Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 5 Issue 1(2013), Pages 71-79.
FINITE NUMBER SUMS IN HIGHER ORDER POWERS OF
HARMONIC NUMBERS
(COMMUNICATED BY TOUFIK MANSOUR)
ANTHONY SOFO
Abstract. We develop a set of identities for finite sums of products of harmonic numbers in higher order and reciprocal binomial coefficients. The results
are analogous to some identities of Euler type.
1. Introduction and preliminaries
Let, as usual,
Hn = γ + ψ (n + 1) =
n
X
1
r=1
r
=
Z∞
1 − tn
dt
1−t
0
be the nth harmonic number and γ denotes the Euler–Mascheroni constant. Let,
also, R, C and N denote, respectively
thesets of real, complex and natural numbers.
w
A generalized binomial coefficient
may be defined by
z
Γ (w + 1)
w
:=
; w, z ∈ C
z
Γ (z + 1) Γ (w − z + 1)
and in the special case when z = n, n ∈ N ∪ {0} , N := {1, 2, 3...} we have
(−1)n (−w)n
w (w − 1) ... (w − n + 1)
w
=
.
:=
n
n!
n!
The familiar gamma function

R∞ −y w−1


e y
dy, ℜ (y) > 0,



0
Γ (w) =

Γ(w+n)

, w ∈ C\ {0, −1, −2, −3, ...} , n ∈ N
 n−1

 Q (w+j)
j=0
2000 Mathematics Subject Classification. 05A10, 05A19.
Key words and phrases. Harmonic numbers, Binomial coefficients and Gamma function,
PolyGamma function, Combinatorial series identities and summation formulas, Partial fraction
approach.
c
2013
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted August 20, 2012. Published February 28, 2013.
71
72
ANTHONY SOFO
and
(w)λ :=


1, λ = 0; w ∈ C\ {0}
Γ (w + λ)
=

Γ (w)
w (w + 1) ... (w + n − 1) , w ∈ C, n ∈ N
, with (0)0 := 1
is known as the Pochhammer symbol. The generalized nth harmonic number in
(r)
power r, Hn , is defined for positive integers n and r as
n
X
1
(−1)r−1 (r−1)
(r−1)
Hn(r) :=
ψ
(n
+
1)
−
ψ
(1)
,
=
mr
(r − 1)!
m =1
where
m+1
ψ (m) (z) = (−1)
m!
∞
X
1
r =0 (z + r)
m+1
m+1
= (−1)
m! ζ (m + 1, z)
are the polygamma functions of order m which are defined by ψ (0) (z) ≡ ψ(z)
and ψ (m) (z) := dm ψ(z)/dz m , m ∈ N and z 6= 0, −1, −2, . . .. Here ψ(z) is the
psi, or digamma, function, given as the logarithmic derivative of the well-known
gamma function Γ(z), i.e. ψ(z) := d log Γ(z)/dz. ζ (α, z) denotes the Hurwitz
zeta function. There also exists the very useful recurrence relation ψ (m) (z + 1) =
(−1)m m!
(m)
(z). We shall provide, in this paper, identities for the general finite
z m+1 + ψ
sums
m
m
m
m
(p)
(p)
(p)
(p)
X
X
X
X
Hn
Hn
H
Hn
and
n
.
,
,
n (n + a)
(n + 1) (n + a)
n+k
n+k
n =1
n =1
n =1 n
n =1
k
k
Analogous results of Euler type for infinite series have been developed by many
authors, see for example [3], [4], [17] and references therein. Many finite versions
of higher order harmonic number sum identities also exist in the literature, for
example [15]
h
p
i
X
6
p
n+1
(−1)
(Hn )3 + 3Hn Hn(2) + 2Hn(3) = 3
n
p
n=0
and by the inversion formula
p
n+1 X
6 (−1)
p
3
= (Hp ) + 3Hp Hp(2) + 2Hp(3) .
n
n3
n=1
Also, after a minor modification, [11]
3 k
2 X
k
(1)
(1)
(2)
(2)
r+1
2
= 2.
1 + 3r Hk−r + Hr−1 + 3r Hk−r − Hr−1 − 1
(−1)
r
r=1
Other finite sum identities can be seen in [1], [18] or [19] where they obtain results
like
3
2m
X
(−1)m (3m)!
n
(−1)k
Hk =
(Hm + 2H2m − H3m ) .
3
k
2 (m!)
k=0
Further work in the summation of harmonic numbers and binomial coefficients has
also been done by Sofo [14]. The works of, [2], [3], [4], [5], [6], [7], [8], [9], [10], [11],
[12], [13],[16], [20], and references therein, also investigate various representations
FINITE NUMBER SUMS
73
of binomial sums and zeta functions in simpler form by the use of the Beta function
and by means of certain summation theorems for hypergeometric series, [17]. The
following results will be useful for later analysis.
Lemma 1.1. Let m and p be a positive integers and a > 0 then
m
(p)
X
a Hn
n (n + a)
n =1
p+1
(p+1)
(p)
= Hm
+ Hm
(Ha−1 + Hm − Hm+a ) + (−1)
+
a−1
X
p+1
(−1)
jp
j =1
(Hj − Hm+j ) +
p−1
X
p−s
(−1)
(p)
Hm Ha−1
(p−s+1)
(s)
Hm
Ha−1 (1.1)
.
s=2
Proof. Consider
m X
n
(p)
X
Hn
1
=
p
n (n + a) n =1
k n (n + a)
n =1
m
X
k =1
since these sums are absolutely convergent we can write
n
m X
X
n =1 k =1
=
m
X
k =1
Now
1
p
k n (n + a)
=
m
m X
X
k =1 n =k
kp
1
n (n + a)
1
[ψ(a + k) − ψ(k) − (ψ(m + a + 1) − ψ(m + 1))] .
a kp
m
X
1
[ψ(a + k) − ψ(k) − (ψ(m + a + 1) − ψ(m + 1))]
a kp
k =1


m
a−1
a−1
X
X
1
1 X 1

=
−
ak p
k
+
j
m
+
1
+
j
j =0
j =0
k =1


m
a−1 X
X
1
1 1
1
1

−
+
−
ak p
k
m + 1 j =1 k + j
m+1+j
=
k =1
(p+1)
Hm
a
=
−
k =1
(p+1)
=
(p+1)
Hm
=
a
Hm
a
a−1 m (p)
Hm
1XX
1
1
+
−
a (m + 1) a j =1
k p (k + j) k p (m + 1 + j)
−
a−1
a−1 m
(p)
(p)
Hm
Hm
1X
1XX
1
−
+
a (m + 1) a j =1 (m + 1 + j) a j =1
k p (k + j)
k =1
a−1 m
(p)
(p)
Hm
Hm
1XX
−
−
(Hm+a − Hm+1 )+
a (m + 1)
a
a j =1
k =1
p+1
p
p−s
X (−1)
(−1)
+
p−1
j
k (k + j) s=2 k s j p−s+1
!
74
ANTHONY SOFO
by partial fraction decomposition, hence
(p)
(p+1)
Hm
a
=
−
a−1
1X
+
a j =1
(p)
Hm
Hm
−
(Hm+a − Hm+1 )
a (m + 1)
a
p+1
(−1)
jp
(Hj + Hm − Hm+j ) +
p
(s)
p−s
X
(−1)
Hm
j p−s+1
s=2
!
.
Therefore
m
(p)
X
a Hn
n (n + a)
n =1
=
(p+1)
Hm
−
(p)
Hm
1
+ Hm+a − Hm+1 − Ha−1
m+1
p−1
a−1
p+1
X
X
(−1)
p−s
(s) (p−s+1)
(−1)
Hm
Ha−1
(H
−
H
)
+
+
j
m+j
p
j
s=2
j =1
p+1
(p)
Hm Ha−1
=
(p+1)
(p)
Hm
+ Hm
(Ha−1 + Hm − Hm+a ) + (−1)
+ (−1)
p+1
+
a−1
X
p+1
(−1)
jp
j =1
(Hj − Hm+j ) +
p−1
X
p−s
(−1)
(p)
Hm Ha−1
(p−s+1)
(s)
Hm
Ha−1
s=2
and the result (1.1) follows.
(α)
Remark. When a = 1, m and p are positive integers, and we define H0
then
m
(p)
(p)
X
Hn
Hm
(p+1)
= Hm
−
n (n + 1)
m+1
n =1
= 0,
(1.2)
and for a = 2
(p)
p−1
m
(p)
p+1
X
X
m 2 + m − 1 Hm
2 Hn
m (−1)
p−s
(s)
(p+1)
(−1)
Hm
.
= Hm
+
+
+
n
(n
+
2)
(m
+
1)
(m
+
2)
m
+
1
n =1
s=2
A related Lemma which will be useful later is the following.
Lemma 1.2. Let m and p be a positive integers and a > 0 then
m
(p)
X
(a − 1) Hn
(n + 1) (n + a)
n =1
p+1
(p)
= Hm
(Ha−1 + Hm+1 − Hm+a ) + (−1)
a−1
X
p+1
(−1)
+
jp
j =1
(Hj − Hm+j ) +
p−1
X
s=2
(p)
Hm Ha−1
(p−s+1)
(s)
(−1)p−s Hm
Ha−1(1.3).
FINITE NUMBER SUMS
75
Proof. Consider
m X
n
(p)
X
1
Hn
=
p
(n + 1) (n + a) n =1
k (n + 1) (n + a)
n =1
m
X
k =1
since these sums are absolutely convergent we can write
m X
m
m X
n
X
X
1
1
=
p (n + 1) (n + a)
p (n + 1) (n + a)
k
k
n =1
k =1 n =k
k =1
=
m
X
1
[ψ(a + k) − ψ(k + 1) − (ψ(m + a + 1) − ψ(m + 2))] .
(a − 1) k p
k =1
Now
m
X
=
1
[ψ(a + k) − ψ(k + 1) − (ψ(m + a + 1) − ψ(m + 2))]
(a − 1) k p
k =1


m
a−1
a−1
X
X
X
1
1
1


−
(a − 1) k p
k
+
j
m
+
1
+
j
j =1
j =1
k =1
=
m
X
k =1
=


a−1 X
1
1
1


−
(a − 1) k p
k+j
m+1+j
j =1
a−1
m XX
1
1
1
−
(a − 1) j =1
k p (k + j) k p (m + 1 + j)
k =1
= −
a−1
a−1
m
(p)
X
XX
1
Hm
1
1
+
(a − 1) j =1 (m + 1 + j) (a − 1) j =1
k p (k + j)
k =1
a−1 X
m
(p)
X
Hm
1
=−
(Hm+a − Hm+1 ) +
(a − 1)
(a − 1) j =1
k =1
by partial fraction decomposition, hence
p+1
p
p−s
X (−1)
(−1)
+
j p−1 k (k + j) s=2 k s j p−s+1
!
(p)
=
−
Hm
(Hm+a − Hm+1 )
(a − 1)
a−1
X
1
+
(a − 1) j =1
p+1
(−1)
jp
(Hj + Hm − Hm+j ) +
p
(s)
p−s
X
(−1)
Hm
s=2
j p−s+1
!
.
Therefore
m
(p)
X
(a − 1) Hn
(n + 1) (n + a)
n =1
p+1
+ (−1)
(p)
Hm Ha−1 +
(p)
= −Hm
(Hm+a − Hm+1 − Ha−1 )
p−1
a−1
p+1
X
X
(−1)
p−s
(s) (p−s+1)
(−1)
Hm
Ha−1
(H
−
H
)
+
j
m+j
p
j
s=2
j =1
76
ANTHONY SOFO
(p)
p+1
(p)
Hm
(Ha−1 + Hm+1 − Hm+a ) + (−1)
=
+
a−1
X
p+1
(−1)
jp
j =1
(Hj − Hm+j ) +
p−1
X
Hm Ha−1
p−s
(−1)
(p−s+1)
(s)
Hm
Ha−1
s=2
and the result (1.3) follows.
Some special cases are noted in the remark.
Remark. When a = 2 and p and m positive integers, then
m
X
(p)
(p)
p+1
(m + 1) Hm
m (−1)
Hn
=
+
(n
+
1)
(n
+
2)
(m
+
2)
m+1
n =1
+
p−1
X
p−s
(−1)
(s)
Hm
.
(1.4)
s=2
2. Two theorems
We now prove the following two theorems.
Theorem 2.1. Let p be a positive integer and k be a positive integer greater than
1, then
m
X
n =1
+
k
X
r
(−1) r
r=2
k
r



(p)
Hn
(p)
= kHm
Hm+1
n+k
k
(p)
p+1
−Hm Hm+r + (−1)
+
(−1)p+1
j =1
jp
Pr−1
(Hj − Hm+j ) +
Proof. Consider the following expansion:
m
X
n =1
(2.1)
(p)
Hm Hr−1 +
Pp
p−s
s=2
(−1)
(s) (p−s+1)
Hm Hr−1


.
m
m
(p)
(p)
(p)
X
X
Hn
k! Hn
k! Hn
=
=
k
Q
(n + 1) (n + 2)k+1
n+k
n =1 (n + 1)
(n + r) n =1
k
r=2
where (α)r is Pochhammer’s symbol given by (α)r = α (α + 1) (α + 2) ... (α + r − 1) , r >
0, (α)0 = 1. Now
m
X
n =1
k
m
(p)
(p)
X
Hn
k! Hn X Ar
=
(n + 1) r =2 (n + r)
n+k
n =1
k
where
r
Ar = lim
n→−r
2 (−1)
n+r
=
k
Q
k!
(n + r)
r =2
k
r
r
2
(2.2)
.
FINITE NUMBER SUMS
77
Now from (2.2) and using Lemma 1.2
X
k
k
m
m
(p)
(p)
X
X
Hn
k! Hn X Ar
k
r
=
(−1) r (r − 1)
r
(n + 1) r =2 (n + r) r =2
(n + 1) (n + r)
n =1
n =1
k
X
=
r
(−1) r
r =2





k
r
(p)
= kHm
Hm+1 +
k
X
(p)
+
Pr−1
r
+
r
(−1) r
r =2
m
X
=
n =1
(p)
Hn
n+k
k
j =1
(−1) r
r =2
k
X
p+1
Hm (Hm+1 − Hm+r ) + (−1)
(−1)p+1
jp
k
r
(Hj − Hm+j ) +
(p)
Pp
(s)
s=2
p+1
−Hm Hm+r + (−1)

(p)
Hm Hr−1
(p−s+1)
(−1)p−s Hm Hr−1
(p)
Hm Hr−1






X
p
r−1
p+1
X
(−1)
k 
p−s
(s) (p−s+1) 
(−1)
Hm
Hr−1
(Hj − Hm+j ) +
p
r
j
s=2
j =1
which is the result (2.1).
Now we consider the following extension of Theorem 2.1.
Theorem 2.2. Let k, m and p be positive integers, then
m
X
n =1
n
h
k
(p)
i
X
Hn
k
(p)
r
p
(p+1)
(p)
(p)
= Hm
+ Hm
Hm +
(−1)
Hm
Hm+r + (−1) Hm Hr−1
r
n+k
r=1
k
(2.3)
−
k
X
r=1
r
(−1)


X
p
r−1
p+1
X
(−1)
k 
p−s
(s) (p−s+1) 
(−1)
Hm
Hr−1
.
(Hj − Hm+j ) +
p
r
j
s=2
j =1
Proof. Consider the following expansion:
m
X
n =1
n
m
m
(p)
(p)
(p)
X
X
k! Hn
k! Hn
Hn
=
=
k
Q
n (n + 1)k+1
n+k
n =1 n
(n + r) n =1
k
r=1
78
ANTHONY SOFO
Now
m
X
n =1
where
m
k
(p)
(p)
X
k! Hn X Br
Hn
=
n
(n + r)
n+k
n =1
r =1
k
Br = lim
n+r
n→−r
k
Q
(n + r)
r+1
(−1)
=
k!
r
k
r
.
r =1
Now from (2.3) and using Lemma 1.1
X
k
k
m
m
(p)
(p)
X
X
Hn
k! Hn X Br
k
r+1
=
(−1)
r
r
n
(n + r) r =1
n (n + r)
r =1
n =1
n =1
=
k
X
r+1
(−1)
r =1
k
r



(p+1)
Hm
+
Pr−1
j =1
(p)
p+1
+ Hm (Hm − Hm+r ) + (−1)
p+1
(−1)
jp
(Hj − Hm+j ) +
Pp
s=2
(p)
Hm Hr−1
p−s
(−1)
(s)
(p−s+1)
Hm Hr−1



(p+1)
(p)
= Hm
+ Hm
Hm
+
k
X
r
(−1)
r =1
=
m
X
n =1
(p)
n
k
r
Hn
n+k
k



(p)
(p)
p
Hm Hm+r + (−1) Hm Hr−1
−
(−1)p+1
j =1
jp
Pr−1
(Hj − Hm+j ) −
which is the result (2.3).
Pp
s=2
p−s
(−1)
(s) (p−s+1)
Hm Hr−1
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
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