Journal of Inequalities in Pure and
Applied Mathematics
AROUND APÉRY’S CONSTANT
WALTHER JANOUS
Ursulinengymnasium
Fürstenweg 86
A-6020 Innsbruck
Austria.
volume 7, issue 1, article 35,
2006.
Received 04 January, 2006;
accepted 18 January, 2006.
Communicated by: A. Lupaş
EMail: walther.janous@tirol.com
Abstract
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2000
Victoria University
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Abstract
In this note we deal with some aspects of Apéry’s constant ζ(3).
2000 Mathematics Subject Classification: 49J40, 90C33, 47H10.
Key words: Apéry’s constant, Harmonic numbers, Infinite sums, Integrals.
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.
Dedicated to Professor Gerd Baron on the occasion of his 65th birthday.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Two Multisums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3
A New Formula for Apéry’s Constant . . . . . . . . . . . . . . . . . . . 8
4
Further Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
5
Two Questions for Further Research . . . . . . . . . . . . . . . . . . . . . 15
References
Around Apéry’s Constant
Walther Janous
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1.
Introduction
Since Apéry’s miraculous proof (1979) that the value ζ(3) of Rieman’s ζ-function
is irrational, Apéry’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 [2].)
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.
Around Apéry’s Constant
Walther Janous
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2.
Two Multisums
Recently in [1] 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.
Around Apéry’s Constant
Walther Janous
Theorem 2.1. For r ≥ 1 the multisum
Sr =
∞
X
∞
X
···
k1 =1
kr
1
k1 · · · kr (k1 + · · · + kr )
=1
attains the value r!ζ(r + 1).
Proof. Firstly we rewrite the multisum as an integral as follows
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Sr =
∞
X
···
k1 =1
∞
X
kr
1
k1 · · · k r
=1
Z
1
xk1 +···+kr −1 dx
0
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that is (upon interchanging of summation and integration),
Z
Sr =
0
1
∞
∞
X
x kr
1 X x k1
···
dx.
x k =1 k1
k
r
k =1
1
Close
r
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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
readily yield the claim.
Subsequently we will deal with a ‘relative’ of Sr , namely the multisum
Tr =
∞
X
k1 =1
···
∞
X
kr
(−1)k1 +···+kr
.
k
1 · · · kr (k1 + · · · + kr )
=1
Around Apéry’s Constant
Walther Janous
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As it will turn out, matters are here more involved. Indeed, we will prove now
J. Ineq. Pure and Appl. Math. 7(1) Art. 35, 2006
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Theorem 2.2. For r ≥ 1,
(2.2) Tr = (−1)r r!ζ(r + 1) −
r
(ln 2)r+1
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.
!
Around Apéry’s Constant
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
Z ln(1/2)
(−t)r
r
Tr = (−1)
(−e−t )dt,
−t − 1
e
0
Walther Janous
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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
− ln 2
tr ekt dt.
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Integration by parts leads to the identity (we suppress integration constants)
!
Z
r
r
X
t
r(r
−
1)...(r
−
m
+
1)
tr ekt dt = ekt
+
(−1)m
tr−m ,
k m=1
k m+1
where k > 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
Around Apéry’s Constant
Walther Janous
!!
.
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Since
∞
X
1
= ln 2,
k
k2
k=1
we finally get the claimed identity (2.2).
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3.
A New Formula for Apéry’s Constant
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
!
∞
3
X
1 ln 2
1
(ln 2)
.
T2 = 2 ζ(3) −
−
+ 3
k
2
3
2
k
k
k=1
Around Apéry’s Constant
Furthermore [2, p. 43], reports
Walther Janous
1
T2 = ζ(3).
4
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Therefore the following holds.
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Theorem 3.1.
(3.1)
8
ζ(3) =
7
∞
(ln 2)3 X 1
+
3
2k
k=1
ln 2
1
+ 3
2
k
k
!
.
This formula should be compared with the following one (see [5])
∞
X
(−1)k+1
2
3
.
ζ(3) = (ln 2) + 4
3
k 3 2k 2k
k
k=1
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4.
•
Further Observations
From S2 + T2 = 94 ζ(3) we infer
X
1
9
2
= ζ(3)
ij(i + j)
4
i,j≥1
i+j even
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
Around Apéry’s Constant
Walther Janous
∞
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
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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
7
1
H2k = ζ(3).
2
(2k + 1)
16
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From these two formulae we get easily
Theorem 4.1.
(4.1)
∞
X
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
Walther Janous
Adding these two identities yields
∞
X
1
Hk = 2ζ(3),
k2
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 EulerMascheroni constant, resp.
Therefore there holds the curious identity
∞
X
(−1)k ψ(k)
k=1
k2
Around Apéry’s Constant
1
π2γ
= ζ(3) +
.
8
12
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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 [3] posed the problem of proving the identity
Z π/4
∞
X
ln(cos x) ln(sin x)
1
=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
that is
∞
∞
X 1
X 1
1
=
−
,
(2n + 1)3
k 3 k=1 (2k)3
k=1
∞
X
n=0
1
7
= ζ(3).
3
(2n + 1)
8
Therefore, the identity under consideration in fact means
Z π/4
ln(cos x) ln(sin x)
·
dx.
ζ(3) = 8
cos x
sin x
0
x)
Letting f (x) = ln(cos
and setting z = π/2 − x we obtain
cos x
Z π/4
Z π/2 π
π
f (x)f
− x dx =
− z f (z)dz
f
2
2
0
π/4
Around Apéry’s Constant
Walther Janous
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whence
π/2
Z
ζ(3) = 4
√
0
ln(cos x) ln(sin x)
·
dx.
cos x
sin x
Next, we substitute sin x = w.
√
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
1
ln( 1 − w) ln( w)
√
· √ √
dw,
· √
ζ(3) = 4
w
2 w 1−w
1−w
0
Walther Janous
that is
ζ(3) =
1
2
1
Z
0
ln(1 − w) ln w
·
dw.
1−w
w
Upon rewriting this as
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1
ζ(3) =
2
Z
0
1
ln(1 − w) ln w
·
dw
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
Around Apéry’s Constant
∞
∞
1 XX 1
ζ(3) =
2 i=1 j=1 ij
Z
0
1
wi−1 (1 − w)j−1 dw.
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Keeping in mind that
Z 1
wi−1 (1 − w)j−1 dw =
0
(i − 1)!(j − 1)!
,
(i + j − 1)!
we arrive at the formula
∞
∞
1
1 XX
ζ(3) =
i+j−1 .
2
2 i=1 j=1 ij
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
·
dz =
z
1−z
Z
0
1
ln(1 − z)
ln(1 − z)dz.
z
Around Apéry’s Constant
Walther Janous
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• Finally, Theorem 4.1 enables us to prove the following finite analogon of
the initial formula (2.1) of the present note, namely
(4.8)
∞ X
i
X
i=1 j=1
1
5
= ζ(3).
ij(i + j)
4
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Indeed, (4.2) and (4.3) imply
∞
X
1
1
1
3
+ ··· +
= ζ(3).
2
k
k+1
2k
4
k=1
However,
1
k2
1
1
+ ··· +
k+1
2k
k
1
1X
=
k j=1 k(k + j)
Around Apéry’s Constant
and
Walther Janous
1
k
k
X
j=1
1
1
=
k(k + j)
k
k
X
j=1
1
j
1
1
−
k k+j
=
1
Hk −
k2
j=1
1
kj(k + j)
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readily lead to
2ζ(3) −
k
∞ X
X
k=1 j=1
as claimed.
k
X
3
1
= ζ(3)
kj(k + j)
4
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5.
Two Questions for Further Research
• The results of Theorem 4.1 may be regarded as special cases of the more
general sums
∞
X
1
H
,
Sa,b =
2 ak−b
(ak
−
b)
k=1
where 0 ≤ b < a are entire numbers.
Problem 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
···
Around Apéry’s Constant
Walther Janous
∞
X
kr
(−1)k1 +···+ks
,
k1 · · · kr (k1 + · · · + kr )
=1
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where r ≥ 1 and 0 ≤ s ≤ r.
Problem 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
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for r ≥ 1 and 0 ≤ s ≤ r.
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References
[1] M. BENCZE, Problem 2984, Crux Math., 29 (2004), 431.
[2] S.R. FINCH, Mathematical Constants, Cambridge Univ. Press, Cambridge
2003.
[3] N. LORD, Problem 89.D, part (b), Math. Gaz., 89 (2005), 115.
[4] A.P. PRUDNIKOV et al., Integrals and Series (Elementary Functions) (in
Russian), Nauka, Moscow 1981.
[5] E.W. WEISSTEIN, Apéry’s Constant, MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/AperysConstant.
html
Around Apéry’s Constant
Walther Janous
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