numbers . Define the partial sum function
A
{\displaystyle A}
A
(
t
)
=
∑
0
≤
n
≤
t
a
n
{\displaystyle A(t)=\sum _{0\leq n\leq t}a_{n}}
for any real number
t
{\displaystyle t}
x
<
y
{\displaystyle x<y}
ϕ
{\displaystyle \phi }
continuously differentiable function on
[
x
,
y
]
{\displaystyle [x,y]}
∑
x
<
n
≤
y
a
n
ϕ
(
n
)
=
A
(
y
)
ϕ
(
y
)
−
A
(
x
)
ϕ
(
x
)
−
∫
x
y
A
(
u
)
ϕ
′
(
u
)
d
u
.
{\displaystyle \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
{\displaystyle A}
ϕ
{\displaystyle \phi }
Taking the left endpoint to be
−
1
{\displaystyle -1}
∑
0
≤
n
≤
x
a
n
ϕ
(
n
)
=
A
(
x
)
ϕ
(
x
)
−
∫
0
x
A
(
u
)
ϕ
′
(
u
)
d
u
.
{\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
(
a
n
)
{\displaystyle (a_{n})}
n
=
1
{\displaystyle n=1}
a
0
=
0
{\displaystyle a_{0}=0}
∑
1
≤
n
≤
x
a
n
ϕ
(
n
)
=
A
(
x
)
ϕ
(
x
)
−
∫
1
x
A
(
u
)
ϕ
′
(
u
)
d
u
.
{\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
x
→
∞
{\displaystyle x\to \infty }
∑
n
=
0
∞
a
n
ϕ
(
n
)
=
lim
x
→
∞
(
A
(
x
)
ϕ
(
x
)
)
−
∫
0
∞
A
(
u
)
ϕ
′
(
u
)
d
u
,
∑
n
=
1
∞
a
n
ϕ
(
n
)
=
lim
x
→
∞
(
A
(
x
)
ϕ
(
x
)
)
−
∫
1
∞
A
(
u
)
ϕ
′
(
u
)
d
u
.
{\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
a
n
=
1
{\displaystyle a_{n}=1}
n
≥
0
{\displaystyle n\geq 0}
A
(
x
)
=
⌊
x
+
1
⌋
{\displaystyle A(x)=\lfloor x+1\rfloor }
∑
0
≤
n
≤
x
ϕ
(
n
)
=
⌊
x
+
1
⌋
ϕ
(
x
)
−
∫
0
x
⌊
u
+
1
⌋
ϕ
′
(
u
)
d
u
.
{\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
a
0
=
0
{\displaystyle a_{0}=0}
a
n
=
1
{\displaystyle a_{n}=1}
n
≥
1
{\displaystyle n\geq 1}
∑
1
≤
n
≤
x
ϕ
(
n
)
=
⌊
x
⌋
ϕ
(
x
)
−
∫
1
x
⌊
u
⌋
ϕ
′
(
u
)
d
u
.
{\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
x
→
∞
{\displaystyle x\to \infty }
∑
n
=
0
∞
ϕ
(
n
)
=
lim
x
→
∞
(
⌊
x
+
1
⌋
ϕ
(
x
)
)
−
∫
0
∞
⌊
u
+
1
⌋
ϕ
′
(
u
)
d
u
,
∑
n
=
1
∞
ϕ
(
n
)
=
lim
x
→
∞
(
⌊
x
⌋
ϕ
(
x
)
)
−
∫
1
∞
⌊
u
⌋
ϕ
′
(
u
)
d
u
,
{\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
ϕ
{\displaystyle \phi }
Riemann–Stieltjes integral :
∑
x
<
n
≤
y
a
n
ϕ
(
n
)
=
A
(
y
)
ϕ
(
y
)
−
A
(
x
)
ϕ
(
x
)
−
∫
x
y
A
(
u
)
d
ϕ
(
u
)
.
{\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
ϕ
{\displaystyle \phi }
summation by parts formula.
If
a
n
=
1
{\displaystyle a_{n}=1}
n
≥
1
{\displaystyle n\geq 1}
ϕ
(
x
)
=
1
/
x
,
{\displaystyle \phi (x)=1/x,}
A
(
x
)
=
⌊
x
⌋
{\displaystyle A(x)=\lfloor x\rfloor }
∑
n
=
1
⌊
x
⌋
1
n
=
⌊
x
⌋
x
+
∫
1
x
⌊
u
⌋
u
2
d
u
.
{\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
H
⌊
x
⌋
{\displaystyle H_{\lfloor x\rfloor }}
edit
Fix a complex number
s
{\displaystyle s}
a
n
=
1
{\displaystyle a_{n}=1}
n
≥
1
{\displaystyle n\geq 1}
ϕ
(
x
)
=
x
−
s
,
{\displaystyle \phi (x)=x^{-s},}
A
(
x
)
=
⌊
x
⌋
{\displaystyle A(x)=\lfloor x\rfloor }
∑
n
=
1
⌊
x
⌋
1
n
s
=
⌊
x
⌋
x
s
+
s
∫
1
x
⌊
u
⌋
u
1
+
s
d
u
.
{\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
ℜ
(
s
)
>
1
{\displaystyle \Re (s)>1}
x
→
∞
{\displaystyle x\to \infty }
ζ
(
s
)
=
s
∫
1
∞
⌊
u
⌋
u
1
+
s
d
u
.
{\displaystyle \zeta (s)=s\int _{1}^{\infty }{\frac {\lfloor u\rfloor }{u^{1+s}}}\,du.}
where
ζ
(
s
)
{\displaystyle \zeta (s)}
Riemann zeta function .
This may be used to derive Dirichlet's theorem that
ζ
(
s
)
{\displaystyle \zeta (s)}
pole with residue 1 at s = 1
Reciprocal of Riemann zeta function
edit
The technique of the previous example may also be applied to other Dirichlet series . If
a
n
=
μ
(
n
)
{\displaystyle a_{n}=\mu (n)}
Möbius function and
ϕ
(
x
)
=
x
−
s
{\displaystyle \phi (x)=x^{-s}}
A
(
x
)
=
M
(
x
)
=
∑
n
≤
x
μ
(
n
)
{\displaystyle A(x)=M(x)=\sum _{n\leq x}\mu (n)}
Mertens function and
1
ζ
(
s
)
=
∑
n
=
1
∞
μ
(
n
)
n
s
=
s
∫
1
∞
M
(
u
)
u
1
+
s
d
u
.
{\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
ℜ
(
s
)
>
1
{\displaystyle \Re (s)>1}
![{\displaystyle [x,y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b7bd6292c6023626c6358bfd3943a031b27d663)
