Identities for the Shifted Harmonic Numbers and Binomial Coefficients

Home , Harmonic number

Filomat 31:19 (2017), 6071–6086 Published by Faculty of Sciences and Mathematics, https://doi.org/10.2298/FIL1719071X University of Nis,ˇ Serbia Available at: http://www.pmf.ni.ac.rs/filomat

Identities for the Shifted Harmonic Numbers and Binomial Coeﬃcients

Ce Xua

aSchool of Mathematical Sciences, Xiamen University, Xiamen 361005, P.R. China

Abstract. We develop new closed form representations of sums of (n+α)th shifted harmonic numbers and reciprocal binomial coeﬃcients through αth shifted harmonic numbers and Riemann zeta function with positive integer arguments. Some interesting new consequences and illustrative examples are considered.

1. Introduction

Let N := 1, 2, 3,... and N− := 1, 2,... be the set of natural numbers and negative integers, { } {− − } respectively, and N0 := N 0 . In this paper we will develop identities, closed form representations of shifted harmonic numbers and∪ { } reciprocal binomial coeﬃcients of the form:

l (m ) l1 (mq) q X∞ 1 (l1, ,lq) (Hn+α ) (Hn+α) W ··· p; m1, , mq; α := ··· ! , (r, k N0, α < N−), (1.1) k,r ··· n + k + r ∈ n=1 np k

(m) for mi, li N (i = 1, 2, , q and q N), p 0, 1 with p + k 2, where Hα stands for the α-th generalized shifted harmonic∈ number··· deﬁned∈ by [24, 26]∈ { } ≥

m 1 (m) ( 1) − (m 1) (m 1) H := − ψ − (α + 1) ψ − (1) , 2 m N, (1.2) α (m 1)! − ≤ ∈ − (1) Hα = Hα := ψ (α + 1) + γ, (1.3) here ψ (z) is digamma function (or called Psi function ) deﬁned by d Γ (z) ψ (z) := (ln Γ (z)) = 0 dz Γ (z) and ψ (z) satisfy the following relations in the forms

X∞ 1 1 ψ (z) = γ + , z < N− := 0, 1, 2 ... , − n + 1 − n + z 0 { − − } n=0

2010 Mathematics Subject Classiﬁcation. Primary 05A10; 05A19; 11B65; 11B83; 11M06; Secondary 33B15; 33D60; 33C20. Keywords. Harmonic numbers; polylogarithm function; binomial coeﬃcients; integral representations; combinatorial series identities; summation formulas. Received: 05 April 2017; Accepted: 24 August 2017 Communicated by Hari M. Srivastava Email address: [email protected] (Ce Xu) C. Xu / Filomat 31:19 (2017), 6071–6086 6072

X∞ ψ(n) (z) = ( 1)n+1n! 1/(z + k)n+1, n N, − ∈ k=0 1 1 1 ψ (x + n) = + + + + ψ (x) , n = 1, 2, 3,.... x x + 1 ··· x + n 1 − From the deﬁnitions of Riemann zeta function and Hurwitz zeta function, we know that

ψ(n) (1) = ( 1)n+1n!ζ (n + 1) , ψ(n) (z) = ( 1)n+1n!ζ (n + 1, z) . − − The Riemann zeta function and Hurwitz zeta function are deﬁned by

X∞ 1 ζ(s):= , (s) > 1, (1.4) ns < n=1 and

X∞ 1 ζ (s, α + 1) := , (s) > 1, α < N− . (1.5) (n + α)s n=1 <

Therefore, in the case of non integer values we may write the generalized shifted harmonic numbers in terms of zeta functions

(m) H = ζ (m) ζ (m, α + 1) , α < N−, 2 m N, (1.6) α − ≤ ∈ and for m = 1,

X∞ 1 1 H H(1) = . (1.7) α ≡ α k − k + α k=1

R∞ t z 1 Here Γ (z) := e− t − dt, (z) > 0 is called gamma function, and γ denotes the Euler-Mascheroni constant 0 < deﬁned by

 n  X 1  γ := lim  ln n = ψ (1) 0.577215664901532860606512.... n  k −  − ≈ →∞ k=1

(n) p The evaluation of the polygamma function ψ ( q ) at rational values of the argument can be explicitly done via a formula as given by Kölbig[15], or Choi and Cvijović[11] in terms of the Polylogarithmic or other special functions. Some specific values are listed in the books [1, 20, 28]. For example, George E. Andrews, Richard Askey and Ranjan Roy [1], or H.M. Srivastava and J. Choi [28] gave the following Gauss’s formula

! [(q 1)/2] ! ! p X− 2kpπ kπ π pπ ψ =2 cos ln 2 sin + rq p γ cot ln q, 0 < p < q; p, q N , (1.8) q q q − − 2 q − ∈ k=1

p where [x] denotes the greatest integer x, and if q is even, rq p = ( 1) ln 2; if q is odd, rq p = 0. For details and historical introductions, please≤ see [1, 3, 4, 11, 15, 20, 28] and− references therein. From (1.2), (1.3) and (1.8), we can obtain some speciﬁc values of shifted harmonic numbers:

8 40 46 H = 2 2 ln 2, H = 2 ln 2, H(2) = 4 2ζ (2) , H(2) = 2ζ (2) , H = 2 ln 2. 1/2 − 3/2 3 − 1/2 − 3/2 9 − 5/2 15 − C. Xu / Filomat 31:19 (2017), 6071–6086 6073

Letting α approach n (n is a positive integer) in (1.6) and (1.7), then the shifted harmonic numbers are reducible to classical harmonic numbers deﬁned by

n n (1) X 1 (m) X 1 Hn H := , H := , n, m N. (1.9) ≡ n j n jm ∈ j=1 j=1 The sum deﬁned in (1.1) is also called Euler type sum, which can be seen as an extension of classical Euler sums. The Euler sums are deﬁned by

(s ) (s ) (s ) X∞ H 1 H 2 H r S := n n ··· n . S,q nq n=1

Here S := (s , s ,..., sr)(r, si N, i = 1, 2,..., r) with s s ... sr and q 2. The quantity 1 2 ∈ 1 ≤ 2 ≤ ≤ ≥ w := s1 + + sr + q is called the weight and the quantity r is called the degree of the sum. Some related results for··· Euler sums may be seen in the works of [10, 12, 14, 16–18, 29–31, 33–35]. For details and historical introductions of Euler sums, please see [2, 5–9, 13]. While there are many results for sums of harmonic numbers (or shifted harmonic numbers) with positive terms, for example we know that [22, 25] ! X∞ H2 k 2 n H(2) k N ! = ζ (2) k 1 + 2 , 2 , n + k k 1 − − (k 1) ≤ ∈ n=1 − k −

(2) X∞ Hn+1/4 140 128 64 69 ! = 152 G ln 2 π ζ (2) , n + 4 − 3 − 3 − 9 − 2 n=1 n 4 where G is Catalans constant, deﬁned by

n 1 X∞ ( 1) − G = − 0.915965.... (2n 1)2 ≈ n=1 − Further work in the summation of harmonic numbers and binomial coefficients has also been done by Sofo [20, 21, 23, 27] and Xu et al. [19, 32]. The motivation for this paper arises from results of Sofo’s papers [26] and [24] with Srivastava. Sofo and Srivastava studied the summation of the product shifted harmonic sums of order p (p = 1, 2) and reciprocal binomial coefficients, of the form, (p) X∞ Hn 1/q − ! n + k n=1 n k and its finite counterpart. In this paper, we develop an approach to evaluate some sums of (1.1). The approach is based on integrals computations. By the approach we prove that the five sums

(1) (2) (3) Wk,r p; m; α , Wk,r p; 1; α , Wk,r p; 1; α and (1,1) (2,1) Wk,r p; 1, 2; α , Wk,r p; 1, 2; α can be expressed as a rational linear combination of products of zeta values and shifted harmonic numbers. Next, we give two lemmas. The following lemmas will be useful in the development of the main theorems. C. Xu / Filomat 31:19 (2017), 6071–6086 6074

Lemma 1.1. For integers k, r 0 and α < N−, then we have ≥ X∞ f (n, α) k α X∞ f (n, α) α r X∞ f (n, α) = − + − , (1.10) (n + r)(n + k) k r (n + k)(n + α) k r (n + r)(n + α) n=1 − n=1 − n=1 where the function f (n, α) satisfy the following relation

lim nβ f (n, α) = c, β > 1, n →∞ − here c is a constant.

Proof. This lemma is almost obvious.

Lemma 1.2. For integer p > 0 and α , 0, 1, 2,..., we have − −

1 p 1 Z X− i 1 α 1 ( 1) − p Hα x − Lip (x) dx = − ζ p + 1 i ( 1) , (1.11) αi − − − αp i 0 =1 where Lip (x) is polylogarithm function deﬁned for x 1 by | | ≤

X∞ xn Lip (x) := , (p) > 1. (1.12) np < n=1

Proof. It is obvious that we can rewrite the integral on the left hand side of (1.11) as

Z1 X∞ α 1 1 x − Li (x) dx = . (1.13) p np (n + α) n 0 =1

By using the partial fraction decomposition

p 1 i 1 1 X− ( 1) − 1 1 1 1 = − ( 1)p , (1.14) np (n + α) αi · np+1 i − − αp n − n + α i=1 − and combining (1.7) with (1.13), we deduce the desired result. This completes the proof of Lemma 1.2.

2. Shifted Harmonic Number Identities

In this section, we will establish some explicit relationships that involve shifted harmonic numbers and the following type sums

(m) X∞ Hn+α , r, k N , r , k, α < N−, m N . (n + r)(n + k) ∈ 0 ∈ n=1

We now prove the following theorems.

Theorem 2.1. For integers m, k 1 and (α) > 0, then we have the following recurrence relation ≥ < m 1 ! m i X− m 1 ( 1) − I (α, m, k) = − (m i 1)! − I (α, i, k) i − − αm i i=0 − C. Xu / Filomat 31:19 (2017), 6071–6086 6075

Xk 1 ! − k m+k j (m+k j) + ( 1) − m + k j 1 !H − I α, 0, j j − − − α j=0 Xk 1 ! − k m+k j ( 1) − m + k j 1 !ζ m + k j I α, 0, j − j − − − − j=0 Xm 1 Xk 1 ! ! − − m 1 k m+k i j (m+k i j) + − ( 1) − − m + k i j 1 !H − − I α, i, j i j − − − − α i=1 j=0 Xm 1 Xk 1 ! ! − − m 1 k m+k i j − ( 1) − − m + k i j 1 !ζ m + k i j I α, i, j . (2.1) − i j − − − − − − i=1 j=0 where I (α, m, k) is deﬁned by the integral

Z1 α 1 m k I (α, m, k) := x − ln xln (1 x)dx. (2.2) − 0 with 1 1 I (α, 0, 0) = , I (α, i, 0) = ( 1)ii! . α − αi+1 Proof. Applying the deﬁnition of Beta function B α, β , we can ﬁnd that

1 Z m+k ∂ B α, β I ( m k) xα 1 mx k ( x)dx α, , := − ln ln 1 = m k , (2.3) − ∂α ∂β β=1 0 where the Beta function is deﬁned by

Z1 α 1 β 1 Γ (α) Γ β B α, β := x − (1 x) − dx = , (α) > 0, β > 0. (2.4) − Γ α + β < < 0 By using (2.4) and the deﬁnition of ψ(x), it is obvious that ∂B α, β = B α, β ψ (α) ψ α + β . ∂α − Therefore, diﬀerentiating m 1 times this equality, we can deduce that − m m 1 ! i X− ∂ B α, β m 1 ∂ B α, β h (m i 1) (m i 1) i = − ψ − − (α) ψ − − α + β . (2.5) ∂αm i ∂αi · − i=0 Since B(α, β) = B(β, α), then we also have

m m 1 ! i X− ∂ B α, β m 1 ∂ B α, β h (m i 1) (m i 1) i = − ψ − − β ψ − − β + α . (2.6) ∂βm i ∂βi · − i=0 Putting β = 1 in (2.6) and combining (1.2), (2.3), we arrive at the conclusion that

m 1 ! X− m i m 1 (m i) I (α, 0, m) = ( 1) − (m i 1)! − I (α, 0, i)H − . (2.7) − − − i α i=0 C. Xu / Filomat 31:19 (2017), 6071–6086 6076

Furthermore, by using (2.5), the following identity is easily derived ! ∂m+kB α, β ∂k ∂mB α, β = ∂αm∂βk ∂βk ∂αm m 1 ! i k X− + m 1 ∂ B α, β h (m i 1) (m i 1) i = − ψ − − (α) ψ − − α + β i ∂αi∂βk · − i=0 k 1 ! j X− B k ∂ α, β (m+k j 1) ψ − − α + β − j ∂βj j=0

m 1 k 1 ! ! i j X− X− + B m 1 k ∂ α, β (m+k i j 1) − ψ − − − α + β . (2.8) − i j ∂αi∂βj i=1 j=0

From (1.2) and (1.5), we know that if β = 1, then we have

(m i 1) (m i 1) m i 1 − m i ψ − − (α) ψ − − (α + 1) = ( 1) ( 1)! m i , (2.9) − − − − α − (m+k j 1) m+k j (m+k j) ψ − − (α + 1) = ( 1) − m + k j 1 ! ζ m + k j H − . (2.10) − − − − − α Hence, taking β = 1 in (2.8), then substituting (2.9) and (2.10) into (2.8) respectively, we can obtain (2.1). The proof of Theorem 2.1 is ﬁnished. From (2.1) and (2.7), we can get the following identities: for α > 0,

Z1 α 1 Hα I (α, 0, 1) = x − ln (1 x) dx = , (2.11) − − α 0 1 Z 2 (2) α 1 2 Hα + Hα I (α, 0, 2) = x − ln (1 x) dx = , (2.12) − α 0 1 Z (2) α 1 Hα ζ (2) Hα I (α, 1, 1) = x − ln x ln (1 x) dx = − , (2.13) − α2 − α 0 1 Z 3 (2) (3) α 1 3 Hα + 3HαHα + 2Hα I (α, 0, 3) = x − ln (1 x) dx = , (2.14) − − α 0 1 Z 2 (2) (3) (2) α 1 2 Hα + Hα ζ (3) Hα ζ (2) Hα I (α, 1, 2) = x − ln xln (1 x) dx = + 2 − + 2 − H . (2.15) − − α2 α α α 0

Theorem 2.2. For integers r, k N0 (r , k) and α > max k, r , we have ∈ { }    2 (2) k  X∞ H + H X H  Hn+α k α  α k α k α+j k  = −  − − −  (n + r)(n + k) k r  α k − j α + j k  n=1 −  − j=1 −     2 (2) Xr H  α r Hα r + Hα r α+j r  + −  − − −  . (2.16) k r  α r − j α + j r  −  − j=1 −  C. Xu / Filomat 31:19 (2017), 6071–6086 6077

1 Proof. Replacing α by n + α in (2.11), then multiplying it by (n + k)− and summing with respect to n, we conclude that Z1 X∞ X∞ Hn+α 1 n+α 1 = x − ln (1 x)dx (n + k)(n + α) − n + k − n n =1 =1 0 1 1 Xk Z Z 1 α+j k 1 α k 1 2 = x − − ln (1 x) dx + x − − ln (1 x) dx. (2.17) j − − j =1 0 0 The relations (2.11), (2.12) and (2.17) yield the following result

2 (2) k X∞ Hn+α Hα k + Hα k X Hα+j k = − − − (α > k) . (2.18) (n + k)(n + α) α k − j α + j k n=1 − j=1 −

Putting f (n, α) = Hn+α in (1.10), and combining (2.18), we may easily deduce the desired result. Corollary 2.3. For integers r, k, m N with r , k and real α > max k, r , we have ∈ { }      2 (2) k   (2) r  X∞ H + H X H   2 X H  Hn+α m k α  α k α k α+j k  α r Hα r + Hα r α+j r  − = −  − − −  + −  − − −  (n + r)(n + k) k r  α k − j α + j k  k r  α r − j α + j r  n=1 −  − j=1 −  −  − j=1 −  m ( ) 1 X Hj+α m Hr Hj+α m Hk − − − − , (2.19) − k r j + α m r − j + α m k − j=1 − − − − where α , m + r j and m + k j j = 1, 2, , m with α m < N−. − − ··· − Proof. By the deﬁnition of Hn+α, we can ﬁnd the following relation

m X 1 Hn+α Hn+α m = . − − j + n + α m j=1 − Hence, using the above identity, we can rewrite the series on the left hand side of (2.19) as

m X∞ Hn+α m X∞ Hn+α X X∞ 1 − = . (2.20) (n + r)(n + k) (n + r)(n + k) − n + j + α m (n + r)(n + k) n=1 n=1 j=1 n=1 − By a direct calculation, we obtain

X∞ 1 n + j + α m (n + r)(n + k) n=1 −     1 X∞ 1 X∞ 1  =   k r  n + j + α m (n + r) − n + j + α m (n + k) − n=1 − n=1 − Hj+α m Hr Hj+α m Hk = − − − − . (2.21) (k r) j + α m r − (k r) j + α m k − − − − − − Substituting (2.16) and (2.21) into (2.20), we can prove (2.19). The proof of Corollary 2.3 is thus completed. From (2.16) and (2.19), we have the following special cases:

X∞ Hn 1/2 2 1 X∞ Hn+1/2 − = + ln 2, = 3 ln 2 1, (n + 1)(n + 2) 3 3 (n + 1)(n + 2) − n=1 n=1 C. Xu / Filomat 31:19 (2017), 6071–6086 6078

X∞ Hn+3/2 14 X∞ Hn+5/2 131 7 = 5 ln 2, = ln 2. (n + 1)(n + 2) 3 − (n + 1)(n + 2) 45 − 3 n=1 n=1

Next, we evaluate the summation of the shifted harmonic sums of order two of the form,

(2) X∞ Hn+α , α > max k, r , k, r N0 , k , r. (n + r)(n + k) { } ∈ n=1

(2) First, we need to obtain the integral representation of Hα . Setting p = 2 in (1.11) gives

Z1 α 1 ζ (2) Hα x − Li (x) dx = . (2.22) 2 α − α2 0

With the help of (2.13), we get

1 1 (2) Z Z Hα α 1 α 1 = x − ln x ln (1 x)dx + x − Li (x) dx, α > 0. (2.23) α − 2 0 0

1 Replacing α by n + α in (2.23), then multiplying it by (n + k)− and summing with respect to n, the result is

1 1 (2) Z Z X∞ X∞ X∞ Hn+α 1 n+α 1 1 n+α 1 = x − ln x ln (1 x)dx + x − Li (x) dx (n + k)(n + α) n + k − n + k 2 n n n =1 =1 0 =1 0 1 1 Z Xk Z α k 1 2 1 α+j k 1 = x − − ln xln (1 x) dx x − − ln x ln (1 x)dx − − − j − j 0 =1 0 1 1 Z Xk Z α k 1 1 α+j k 1 x − − ln (1 x) Li (x) dx x − − Li (x)dx. (2.24) − − 2 − j 2 j 0 =1 0

We note that by using (2.11), the following identities ar easily derived

Z1 X∞ α 1 Hn+α x − ln (1 x) Li (x) dx = − 2 − n2 (n + α) n 0 =1 1 X∞ H 1 X∞ H = n+α n+α . (2.25) α n (n + α) − α n2 n=1 n=1

On the other hand, from (2.18), setting k = 0, we have

2 (2) X∞ Hn+α Hα + Hα = , α < N− 0 . (2.26) n (n + α) α ∪ { } n=1

Hence, combining (2.13), (2.15), (2.22) and (2.24)-(2.26), we obtain

(2) (3) X∞ Hn+α 1 X∞ Hn+α k ζ (3) Hα k = − 2 − − (n + k)(n + α) α k n2 − α k n=1 − n=1 − C. Xu / Filomat 31:19 (2017), 6071–6086 6079

(2) k H(2) ζ (2) Hα k X α+j k 2 − − Hα k − , (α > k). (2.27) − α k − − j α + j k − j=1 − (2) Letting f (n, α) = Hn+α in (1.10), then substituting (2.27) into (1.10), we get the following Theorem. Theorem 2.4. For integers r, k 0 and real α > k > r, then we have ≥  r H(2) k H(2)   P α+j r P α+j k   (r α) − (k α) −   j(α+j r) j(α+j k)  (2)  − j=1 − − − j=1 −  X∞   Hn+α 1  k r  = P Hα+j k (3) (2) . (2.28)  − −  (n + r)(n + k) k r  2 + 2Hα r + Hα kζ (2) + 2Hα rHα r  n=1 −  − j=1 (α+j k) − − − −   −   (3) (2)  2Hα k Hα rζ (2) 2Hα kHα k − − − − − − − Proof. From (1.10) and (2.27), we arrive at the conclusion that

(2) r H(2) k H(2) X∞ Hn+α 1 X∞ Hn+α r Hn+α k r α X α+j r k α X α+j k = − − − + − − − − (n + r)(n + k) k r n2 k r j α + j r − k r j α + j k n=1 − n=1 − j=1 − − j=1 − ( (3) (2) ) 2 Hα r + Hα kζ (2) + Hα rHα r + − − . (2.29) k r −(3) −(2) Hα k Hα rζ (2) Hα kHα k − − − − − − − − By the deﬁnition of shifted harmonic number, and using (1.14), we conclude that

k r (m) (m) X− 1 Hα r Hα k = m , − − − j + α k j=1 − and for 2 p N, 0 r < k N, ≤ ∈ ≤ ∈ k r X∞ Hn+α r Hn+α k X− X∞ 1 − − − = np np n + j + α k n=1 j=1 n=1 − k r p 1 k r X X− i 1 X − ( 1) − p 1 − Hj+α k = ζ p + 1 i + ( 1) − − − i p j + α k − − j + α k j=1 i=1 − j=1 − p 1 k r X− X i 1 (i) (i) p 1 − Hj+α k − − − = ( 1) ζ p + 1 i Hα r Hα k + ( 1) p . (2.30) − − − − − − j + α k i=1 j=1 − Taking p = 2 in (2.30) and substituting it into (2.29), by a simple calculation, we obtain the desired result. Corollary 2.5. For integers r, k, m N and α > k > r, then ∈  r H(2) k H(2)   P α+j r P α+j k   (r α) − (k α) −   j(α+j r) j(α+j k)  (2)  − j=1 − − − j=1 −  X∞   Hn+α m 1  k r  − = P Hα+j k (3) (2)  − −  (n + r)(n + k) k r  2 + 2Hα r + Hα kζ (2) + 2Hα rHα r  n=1 −  − j=1 (α+j k) − − − −   −   (3) (2)  2H Hα rζ (2) 2Hα kH − α k − − − − α k  − ( ) −  m m ζ(2) H 2  P Hα+j m Hr P α+j m   − − − −   2   (α+j m r) α+j m r  1  j=1 − − − j=1 − −  (2) . (2.31)  m m ζ(2) H  − k r  P Hα+j m Hk P α+j m   − − − −  −  2 +   (α+j m k) α+j m k   − j=1 − − j=1 − −  where α , m + r j, m + k j j = 1, 2, , m and α m < N−. − − ··· − C. Xu / Filomat 31:19 (2017), 6071–6086 6080

Proof. By a similar argument as in the proof of Corollary 2.3, we have

(2) (2) m X∞ Hn+α m X∞ Hn+α X X∞ 1 − = . (2.32) 2 (n + r)(n + k) (n + r)(n + k) − n + j + α m (n + k)(n + r) n=1 n=1 j=1 n=1 − On the other hand, we easily obtain the result

X∞ 1 2 n=1 n + j + α m (n + k)(n + r)  −  1 X∞  1 1  =    2 2  k r  n + j + α m (n + r) − n + j + α m (n + k) − n=1 − −    1 P∞ 1 1   2 n+r n+j+α m   (j+α m r) n=1 − −   − −   1 P∞ 1   j+α m r 2  1  − − − n=1 (n+j+α m)  =  −  k r  1 P∞ 1 1  −  2 n+k n+j+α m   −(j+α m k) n=1 − −   − −   1 P∞ 1   + 2   j+α m k (n+j+α m)  − − n=1 −  ( )  m m ζ(2) H 2  P Hα+j m Hr P α+j m   − − − −   2   (α+j m r) α+j m r  1  j=1 − − − j=1 − −  = (2) . (2.33)  m m ζ(2) H  k r  P Hα+j m Hk P α+j m   − − − −  −  2 +   (α+j m k) α+j m k   − j=1 − − j=1 − −  Substituting (2.28) and (2.33) into (2.32) respectively, we deduce (2.31). This completes the proof of Corollary 2.5. Theorem 2.6. For integers r, k 0 and α > k > r, then we have ≥  (2)  X 2 (2) H3 + 3H H(2) + 2H(3) Xk H2 + H  ∞ Hn+α + Hn+α k α  α k α k α k α k α+j k α+j k  = −  − − − − − −  (n + r)(n + k) k r  α k − j α + j k  n=1 −  − j=1 −   (2)   3 (2) (3) Xr H2 + H  α r Hα r + 3Hα rHα r + 2Hα r α+j r α+j r  + −  − − − − − −  . (2.34) k r  α r − j α + j r  −  − j=1 −  Proof. From (2.12) and (2.14), we know that

1 (2) Z X∞ 2 X∞ Hn+α + Hn+α 1 n+α 1 2 = x − ln (1 x)dx (n + k)(n + α) n + k − n n =1 =1 0 1 1 Z Xk Z α k 1 3 1 α+j k 1 2 = x − − ln (1 x)dx x − − ln (1 x)dx − − − j − j 0 =1 0 3 (2) (3) k H2 + H(2) Hα k + 3Hα kHα k + 2Hα k X α+j k α+j k = − − − − − − . (2.35) α k − j α + j k − j=1 −

2 (2) Setting f (n, α) = Hn+α + Hn+α in (1.10) and combining (2.35), the result is (2.34). Similarly to the proof of Theorem 2.6, we have the following similar result. C. Xu / Filomat 31:19 (2017), 6071–6086 6081

Theorem 2.7. For integers r, k 0 and α > k > r, then the following identity holds: ≥ 3 (2) (3) X∞ Hn+α + 3Hn+αHn+α + 2Hn+α (n + r)(n + k) n=1 2  4 2 (2) (3) (2) (4)  H + 6H H + 8Hα kH + 3 H + 6H   α k α k α k − α k α k α k   − − − − − −  k α  α k   −  = −  k 3 (2) (3)  k r  X H + 3Hα+j kH + 2H  −  α+j k − α+j k α+j k   − − −  − j α + j k  j=1 − 2  4 2 (2) (3) (2) (4)  Hα r + 6Hα rHα r + 8Hα rHα r + 3 Hα r + 6Hα r   − − − − − − −    α r  α r   −  + −  r 3 (2) (3)  . (2.36) k r  X H + 3Hα+j rH + 2H  −  α+j r − α+j r α+j r   − − −  − j α + j r  j=1 − Proof. Applying the same arguments as in the proof of Theorem 2.6, we may easily deduce

1 1 (2) (3) Z k Z X∞ 3 X Hn+α + 3Hn+αHn+α + 2Hn+α α k 1 4 1 α+j k 1 3 = x − − ln (1 x)dx + x − − ln (1 x)dx. (2.37) (n + α)(n + k) − j − n j =1 0 =1 0 Setting m = 4 in (2.7) and combining (1.10), (2.11), (2.12) with (2.14) we obtain

2 1 4 2 (2) (3) (2) (4) Z H + 6H H + 8HαH + 3 H + 6H α 1 4 α α α α α α x − ln (1 x)dx = . (2.38) − α 0 Therefore, by using (2.14), (2.37) and (2.38) yields the desired result. This completes the proof Theorem 2.7. Substituting (2.28) into (2.34), we can obtain the following Corollary. Corollary 2.8. For integers r, k 0 and α > k > r, then we have the quadratic sums ≥  3 (2)   Hα r + Hα rHα r + Hα rζ (2)   − −   −3 −(2)   H Hα kH Hα kζ (2)   − α k − − α k − −  X 2  − r H2 − k r  ∞ Hn+α 1  P α+j r P Hα+j k   − − −  = + (r α) + 2 . (2.39)  j α+j r  (n + r)(n + k) k r  − j=1 ( ) j=1 (α+j k)  n=1 −  − −   k H2   P α+j k   (k α) −   j(α+j k)   − − j=1 −  Further, using the deﬁnition of shifted harmonic number, by a simple calculation, the following relations are easily derived ∂ H(m) = m ζ (m + 1) H(m+1) , ∂α n+α − n+α m ∂ m+1 (m+1) (Hn ) = ( 1) m! ζ (m + 1) H , ∂αm +α − − n+α ∂ 2 (2) H = 2ζ (2) Hn 2Hn H , ∂α n+α +α − +α n+α ∂ H3 = 3ζ (2) H2 3H2 H(2) . ∂α n+α n+α − n+α n+α Hence, from Theorems 2.2, 2.4, 2.7 and Corollary 2.8 with the above relations, we can get the Theorem 2.9. C. Xu / Filomat 31:19 (2017), 6071–6086 6082

Theorem 2.9. For positive integers k, r, m and real α > max k, r with k , r. Then the linear, quadratic and cubic sums { }

(m) (2) 3 2 (2) X∞ H X∞ Hn+αH X∞ H X∞ H H n+α , n+α , n+α , n+α n+α (n + r)(n + k) (n + r)(n + k) (n + r)(n + k) (n + r)(n + k) n=1 n=1 n=1 n=1 can be expressed in terms of shifted harmonic numbers and zeta values. As simple example is as follows: Corollary 2.10. For integers r, k 0 and α > k > r, then we have ≥  r H(3) k H(3)   P α+j r P α+j k   (r α) − (k α) −   j(α+j r) j(α+j k)   − j=1 − − − j=1 −   (2)  (3)  k r H k r   1 P− α+j k P− Hα+j k (4) (4)  X∞ Hn+α 1  − −   2 2 3 + 3 Hα r Hα k  =  − (α+j k) − (α+j k) − −  . (n + r)(n + k) k r  j=1 − j=1 − −  n=1 −  (2) (2)   + H Hα r ζ (2) + (Hα k Hα r) ζ (3)   α k − −   − −2 − 2 −   (2) (2) (3) (3)   + Hα r Hα k + 2 Hα rHα r Hα kHα k  − − − − − − − −

3. Some Results on Shifted Harmonic Number Sums

(l1, ,lq) In this section, we give some closed form of sums W ··· p; m , , mq; α through shifted harmonic k,r 1 ··· numbers and zeta values. First, we consider the partial fraction decomposition

m 1 X Aj m = , (3.1) Q n + aj (n + ai) j=1 i=1 where n n + aj Y 1 − Aj = lim m = ai aj . n aj Q − →− (n + ai) i=1,i,j i=1

Letting m = k and ai = r + i in (3.1), we have

k ! 1 k! X j+1 k 1 ! = = ( 1) j , k, r N0 . (3.2) n + k + r Qk − j n + r + j ∈ (n + r + i) j=1 k i=1 On the other hand, we can ﬁnd that 1 k ! = ! n + k + r n + k + r (n + r + 1) k k 1 − k 1 ! k X− k 1 1 = ( 1)j+1 j − (n + r + 1) − j n + r + 1 + j j=1 k 1 ! X− j+1 k 1 1 =k ( 1) j − , r N0, k N. (3.3) − j (n + r + 1) n + r + 1 + j ∈ ∈ j=1 C. Xu / Filomat 31:19 (2017), 6071–6086 6083

From identities (2.16), (2.28), (2.39), (3.2) and (3.3), we obtain the following new results: for α > k + r, 2 k N, r N , ≤ ∈ ∈ 0

(1) X∞ Hn+α Wk,r (0; 1; α) = ! n=1 n + k + r k k 1 ! X− j+1 k 1 X∞ Hn+α = k ( 1) j − − j (n + r + 1) n + r + 1 + j j=1 n=1  2 (2) 2 (2)   Hα r 1 + Hα r 1 Hα r 1 j Hα r 1 j   − − − − − − − − − − − −  k 1 !  r+1+j  X−  P Hα+i r 1 j  j+1 k 1  r + 1 + j α − − −  = k ( 1) −  i(α+i r 1 j) . (3.4) − j  − − i=1 − − −  j=1  r+1   P Hα+i r 1   + (r + 1 α) i(α+i −r− 1)  − i=1 − − (2) (1) X∞ Hn+α Wk,r (0; 2; α) = ! n=1 n + k + r k

k 1 ! (2) X− j+1 k 1 X∞ Hn+α = k ( 1) j − − j (n + r + 1) n + r + 1 + j j=1 n=1

 r+1 H(2)   P α+i r 1   (r + 1 α) − −   − i(α+i r 1)   i=1 − −   r+1+j H(2)   P α+i r 1 j   r + 1 + j α − − −  Xk 1 !  i(α+i r 1 j)  − j+1 k 1  − − i=1 − − −  = k ( 1) −  j , (3.5) j  P Hα+i r 1 j  −  − − −  j=1  2   (α+i r 1 j)   − i=1 − − −   (3) (2)   +2H + Hα r 1 jζ (2) + 2Hα r 1H   α r 1 − − − − − α r 1   (3−) − −(2−)  2Hα r 1 j Hα r 1ζ (2) 2Hα r 1 jHα r 1 j − − − − − − − − − − − − − − 2 (2) X∞ Hn+α Wk,r (0; 1; α) = ! n=1 n + k + r k

k 1 ! 2 X− j+1 k 1 X∞ Hn+α = k ( 1) j − − j (n + r + 1) n + r + 1 + j j=1 n=1  3 (2)   H + Hα r 1H + Hα r 1ζ (2)   α r 1 − − α r 1 − −   −3 − − − (2)   Hα r 1 j Hα r 1 jHα r 1 j Hα r 1 jζ (2)   − − − − − − − − − − − − − − −   r+1 H2   P α+i r 1  k 1 !  + (r + 1 α) − −  X− k 1  i(α+i r 1)  = k ( 1)j+1 −  − i=1 − −  . (3.6) j  j  −  P Hα+i r 1 j  j=1  − − −   + 2   (α+i r 1 j)   i=1 − − −   r+1+j H2   P α+i r 1 j   r + 1 + j α − − −   i(α+i r 1 j)  − − i=1 − − − C. Xu / Filomat 31:19 (2017), 6071–6086 6084 and for k N and α > k, ∈

(1) X∞ Hn+α W (1; 1; α) = ! k,0 n + k n=1 n k k ! X j+1 k X∞ Hn+α = ( 1) j − j n n + j j=1 n=1  2 (2) 2 (2)  k !  H + Hα H H  X k  α − α j − α j  = ( 1)j+1  j − −  , (3.7)  P Hα+i j  − j  j α −  j=1  i(α+i j)  − − i=1 − (2) (1) X∞ Hn+α W (1; 2; α) = ! k,0 n + k n=1 n k k ! (2) X j+1 k X∞ Hn+α = ( 1) j − j n n + j j=1 n=1  (3) (2)   2Hα + Hα jζ (2) + 2HαHα   −   (3) (2)   2Hα j Hαζ (2) 2Hα jHα j  k !  − − − − − −  X k  j H(2)  j+1  P α+i j  = ( 1)  j α − , (3.8) − j  i(α+i j)  j=1  − − i=1 −   j   P Hα+i j   −   2   (α+i j)  − i=1 − 2 (2) X∞ Hn+α W (1; 1; α) = ! k,0 n + k n=1 n k

k ! 2 X j+1 k X∞ Hn+α = ( 1) j − j n n + j j=1 n=1    H3 H H(2) H ( )   α + α α + αζ 2   3 (2)   H Hα jH Hα jζ (2)  k !  − α j − − α j − −  X k  j − −  j+1  P Hα+i j  = ( 1) − . (3.9)  + 2  − j  (α+i j)  j=1  i=1 −   j H2   P α+i j   j α −   i(α+i j)  − − i=1 − Hence, from Corollary 2.8, Theorem 2.9 and formulas (3.2), (3.3), we obtain the following description of (l1, ,lq) W ··· p; m , , mq; α . k,r 1 ··· Theorem 3.1. For positive integers k, r, m and real α (α > k > r) with p = 0, 1, then the sums

(1) (2) (3) Wk,r p; m; α , Wk,r p; 1; α , Wk,r p; 1; α and

(1,1) (2,1) Wk,r p; 1, 2; α , Wk,r p; 1, 2; α can be expressed in terms of shifted harmonic numbers and ordinary zeta values. C. Xu / Filomat 31:19 (2017), 6071–6086 6085

At the end of this section we give a explicit formula of sums associated with shifted harmonic numbers. We deﬁne the parametric polylogarithm function by the series

X∞ xn Lip,α (x) := , x ( 1, 1) , (p) > 1, α < N−. (n + α)p ∈ − < n=1

Next, we consider the following integral

Z1 r 1 x − Lip (x) Lim (x)dx, p, m N, α, β, r < N−. ,α ,β ∈ 0

First, using integration by parts, the following identity is easily derived

1 p 1 Z X− i 1 r 1 ( 1) − p 1 Hr Hα x − Lip,α (x)dx = − ζ p + 1 i, α + 1 + ( 1) − − . (3.10) i − − (r )p i (r α) α 0 =1 − − We note that

Z1 Z1 X∞ r 1 1 n+r 1 x − Lip,α (x) Lim,β (x)dx = x − Lim,β (x)dx (n + )p n α 0 =1 0 Z1 X∞ 1 n+r 1 = m x − Lip,α (x)dx. (3.11) n n + β =1 0

Substituting (3.11) into (3.10), we can deduce that

m 1 X∞ Hn+r p 1 X∞ Hn+r ( 1) − p m ( 1) − m p − (n + α) n + r β − − n + β (n + r α) n=1 − n=1 − m 1 X∞ 1 p 1 X∞ 1 = ( 1) − Hβ p m ( 1) − Hα m p − (n + α) n + r β − − n + β (n + r α) n=1 − n=1 − p 1 X− X i 1 ∞ 1 ( 1) − p 1 i 1 + ζ + , α + m i − − n + β (n + r α) i=1 n=1 − Xm 1 X − i 1 ∞ 1 ( 1) − ζ m + 1 i, β + 1 . (3.12) p i − − − (n + α) n + r β i=1 n=1 − Putting r = 2α, α = β in (3.12), we have that

n m 1 p 1o X∞ Hn+2α ( 1) − ( 1) − − − − p+m n=1 (n + α) p 1 X− n m 1 p 1o i 1 = ( 1) − ( 1) − H ζ p + m, α + 1 + ( 1) − ζ p + 1 i, α + 1 ζ (m + i, α + 1) − − − α − − i=1 Xm 1 − i 1 ( 1) − ζ m + 1 i, β + 1 ζ p + i, α + 1 . (3.13) − − − i=1 C. Xu / Filomat 31:19 (2017), 6071–6086 6086

From (3.13), we can get some speciﬁc cases

X∞ H 1 n+2α = H ζ (3, α + 1) + ζ2 (2, α + 1) , 3 α 2 n=1 (n + α) X∞ H 1 n+2α = H ζ (5, α + 1) + ζ (2, α + 1) ζ (4, α + 1) ζ2 (3, α + 1) , 5 α − 2 n=1 (n + α) X∞ H 1 n+2α = H ζ (7, α + 1) + ζ (2, α + 1) ζ (6, α + 1) + ζ2 (4, α + 1) ζ (3, α + 1) ζ (5, α + 1) . 7 α 2 − n=1 (n + α)

Acknowledgments. The authors would like to thank the anonymous referee for his/her helpful com- ments, which improve the presentation of the paper.

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