Parametric harmonic number sum identities

0.16 MB·

by  Ce Xu

Parametric harmonic number sum identities
Ce Xu∗
7
School of Mathematical Sciences, Xiamen University
1
0 Xiamen 361005,P.R. China
2

r
a Abstract WeestablishsomeidentitiesofEulerrelatedsums. Byusingtheidentities,wediscuss
M
theclosedformrepresentationsofsumsofharmonicnumbersandreciprocalparametricbinomial
9 coefficients throughparametricharmonicnumbers,shiftedharmonicnumbersandRiemannzeta
2 function with positive integer arguments. In particular we investigate products of quadratic

harmonic numbers and reciprocal parametric binomial coefficients. Some illustrative special
]
T cases as well as immediate consequences of the main results are also considered.
N
Keywords Euler sums; harmonic numbers; binomial coefficients; polylogarithms.
.
h
t AMS Subject Classifications (2010): 05A10, 05A19, 11B65, 11B83, 11M06, 33D60, 33C20.
a
m
1 Introduction
[

2
v Let R and C denote respectively, the sets of real and complex numbers and let N = {1,2,3,...}
6 be the set of natural numbers, and N := N∪{0}. The classical harmonic numbers of order s
2 0
are given by
7
n
3 1
0 Hn(s) := js (n ∈ N, s ∈ C) (1.1)
.
1 Xj=1
0 and
17 n 1 1 1−tn
H := H(1) = = dt = γ+ψ(n+1) (n ∈ N), (1.2)
v: n n j 1−t
j=1 Z
i X 0
X
where γ denotes the Euler-Mascheroni constant, defined by
r
a n
1
γ := lim −lnn = −ψ(1) ≈ 0.577215664901532860606512...,
n→∞ k !
k=1
X
and ψ(z) denotes the digamma function (or called Psi function ) which is defined as the loga-
rithmic derivative of the well known gamma function:
d Γ′(z)
ψ(z) := (lnΓ(z)) = . (1.3)
dz Γ(z)
In general, the polygamma function of order m is defined as the (m+1)-th derivative of the
logarithm of the gamma function:
dm dm+1
ψ(m)(z) := ψ(z) = ln(Γ(z)) (m ∈ N ;z ∈ C\Z− := {0,−1,−2,...}). (1.4)
dzm dzm+1 0 0
∗Corresponding author. Email: [email protected]
1
For any s ∈ C\{0}, we assume that
(s) (0)
H := 0, H := 0, H := 1.
0 0 0
Equation (1.1) can be written in the following form:
H(s) = ζ(s)−ζ(s,n+1) (ℜ(s)> 1;n ∈N)
n
by recalling the well-known (easily derivable) relationship between the Riemann Zeta function
ζ(s) (or called Euler-Riemann zeta function) and the Hurwitz (or generalized) Zeta function
ζ(s,a) (see [3,6,7])
n
1
ζ(s) =ζ(s,n+1)+ (n ∈ N ). (1.5)
js 0
j=1
X
The Riemann Zeta function or Euler-Riemann zeta function (for more details, see for instance,
[1,3,5–7,28])
∞
1
ζ(s):= (ℜ(s)> 1)
ns
n=1
X
is probablythemost important, fascinating, challenging andmysterious object of modernmath-
ematics, inspiteofits uttersimplicity. Thisfunctionisdefinedover thecomplexplaneandplays
a pivotal role in analytic number theory having applications in physics, probability theory, ap-
plied statistics and other fields of mathematics. There is an enormous amount of literature on
the Riemann zeta function.
A well-known (and potentially useful) relationship between the Polygamma functions ψ(m)(z)
and the generalized Zeta function (or Hurwitz Zeta function) ζ(s,a) is given by
∞
1
ψ(m)(z) = (−1)m+1m! = (−1)m+1m!ζ(m+1,z). (1.6)
(n+z)m+1
n=0
X
It is also easy to have the following expression
ψ(m)(n+1)−ψ(m)(1) = (−1)m+1m!H(m+1) (m,n ∈ N ),
n 0
(m)
which immediately yields another expression for H as follows
n
(−1)m−1
H(m) = ψ(m−1)(n+1)−ψ(m−1)(1) (m ∈ N;n ∈N ). (1.7)
n (m−1)! 0
n o
Replacing n by α (α ∈ C\Z−;Z− := {−1,−2,...}) in above equation (1.2) and (1.7), we can
define the generalized α-th shifted harmonic number of order m by
(−1)m−1
H(m) := ψ(m−1)(α+1)−ψ(m−1)(1) , 2≤ m ∈N, (1.8)
α (m−1)!
(cid:16) (cid:17)
and
∞
1 1
H = H(1) := ψ(α+1)+γ = − . (1.9)
α α k k+α
k=1(cid:18) (cid:19)
X
2
It is easy to otain the following expression by using the relation (1.6) and the definition of
Hurwitz Zeta function ζ(s,a+1)
n
1
H(m) = ζ(m)−ζ(m,α+1), H(m) = H(m) + , 2 ≤ m ∈ N. (1.10)
α n+α α (j +α)m
j=1
X
The generalized harmonic functions (also called parametric harmonic numbers or partial sums
of Hurwitz Zeta function) are defined by (see [12,19,23])
n
1
H(s)(a) := (n ∈N;s ∈ C;a ∈C\Z−), (1.11)
n (j +a)s
j=1
X
(s) (s)
so that, obviously, H (0) = H . By using the equation (1.6), we may write the generalized
n n
(m)
harmonic numbers, H (a) , in terms of polygamma functions
n
(−1)m−1
H(m)(a) = ψ(m−1)(n+a+1)−ψ(m−1)(a+1) (m ∈N;n ∈ N ). (1.12)
n (m−1)! 0
n o
a
A generalized binomial coefficient (a,b ∈ C) is defined, in terms of the familiar (Euler’s)
b
(cid:18) (cid:19)
gamma function, by
a Γ(a+1)
:= , a,b ∈C, (1.13)
b Γ(b+1)Γ(a−b+1)
(cid:18) (cid:19)
which, in the special case when b = n,n ∈ N , yields
0
a a a(a−1)···(a−n+1)
:= 1, := , n ∈ N.
0 n n!
(cid:18) (cid:19) (cid:18) (cid:19)
Here we are interested in evaluating closed form and integral representations of Euler type sums
containing both harmonic numbers and binomial coefficients. In this paper we will develop
identities, closed form representations of harmonic numbers and reciprocal binomial coefficients
of the form:
∞ H(m1)H(m2)···H(mr)
W(a,b)(m ,m ,··· ,m ,p):= n n n (a,b ∈ C\Z−) (1.14)
k 1 2 r n+k+b
n=1 (n+a)p
X k
(cid:18) (cid:19)
for r = 1,p ∈ {0,1} and r = 2,m = 1,p ∈ {0,1}, where p ∈ N ,k,r,m ∈ N,i =1,2,...,r and
i 0 i
(a,b)
p+ k > 1. When a = b = 0, then the parametric Euler type sum W (m ,m ,··· ,m ,p)
k 1 2 r
reduces to the classical Euler type sum W (m ,m ,··· ,m ,p) which is defined by (see [30])
k 1 2 r
∞ H(m1)H(m2)···H(mr)
W (m ,m ,··· ,m ,p) := n n n , (p ∈ N ,k,r,m ∈ N) (1.15)
k 1 2 r 0 i
n+k
n=1 np
X k
(cid:18) (cid:19)
with p+k > 1. While there are many results for sums of harmonic numbers with positive terms
W (m ,m ,··· ,m ,p), for example we know that from [22] and [30]
k 1 2 r
k 2
W (1,1,0) = ζ(2)−H(2) + , 2 ≤ k ∈ N,
k k−1 k−1 (k−1)2
(cid:18) (cid:19)
3
H3+3H H(2)+2H(3)
k 3ζ(3)+ r r r r
k 3
Wk(1,1,1) = Xr=1(−1)r+1(cid:18)r(cid:19)−Hr2+rHr(2) −r−1 Hi2i +ζ(2)Hr−1, k ∈ N.
i=1
X
 
 
Therehasrecentlybeenrenewedinterestinthestudyofseriesinvolvingbinomialcoefficientsand
a number of authors have obtained either closed form representation or integral representation
for some particular cases of these series. The interested reader is referred to [12,13,20–28,30],
and references therein. For results on alternating quadratic harmonic number sums see [25].
When k = 0 in (1.15), then the Euler type sum W (m ,m ,··· ,m ,p) reduces to the classical
k 1 2 r
Euler sum Sm,p (m := {m1,m2,...,mr}) defined by (see [14])
∞ H(m1)H(m2)···H(mr)
n n n
Sm,p := np . (1.16)
n=1
X
There are many works investigating Euler sums Sm,p involving harmonic numbers, see for ex-
ample [4,8–10,14,15,18,29], and references therein. There are fewer results for sums of the type
(1.14). In this paper, we will prove that all Euler type sums
(a,b) (a,a) (a,a)
W (1,p), W (m,p), W (1,1,p)
k k k
for p = 0,1 can be expressed as a rational linear combination of products of zeta values, shifted
harmonic numbers and parametric harmonic numbers. Finally, we use certain integral represen-
tations to evaluate several series with alternating harmonic numbers. The alternating harmonic
numbers are defined by
H(s) := n (−1)j−1 (n ∈ N, s ∈ C) (1.17)
n js
j=1
X
(a,b)
and we define the alternating harmonic number sums W (m¯,p) by the series
k
∞ (m)
H
W(a,b)(m,p) := n (m,k ∈ N;p ∈N ;a,b ∈ C\Z−) (1.18)
k n+k+b 0
n=1(n+a)p
X k
(cid:18) (cid:19)
with p+k > 1.
The following lemmas will be useful in the development of the main theorems.
Lemma 1.1 For m ∈N, a,b ∈ C\Z− and x ∈ (−1,1). Then the following identity holds:
x
m−1 (−1)k−1xn+b
H (t,a)tn+b−1dt = H (x,a)
m k m+1−k
(n+b)
Z0 Xk=1
(−1)m−1 n xk+a+b
+ xn+bH (x,a)+ −H (x,a+b) , (1.19)
m 1 1
(n+b) k+a+b
( )
k=1
X
where the function H (x,a) is defined by
m
∞ xn+a
H (x,a) := , ℜ(s)> 1, a ∈/ Z−, x ∈ [−1,1], (1.20)
s s
(n+a)
n=1
X
4
∞ xn+a
with H (x,a) := , x ∈ [−1,1).
1
n+a
n=1
X
Proof. By using integration by parts, we have the following recurrence relation
x x
xn+b 1
H (t,a)tn+b−1dt = H (x,a)− H (t,a)tn+b−1dt = ···
m m m−1
n+b n+b
Z Z
0 0
x
m−1 (−1)k−1xn+b (−1)m−1
= H (x,a)+ H (t,a)tn+b−1dt.
(n+b)k m+1−k (n+b)m−1 1
k=1 Z
X 0
(1.21)
By simple calculation, the second term integral on the right hand side is equal to
x
1 n xk+a+b
H (t,a)tn+b−1dt = xn+bH (x,a)+ −H (x,a+b) . (1.22)
1 1 1
n+b k+a+b
( )
Z0 Xk=1
Substituting (1.22) into (1.21), we may easily deduce the desired result. (cid:3)
Corollary 1.2 For m ∈ N, b ∈C\Z− and x ∈ (−1,1), we have
x
m−1 (−1)i−1 (−1)m−1 n xj+b
Li (t)tn+b−1dt = xn+bLi (x)+
m (n+b)i m+1−i (n+b)m j +b
Z i=1 j=1
0 X X
(−1)m−1
+ xn+bLi (x)−H (x,b) , (1.23)
m 1 1
(n+b)
n o
where Li (x) denotes the polylogarithm function which is defined by
s
∞ xn
Li (x) = , ℜ(s) ≥ 1,−1 ≤ x < 1,
s ns
n=1
X
with Li (x) = −ln(1−x). Obviously, Li (x) = H (x,0).
1 s s
Proof. Setting a = 0 in (1.19), the result is (1.23). (cid:3)
Lemma 1.3 ( [31]) Let s be a integer with s ≥ 1 and a ∈/ Z−, then we have
n−1xn−j n−1yn−j
yn +xn
∞ j j s
j=1 j=1
X X =sLi (a,xy)− Li (a,x)Li (a,y)
s s+1 j s+1−j
(n+a)
n=1 j=1
X X
+Li (a,xy)(Li (x)+Li (y)). (1.24)
s 1 1
where x,y ∈ [−1,1) and the parametric polylogarithm function Li (a,x) is defined by
s
∞ xn
Li (a,x) = , ℜ(s)≥ 1,−1 ≤ x < 1.
s s
(n+a)
n=1
X
5
Proof. Takingx = y = q andb = ainTheorem2.2inthereference[31], thenlettingq approach
1, we obtain formula (1.24). (cid:3)
Putting x = y in above equation, we arrive at the conclusion that (s ≥ 2)
∞ xn n−1xn−j s
= Li a,x2 +Li a,x2 Li (x)−Li (a,x)Li (a,x)
s s+1 s 1 s 1
(n+a) j 2
n=1 j=1
X X (cid:0) (cid:1) (cid:0) (cid:1)
s−1
1
− Li (a,x)Li (a,x), (1.25)
j s+1−j
2
j=2
X
Note that when s > 1, lim Li (a,1) = ζ(s,a+1). When x approach 1, we arrive at the
s
x→1
conclusion that
lim Li a,x2 Li (x)−Li (a,x)Li (a,x) = ζ(s,a+1)H . (1.26)
s 1 s 1 a
x→1
(cid:8) (cid:0) (cid:1) (cid:9)
Hence, letting x → 1 in (1.25), we obtain
∞ s−2
H s 1
n
= ζ(s+1,a+1)− ζ(s−j,a+1)ζ(j +1,a+1)
s
(n+a) 2 2
n=1 j=1
X X
∞
1
+ζ(s,a+1)H + . (1.27)
a s
n(n+a)
n=1
X
By using partial fraction decomposition, we get
∞ s
1 H ζ(j,a+1)
= a − (s ∈N;a ∈ C\Z−). (1.28)
n(n+a)s as as+1−j 0
n=1 j=2
X X
∞
H
Therefore, we know that the parametric Euler sums n (2 ≤ s ∈ N) can be expressed
s
(n+a)
n=1
X
in term of parametric harmonic numbers and shifted harmonic numbers.
2 Identities for Euler related sums
We now prove the following theorems.
Theorem 2.1 For m ∈ N and a > 0, we have the recurrence relation
m−1
Y (a)
Y (a) =(m−1)! i H(m−i), Y (a) = 1, (2.1)
m i! a 0
i=0
X
where Y (a) is defined by the following integral
m
1
Y (a)
m := (−1)m xa−1lnm(1−x)dx. (2.2)
a
Z
0
6
Proof. Applying the definition of Beta function, we know that for a > 0,m ∈ N,
1
∂mB(a,b)
xa−1lnm(1−x)dx = lim
b→1 ∂bm
Z
0
m−1 m−1 ∂iB(a,b)
= ψ(m−i−1)(1)−ψ(m−i−1)(a+1) , (2.3)
i ∂bi
Xi=0 (cid:18) (cid:19) (cid:12)(cid:12)b=1(cid:16) (cid:17)
(cid:12)
where the Beta function is defined by (cid:12)
1
Γ(a)Γ(b)
B(a,b) := xa−1(1−x)b−1dx = , ℜ(a)> 0,ℜ(b) > 0.
Γ(a+b)
Z
0
(m)
By using the notation of H , then (2.3) can be written as
a
1 m−1 1
m−1
xa−1lnm(1−x)dx= (−1)m−i(m−i−1)! xa−1lni(1−x)dx H(m−i).
i   a
Z i=0 (cid:18) (cid:19) Z
0 X 0
  (2.4)
Then with the help of formula (2.2) we may easily deduce the result. (cid:3)
From the recurrence relation (2.1), we obtain the following identities
1
H
xa−1ln(1−x)dx = − a, (2.5)
a
Z
0
1
H2+H(2)
xa−1ln2(1−x)dx = a a , (2.6)
a
Z
0
1
H3+3H H(2)+2H(3)
xa−1ln3(1−x)dx = − a a a a , (2.7)
a
Z
0
2
1 H4+6H2H(2)+8H H(3)+3 H(2) +6H(4)
a a a a a a a
xa−1ln4(1−x)dx = . (2.8)
a (cid:16) (cid:17)
Z
0
Theorem 2.2 For a,b ∈ C\Z− with a 6= b. Then the following identity holds:
0
∞ H 1 H H H2−H2 ζ(2,a+1)−ζ(2,b+1)
n = a − b + a b + . (2.9)
(n+a)(n+b) b−a a b 2(b−a) 2(b−a)
n=1 (cid:18) (cid:19)
X
Proof. First we note that we may rewrite the series on the left hand side of (2.9) as
∞ H 1 ∞ xn xn n xn−j
n
= lim − . (2.10)
(n+a)(n+b) b−a x→1 n+a n+b  j 
Xn=1 Xn=1(cid:18) (cid:19) Xj=1 
 
 
7
On the other hand, from Lemma 1.3, we deduce that
∞ xn n−1xn−j 1 1
= Li (a,x2)Li (x)+ Li (a,x2)− (Li (a,x))2. (2.11)
1 1 2 1
n+a j 2 2
n=1 j=1
X X
Substituting (2.11) into (2.10), by simple calculation, the result is
∞ ∞ ∞
H 1 1 1 1
n
= + −
(n+a)(n+b) n(n+a)(n+b) 2(b−a) (n+a)2 (n+b)2
n=1 n=1 n=1(cid:26) (cid:27)
X X X
∞ ∞ ∞
1 1 1 1
+ a +b . (2.12)
2 (n+a)(n+b) n(n+a) n(n+b)
! !
n=1 n=1 n=1
X X X
Using the definitions of shifted harmonic number and Hurwitz Zeta function we obtain the
desired result. (cid:3)
Theorem 2.3 For m,k ∈ N and a ∈ C\Z−. Then
∞ m−1
1
+ (−1)j−1ζ(m+1−j)H(j) (a)
∞ Hn(m) 1 n=1nm(n+a) j=1 k−1 
= X X , (2.13)
 
Xn=1(n+a)(n+a+k) k +(−1)m−1H H(m)(a)+(−1)m−1k−1 Hi(1)(a) 
a k−1 (i+a)m
 Xi=1 
 
 
when a ∈C\Z−, we have  
0
1 ∞ m−1
1 ζ(m+1−l) H
xa−1Li (x)dx = = (−1)l−1 +(−1)m−1 a. (2.14)
m nm(n+a) al am
Z0 nX=1 Xl=1
Proof. By the definition of polylogarithm function and Cauchy product formula, we can verify
that
∞
Li (t)
m = H(m)tn, t ∈ (−1,1). (2.15)
1−t n
n=1
X
Multiplying (2.15) by ta−1−ta+k−1 and integrating over (0,1), we obtain
∞ (m) 1 k−1 1
H
k n = xa−1Li (x)dx+ xa+i−1Li (x)dx. (2.16)
m m
(n+a)(n+a+k)
n=1 Z i=1Z
X 0 X0
In Corollary 1.2, letting x → 1 and noting that
lim xn+aLi (x)−H (x,a) = H ,
1 1 a
x→1
(cid:8) (cid:9)
we arrive at the conclusion that
1 m−1 (−1)j−1 H H(1)(a)
xa+n−1Li (x)dx = ζ(m+1−j)+(−1)m−1 a +(−1)m−1 n .
m (n+a)j (n+a)m (n+a)m
Z j=1
0 X
(2.17)
Combining (2.14), (2.16) and (2.17), the result is (2.13). (cid:3)
From Theorem 2.3, we can get the following corollary.
8
Corollary 2.4 For integers m,k,r ∈ N and k 6=r, we have
∞ (m) m−1 k−1
H 1 H
n = ζ(m+1)+ (−1)j−1ζ(m+1−j)H(j) +(−1)m−1 i . (2.18)
n(n+k) k  k−1 im
Xn=1  Xj=1 Xi=1 
 m−1 
(−1)j−1ζ(m+1−j) H(j) −H(j)
∞ (m) k−1 r−1
(n+Hr)n(n+k) = k−1 r Xj=1 k−1 r−1(cid:16) (cid:17). (2.19)
 H H 
nX=1 +(−1)m−1 i − i 
im im
!
i=1 i=1
 X X 
 
 
Putting m = 1,2 in (2.13), we givethe following identities 
2
(1) (2)
∞ (1) H (a) +H (a)
Hn = 1 H H(m)(a−1)− Hk (a) + k k , (2.20)
(n+a)(n+a+k) k  a k k+a (cid:16) (cid:17)2 
nX=1  
 
∞ (2)  k−1 (1) 
H 1 H (a)
n = ζ(2)H(1)(a−1)−H H(2)(a−1)− i . (2.21)
(n+a)(n+a+k) k ( k a k (i+a)2)
n=1 i=1
X X
Theorem 2.5 For k ∈ N and a > 0, we have
(1) (2)
ζ(2)H (a−1)−H H (a−1)
∞ H2 1 k a k
n =  k−1 H(1)(a) k H2 +H(2) . (2.22)
(n+a)(n+a+k) k − i + a+j−1 a+j−1
nX=1  (i+a)2 a+j −1 
i=1 j=1
X X
 
 
Proof. From [17], we know that thegenerating function of Stirling numbers of the first kind is
given by
∞ xn
lnm+1(1−x) = (−1)m+1(m+1)! s(n,m+1) ,m ∈ N ,x ∈ [−1,1). (2.23)
0
n!
n=m+1
X
where s(n,k) denotes the (unsigned) Stirling number of the first kind (see [17]), and we have
(n−1)!
s(n,1) = (n−1)!,s(n,2) = (n−1)!H ,s(n,3) = H2 −H(2) ,
n−1 2 n−1 n−1
(n−1)! h i
s(n,4) = H3 −3H H(2) +2H(3) ,
6 n−1 n−1 n−1 n−1
(n−1)!h i
s(n,5) = H4 −6H(4) −6H2 H(2) +3(H(2) )2+8H H(3) .
24 n−1 n−1 n−1 n−1 n−1 n−1 n−1
h i
Hence, we obtain the following formula
∞ H(2)−H2
ln3(1−x)= 3 n nxn+1. (2.24)
n+1
n=1
X
9
Differentiating this equality above, we deduce that
ln2(1−x) ∞
= H2−H(2) xn. (2.25)
1−x n n
Xn=1(cid:16) (cid:17)
Multiplying (2.15) by xa−1−xa+k−1 and integrating over (0,1), yields
∞ H2−H(2) k 1
k n n = xa+j−2ln2(1−x)dx. (2.26)
(n+a)(n+a+k)
n=1 j=1Z
X X0
Combining (2.6) with (2.26), we get
∞ H2−H(2) 1 k H2 +H(2)
n n = a+j−1 a+j−1, (2.27)
(n+a)(n+a+k) k a+j −1
n=1 j=1
X X
which together with (2.21) gives (2.22) and finishes the proof of Theorem 2.5. (cid:3)
3 The main results
(a,b)
In this section, we give some closed form sums of W (m ,m ,··· ,m ,p) through parametric
k 1 2 r
harmonic numbers, shifted harmonic numbers and zeta values. First, we consider the expansion
k
1 A
= j (k ∈ N ;a ∈C\Z−) (3.1)
0 i
k n+a
j
(n+a ) j=1
i X
i=1
Q
where
k
n+a
A = lim j = (a −a )−1. (3.2)
j i j
n→−aj k
(n+a ) i=1,i6=j
i Y
i=1
Q
Taking a =a+i in (3.2) we obtain
i
r k
A = (−1)r+1 , (3.3)
r
k! r
(cid:18) (cid:19)
k
1 r k 1
= (−1)r+1 . (3.4)
k k! r n+a+r
(n+a+i) r=1 (cid:18) (cid:19)
X
i=1
Q
Furthermore, by using the equation (3.4) and the definition of binomial coefficient, we have the
following expansions
k
1 k 1
= (−1)r+1r (k ∈ N ;a ∈ C\Z−), (3.5)
n+k+a r n+a+r 0
r=1 (cid:18) (cid:19)
k X
(cid:18) (cid:19)
10