# Journal of Inequalities in Pure and Applied Mathematics ```Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 35, 2006
AROUND AP&Eacute;RY’S CONSTANT
WALTHER JANOUS
U RSULINENGYMNASIUM
F &Uuml;RSTENWEG 86
A-6020 I NNSBRUCK
AUSTRIA .
walther.janous@tirol.com
Received 04 January, 2006; accepted 18 January, 2006
Communicated by A. Lupaş
Dedicated to Professor Gerd Baron on the occasion of his 65th birthday.
A BSTRACT. In this note we deal with some aspects of Ap&eacute;ry’s constant ζ(3).
Key words and phrases: Ap&eacute;ry’s constant, Harmonic numbers, Infinite sums, Integrals.
2000 Mathematics Subject Classification. 49J40, 90C33, 47H10.
1. I NTRODUCTION
Since Ap&eacute;ry’s miraculous proof (1979) that the value ζ(3) of Rieman’s ζ-function is irrational, Ap&eacute;ry’s constant ζ(3) has been the focus of attention for many mathematicians. (An
extensive list of results and references are found in Section 1.6 of the highly recommended
encyclopedic book .)
It is the purpose of this note to extend some of these results. Thereby we will also obtain a
new infinite sum rapidly converging to ζ(3).
At the end of this note we raise two questions for further investigation.
ISSN (electronic): 1443-5756
This research is partially supported by grant 04-39-3265-2/03 (11.XII.2003) of the Federal Ministry of Education and Sciences, Bosnia and
Herzegowina.
020-06
2
WALTHER JANOUS
2. T WO M ULTISUMS
Recently in  the proof of
∞
∞ X
X
(2.1)
1
= 2ζ(3),
ij(i + j)
i=1 j=1
where ζ(3) = 1.202056903..., was posed as a problem. Although this result was published
earlier (see [2, p. 43]) it is worthwhile reconsidering in the following more general way.
Theorem 2.1. For r ≥ 1 the multisum
∞
X
Sr =
∞
X
&middot;&middot;&middot;
k1 =1
kr
1
k1 &middot; &middot; &middot; kr (k1 + &middot; &middot; &middot; + kr )
=1
attains the value r!ζ(r + 1).
Proof. Firstly we rewrite the multisum as an integral as follows
Z 1
∞
∞
X
X
1
Sr =
&middot;&middot;&middot;
xk1 +&middot;&middot;&middot;+kr −1 dx
k
&middot;
&middot;
&middot;
k
1
r 0
k =1
k =1
r
1
that is (upon interchanging of summation and integration),
Z 1 X
∞
∞
X
x k1
x kr
1
Sr =
&middot;&middot;&middot;
dx.
k
r
0 x k =1 k1
k =1
r
1
Due to
∞
X
xj
j=1
= − ln(1 − x),
j
we get
r
Z
1
ln(1 − x)r
dx.
x
Sr = (−1)
0
Substituting x = 1 − t yields
r
Z
Sr = (−1)
0
1
ln(t)r
dt.
1−t
This and the known result ([2, p. 47])
Z 1
ln(t)r
dt = (−1)r r!ζ(r + 1)
1
−
t
0
Subsequently we will deal with a ‘relative’ of Sr , namely the multisum
Tr =
∞
X
k1 =1
&middot;&middot;&middot;
∞
X
kr
(−1)k1 +&middot;&middot;&middot;+kr
.
k
1 &middot; &middot; &middot; kr (k1 + &middot; &middot; &middot; + kr )
=1
As it will turn out, matters are here more involved. Indeed, we will prove now
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
http://jipam.vu.edu.au/
A ROUND A P&Eacute;RY ’ S C ONSTANT
3
Theorem 2.2. For r ≥ 1,
(2.2)
r
(ln 2)r+1
r+1
Tr = (−1)r r!ζ(r + 1) −
∞ X
r
X
r(r − 1)...(r − m + 1)
−
(ln 2)r−m
k
m+1
2 k
k=1 m=1
!
holds.
Proof. Proceeding as in the previous proof we get
Z 1
ln(1 + x)r
r
Tr = (−1)
dx.
x
0
Substitution of x = e−t − 1 yields
r
ln(1/2)
Z
Tr = (−1)
0
(−t)r
(−e−t )dt,
e−t − 1
that is,
Z
0
Tr =
ln(1/2)
Upon expanding
1
1−et
tr
dt.
1 − et
as a geometric series we arrive at
Tr =
∞ Z
X
k=0
0
tr ekt dt.
− ln 2
Integration by parts leads to the identity (we suppress integration constants)
Z
tr ekt dt = ekt
!
r
tr X
r(r
−
1)...(r
−
m
+
1)
+
(−1)m
tr−m ,
k m=1
k m+1
where k &gt; 0.
Therefore a straightforward simplification yields
∞
r
Tr = (−1)
(ln 2)r+1 X r!
+
r+1
k r+1
k=1
∞
X
1
−
2k
k=1
r
(ln 2)r X r(r − 1)...(r − m + 1)
+
(ln 2)r−m
m+1
k
k
m=1
!!
.
Since
∞
X
1
= ln 2,
k2k
k=1
we finally get the claimed identity (2.2).
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
http://jipam.vu.edu.au/
4
WALTHER JANOUS
3. A N EW F ORMULA
FOR
A P&Eacute;RY ’ S C ONSTANT
Theorem 2.2 enables us to obtain a new way to express ζ(3) by a fast converging series.
Indeed, letting r = 2 we get
∞
(ln 2)3 X 1
T2 = 2 ζ(3) −
−
3
2k
k=1
ln 2
1
+ 3
2
k
k
!
.
Furthermore [2, p. 43], reports
1
T2 = ζ(3).
4
Therefore the following holds.
Theorem 3.1.
(3.1)
∞
(ln 2)3 X 1
+
3
2k
k=1
8
ζ(3) =
7
ln 2
1
+ 3
2
k
k
!
.
This formula should be compared with the following one (see )
∞
X
2
(−1)k+1
.
ζ(3) = (ln 2)3 + 4
3 2k 2k
3
k
k
k=1
4. F URTHER O BSERVATIONS
•
From S2 + T2 = 94 ζ(3) we infer
X
2
i,j≥1
i+j even
1
9
= ζ(3)
ij(i + j)
4
that is (we put i + j = 2k),
∞ 2k−1
X
X
k=1 j=1
i.e.
1
9
= ζ(3),
2(2k − j)jk
8
∞
2k−1
X
1 X 1 1
1
9
+
= ζ(3).
2k j=1 2k j 2k − j
8
k=1
This can be summarized as
∞
X
1
9
H2k−1 = ζ(3),
2
k
4
k=1
where
Hn =
n
X
1
j=1
j
denotes the n-th harmonic number.
In a similar way S2 − T2 = 74 ζ(3) implies the formula
∞
X
k=1
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
1
7
H2k = ζ(3).
2
(2k + 1)
16
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A ROUND A P&Eacute;RY ’ S C ONSTANT
5
From these two formulae we get easily
Theorem 4.1.
∞
X
(4.1)
k=1
1
21
H2k−1 = ζ(3)
2
(2k − 1)
16
and
∞
X
(4.2)
k=1
1
11
H2k = ζ(3).
2
(2k)
16
∞
X
1
H = 2ζ(3),
2 k
k
k=1
(4.3)
a result already known to L. Euler.
•
[4, p. 499, item 2.6.9.14] reads
Z 1
∞
X
ln(1 + x)2
(−1)k ψ(k) π 2 γ
T2 =
dx = 2
−
,
x
k2
6
0
k=1
where ψ(z) = (ln Γ(z))0 and γ denote the digamma function and the Euler-Mascheroni constant, resp.
Therefore there holds the curious identity
∞
X
(−1)k ψ(k)
k2
k=1
1
π2γ
= ζ(3) +
.
8
12
Because of ψ(k) = −γ + Hk−1 it reads in equivalent form
∞
X
(−1)k
(4.4)
k=1
•
k2
5
Hk = ζ(3)
8
Recently  posed the problem of proving the identity
Z π/4
∞
X
1
ln(cos x) ln(sin x)
=7
dx.
3
(2n
+
1)
cos
x
sin
x
0
n=0
We show that it implies a remarkable result for two doublesums.
Indeed, we firstly note
∞
X
n=0
∞
∞
X 1
X 1
1
=
−
,
(2n + 1)3
k 3 k=1 (2k)3
k=1
that is
∞
X
n=0
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
1
7
= ζ(3).
3
(2n + 1)
8
http://jipam.vu.edu.au/
6
WALTHER JANOUS
Therefore, the identity under consideration in fact means
Z π/4
ln(cos x) ln(sin x)
ζ(3) = 8
&middot;
dx.
cos x
sin x
0
Letting f (x) =
ln(cos x)
cos x
and setting z = π/2 − x we obtain
Z π/2 π
π/4
π
f (x)f
− x dx =
f
− z f (z)dz
2
2
0
π/4
Z
whence
Z
ζ(3) = 4
√
Next, we substitute sin x = w.
π/2
0
ln(cos x) ln(sin x)
&middot;
dx.
cos x
sin x
√
From cos xdx = 2√1 w dw and cos x = 1 − w we get dx = 2√w√1 1−w dw.
This in turn yields
√
√
Z 1
ln( 1 − w) ln( w)
1
√
ζ(3) = 4
&middot; √
&middot; √ √
dw,
w
1−w
2 w 1−w
0
that is
1
ζ(3) =
2
Z
0
1
ln(1 − w) ln w
&middot;
dw.
1−w
w
Upon rewriting this as
Z
1 1 ln(1 − w) ln w
ζ(3) =
&middot;
dw
2 0
w
1−w
and developing the two factors of the integrand we get
!
!
Z
∞
∞
X
X
wi−1
(1 − w)j−1
1 1
ζ(3) =
−
−
dw,
2 0
i
j
i=1
j=1
that is
∞
∞
1 XX 1
ζ(3) =
2 i=1 j=1 ij
1
Z
wi−1 (1 − w)j−1 dw.
0
Keeping in mind that
1
Z
wi−1 (1 − w)j−1 dw =
0
(i − 1)!(j − 1)!
,
(i + j − 1)!
we arrive at the formula
∞
∞
1 XX
1
.
ζ(3) =
2 i=1 j=1 ij 2 i+j−1
j
(4.5)
Equation (4.5) and ζ(3) = 12 S2 give the two noteworthy identities
∞ X
∞
X
(4.6)
1
ij 2 i+j−1
j
i=1 j=1
∞ X
∞
X
=
i=1 j=1
1
ij(i + j)
and
Z
(4.7)
0
1
ln(1 − z) ln z
&middot;
dz =
z
1−z
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
Z
0
1
ln(1 − z)
ln(1 − z)dz.
z
http://jipam.vu.edu.au/
A ROUND A P&Eacute;RY ’ S C ONSTANT
•
7
Finally, Theorem 4.1 enables us to prove the following finite analogon of the initial for-
mula (2.1) of the present note, namely
∞ X
i
X
(4.8)
i=1 j=1
1
5
= ζ(3).
ij(i + j)
4
Indeed, (4.2) and (4.3) imply
∞
X
1
1
3
1
+ &middot;&middot;&middot; +
= ζ(3).
2
k
k+1
2k
4
k=1
However,
1
k2
1
1
+ &middot;&middot;&middot; +
k+1
2k
k
1X
1
=
k j=1 k(k + j)
and
k
k
1X
1
1X1
=
k j=1 k(k + j)
k j=1 j
1
1
−
k k+j
k
=
X
1
1
H
−
k
k2
kj(k + j)
j=1
2ζ(3) −
∞ X
k
X
k=1 j=1
1
3
= ζ(3)
kj(k + j)
4
as claimed.
5. T WO Q UESTIONS FOR F URTHER R ESEARCH
•
The results of Theorem 4.1 may be regarded as special cases of the more general sums
Sa,b =
∞
X
k=1
1
Hak−b ,
(ak − b)2
where 0 ≤ b &lt; a are entire numbers.
Problem 5.1. Determine Sa,b for a ≥ 3 in terms of ’familiar’ expressions.
•
Let, in analogy to Sr and Tr , Ur,s denote the multisum
Ur,s =
∞
X
k1 =1
&middot;&middot;&middot;
∞
X
kr
(−1)k1 +&middot;&middot;&middot;+ks
,
k
1 &middot; &middot; &middot; kr (k1 + &middot; &middot; &middot; + kr )
=1
where r ≥ 1 and 0 ≤ s ≤ r.
Problem 5.2. Determine Ur,s in the spirit of Theorem 2.2.
In other words evaluate the integrals
Z 1
ln(1 − x)s ln(1 + x)r−s
Ir,s =
dx
x
0
for r ≥ 1 and 0 ≤ s ≤ r.
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
http://jipam.vu.edu.au/
8
WALTHER JANOUS
R EFERENCES
 M. BENCZE, Problem 2984, Crux Math., 29 (2004), 431.
 S.R. FINCH, Mathematical Constants, Cambridge Univ. Press, Cambridge 2003.
 N. LORD, Problem 89.D, part (b), Math. Gaz., 89 (2005), 115.
 A.P. PRUDNIKOV et al., Integrals and Series (Elementary Functions) (in Russian), Nauka, Moscow
1981.
 E.W. WEISSTEIN, Ap&eacute;ry’s Constant, MathWorld–A Wolfram Web Resource. http://
mathworld.wolfram.com/AperysConstant.html
J. Inequal. Pure and Appl. Math., 7(1) Art. 35, 2006
http://jipam.vu.edu.au/
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