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 har- monic 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
r=1
1 r =
∞
Z
0
1−tn 1−t dt
be the nth harmonic number andγ denotes the Euler–Mascheroni constant. Let, also,R, CandNdenote, respectively the sets of real, complex and natural numbers.
A generalized binomial coefficient w
z
may be defined by w
z
:= Γ (w+ 1)
Γ (z+ 1) Γ (w−z+ 1); w, z∈C
and in the special case when z=n, n∈N∪ {0}, N:={1,2,3...}we have w
n
:= w(w−1)...(w−n+ 1)
n! =(−1)n(−w)n n! . The familiar gamma function
Γ (w) =
∞
R
0
e−yyw−1dy, ℜ(y)>0,
Γ(w+n)
n−Q1 j=0
(w+j)
, w∈C\ {0,−1,−2,−3, ...}, n∈N
2000Mathematics 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 Prishtin¨es, Prishtin¨e, Kosov¨e.
Submitted August 20, 2012. Published February 28, 2013.
71
and
(w)λ:=Γ (w+λ) Γ (w) =
1, λ= 0; w∈C\ {0}
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 powerr,Hn(r), is defined for positive integersnandras
Hn(r):=
n
X
m=1
1
mr = (−1)r−1 (r−1)!
ψ(r−1)(n+ 1)−ψ(r−1)(1) ,
where
ψ(m)(z) = (−1)m+1m!
∞
X
r=0
1
(z+r)m+1 = (−1)m+1m! ζ(m+ 1, z)
are the polygamma functions of order m which are defined by ψ(0)(z) ≡ ψ(z) and ψ(m)(z) := dmψ(z)/dzm, 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) := dlog Γ(z)/dz. ζ(α, z) denotes the Hurwitz zeta function. There also exists the very useful recurrence relationψ(m)(z+ 1) =
(−1)mm!
zm+1 +ψ(m)(z). We shall provide, in this paper, identities for the general finite sums
m
X
n=1
Hn(p)
n(n+a),
m
X
n=1
Hn(p)
(n+ 1) (n+a),
m
X
n=1
Hn(p)
n+k k
and
m
X
n=1
Hn(p)
n
n+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]
p
X
n=0
(−1)n+1 p
n h
(Hn)3+ 3HnHn(2)+ 2Hn(3)i
= 6 p3 and by the inversion formula
p
X
n=1
6 (−1)n+1 n3
p n
= (Hp)3+ 3HpHp(2)+ 2Hp(3). Also, after a minor modification, [11]
k
X
r=1
(−1)r+1 k
r 3
1 + 3r2
Hk−r(2) +Hr−1(2) +
3r
Hk−r(1) −Hr−1(1)
−12
= 2.
Other finite sum identities can be seen in [1], [18] or [19] where they obtain results
like 2m
X
k=0
(−1)k n
k 3
Hk =(−1)m(3m)!
2 (m!)3 (Hm+ 2H2m−H3m).
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
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 mandpbe a positive integers and a >0 then
m
X
n=1
a Hn(p)
n(n+a) = Hm(p+1)+Hm(p)(Ha−1+Hm−Hm+a) + (−1)p+1HmHa−(p)1 +
a−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p−1
X
s=2
(−1)p−sHm(s)Ha−1(p−s+1)(1.1).
Proof. Consider
m
X
n=1
Hn(p)
n(n+a)=
m
X
n=1 n
X
k=1
1 kp n (n+a) since these sums are absolutely convergent we can write
m
X
n=1 n
X
k=1
1
kp n (n+a) =
m
X
k=1 m
X
n=k
1 kp n (n+a)
=
m
X
k=1
1
a kp[ψ(a+k)−ψ(k)−(ψ(m+a+ 1)−ψ(m+ 1))]. Now
m
X
k=1
1
a kp [ψ(a+k)−ψ(k)−(ψ(m+a+ 1)−ψ(m+ 1))]
=
m
X
k=1
1 akp
a−1
X
j=0
1 k+j −
a−1
X
j=0
1 m+ 1 +j
=
m
X
k=1
1 akp
1 k − 1
m+ 1 +
a−1
X
j=1
1
k+j − 1 m+ 1 +j
= Hm(p+1)
a − Hm(p)
a(m+ 1) +1 a
a−1
X
j=1 m
X
k=1
1
kp(k+j)− 1 kp(m+ 1 +j)
= Hm(p+1)
a − Hm(p)
a(m+ 1)−1 a
a−1
X
j=1
Hm(p)
(m+ 1 +j)+1 a
a−1
X
j=1 m
X
k=1
1 kp(k+j)
= Hm(p+1)
a − Hm(p)
a(m+ 1)−Hm(p)
a (Hm+a−Hm+1)+1 a
a−1
X
j=1 m
X
k=1
(−1)p+1 jp−1k(k+j)+
p
X
s=2
(−1)p−s ksjp−s+1
!
by partial fraction decomposition, hence
= Hm(p+1)
a − Hm(p)
a(m+ 1) −Hm(p)
a (Hm+a−Hm+1) +1
a
a−1
X
j=1
(−1)p+1
jp (Hj+Hm−Hm+j) +
p
X
s=2
(−1)p−sHm(s)
jp−s+1
! .
Therefore
m
X
n=1
a Hn(p)
n(n+a) = Hm(p+1)−Hm(p) 1
m+ 1+Hm+a−Hm+1−Ha−1
+ (−1)p+1HmHa−(p)1+
a−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p−1
X
s=2
(−1)p−sHm(s)Ha−(p−s+1)1
= Hm(p+1)+Hm(p)(Ha−1+Hm−Hm+a) + (−1)p+1HmHa−1(p)
+
a−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p−1
X
s=2
(−1)p−sHm(s)Ha−1(p−s+1)
and the result (1.1) follows.
Remark. When a = 1, m and p are positive integers, and we define H0(α) = 0, then
m
X
n=1
Hn(p)
n(n+ 1) =Hm(p+1)− Hm(p)
m+ 1 (1.2)
and for a= 2
m
X
n=1
2 Hn(p)
n(n+ 2) =Hm(p+1)+ m2+m−1 Hm(p)
(m+ 1) (m+ 2) +m(−1)p+1 m+ 1 +
p−1
X
s=2
(−1)p−sHm(s). A related Lemma which will be useful later is the following.
Lemma 1.2. Let mandpbe a positive integers and a >0 then
m
X
n=1
(a−1) Hn(p)
(n+ 1) (n+a) = Hm(p)(Ha−1+Hm+1−Hm+a) + (−1)p+1HmHa−(p)1 +
a−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p−1
X
s=2
(−1)p−sHm(s)Ha−1(p−s+1)(1.3).
Proof. Consider
m
X
n=1
Hn(p)
(n+ 1) (n+a)=
m
X
n=1 n
X
k=1
1
kp (n+ 1) (n+a) since these sums are absolutely convergent we can write
m
X
n=1 n
X
k=1
1
kp (n+ 1) (n+a) =
m
X
k=1 m
X
n=k
1
kp (n+ 1) (n+a)
=
m
X
k=1
1
(a−1) kp[ψ(a+k)−ψ(k+ 1)−(ψ(m+a+ 1)−ψ(m+ 2))]. Now
m
X
k=1
1
(a−1) kp[ψ(a+k)−ψ(k+ 1)−(ψ(m+a+ 1)−ψ(m+ 2))]
=
m
X
k=1
1 (a−1)kp
a−1
X
j=1
1 k+j −
a−1
X
j=1
1 m+ 1 +j
=
m
X
k=1
1 (a−1)kp
a−1
X
j=1
1
k+j − 1 m+ 1 +j
= 1
(a−1)
a−1
X
j=1 m
X
k=1
1
kp(k+j)− 1 kp(m+ 1 +j)
= − 1
(a−1)
a−1
X
j=1
Hm(p)
(m+ 1 +j)+ 1 (a−1)
a−1
X
j=1 m
X
k=1
1 kp(k+j)
=− Hm(p)
(a−1)(Hm+a−Hm+1) + 1 (a−1)
a−1
X
j=1 m
X
k=1
(−1)p+1 jp−1k(k+j)+
p
X
s=2
(−1)p−s ksjp−s+1
!
by partial fraction decomposition, hence
= − Hm(p)
(a−1)(Hm+a−Hm+1)
+ 1
(a−1)
a−1
X
j=1
(−1)p+1
jp (Hj+Hm−Hm+j) +
p
X
s=2
(−1)p−sHm(s)
jp−s+1
! .
Therefore
m
X
n=1
(a−1) Hn(p)
(n+ 1) (n+a) = −Hm(p)(Hm+a−Hm+1−Ha−1)
+ (−1)p+1HmHa−(p)1+
a−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p−1
X
s=2
(−1)p−sHm(s)Ha−(p−s+1)1
= Hm(p)(Ha−1+Hm+1−Hm+a) + (−1)p+1HmHa−(p)1
+
a−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p−1
X
s=2
(−1)p−sHm(s)Ha−(p−s+1)1
and the result (1.3) follows.
Some special cases are noted in the remark.
Remark. Whena= 2 andpandmpositive integers, then
m
X
n=1
Hn(p)
(n+ 1) (n+ 2) = (m+ 1)Hm(p)
(m+ 2) +m(−1)p+1 m+ 1 +
p−1
X
s=2
(−1)p−sHm(s). (1.4)
2. Two theorems We now prove the following two theorems.
Theorem 2.1. Let pbe a positive integer and kbe a positive integer greater than 1, then
m
X
n=1
Hn(p)
n+k k
=kHm(p) Hm+1 (2.1)
+
k
X
r=2
(−1)rr k
r
−Hm(p)Hm+r+ (−1)p+1HmHr(p)−1+ +Pr−1
j=1 (−1)p+1
jp (Hj−Hm+j) +Pp
s=2(−1)p−sHm(s)Hr−1(p−s+1)
.
Proof. Consider the following expansion:
m
X
n=1
Hn(p)
n+k k
=
m
X
n=1
k!Hn(p)
(n+ 1)
k
Q
r=2
(n+r)
=
m
X
n=1
k!Hn(p)
(n+ 1) (n+ 2)k+1
where (α)ris 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
Hn(p)
n+k k
=
m
X
n=1
k!Hn(p)
(n+ 1)
k
X
r=2
Ar
(n+r) (2.2)
where
Ar= lim
n→−r
n+r
k
Q
r=2
(n+r)
=2 (−1)r k!
k r
r 2
.
Now from (2.2) and using Lemma 1.2
m
X
n=1
k!Hn(p)
(n+ 1)
k
X
r=2
Ar
(n+r) =
k
X
r=2
(−1)r r(r−1) k
r m
X
n=1
Hn(p)
(n+ 1) (n+r)
=
k
X
r=2
(−1)r r k
r
Hm(p)(Hm+1−Hm+r) + (−1)p+1HmHr−(p)1
+Pr−1 j=1
(−1)p+1
jp (Hj−Hm+j) +Pp
s=2(−1)p−sHm(s)Hr−(p−s+1)1
= kHm(p) Hm+1+
k
X
r=2
(−1)r r k
r
−Hm(p)Hm+r+ (−1)p+1HmHr−(p)1
+
k
X
r=2
(−1)r r k
r
r−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p
X
s=2
(−1)p−sHm(s)Hr−1(p−s+1)
=
m
X
n=1
Hn(p)
n+k k
which is the result (2.1).
Now we consider the following extension of Theorem 2.1.
Theorem 2.2. Let k, m andpbe positive integers, then
m
X
n=1
Hn(p)
n
n+k k
=Hm(p+1)+Hm(p)Hm+
k
X
r=1
(−1)r k
r h
Hm(p)Hm+r+ (−1)pHmHr−(p)1
i
(2.3)
−
k
X
r=1
(−1)r k
r
r−1
X
j=1
(−1)p+1
jp (Hj−Hm+j) +
p
X
s=2
(−1)p−sHm(s)Hr(p−s+1)−1
.
Proof. Consider the following expansion:
m
X
n=1
Hn(p)
n
n+k k
=
m
X
n=1
k! Hn(p)
n
k
Q
r=1
(n+r)
=
m
X
n=1
k!Hn(p)
n(n+ 1)k+1
Now
m
X
n=1
Hn(p)
n+k k
=
m
X
n=1
k! Hn(p)
n
k
X
r=1
Br
(n+r) where
Br= lim
n→−r
n+r
k
Q
r=1
(n+r)
=(−1)r+1r k!
k r
.
Now from (2.3) and using Lemma 1.1
m
X
n=1
k!Hn(p)
n
k
X
r=1
Br
(n+r) =
k
X
r=1
(−1)r+1 r k
r m
X
n=1
Hn(p)
n(n+r)
=
k
X
r=1
(−1)r+1 k
r
Hm(p+1)+Hm(p)(Hm−Hm+r) + (−1)p+1HmHr−(p)1
+Pr−1 j=1
(−1)p+1
jp (Hj−Hm+j) +Pp
s=2(−1)p−sHm(s)Hr(p−s+1)−1
= Hm(p+1)+Hm(p)Hm
+
k
X
r=1
(−1)r k
r
Hm(p)Hm+r+ (−1)pHmHr−1(p)
−Pr−1 j=1
(−1)p+1
jp (Hj−Hm+j) −Pp
s=2(−1)p−sHm(s)Hr−1(p−s+1)
=
m
X
n=1
Hn(p)
n
n+k k
which is the result (2.3).
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