Half integer values of order-two harmonic numbers sums
Half integer values of harmonic numbers and reciprocal binomial coefficients sums are investigated in this paper. Closedform representations and integral expressions are developed for the infinite series.
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| description | Half integer values of harmonic numbers and reciprocal binomial coefficients sums are investigated in this paper. Closedform
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UDC 517.9
A. Sofo (Victoria Univ., Melbourne City, Australia)
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS
НАПIВЦIЛI ЗНАЧЕННЯ СУМ ГАРМОНIЧНИХ ЧИСЕЛ ДРУГОГО ПОРЯДКУ
Half integer values of harmonic numbers and reciprocal binomial coefficients sums are investigated in this paper. Closed-
form representations and integral expressions are developed for the infinite series.
Вивчаються напiвцiлi значення сум гармонiчних чисел та обернених бiномiальних коефiцiєнтiв. Для нескiнченних
рядiв отримано зображення в замкненiй формi та iнтегральнi вирази.
1. Introduction and preliminaries. It is known that the harmonic number Hn has the usual
definition
Hn =
n\sum
r=1
1
r
=
\infty \sum
j=1
n
j(j + n)
=
1\int
0
1 - xn
1 - x
dx (H0 := 0) (1.1)
for n \in \BbbN where \BbbN := \{ 1, 2, 3, . . .\} is the set of positive integers, and \BbbN 0 := \BbbN \cup \{ 0\} . An unusual,
but intriguing representation has recently been given by Ciaurri et al. [5], as
Hn = \pi
1\int
0
\biggl(
x - 1
2
\biggr) \left( \mathrm{c}\mathrm{o}\mathrm{s}
\biggl(
(4n+ 1)\pi x
2
\biggr)
- \mathrm{c}\mathrm{o}\mathrm{s}
\Bigl( \pi x
2
\Bigr)
\mathrm{s}\mathrm{i}\mathrm{n}
\Bigl( \pi x
2
\Bigr)
\right) dx.
Let \BbbR and \BbbC denote, respectively the sets of real and complex numbers. We define harmonic numbers
at half integer values as Hn - 1
2
, which may be expressed in terms of the digamma (or Psi) function
\psi (z), z \in \BbbR , and the Euler – Mascheroni constant, \gamma as Hn - 1
2
= \gamma + \psi
\biggl(
n+
1
2
\biggr)
. The digamma
function is defined by
\psi (z) :=
d
dz
\{ \mathrm{l}\mathrm{o}\mathrm{g} \Gamma (z)\} =
\Gamma \prime (z)
\Gamma (z)
or \mathrm{l}\mathrm{o}\mathrm{g} \Gamma (z) =
z\int
1
\psi (t) dt.
The Lerch transcendent \Phi (z, t, a) =
\sum \infty
m=0
zm
(m+ a)t
is defined for | z| < 1 and \Re (a) > 0 and
satisfies the recurrence
\Phi (z, t, a) = z \Phi (z, t, a+ 1) + a - t.
The Lerch transcendent generalizes the Hurwitz zeta function, \zeta (t, a) at z = 1,
\Phi (1, t, a) = \zeta (t, a) =
\infty \sum
m=0
1
(m+ a)t
and the polylogarithm, or de Jonquière’s function, when a = 1,
c\bigcirc A. SOFO, 2016
1418 ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS 1419
Lit (z) :=
\infty \sum
m=1
zm
mt
, t \in \BbbC , when | z| < 1; \Re (t) > 1 when | z| = 1.
Moreover
1\int
0
Lit (px)
x
dx =
\left\{ \zeta (1 + t) for p = 1,
(2 - r - 1) \zeta (1 + t) for p = - 1,
where \zeta (s) denotes the Riemann zeta function defined by
\zeta (s) =
\infty \sum
m=1
1
ms
, \Re (s) > 1.
A generalized binomial coefficient
\biggl(
\lambda
\mu
\biggr)
, \lambda , \mu \in \BbbC , is defined, in terms of the familiar (Euler’s)
gamma function, by \biggl(
\lambda
\mu
\biggr)
:=
\Gamma (\lambda + 1)
\Gamma (\mu + 1)\Gamma (\lambda - \mu + 1)
, \lambda , \mu \in \BbbC ,
which, in the special case when \mu = n, n \in \BbbN 0, yields\biggl(
\lambda
0
\biggr)
:= 1 and
\biggl(
\lambda
n
\biggr)
:=
\lambda (\lambda - 1) . . . (\lambda - n+ 1)
n!
=
( - 1)n ( - \lambda )n
n!
, n \in \BbbN ,
where (\lambda )\nu , \lambda , \nu \in \BbbC , is the Pochhammer symbol defined, also in terms of the gamma function, by
(\lambda )\nu :=
\Gamma (\lambda + \nu )
\Gamma (\lambda )
=
\left\{ 1, \nu = 0, \lambda \in \BbbC \setminus \{ 0\} ,
\lambda (\lambda + 1) . . . (\lambda + n - 1), \nu = n \in \BbbN , \lambda \in \BbbC ,
it being understood conventionally that (0)0 := 1 and assumed that the \Gamma -quotient exists. A gene-
ralized harmonic number H(m)
n of order m is defined, for positive integers n and m, as follows:
H(m)
n :=
n\sum
r=1
1
rm
, m, n \in \BbbN , and H
(m)
0 := 0, m \in \BbbN ,
and
\psi (n)(z) :=
dn
dzn
\{ \psi (z)\} =
dn+1
dzn+1
\{ \mathrm{l}\mathrm{o}\mathrm{g} \Gamma (z)\} , n \in \BbbN 0.
In the case of non integer values of the argument z =
r
q
, we may write the generalized harmonic
numbers, H(\alpha +1)
z , in terms of polygamma functions
H
(\alpha +1)
r
q
= \zeta (\alpha + 1) +
( - 1)\alpha
\alpha !
\psi (\alpha )
\biggl(
r
q
+ 1
\biggr)
,
r
q
\not = \{ - 1, - 2, - 3, . . .\} ,
where \zeta (z) is the zeta function. We also define
H r
q
= H
(1)
r
q
= \gamma + \psi
\biggl(
r
q
+ 1
\biggr)
.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
1420 A. SOFO
The evaluation of the polygamma function \psi (\alpha )
\Bigl( r
a
\Bigr)
at rational values of the argument can be
explicitly done via a formula as given by Kölbig [7], or Choi and Cvijovic [2] in terms of the
polylogarithmic or other special functions. Some specific values are given as
H
(2)
3
2
=
40
9
- 2\zeta (2), H
(2)
1
2
= 4 - 2\zeta (2), H
(3)
3
2
=
224
27
- 6\zeta (3).
Many others are listed in the books [13, 19, 20]. In this paper we will develop identities, closed form
representations of alternating half integer harmonic numbers and reciprocal binomial coefficients of
the form
Wk (p) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
np
\biggl(
n+ k
k
\biggr) (1.2)
for p = 0 and 1. While there are many results for sums of harmonic numbers with positive terms,
see, for example, [1, 3, 4, 6, 8 – 12, 14, 16 – 18, 21 – 23] and references therein. There are fewer
results for sums of the type (1.2).
The following lemma will be useful in the development of the main theorems.
Lemma 1.1. Let r be a positive integer. Then for p \in \BbbN
r\sum
j=1
( - 1)j
jp
=
1
2p
\biggl(
H
(p)
[ r2 ]
+H
(p)
[ r - 1
2 ]
\biggr)
- H
(p)
2[ r+1
2 ] - 1
. (1.3)
For p = 1
r\sum
j=1
( - 1)j
j
= H[ r2 ]
- Hr
where [x] is the integer part of x. We also have the known results, for 0 < t \leq 1
\mathrm{l}\mathrm{n}2 (1 + t) = 2
\infty \sum
n=1
( - t)n+1Hn
n+ 1
and when t = 1
\mathrm{l}\mathrm{n}2 2 = 2
\infty \sum
n=1
( - 1)n+1Hn
n+ 1
= \zeta (2) - 2Li2
\biggl(
1
2
\biggr)
, (1.4)
t \mathrm{l}\mathrm{n} (1 + t) =
\infty \sum
n=1
( - t)n+1
n
,
hence
\mathrm{l}\mathrm{n} 2 =
\infty \sum
n=1
( - 1)n+1
n
=
\infty \sum
n=1
1
n2n
=
1
2
\infty \sum
n=1
Hn
2n
, (1.5)
B(0) :=
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n
= 7\zeta (3) - 2\pi G - 2 \mathrm{l}\mathrm{n} 2\zeta (2). (1.6)
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS 1421
Proof. The proof of (1.3) is given in the paper [15].
Firstly, (1.4) and (1.5) are standard known results. Next from the definition, for p \in \BbbN 0
H(p+1)
n =
( - 1)p
p!
1\int
0
\mathrm{l}\mathrm{n}p x
1 - x
(1 - xn) dx (1.7)
we can write
B(0) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n
=
1\int
0
\mathrm{l}\mathrm{o}\mathrm{g} (x)
1 - x
\infty \sum
n=1
( - 1)n+1
\Bigl(
1 - xn -
1
2
\Bigr)
n
dx =
=
1\int
0
\mathrm{l}\mathrm{o}\mathrm{g} (x)
1 - x
\biggl(
\mathrm{l}\mathrm{n} (1 + x)\surd
x
- \mathrm{l}\mathrm{n} 2
\biggr)
dx =
= 7\zeta (3) - 2\pi G - 2 \mathrm{l}\mathrm{n} 2\zeta (2),
where G :=
\sum \infty
n=0
( - 1)n
(2n+ 1)2
\approxeq 0.91596 is Catalan’s constant.
Lemma 1.1 is proved.
Lemma 1.2. Let r be a positive integer, then we have the recurrence relation
B(r) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n+ r
= - B (r - 1) - 2\pi
(2r - 1)2
+
8G
2r - 1
- 2\zeta (2)
r
-
- 4( - 1)r
(2r - 1)2
\Bigl(
\mathrm{l}\mathrm{n} 2 +H[ r - 1
2 ] - Hr - 1
\Bigr)
(1.8)
with solution
B(r) = ( - 1)rB(0) - 2( - 1)r\zeta (2)
\Bigl(
H
[ r2 ]
- Hr
\Bigr)
-
- ( - 1)r \mathrm{l}\mathrm{n} 2
\biggl(
2\zeta (2) +H
(2)
r - 1
2
\biggr)
+
+8( - 1)rG
r\sum
j=1
( - 1)j
2j - 1
- 2( - 1)r\pi
r\sum
j=1
( - 1)j
(2j - 1)2
-
- 4( - 1)r
r\sum
j=1
\Bigl(
H[ j - 1
2 ] - Hj - 1
\Bigr)
(2j - 1)2
(1.9)
and B(0) =
\sum \infty
n=1
( - 1)n+1H
(2)
n - 1
2
n
= 7\zeta (3) - 2\pi G - 2 \mathrm{l}\mathrm{n} 2\zeta (2).
Proof. By a change of counter
B(r) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n+ r
=
\infty \sum
n=2
( - 1)nH
(2)
n - 3
2
n+ r - 1
=
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
1422 A. SOFO
=
\infty \sum
n=2
( - 1)n
n+ r - 1
\Biggl(
H
(2)
n - 1
2
-
\biggl(
2
2n - 1
\biggr) 2
\Biggr)
=
= -
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n+ r - 1
+
\infty \sum
n=1
4( - 1)n+1
(2n - 1)2 (n+ r - 1)
+
H
(2)
1
2
r
- 4
r
=
= - B (r - 1) +
1
r
\biggl(
H
(2)
1
2
- 4
\biggr)
+ 4
\infty \sum
n=1
( - 1)n+1
(2r - 1)2
\left(
2 (2r - 1)
(2n - 1)2
- 2
2n - 1
+
1
n+ r - 1
\right) =
= - B(r - 1) - 2\zeta (2)
r
- \pi
2(2r - 1)2
+
8G
2r - 1
+
+ 4
\infty \sum
n=1
( - 1)n+1
(2r - 1)2(n+ r - 1)
.
From Lemma 1.1 and using the known results
B(r) = - B (r - 1) - 2\zeta (2)
r
- \pi
2 (2r - 1)2
+
8G
2r - 1
+
+
4( - 1)r
(2r - 1)2
\Biggl( \infty \sum
n=1
( - 1)n+1
n
-
r - 1\sum
n=1
( - 1)n+1
n
\Biggr)
which, for r \geq 1, results in the recurrence relation
B(r) +B (r - 1) = - 2\pi
(2r - 1)2
+
8G
2r - 1
- 2\zeta (2)
r
-
- 4( - 1)r
(2r - 1)2
\Bigl(
\mathrm{l}\mathrm{n} 2 +H[ r - 1
2 ] - Hr - 1
\Bigr)
.
By the subsequent reduction of the B(r), B(r - 1), B(r - 2), . . . , B(1) terms in (1.8), we arrive at
the identity (1.9).
Lemma 1.2 is proved.
Example 1.1.
B (5) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n+ 5
= - 184939
297675
- 182738\pi
99225
- 47
30
\zeta (2) - 7\zeta (3) +
+
469876 \mathrm{l}\mathrm{n} 2
99225
+ 2 \mathrm{l}\mathrm{n} 2\zeta (2) + 2\pi G+
2104G
315
.
It is of some interest to note that B(r) may be expanded in a slightly different way so that it
gives rise to another unexpected harmonic series identity. This is pursued in the next lemma.
Lemma 1.3. For r \in \BbbN , we have the identity
V (r) =
\infty \sum
n=1
H
(2)
2n - 1
2
(2n+ r) (2n+ r - 1)
=
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS 1423
= B(r) +
1
2r - 1
(3\zeta (2) - 4G) +
1
(2r - 1)2
\Bigl(
\pi - 6 \mathrm{l}\mathrm{n} 2 - 2H r - 1
2
\Bigr)
.
For r = 0
V (0) =
\infty \sum
n=1
H
(2)
2n - 1
2
2n(2n - 1)
= B(0) + 4G - 3\zeta (2) + \pi - 6 \mathrm{l}\mathrm{n} 2 - 2H - 1
2
=
=
1\int
0
\mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
\Biggl(
\mathrm{l}\mathrm{n} 2 -
\surd
x \mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h} - 1 (x) -
\mathrm{l}\mathrm{n}
\bigl(
1 - x2
\bigr)
2
\surd
x
\Biggr)
dx =
= 7\zeta (3) - 2\pi G - 2 \mathrm{l}\mathrm{n} 2\zeta (2) + 4G - 3\zeta (2) + \pi - 2 \mathrm{l}\mathrm{n} 2.
Proof. By expansion
B(r) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n+ r
=
\infty \sum
n=1
\left( H
(2)
2n - 3
2
2n+ r - 1
-
H
(2)
2n - 1
2
2n+ r
\right) =
=
\infty \sum
n=1
\left( H
(2)
2n - 1
2
(2n+ r - 1) (2n+ r)
- 4
(4n - 1)2 (2n+ r - 1)
\right) ,
and by rearrangement
\infty \sum
n=1
H
(2)
2n - 1
2
(2n+ r - 1) (2n+ r)
= B(r) +
\infty \sum
n=1
4
(4n - 1)2 (2n+ r - 1)
=
= B(r) +
1
2r - 1
(3\zeta (2) - 4G) +
1
(2r - 1)2
\Bigl(
\pi - 6 \mathrm{l}\mathrm{n} 2 - 2H r - 1
2
\Bigr)
,
where B(r) is given by (1.9).
From (1.1) we can write
\infty \sum
n=1
H
(2)
2n - 1
2
2n (2n - 1)
= -
1\int
0
\mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
\infty \sum
n=1
\Bigl(
1 - x2n -
1
2
\Bigr)
2n(2n - 1)
dx =
=
1\int
0
\mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
\Biggl(
\surd
x \mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h} - 1 (x) +
\mathrm{l}\mathrm{n}
\bigl(
1 - x2
\bigr)
2
\surd
x
- \mathrm{l}\mathrm{n} 2
\Biggr)
dx =
= 7\zeta (3) - 2\pi G - 2 \mathrm{l}\mathrm{n} 2\zeta (2) + 4G - 3\zeta (2) + \pi - 2 \mathrm{l}\mathrm{n} 2 =
= B(0) + 4G - 3\zeta (2) + \pi - 6 \mathrm{l}\mathrm{n} 2 - 2H - 1
2
.
Lemma 1.3 is proved.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
1424 A. SOFO
Remark 1.1. Since we have the relation
H
(2)
n - 1
2
= 4H
(2)
2n - H(2)
n - 2\zeta (2)
it is possible to obtain identities for p = 0, 1 and k \geq 1
Xk (p) =
\infty \sum
n=1
( - 1)n+1H
(2)
2n
np
\biggl(
n+ k
k
\biggr) .
The next three theorems relate the main results of this investigation, namely the closed form and
integral representation of (1.2).
2. Closed form and integral identities. We now prove the following theorems.
Theorem 2.1. Let k be real positive integer. Then from (1.2) with p = 0 we have
Wk(0) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2\biggl(
n+ k
k
\biggr) =
k\sum
r=1
( - 1)1+rr
\biggl(
k
r
\biggr)
B(r), (2.1)
where B(r) is given by (1.9).
Proof. Consider the expansion
Wk(0) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2\biggl(
n+ k
k
\biggr) =
\infty \sum
n=1
( - 1)n+1k!H
(2)
n - 1
2
(n+ 1)k
=
=
\infty \sum
n=1
( - 1)n+1k!H
(2)
n - 1
2
k\sum
r=1
Mr
n+ r
,
where
Mr = \mathrm{l}\mathrm{i}\mathrm{m}
n\rightarrow - r
\left\{ n+ r\prod k
r=1
n+ r
\right\} =
( - 1)r+1r
k!
\biggl(
k
r
\biggr)
.
Hence,
Wk(0) =
k\sum
r=1
( - 1)r+1r
\biggl(
k
r
\biggr) \infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n+ r
=
=
k\sum
r=1
( - 1)r+1r
\biggl(
k
r
\biggr)
B(r).
Theorem 2.1 is proved.
The other case of Wk(1) can be evaluated in a similar fashion. We list the results in the next
corollary.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS 1425
Corollary 2.1. Under the assumptions of Theorem 2.1, we have
Wk(1) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
n
\biggl(
n+ k
k
\biggr) = B(0) -
k\sum
r=1
( - 1)r+1
\biggl(
k
r
\biggr)
B(r) =
= 7\zeta (3) - 2\pi G - 2 \mathrm{l}\mathrm{n} 2\zeta (2) -
k\sum
r=1
( - 1)r+1
\biggl(
k
r
\biggr)
B(r). (2.2)
Proof. The proof follows directly from Theorem 2.1 and using the same technique.
It is possible to represent the alternating harmonic number sums (2.1), (2.2) and (1.8) in terms of
an integral, which is developed in the next theorem.
Theorem 2.2. Let k be a positive integer. Then we have
k
1 + k
1\int
0
\surd
x \mathrm{l}\mathrm{o}\mathrm{g} x
(1 - x) (1 + x)
2F1
\Biggl[
1, 1
2 + k
\bigm| \bigm| \bigm| \bigm| \bigm| - x
\Biggr]
dx = (2.3)
= 2G - \zeta (2) - k\zeta (2)
2 (1 + k)
2F1
\biggl[
1, 1
2 + k
\bigm| \bigm| \bigm| \bigm| - 1
\biggr]
- Wk(0), (2.4)
where Wk(0) is given by (2.1) and 2F1
\biggl[
\cdot , \cdot
\cdot
\bigm| \bigm| \bigm| \bigm| z\biggr] is the classical Gauss hypergeometric function.
Proof. From (1.7)
H(2)
n = -
1\int
0
\mathrm{l}\mathrm{n}x
1 - x
(1 - xn) dx
we can therefore write
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2\biggl(
n+ k
k
\biggr) = -
1\int
0
\mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
\infty \sum
n=1
( - 1)n+1
\Bigl(
1 - xn -
1
2
\Bigr)
\biggl(
n+ k
k
\biggr) dx =
=
k\sum
r=1
( - 1)1+rr
\biggl(
k
r
\biggr)
B(r) =Wk(0),
hence,
1\int
0
1
1 - x
\left(
k
2 (1 + k)
2F1
\Biggl[
1, 1
2 + k
\bigm| \bigm| \bigm| \bigm| \bigm| - 1
\Biggr]
- (1 -
\surd
x)
2
2 (1 + x)
- k
\surd
x
(1 + k) (1 + x)
2F1
\Biggl[
1, 1
2 + k
\bigm| \bigm| \bigm| \bigm| \bigm| - x
\Biggr]
\right) dx =
k\sum
r=1
( - 1)1+rr
\Biggl(
k
r
\Biggr)
B(r).
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
1426 A. SOFO
Therefore, (2.3) and (2.4) follows.
Theorem 2.2 is proved.
A similar integral representation can be evaluated for Wk(1) and B(r), the results is recorded in
the next theorem.
Theorem 2.3. Let the conditions of Theorem 2.2 hold. Then we have
1
1 + k
1\int
0
\surd
x \mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
2F1
\Biggl[
1, 1
2 + k
\bigm| \bigm| \bigm| \bigm| \bigm| - x
\Biggr]
dx =
=Wk(1) -
\zeta (2)
1 + k
2F1
\Biggl[
1, 1
2 + k
\bigm| \bigm| \bigm| \bigm| \bigm| - 1
\Biggr]
. (2.5)
Also for B(r)
1\int
0
\surd
x \mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
\Phi ( - x, 1, 1 + r) dx = B(r) - \zeta (2)
2
\Bigl(
H r
2
- H r - 1
2
\Bigr)
, (2.6)
where \Phi ( - x, 1, 1 + r) is the Lerch transcendent.
Proof. The proof follows the same pattern as that employed in Theorem 2.2.
Example 2.1. From (2.6), for r = 2,
1\int
0
\surd
x \mathrm{l}\mathrm{n}x
1 - x
\Phi ( - x, 1, 3) dx = B(2) - \zeta (2)
2
(2 \mathrm{l}\mathrm{n} 2 - 1) ,
which reduces to
\mathrm{l}\mathrm{i}\mathrm{m}
\epsilon - \rightarrow 0
1\int
\epsilon
\mathrm{l}\mathrm{n}x
x
5
2 (1 - x)
\Bigl(
\mathrm{l}\mathrm{n} (1 + x) - x
\Bigl(
1 - x
2
\Bigr) \Bigr)
dx =
= 7\zeta (3) - 2
\biggl(
\pi +
8
3
\biggr)
G+ 3
\biggl(
1
2
- \mathrm{l}\mathrm{n} 2
\biggr)
\zeta (2) - 40
9
\mathrm{l}\mathrm{n} 2 +
16
9
\pi +
4
9
.
From (2.5), for k = 2,
1
3
1\int
0
\surd
x \mathrm{l}\mathrm{o}\mathrm{g} x
1 - x
2F1
\Biggl[
1, 1
4
\bigm| \bigm| \bigm| \bigm| \bigm| - x
\Biggr]
dx =W2(1) -
\zeta (2)
2
2F1
\Biggl[
1, 1
4
\bigm| \bigm| \bigm| \bigm| \bigm| - 1
\Biggr]
,
which reduces to
\mathrm{l}\mathrm{i}\mathrm{m}
\epsilon - \rightarrow 0
1\int
\epsilon
\mathrm{l}\mathrm{n}x
x
5
2 (1 - x)
\Bigl(
(1 + x)2 \mathrm{l}\mathrm{n} (1 + x) - x
\Bigr)
dx =
= 28\zeta (3) + 3 (1 - 4 \mathrm{l}\mathrm{n} 2) \zeta (2) - 8
\biggl(
\pi +
8
3
\biggr)
G - 112
9
\mathrm{l}\mathrm{n} 2 +
52
9
\pi +
4
9
.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS 1427
From (2.4), for k = 2,
2
3
1\int
0
\surd
x \mathrm{l}\mathrm{o}\mathrm{g} x
1 - x (1 + x)
2F1
\Biggl[
1, 1
4
\bigm| \bigm| \bigm| \bigm| \bigm| - x
\Biggr]
dx =
= 2G - \zeta (2) - \zeta (2)
3
2F1
\Biggl[
1, 1
4
\bigm| \bigm| \bigm| \bigm| \bigm| - 1
\Biggr]
- W2(0)
which reduces to
\mathrm{l}\mathrm{i}\mathrm{m}
\epsilon - \rightarrow 0
1\int
\epsilon
\mathrm{l}\mathrm{n}x
x
5
2 (1 - x)
\biggl(
(1 + x) \mathrm{l}\mathrm{n}(1 + x) - x
1 + x
\biggr)
dx =
= 14\zeta (3) + 3
\biggl(
1
2
- 2 \mathrm{l}\mathrm{n} 2
\biggr)
\zeta (2) -
\biggl(
46
3
+ 4\pi
\biggr)
G - 76
9
\mathrm{l}\mathrm{n} 2 +
34
9
\pi +
4
9
.
The Wolfram online integrator yields no solution to these integrals.
Remark 2.1. It appears that, for r \in \BbbN 0, p \geq 3
Y (p, r) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
(n+ r)p
may not have a closed form solution, in terms of some common special functions. Remarkably,
however, the sum of two consecutive terms of Y (p, r) does have a closed form solution, this result
is pursued in the next lemma.
Lemma 2.1. For r \in \BbbN , p \in \BbbN
Y (p, r) + Y (p, r + 1) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
\biggl(
1
(n+ r)p
+
1
(n+ r + 1)p
\biggr)
=
= 4
\biggl(
2
2r + 1
\biggr) p
G - p\pi
2
\biggl(
2
2r + 1
\biggr) p+1
- 2\zeta (2)
(r + 1)p
+
+ ( - 1)rp
\biggl(
2
2r + 1
\biggr) p+1 \Bigl(
\mathrm{l}\mathrm{n} 2 +H[ r2 ]
- Hr
\Bigr)
+
+( - 1)r
p\sum
j=2
(p+ 1 - j)
\biggl(
2
2r + 1
\biggr) p+2 - j \bigl(
1 - 21 - j
\bigr)
\zeta (j) +
+ ( - 1)r
p\sum
j=2
(p+ 1 - j)
\biggl(
2
2r + 1
\biggr) p+2 - j \biggl( 1
2j
\biggl(
H
(j)
[ r2 ]
+H
(j)
[ r - 1
2 ]
\biggr)
- H
(j)
2[ r+1
2 ] - 1
\biggr)
, (2.7)
and for r = 0
Y (p, 0) + Y (p, 1) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
\biggl(
1
np
+
1
(n+ 1)p
\biggr)
=
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
1428 A. SOFO
= 2p+2G - 2pp\pi + 2p+1p \mathrm{l}\mathrm{n} 2 +
\bigl(
2p - 1 (p - 1) - 2
\bigr)
\zeta (2)+
+
p\sum
j=3
(p+ 1 - j)2 - j
\bigl(
1 - 21 - j
\bigr)
\zeta (j).
Proof. Consider
Y (p, r) + Y (p, r + 1) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
\biggl(
1
(n+ r)p
+
1
(n+ r + 1)p
\biggr)
and by a change of counter in the second sum we can write
Y (p, r) + Y (p, r + 1) =
\infty \sum
n=1
( - 1)n+1H
(2)
n - 1
2
(n+ r)p
- 2\zeta (2)
(r + 1)p
-
-
\infty \sum
n=1
( - 1)n+1
(n+ r)p
\biggl(
H
(2)
n - 1
2
- 4
(2n - 1)2
\biggr)
=
=
\infty \sum
n=1
4( - 1)n+1
(2n - 1)2(n+ r)p
- 2\zeta (2)
(r + 1)p
=
=
\infty \sum
n=1
( - 1)n+1
\left(
\biggl(
2
2r + 1
\biggr) p 4
(2n - 1)2
-
\biggl(
2
2r + 1
\biggr) p+1 2p
2n - 1
+
+
\biggl(
2
2r + 1
\biggr) p+1 p
n+ r
+
p\sum
j=2
\biggl(
2
2r + 1
\biggr) p+2 - j p+ 1 - j
(n+ r)j
\right) - 2\zeta (2)
(r + 1)p
=
= 4
\biggl(
2
2r + 1
\biggr) p
G - p\pi
2
\biggl(
2
2r + 1
\biggr) p+1
- 2\zeta (2)
(r + 1)p
+
+ ( - 1)rp
\biggl(
2
2r + 1
\biggr) p+1 \Bigl(
\mathrm{l}\mathrm{n} 2 +H[ r2 ]
- Hr
\Bigr)
+
+ ( - 1)r
p\sum
j=2
(p+ 1 - j)
\biggl(
2
2r + 1
\biggr) p+2 - j \bigl(
1 - 21 - j
\bigr)
\zeta (j) +
+( - 1)r
p\sum
j=2
(p+ 1 - j)
\biggl(
2
2r + 1
\biggr) p+2 - j \biggl( 1
2j
\biggl(
H
(j)
[ r2 ]
+H
(j)
[ r - 1
2 ]
\biggr)
- H
(j)
2[ r+1
2 ] - 1
\biggr)
,
and a rearrangement leads to (2.7). The case of r = 0 follows.
Lemma 2.1 is proved.
Example 2.2.
Y (5, 2) + Y (5, 3) =
128
3125
G - 32
3125
\pi +
64
3125
\mathrm{l}\mathrm{n} 2 - 9579
25000
+
+
9302
759375
\zeta (2) +
36
625
\zeta (3) +
14
125
\zeta (4) +
3
20
\zeta (5),
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
HALF INTEGER VALUES OF ORDER-TWO HARMONIC NUMBERS SUMS 1429
Y (5, 0) + Y (5, 1) = 128G - 160\pi + 320 \mathrm{l}\mathrm{n} 2 +
+ 62\zeta (2) + 36\zeta (3) + 14\zeta (4) +
15
4
\zeta (5).
References
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Received 23.06.15,
after revision — 26.03.16
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 10
|
| id | umjimathkievua-article-1929 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:15:24Z |
| publishDate | 2016 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/1a/a553490a31fdaabfafc134434d72fc1a.pdf |
| spelling | umjimathkievua-article-19292019-12-05T09:31:57Z Half integer values of order-two harmonic numbers sums Напiвцiлi значення сум гармонiчних чисел другого порядку Sofo, A. Софо, А. Half integer values of harmonic numbers and reciprocal binomial coefficients sums are investigated in this paper. Closedform representations and integral expressions are developed for the infinite series. Вивчаються напiвцiлi значення сум гармонiчних чисел та обернених бiномiальних коефiцiєнтiв. Для нескiнченних рядiв отримано зображення в замкненiй формi та iнтегральнi вирази. Institute of Mathematics, NAS of Ukraine 2016-10-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1929 Ukrains’kyi Matematychnyi Zhurnal; Vol. 68 No. 10 (2016); 1418-1429 Український математичний журнал; Том 68 № 10 (2016); 1418-1429 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1929/911 Copyright (c) 2016 Sofo A. |
| spellingShingle | Sofo, A. Софо, А. Half integer values of order-two harmonic numbers sums |
| title | Half integer values of order-two harmonic numbers sums |
| title_alt | Напiвцiлi значення сум гармонiчних чисел другого порядку |
| title_full | Half integer values of order-two harmonic numbers sums |
| title_fullStr | Half integer values of order-two harmonic numbers sums |
| title_full_unstemmed | Half integer values of order-two harmonic numbers sums |
| title_short | Half integer values of order-two harmonic numbers sums |
| title_sort | half integer values of order-two harmonic numbers sums |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1929 |
| work_keys_str_mv | AT sofoa halfintegervaluesofordertwoharmonicnumberssums AT sofoa halfintegervaluesofordertwoharmonicnumberssums AT sofoa napivciliznačennâsumgarmoničnihčiseldrugogoporâdku AT sofoa napivciliznačennâsumgarmoničnihčiseldrugogoporâdku |