# Sums of Reciprocals of Triple Binomial Coefficients

Authors: A. Sofo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

---

## Abstract

We investigate the integral representation of infinite sums involving
the reciprocals of triple binomial coefficients. We also recover some wellknown
properties ofζ(3) and extend the range of results given by other authors.


---

## Body

## 1. Introduction

In this paper, we investigate the summation of the
reciprocal of triple products of combinatorial coefficients. In particular, we
develop integral representations for(1.1)
∑
n
=
0
∞
1
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
,
∑
n
=
0
∞
(
n
+
m
−
1
n
)
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
,and their alternating series
counterparts.For the representation of sums of reciprocals of
single and double binomial coefficients, one may refer to some results in the
papers [1–3], see also [4].For designated cases of the parameter values(
a
,
b
,
c
,
j
,
k
,
l
,
m
), various particular sums may be expressed in terms of ζ
(
2
) and ζ
(
3
).  For many
interesting properties of the Zeta function, the reader is referred to
[5].The representation of sums in terms of integrals is
extremely useful because it allows one to estimate bounds on the sums in cases
they cannot be written in closed form. Convexity properties for sums may also
be investigated.Apéry's [6], see also Beukers [7], proof of the irrationality of ζ
(
3
) uses an
elementary and quite complicated construction of the approximants α
n
/
β
n
∈
Q to this number
based on a recurrence relation. The integral representation(1.2)
∭
0
1
{
x
(
1
−
x
)
y
(
1
−
y
)
z
(
1
−
z
)
}
n
(
1
−
(
1
−
x
y
)
z
)
n
+
1
d
x
d
y
d
z
=
2
β
n
ζ
(
3
)
−
2
α
nfor the sequence {
α
n
,
β
n
} was proposed.It is important to note that other integral
representations ofζ
(
3
) are available
in terms of both single and double integrals. Guillera and Sondow [8] list a number of them
including the classical results(1.3)
∬
0
1
−
ln
(
x
y
)
1
−
x
y
d
x
d
y
=
2
ζ
(
3
)
,
∬
0
1
ln
(
2
−
x
y
)
1
−
x
y
d
x
d
y
=
5
8
ζ
(
3
)
.In a recent paper, Muzaffar
[9] also obtained some
results of the combinatorial type(1.4)
∑
n
=
0
∞
(
2
n
n
)
(
2
n
+
1
1
)
(
2
n
+
k
+
1
1
)
(
2
n
+
2
k
n
+
k
)
=
α
k
π
2
+
β
kby utilising the power series
expansion of (
sin
−
1
x
)
q, and (
α
k
,
β
k
) are constants
depending on k
≥
0. In this paper,
we complement and extend some of the results given by Muzaffar.There are some identities in the literature involving
reciprocals of triple products of combinatorial coefficients, one
prominent
identity is the Dougall
identity, see [10] or
[11],(1.5)
1
+
2
∑
n
=
1
∞
(
−
1
)
n
(
a
n
)
(
b
n
)
(
c
n
)
(
n
+
a
a
)
(
n
+
b
b
)
(
n
+
c
c
)
=
(
a
+
b
+
c
b
)
(
a
+
b
b
)
(
b
+
c
b
)for R
(
a
+
b
+
c
)
>
−
1.


## 2. The Main Results

In this section, we develop integral identities for reciprocals
of triple products of binomial coefficients.Theorem 2.1.
Fora
,
b, and c positive real
numbers and j
,
k
,
l
≥
0
, then (2.1)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
∑
n
=
0
∞
1
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
(2.2)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
1
−
X
Y
Z
d
x
d
y
d
z
(2.3)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
1
+
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
X
Y
Z
(
(
X
Y
Z
)
2
+
4
X
Y
Z
+
1
)
d
x
d
y
d
z
x
y
z
(
1
−
X
Y
Z
)
4
(2.4)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
j
+
k
+
l
+
1
F
j
+
k
+
l
[
1
,
1
a
,
2
a
,
…
,
j
a
,
1
b
,
2
b
,
…
,
k
b
,
1
c
,
2
c
,
…
,
l
c
a
+
1
a
,
a
+
2
a
,
…
,
a
+
j
a
,
b
+
1
b
,
b
+
2
b
,
…
,
b
+
k
b
,
c
+
1
c
,
c
+
2
c
,
…
,
c
+
l
c
|
1
]
,and similarly (2.5)
T
(
a
,
b
,
c
,
j
,
k
,
l
)
=
∑
n
=
0
∞
(
−
1
)
n
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
(2.6)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
1
+
X
Y
Z
d
x
d
y
d
z
(2.7)
=
1
−
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
X
Y
Z
(
(
X
Y
Z
)
2
−
4
X
Y
Z
+
1
)
d
x
d
y
d
z
x
y
z
(
1
+
X
Y
Z
)
4
(2.8)
=
j
+
k
+
l
+
1
F
j
+
k
+
l
[
1
,
1
a
,
2
a
,
…
,
j
a
,
1
b
,
2
b
,
…
,
k
b
,
1
c
,
2
c
,
…
,
l
c
a
+
1
a
,
a
+
2
a
,
…
,
a
+
j
a
,
b
+
1
b
,
b
+
2
b
,
…
,
b
+
k
b
,
c
+
1
c
,
c
+
2
c
,
…
,
c
+
l
c
|
−
1
]
,where(2.9)
X
Y
Z
=
x
a
y
b
z
c
.Proof.
Consider (2.1):(2.10)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
∑
n
=
0
∞
1
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
=
j
k
l
∑
n
=
0
∞
Γ
(
a
n
+
1
)
Γ
(
j
)
Γ
(
b
n
+
1
)
Γ
(
k
)
Γ
(
c
n
+
1
)
Γ
(
l
)
Γ
(
a
n
+
j
+
1
)
Γ
(
b
n
+
k
+
1
)
Γ
(
c
n
+
l
+
1
)
=
j
k
l
∑
n
=
0
∞
B
(
a
n
,
j
+
1
)
B
(
b
n
,
k
+
1
)
B
(
c
n
,
l
+
1
)
,where Γ
(
⋅
) is the
classical Gamma function and B
(
⋅
,
⋅
) is the Beta
function. It holds that
(2.11)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
j
k
l
∑
n
=
0
∞
∫
x
=
0
1
(
1
−
x
)
j
−
1
x
a
n
d
x
∫
y
=
0
1
(
1
−
y
)
k
−
1
y
b
n
d
y
∫
z
=
0
1
(
1
−
z
)
l
−
1
z
c
n
d
z
=
j
k
l
∫
x
=
0
1
∫
y
=
0
1
∫
z
=
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
∑
n
=
0
∞
(
x
a
y
b
z
c
)
n
d
x
d
y
d
zby an allowable change of
integral and sum, and hence we have
(2.12)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
1
−
X
Y
Z
d
x
d
y
d
z
,which is the result
(2.2).
To prove identity (2.3), consider (2.1) and expand as
follows:(2.13)
S
(
a
,
b
,
c
,
j
,
k
,
l
)
=
∑
n
=
0
∞
a
b
c
n
3
Γ
(
a
n
)
Γ
(
j
+
1
)
Γ
(
b
n
)
Γ
(
k
+
1
)
Γ
(
c
n
)
Γ
(
l
+
1
)
Γ
(
a
n
+
j
+
1
)
Γ
(
b
n
+
k
+
1
)
Γ
(
c
n
+
l
+
1
)
=
∑
n
=
0
∞
a
b
c
n
3
B
(
j
+
1
,
a
n
)
B
(
k
+
1
,
b
n
)
B
(
k
+
1
,
c
n
)
=
1
+
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
x
y
z
∑
n
=
1
∞
n
3
(
x
a
y
b
z
c
)
n
d
x
d
y
d
z
=
1
+
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
x
y
z
(
1
−
X
Y
Z
)
4
X
Y
Z
(
(
X
Y
Z
)
2
+
4
X
Y
Z
+
1
)
d
x
d
y
d
zwhich is the result (2.3). The
results (2.6) and (2.7) may be obtained in a similar fashion and therefore will
not be pursued here.The hypergeometric representation (2.4) and
(2.8) can
be obtained by the consideration of the ratio of successive terms (2.1) and
(2.5), respectively.We may also
note that from known properties of the hypergeometric function, we may write,
from (2.4) and
(2.8),(2.14)
j
+
k
+
l
+
1
F
j
+
k
+
l
[
1
,
1
a
,
2
a
,
…
,
j
a
,
1
b
,
2
b
,
…
,
k
b
,
1
c
,
2
c
,
…
,
l
c
a
+
1
a
,
a
+
2
a
,
…
,
a
+
j
a
,
b
+
1
b
,
b
+
2
b
,
…
,
b
+
k
b
,
c
+
1
c
,
c
+
2
c
,
…
,
c
+
l
c
|
1
]
=
a
+
b
+
c
+
1
F
a
+
b
+
c
[
1
,
1
,
1
,
1
,
1
a
,
2
a
,
…
,
a
−
1
a
,
1
b
,
2
b
,
…
,
b
−
1
b
,
1
c
,
2
c
,
…
,
c
−
1
c
j
+
1
a
,
j
+
2
a
,
…
,
j
+
a
a
,
k
+
1
b
,
k
+
2
b
,
…
,
k
+
b
b
,
l
+
1
c
,
l
+
2
c
,
…
,
l
+
c
c
|
1
]
,
j
+
k
+
l
+
1
F
j
+
k
+
l
[
1
,
1
a
,
2
a
,
…
,
j
a
,
1
b
,
2
b
,
…
,
k
b
,
1
c
,
2
c
,
…
,
l
c
a
+
1
a
,
a
+
2
a
,
…
,
a
+
j
a
,
b
+
1
b
,
b
+
2
b
,
…
,
b
+
k
b
,
c
+
1
c
,
c
+
2
c
,
…
,
c
+
l
c
|
−
1
]
=
a
+
b
+
c
+
1
F
a
+
b
+
c
[
1
,
1
,
1
,
1
,
1
a
,
2
a
,
…
,
a
−
1
a
,
1
b
,
2
b
,
…
,
b
−
1
b
,
1
c
,
2
c
,
…
,
c
−
1
c
j
+
1
a
,
j
+
2
a
,
…
,
j
+
a
a
,
k
+
1
b
,
k
+
2
b
,
…
,
k
+
b
b
,
l
+
1
c
,
l
+
2
c
,
…
,
l
+
c
c
|
−
1
]
Examples
Example 3.1.
It holds
that(3.1)
S
(
1,1,1,1,1,1
)
=
∑
n
=
0
∞
1
(
n
+
1
)
3
=
ζ
(
3
)
=
∭
0
1
d
x
d
y
d
z
1
−
x
y
z
=
4
F
3
[
1,1,1,1
2,2,2
|
1
]
=
1
+
∭
0
1
(
1
−
x
)
(
1
−
y
)
(
1
−
z
)
(
(
x
y
z
)
2
+
4
x
y
z
+
1
)
(
1
−
x
y
z
)
4
d
x
d
y
d
z
.
Other integral representations ofζ
(
3
) do exist, some
of which are as follows.
Finch [12] gave the
expression(3.2)
(
−
1
)
n
n
!
ζ
(
n
+
1
)
=
∫
x
=
0
1
{
ln
(
x
)
}
n
1
−
x
d
x
=
∫
x
=
0
1
{
ln
(
1
−
x
)
}
n
x
d
x
.
Lord [13]
posed the problem to show that(3.3)
S
(
2,2,2,1,1,1
)
=
∑
n
=
0
∞
1
(
2
n
+
1
)
3
=
7
8
ζ
(
3
)
=
∫
x
=
0
π
/
4
ln
(
cos
(
x
)
)
ln
(
sin
(
x
)
)
cos
(
x
)
sin
(
x
)
d
x
=
∭
0
1
d
x
d
y
d
z
1
−
x
2
y
2
z
2
=
1
+
8
∭
0
1
(
1
−
x
)
(
1
−
y
)
(
1
−
z
)
x
y
z
(
(
x
y
z
)
4
+
4
(
x
y
z
)
2
+
1
)
(
1
−
x
2
y
2
z
2
)
4
d
x
d
y
d
z
=
4
F
3
[
1
2
,
1
2
,
1
2
,1
3
2
,
3
2
,
3
2
|
1
]
,the last three expressions are
directly from
(2.2),
(2.3), and
(2.4), respectively.
Nan-Yue and  Williams [14] also gave(3.4)
ζ
(
3
)
=
−
5
∫
x
=
0
ln
(
ϕ
2
)
x
ln
(
2
sinh
(
x
2
)
)
d
x
,where ϕ = golden ratio
= (
1
+
5
)
/
2.
Example 3.2.
It holds
that(3.5)
S
(
4
,
2
,
3
,
j
,
k
,
l
)
=
∑
n
=
0
∞
1
(
4
n
+
j
j
)
(
2
n
+
k
k
)
(
3
n
+
l
l
)
=
∑
n
=
0
∞
j
!
k
!
l
!
∏
r
=
1
j
(
4
n
+
r
)
∏
r
=
1
k
(
2
n
+
r
)
∏
r
=
1
l
(
3
n
+
r
)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
d
x
d
y
d
z
1
−
x
4
y
2
z
3
=
10
F
9
[
1
,
1
,
1
,
1
,
3
4
,
2
3
,
1
2
,
1
2
,
1
3
,
1
4
j
+
1
4
,
j
+
2
4
,
j
+
3
4
j
+
4
4
,
k
+
1
2
,
k
+
2
2
,
l
+
1
3
,
l
+
2
3
,
l
+
3
3
|
1
]
=
α
1
+
α
2
π
+
α
3
ζ
(
2
)
+
α
4
ln
(
2
)
+
α
5
ln
(
3
)
+
α
6
ζ
(
3
)
.For j
=
5
,
k
=
6,  and l
=
6,   we have the
values (3.6)
α
1
=
495762799093
17
⋅
11
⋅
7
2
⋅
3
2
⋅
2
4
,
α
2
=
(
167
⋅
2
25
17
⋅
13
⋅
11
⋅
7
2
⋅
3
2
−
23
⋅
5
⋅
3
16
3
17
⋅
13
⋅
11
⋅
7
2
⋅
2
4
)
,
α
3
=
1709
⋅
5
⋅
2
7
,
α
4
=
−
755357
⋅
2
19
17
⋅
13
⋅
11
⋅
7
2
⋅
3
2
,
α
5
=
43
⋅
23
⋅
5
⋅
3
16
17
⋅
13
⋅
11
⋅
7
2
⋅
2
4
,
α
6
=
5
3
⋅
3
⋅
2
2
.
Example 3.3.
For the
alternating case,(3.7)
T
(
1,1,1,1,1,1
)
=
∑
n
=
0
∞
(
−
1
)
n
(
n
+
1
)
3
=
3
4
ζ
(
3
)
=
∭
0
1
d
x
d
y
d
z
1
+
x
y
z
=
4
F
3
[
1,1,1,1
2,2,2
|
−
1
]
=
1
−
∭
0
1
(
1
−
x
)
(
1
−
y
)
(
1
−
z
)
(
(
x
y
z
)
2
−
4
x
y
z
+
1
)
(
1
+
x
y
z
)
4
d
x
d
y
d
z
.
Example 3.4.
It holds
that(3.8)
T
(
2
/
3
,
2
/
3
,
5
/
6
,
2
,
4
,
3
)
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
/
3
+
2
2
)
(
2
n
/
3
+
4
4
)
(
5
n
/
6
+
3
3
)
=
∑
n
=
0
∞
(
−
1
)
n
2
5
3
11
(
n
+
3
)
2
(
n
+
6
)
(
2
n
+
3
)
2
(
2
n
+
9
)
(
5
n
+
6
)
(
5
n
+
12
)
(
5
n
+
18
)
=
24
∭
0
1
(
1
−
x
)
(
1
−
y
)
3
(
1
−
z
)
2
d
x
d
y
d
z
1
+
x
2
/
3
y
2
/
3
z
5
/
6
=
10
F
9
[
1
,
3
2
,
3
2
,
3
,
3
,
9
2
,
6
,
6
5
,
12
5
,
18
5
5
2
,
5
2
,
4
,
4
,
11
2
,
11
5
,
17
5
,
23
5
,7
|
−
1
]
=
70663
⋅
1669
13
⋅
11
⋅
7
⋅
5
⋅
3
⋅
2
+
3
3
⋅
2
4
ζ
(
2
)
+
3
2
⋅
2
10
7
G
+
109
⋅
5
⋅
3
⋅
2
9
11
⋅
7
2
ln
2
−
2
8
⋅
3
4
11
⋅
7
2
π
+
(
113
⋅
5
6
⋅
5
11
⋅
7
2
⋅
2
2
−
271
⋅
5
5
11
⋅
7
2
⋅
2
3
)
ln
(
α
)
−
(
113
⋅
5
6
⋅
5
11
⋅
7
2
⋅
2
2
+
271
⋅
5
5
11
⋅
7
2
⋅
2
3
)
ln
(
ϕ
)
−
(
5
4
⋅
3
⋅
5
⋅
ϕ
α
5
11
⋅
2
+
37
⋅
5
4
⋅
5
⋅
α
ϕ
5
7
2
)
π
,where G is Catalan's
constant, ϕ is the golden
ratio, and α = silver ratio
= (
5
−
1
)
/
2
.Now consider the following theorem, which is a
generalisation of Theorem2.1.Theorem 3.5.
Fora
,
b
,
c, and m positive real
numbers and j
,
k
,
l
≥
0 with j
+
k
+
l
≥
m
, then (3.9)
Q
(
a
,
b
,
c
,
j
,
k
,
l
,
m
)
=
∑
n
=
0
∞
(
n
+
m
−
1
n
)
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
(3.10)
=
∑
n
=
0
∞
j
!
k
!
l
!
(
n
+
1
)
m
−
1
(
m
−
1
)
!
(
a
n
+
1
)
j
(
b
n
+
1
)
k
(
c
n
+
1
)
l
(3.11)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
(
1
−
X
Y
Z
)
m
d
x
d
y
d
z
(3.12)
=
1
+
m
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
X
Y
Z
(
(
X
Y
Z
)
2
+
(
3
m
+
1
)
X
Y
Z
+
1
)
d
x
d
y
d
z
x
y
z
(
1
−
X
Y
Z
)
m
+
3
(3.13)
=
j
+
k
+
l
+
1
F
j
+
k
+
l
[
m
,
1
a
,
2
a
,
…
,
j
a
,
1
b
,
2
b
,
…
,
k
b
,
1
c
,
2
c
,
…
,
l
c
a
+
1
a
,
a
+
2
a
,
…
,
a
+
j
a
,
b
+
1
b
,
b
+
2
b
,
…
,
b
+
k
b
,
c
+
1
c
,
c
+
2
c
,
…
,
c
+
l
c
|
1
]
,
(3.14)
R
(
a
,
b
,
c
,
j
,
k
,
l
,
m
)
=
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
(3.15)
=
∑
n
=
0
∞
(
−
1
)
n
j
!
k
!
l
!
(
n
+
1
)
m
−
1
(
m
−
1
)
!
(
a
n
+
1
)
j
(
b
n
+
1
)
k
(
c
n
+
1
)
l
(3.16)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
(
1
+
X
Y
Z
)
m
d
x
d
y
d
z
(3.17)
=
1
−
m
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
X
Y
Z
(
(
X
Y
Z
)
2
−
(
3
m
+
1
)
X
Y
Z
+
1
)
d
x
d
y
d
z
x
y
z
(
1
+
X
Y
Z
)
m
+
3
(3.18)
=
j
+
k
+
l
+
1
F
j
+
k
+
l
[
m
,
1
a
,
2
a
,
…
,
j
a
,
1
b
,
2
b
,
…
,
k
b
,
1
c
,
2
c
,
…
,
l
c
a
+
1
a
,
a
+
2
a
,
…
,
a
+
j
a
,
b
+
1
b
,
b
+
2
b
,
…
,
b
+
k
b
,
c
+
1
c
,
c
+
2
c
,
…
,
c
+
l
c
|
−
1
]
,where X
Y
Z is given by
(2.9) and(3.19)
(
p
)
α
=
p
(
p
+
1
)
⋯
(
p
+
α
−
1
)
=
Γ
(
p
+
α
)
Γ
(
p
)is Pochhammer's symbol.Proof.
Consider (3.14):(3.20)
R
(
a
,
b
,
c
,
j
,
k
,
l
,
m
)
=
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
=
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
a
b
c
n
3
Γ
(
a
n
)
Γ
(
j
+
1
)
Γ
(
b
n
)
Γ
(
k
+
1
)
Γ
(
c
n
)
Γ
(
l
+
1
)
Γ
(
a
n
+
j
+
1
)
Γ
(
b
n
+
k
+
1
)
Γ
(
c
n
+
l
+
1
)
=
∑
n
=
0
∞
(
−
1
)
n
a
b
c
n
3
(
n
+
m
−
1
n
)
B
(
a
n
,
j
+
1
)
B
(
b
n
,
k
+
1
)
B
(
c
n
,
l
+
1
)
=
1
+
a
b
c
∑
n
=
1
∞
(
−
1
)
n
n
3
(
n
+
m
−
1
n
)
∫
0
1
(
1
−
x
)
j
−
1
x
a
n
d
x
∫
0
1
(
1
−
y
)
k
−
1
y
b
n
d
y
∫
0
1
(
1
−
z
)
l
−
1
z
c
n
d
z
.By  an allowable change of
integral and sum, we have(3.21)
R
(
a
,
b
,
c
,
j
,
k
,
l
,
m
)
=
1
+
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
x
y
z
∑
n
=
1
∞
(
−
1
)
n
(
n
+
m
−
1
m
−
1
)
n
3
(
x
a
y
b
z
c
)
n
d
x
d
y
d
z
=
1
−
m
a
b
c
∭
0
1
(
1
−
x
)
j
(
1
−
y
)
k
(
1
−
z
)
l
X
Y
Z
(
(
X
Y
Z
)
2
−
(
3
m
+
1
)
X
Y
Z
+
1
)
x
y
z
(
1
+
X
Y
Z
)
m
+
3
d
x
d
y
d
zwhich is the result
(3.17).
To arrive at the result (3.16), consider(3.22)
R
(
a
,
b
,
j
,
c
,
k
,
l
,
m
)
=
j
k
l
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
Γ
(
a
n
+
1
)
Γ
(
j
)
Γ
(
b
n
+
1
)
Γ
(
k
)
Γ
(
c
n
+
1
)
Γ
(
l
)
Γ
(
a
n
+
j
+
1
)
Γ
(
b
n
+
k
+
1
)
Γ
(
c
n
+
l
+
1
)
=
j
k
l
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
B
(
a
n
+
1
,
j
)
B
(
b
n
+
1
,
k
)
B
(
c
n
+
1
,
l
)
=
j
k
l
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
∫
0
1
(
1
−
x
)
j
−
1
x
a
n
d
x
∫
0
1
(
1
−
y
)
k
−
1
y
b
n
d
y
∫
0
1
(
1
−
z
)
l
−
1
z
c
n
d
z
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
∑
n
=
0
∞
(
−
1
)
n
(
n
+
m
−
1
n
)
(
x
a
y
b
z
c
)
n
d
x
d
y
d
zby an allowable change of sum and
integral, hence(3.23)
R
(
a
,
b
,
j
,
c
,
k
,
l
,
m
)
=
j
k
l
∭
0
1
(
1
−
x
)
j
−
1
(
1
−
y
)
k
−
1
(
1
−
z
)
l
−
1
(
1
+
X
Y
Z
)
m
d
x
d
y
d
zwhich is the result (3.16).The hypergeometric
representations (3.13) and
(3.18) can be obtained
by the consideration of the ratio of successive terms (3.9)
and (3.14),
respectively.In the case whenm
=
1
, Theorem 3.5
reduces to Theorem 2.1.Examples
Example 4.1.
It holds
that(4.1)
Q
(
4,3,2,5,3,6,11
)
=
∑
n
=
0
∞
(
n
+
10
n
)
(
4
n
+
5
5
)
(
3
n
+
3
3
)
(
2
n
+
6
6
)
=
90
∭
0
1
(
1
−
x
)
4
(
1
−
y
)
2
(
1
−
z
)
5
(
1
−
x
4
y
3
z
2
)
11
d
x
d
y
d
z
=
1
+
264
∭
0
1
(
1
−
x
)
5
(
1
−
y
)
3
(
1
−
z
)
6
x
3
y
2
z
(
x
8
y
6
z
4
+
34
x
4
y
3
z
2
+
1
)
d
x
d
y
d
z
(
1
−
x
4
y
3
z
2
)
14
=
10
F
9
[
11
,
1
,
1
,
1
,
3
4
,
2
3
,
1
2
,
1
2
,
1
3
,
1
4
4
,
7
2
,
9
4
,
2
,
2
,
5
3
,
7
4
,
3
2
,
4
3
|
1
]
=
3413
⋅
43
⋅
5
⋅
3
2
14
ζ
(
2
)
+
(
1931
⋅
509
2
12
−
90379
3
7
⋅
3
⋅
2
5
)
π
−
1459
⋅
1231
7
⋅
3
2
⋅
2
11
−
379880779
7
⋅
3
⋅
2
10
ln
2
+
22567
⋅
3
2
7
⋅
2
5
ln
3.
Example 4.2.
It holds
that(4.2)
R
(
1
/
4
,
1
/
6
,
1
/
2
,
5
,
3
,
7
,
14
)
:
=
∑
n
=
0
∞
(
−
1
)
n
(
n
+
13
n
)
(
n
/
4
+
5
5
)
(
n
/
6
+
3
3
)
(
n
/
2
+
7
7
)
=
∑
n
=
0
∞
(
−
1
)
n
5
!
3
!
7
!
(
n
+
1
)
13
13
!
(
n
/
4
+
1
)
5
(
n
/
6
+
1
)
3
(
n
/
2
+
1
)
7
=
105
∭
0
1
(
1
−
x
)
4
(
1
−
y
)
2
(
1
−
z
)
6
(
1
+
x
1
/
4
y
1
/
6
z
1
/
2
)
14
d
x
d
y
d
z
=
1
+
7
24
∭
0
1
(
1
−
x
)
5
(
1
−
y
)
3
(
1
−
z
)
7
×
(
x
1
/
2
y
1
/
3
z
+
43
x
1
/
4
y
1
/
6
z
1
/
2
+
1
)
x
3
/
4
y
5
/
6
z
1
/
2
(
1
−
x
1
/
4
y
1
/
6
z
1
/
2
)
17
d
x
d
y
d
z
=
16
F
15
[
2,4,4,6,6,8,8,10,12,12,12,14,14,16,18,20
3,5,5,7,7,9,9,11,13,13,13,15,17,19,21
|
−
1
]
=
7
⋅
5
⋅
3
3
⋅
2
4
13
ζ
(
2
)
−
16231
⋅
11
⋅
3
7
⋅
5
⋅
2
3
.


## 3. Conclusion

We have provided triple integral identities for sums
of the reciprocal of triple binomial coefficients. In doing so, we have
recovered the standard representation forζ
(
3
) and have
generalised and extended some results published previously by other authors.In another forum, we will extend our results to
consider binomial coefficients of the form(5.1)
∑
n
=
0
∞
n
s
(
n
+
m
−
1
n
)
(
a
n
+
j
j
)
(
b
n
+
k
k
)
(
c
n
+
l
l
)
,
∑
n
=
0
∞
(
n
+
m
−
1
n
)
(
a
n
+
j
b
n
)
(
c
n
+
k
d
n
)
(
p
n
+
j
q
n
)
.



---

*Source: 794181-2008-04-01.xml*