1
Journal of Integer Sequences, Vol. 9 (2006),
Article 06.2.3
2
3
47
6
23 11
Evaluations of Some Variant Euler Sums
Hongwei Chen
Department of Mathematics
Christopher Newport University
Newport News, VA 23606
hchen@cnu.edu
Abstract
In this note we present some elementary methods for the summation of certain
Euler sums with terms involving 1 + 1/3 + 1/5 + · · · + 1/(2k − 1).
1
Introduction
In the last decade, based on extensive experimentation with computer algebraic systems,
a large class of Euler sums have been explicitly evaluated in terms of the Riemann zeta
function ζ(k). For example, let
Hk = 1 +
1
1
+ ··· + .
2
k
Then
∞
X
1
Hk = 2 ζ(3),
k2
k=1
∞
X
k=1
1
π2
H
=
ζ(3)
−
ln 2,
k
2k k 2
12
∞
X
1 2
17
Hk =
ζ(4),
2
k
4
k=1
and
∞
X
(−1)k−1
5
H
=
ζ(3).
k
2
k
8
k=1
1
More details can be found in [1, 2, 3, 4] In particular, Borwein and Bradley [3] collected 32
beautiful proofs of the first sum above.
Motivated by the above results, in this note, replacing Hk by
hk = H2k −
1
1
1
Hk = 1 + + · · · +
,
2
3
2k − 1
(1)
we study the following variant Euler sums
∞
X
ak h k
k=1
where the ak are relatively simple function of k.
2
The Main Results
We begin to derive some series involving hk . Since
− ln(1 − x) =
replacing x by −x gives
Z
x
0
∞
X xk
dt
=
,
1 − t k=1 k
∞
X
(−1)k−1 xk
ln(1 + x) =
.
k
k=1
Averaging these two series gives us
µ
¶ X
∞
1+x
1
1
ln
=
x2k−1 .
2
1−x
2k − 1
k=1
(2)
In term of the Cauchy product and partial fractions, we have
µ
¶
¶
∞ µ
X
1 2 1+x
1
1
1
ln
=
+
+ ··· +
x2k
4
1−x
(2k
−
1)
·
1
(2k
−
3)
·
3
1
·
(2k
−
1)
k=1
·µ
¶ µ
¶
µ
¶¸
∞
X
1
1
1
1
1
1
1
+
+
+
+ ··· +
+
x2k
=
2k
2k
−
1
1
2k
−
3
3
1
2k
−
1
k=1
¶ 2k
∞ µ
X
1
1
x
=
.
1 + + ··· +
3
2k − 1
k
k=1
Noting that hk is given by (1), we have
µ
¶
∞
X
hk 2k 1 2 1 + x
x = ln
.
k
4
1−x
k=1
2
(3)
This enables us to evaluate a wide variety of interesting series via specialization, differentiation and integration.
First, setting x = 1/2, we find
For x =
√
∞
X
hk
1 2
=
ln 3.
2k k
2
4
k=1
(4)
2/2,
∞
X
√
1 2
hk
=
ln
(3
+
2
2).
kk
2
4
k=1
(5)
√
Putting x = ( 5 − 1)/2 = φ, the golden ratio, we get
∞
X
√
hk 2k 1 2
φ = ln (2 + 5).
k
4
k=1
(6)
√
√
Furthermore, for any α ≥ 2, putting x = ( 5+1)/2α and x = ( 5−1)/2α in (3) respectively,
we get
Ã√
!2k
Ã
√ !
∞
X
1 2 (2α + 1) + 5
hk
5+1
√
= ln
(7)
2k k
α
2
4
(2α
−
1)
−
5
k=1
and
∞
X
hk
α2k k
k=1
Ã√
5−1
2
!2k
1
= ln2
4
Ã
√ !
(2α − 1) + 5
√
.
(2α + 1) − 5
(8)
Recalling the Fibonacci numbers which are defined by
F1 = 1, F2 = 1, Fk = Fk−1 + Fk−2 for k ≥ 2
and Binet’s formula
Ã
1
Fk = √ 
5
√
5+1
2
!k
combining (7) and (8), we find
−
Ã

√ !k
1− 5 
,
2
!
√
√
µ 2
¶ Ã 2
∞
X
hk
5
α +α 5+1
α +α−1
√
F2k =
ln
.
ln
2k
2
2−α 5+1
α
k
20
α
−
α
−
1
α
k=1
(9)
In particular, for α = 2
√
∞
X
√
hk
5
F
=
ln
5
ln(9
+
4
5).
2k
2k k
2
4
k=1
3
(10)
Another step along this path is to change variables. Setting x = cos θ in (3) leads to
∞
X
hk
cos2k θ = ln2 (cot(x/2)) .
k
k=1
(11)
Integrating both sides from 0 to π, and using
¶
µ
Z π
π
2k
2k
cos θ dθ = 2k
k
2
0
and
Z
we find
π
ln2 (cot(x/2)) dθ =
0
π3
,
4
µ
¶
∞
X
π2
hk
2k
= .
k
22k k
4
k=1
(12)
This adds another interesting series to Lehmer’s list [6].
Next, for 0 < x < 1, differentiating (3), then multiplying both sides by x, we obtain
µ
¶
∞
X
1+x
x
2k
ln
.
(13)
hk x =
2)
2(1
−
x
1
−
x
k=1
Setting x = 1/2, we get
For x =
∞
X
1
hk
= ln 3.
2k
2
3
k=1
√
2/2,
(14)
√
∞
X
√
2
hk
=
ln(3
+
2
2).
2k
2
k=1
(15)
√
∞
X
√
√
hk
5
F
=
(10
ln(5
+
2
5)
+
3
5 ln 5 − 5 ln 5).
2k 2k
2
50
k=1
(16)
Similar to (10), we have
Finally, for 0 < x ≤ 1, dividing both sides of (3) by x and integrating from 0 to x, we
obtain
µ
¶
Z
∞
X
hk 2k 1 x 1 2 1 + t
x =
ln
dt.
(17)
2
k
2
t
1
−
t
0
k=1
Using the substitution u = (1 − x)/(1 + x) and integration by parts, we get
Z 1
∞
X
hk 2k
ln2 u
x
=
du
2
k2
(1−x)/(1+x) 1 − u
k=1
µ
¶ Z 1
µ
¶
1−x
1−u
1
ln u
2
=
ln x ln
+
ln
du.
2
1+x
1+u
(1−x)/(1+x) u
4
In view of (2), we have
Z
1
(1−x)/(1+x)
Since
Z
we find
µ
ln u
ln
u
1−u
1+u
u2k ln u du =
¶
du = −2
∞
X
k=0
1
2k + 1
Z
1
u2k ln u du.
(1−x)/(1+x)
1
1
u2k+1 ln u −
u2k+1 + C,
2k + 1
(2k + 1)2
¶
µ
¶ ∞
¶2k+1
µ
µ
∞
X
1−x X
1
hk 2k 1
1−x
1−x
2
+ 2 ln
x = ln x ln
k2
2
1+x
1 + x k=0 (2k + 1)2 1 + x
k=1
+2
∞
X
k=0
¶2k+1
µ
∞
X
1
1
1−x
−2
.
(2k + 1)3
(2k + 1)3 1 + x
k=0
(18)
In terms of the polylogarithm function [5]
∞
X
xn
,
Lin (x) =
kn
k=1
and noting that
∞
X
k=0
and
∞
X
k=0
we finally obtain
1
xn
= (Lin (x) − Lin (−x))
n
(2k + 1)
2
∞
∞
X 1
X 1
1
7
=
−
=
ζ(3),
3
3
(2k + 1)3
k
(2k)
8
k=0
k=0
µ
¶
∞
X
1−x
1
hk 2k 7
2
x = ζ(3) + ln x ln
2
k
4
2
1+x
k=1
µ
µ
µ
µ
µ
¶µ
¶
¶¶ µ
¶
¶¶
1−x
x−1
1−x
x−1
1−x
Li2
− Li2
− Li3
− Li3
.
+ ln
1+x
1+x
1+x
1+x
1+x
Setting x = 1, we get
∞
X
7
hk
=
ζ(3).
2
k
4
k=1
For x = 1/3,
∞
X
k=1
7
1
hk
=
ζ(3) − ln 3 ln3 2
2
k
8
2
32k
+
1 3
ln 2 + ln 2 Li2 (−1/2) + Li3 (−1/2),
3
5
(19)
where we have used
π2 1 2
− ln 2;
12 2
7
1
π2
Li3 (1/2) =
ζ(3) + ln3 2 −
ln 2.
8
6
12
Li2 (1/2) =
Moreover, noting that
hk =
k Z
X
i=1
1
x
0
2(i−1)
dt =
Z
1
(
0
k
X
x
2(i−1)
) dt =
i=1
Z
0
1
1 − x2k
dx
1 − x2
and rewriting (8) as
µ
¶
Z
∞
X
1 1 1 2 1+t
hk
2k
dt,
(1 − x ) =
ln
2
k
2
t
1
−
t
x
k=1
we have
Z
∞
∞
X
X
hk 1 1 − x2k
h2k
=
dx
k2
k 2 0 1 − x2
k=1
k=1
¶ ¶
µ
Z 1
Z µ
1 1
1 2 1+t
1
=
ln
dt dx.
2 0
1 − x2 x t
1−t
Exchanging the order of the integration, we get
¶
µ
¶Z t
Z µ
∞
X
1 1 1 2 1+t
1
h2k
=
dx dt.
ln
2
k2
2 0
t
1−t
0 1−x
k=1
µ
¶
Z
1 1 1 3 1+t
=
ln
dt.
4 0 t
1−t
Using the substitution x = (1 − t)/(1 + t) and the well-known fact that
Z 1
6
xk ln3 x dx = −
,
(k + 1)3
0
we find
Z
∞
X
1 1 ln3 x
h2k
=−
dx
2
2
k
2
1
−
x
0
k=1
∞ Z
∞
X
1 X 1 2k 3
45
1
=−
=
x ln x dx = 3
ζ(4)
4
2 k=0 0
(2k + 1)
16
k=0
Another path out of (3) is to bring in complex variables. Since
µ
¶
1
1
1+z
−1
−1
tan (iz) = tanh z = ln
i
2
1−z
6
(20)
Replacing x by ix in (3), we obtain
∞
X
(−1)k−1 hk 2k
x = (tan−1 x)2 .
k
k=1
√ √
This series may be evaluated at values such as x = 2 − 3, 3/3, 1 explicitly:
√
∞
X
(−1)k−1 hk (2 − 3)2k
π2
3
=
=
ζ(2),
k
144
72
k=1
∞
X
π2
1
(−1)k−1 hk
=
= ζ(2),
k
3 k
36
6
k=1
∞
X
π2
3
(−1)k−1 hk
=
= ζ(2).
k
16
8
k=1
(21)
(22)
(23)
(24)
Similarly, applying differentiation and integration to (21), we deduce the corresponding
formulas
∞
X
x
(−1)k−1 hk x2k =
tan−1 x,
(25)
2
1
+
x
k=1
Z x
∞
X (−1)k−1 hk
(tan−1 t)2
2k
x
=
2
dt.
(26)
2
k
t
0
k=1
In particular, we find
√
∞
X
(−1)k−1 hk
3
=
π,
k
3
24
k=1
∞
X
(−1)k−1 hk
7
= G π − ζ(3),
2
k
4
k=1
(27)
(28)
where G is the Catalan’s constant which is defined by
G=
∞
X
k=0
(−1)k
.
(2k + 1)2
Finally, following the excellent suggestion of an anonymous referee, recalling that
hk = H2k −
1
Hk ,
2
(29)
we find from (19)
∞
∞
∞
X
X
1
1
1 X 1
11
H
=
h
+
Hk =
ζ(3).
2k
k
2
2
2
k
k
2
k
4
k=1
k=1
k=1
7
(30)
Furthermore, in terms of the multiple series [7]
∞
∞ X
X
i=1 j=1
∞
∞ X
X
1
(−1)i+j
1
= 2ζ(3),
= ζ(3),
ij(i + j)
ij(i + j)
4
i=1 j=1
the difference gives
X
i,j>1,i+j=odd
7
1
= ζ(3).
ij(i + j)
8
Setting i + j = 2k + 1 and using partail fractions, we have
X
i,j>1,i+j=odd
2k
∞ X
X
1
1
=
ij(i + j)
j(2k + 1 − j)(2k + 1)
k=1 j=1
=
=
∞
X
k=1
∞
X
k=1
Thus,
∞
X
k=1
Subsequently, we have
∞
X
k=1
∞
X
k=1
¶
2k µ
X
1
1
1
+
(2k + 1)2 j=1 j 2k + 1 − j
1
2H2k .
(2k + 1)2
7
1
H2k =
ζ(3).
2
(2k + 1)
16
(31)
∞
X
1
1
H
=
H2k +
2k
2
(2k − 1)2
(2k
+
1)
k=1
∞
X
21
1
1
1
+
=
ζ(3) + (π 2 − 8 ln 2).
3
2
(2k − 1)
2k(2k − 1)
16
8
k=1
(32)
From this and the known result [1]
∞
X
k=1
we finally get
1
1
Hk = (π 2 − π 2 ln 2 − 8 ln 2 + 7ζ(3)),
2
(2k − 1)
4
∞
X
k=1
3
7
3
1
hk =
ζ(3) + ζ(2) ln 2.
2
(2k − 1)
16
4
(33)
Acknowledgments
The author would like to thank the referee and the editor for their thoughtful comments and
suggestions for improving the original version of the manuscript.
8
References
[1] D. Bailey, J. Borwein and R. Girgensohn, Experimental evaluation of Euler sums, Experimental Math., 3 (1994), 17–30.
[2] J. Borwein, D. Bailey and R. Girgensohn, Experimentation in Mathematics, A. K. Peters,
2004.
[3] J. Borwein and D. V. Bradley, Thirty-two Goldbach variations, preprint. Available at
http://users.cs.dal.ca/~jborwein/32goldbach.pdf.
[4] J. Borwein and I. J. Zucker and J. Boersma, The evaluation of character Euler double
sums, preprint. Available at http://users.cs.dal.ca/~jborwein/bzb7.pdf.
[5] L. Lewin, Polylogarithms and Associated Functions, Elsevier North Holland, New YorkAmsterdam, 1981.
[6] D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math.
Monthly, 92 (1985), 449–457.
[7] D. Zagier, Values of zeta functions and their applications, First European Congress of
Mathematics, Vol. 2, Paris, Birkhauser, 1994, pp. 497–512.
2000 Mathematics Subject Classification: Primary 40C15; Secondary 40A25, 11M41, 11M06.
Keywords: Euler sums, Riemann zeta function, closed forms.
(Concerned with sequences A000045, A000984, A001008, and A005408.)
Received February 9 2006; revised version received April 21 2006. Published in Journal of
Integer Sequences, May 18 2006.
Return to Journal of Integer Sequences home page.
9