The Australian Journal of Mathematical Analysis and Applications AJMAA Volume 16, Issue 1, Article 14, pp. 1-14, 2019 CUBIC ALTERNATING HARMONIC NUMBER SUMS ANTHONY SOFO Received 12 September, 2018; accepted 23 January, 2019; published 27 May, 2019. VICTORIA UNIVERSITY,COLLEGE OF ENGINEERING AND SCIENCE,MELBOURNE CITY,AUSTRALIA. [email protected] ABSTRACT. We develop new closed form representations of sums of cubic alternating harmonic numbers and reciprocal binomial coefficients. We also identify a new integral representation for the ζ (4) constant. Key words and phrases: Polylogarithm function, Alternating cubic harmonic numbers, Combinatorial series identities, Sum- mation formulas, Partial fraction approach, Binomial coefficients. 2000 Mathematics Subject Classification. Primary 05A19, 33C20. Secondary 11M06, 11B83. ISSN (electronic): 1449-5910 c 2019 Austral Internet Publishing. All rights reserved. 2 ANTHONY SOFO 1. INTRODUCTION AND PRELIMINARIES In this paper we will develop identities, closed form representations of alternating cubic har- monic numbers and reciprocal binomial coefficients, including integral representations, of the form: ∞ n+1 X (−1) H3 (1.1) Ω(k, p) = n , n + k n =1 np k th for p = 0, 1 and k ∈ N0. Here, the n harmonic number n X 1 Z 1 1 − tn (1.2) H = = γ + ψ (n + 1) = dt, H := 0 n r 1 − t 0 r=1 0 and as usual, γ denotes the Euler-Mascheroni constant and ψ(z) is the Psi (or Digamma) func- tion defined by d Γ0(z) Z z ψ(z) := {log Γ(z)} = or log Γ(z) = ψ(t) dt. dz Γ(z) 1 For sums of harmonic numbers with positive terms [10], [27], [28] and [29] have given many results, including sums of the form ∞ X H3 n . n + k n =1 np k Other results are given by [5] and [17]. Let R and C denote, respectively the sets of real and complex numbers and let N := {1, 2, 3, ···} be the set of positive integers, and N0 := N ∪ {0} . λ A generalized binomial coefficient µ (λ, µ ∈ C) is defined, in terms of the familiar gamma function, by λ Γ(λ + 1) := , λ, µ ∈ . µ Γ(µ + 1) Γ (λ − µ + 1) C The Pochhammer symbol (λ)ν (λ, ν ∈ C) is also defined in terms of the gamma function, by 1 (ν = 0; λ ∈ \{0}) Γ(λ + ν) C (λ) := = . ν Γ(λ) λ(λ + 1) ··· (λ + n − 1) (ν = n ∈ N; λ ∈ C), (m) A generalized harmonic number Hn of order m is defined, for positive integers n and m, as follows: n X 1 (m) H(m) := , m, n ∈ and H := 0, m ∈ . n rm N 0 N r=1 In the case of non-integer values of n such as (for example) a value ρ ∈ R, the generalized (m+1) harmonic numbers Hρ may be defined, in terms of the polygamma functions dn dn+1 ψ(n)(z) := {ψ(z)} = {log Γ(z)}, n ∈ , dzn dzn+1 N0 by (−1)m (1.3) H(m+1) = ζ (m + 1) + ψ(m) (ρ + 1) ,H(m+1) = 0 ρ m! 0 (ρ ∈ R \ {−1, −2, −3, ···} ; m ∈ N) , AJMAA, Vol. 16, No. 1, Art. 14, pp. 1-14, 2019 AJMAA CUBIC ALTERNATING HARMONIC NUMBER SUMS 3 (α) r where ζ (z) is the Riemann zeta function. The evaluation of the polygamma function ψ a 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 listed in the books[15], [21] and [22]. The polylogarithm or de-Jonquière function Lip (z), is defined as, ∞ X zn Li (z) := , p ∈ when |z| < 1; < (p) > 1 when |z| = 1. p np C n =1 Some results for sums of alternating and non-alternating harmonic numbers may be seen in the works of [3], [4], [12], [13], [14], [16], [17], [18], [19], [20], [23], [24], [26] and references therein. Some explicit, and closely related results may also be seen in the well presented papers [9] and [25]. The following lemma will be useful in the development of the main theorems. Lemma 1.1. Let r be a positive integer and p ∈ N. Then: r j X (−1) 1 (p) (p) (p) (1.4) F (p, r) = p = p H r + H r−1 − H r+1 j 2 [ 2 ] [ 2 ] 2[ 2 ]−1 j =1 where [x] is the integer part of x, and when p = 1, r j X (−1) (1.5) F (1, r) = = H r − Hr. j [ 2 ] j =1 For p = 2, r j X (−1) 1 (2) (2) 1 (1.6) F (2, r) = 2 = H r − H r+1 1 − ζ (2) . j 4 [ 2 ] [ 2 ]− 2 2 j =1 Proof. The proof is given in the paper [11]. Lemma 1.2. The following identity holds, ∞ n+1 X (−1) H3 5 3 1 9 (1.7) X (0) = n = ζ (4) + ζ (2) ln2 2 − ln4 2 − ζ (3) ln 2, n 8 4 4 8 n =1 1 2 Z log (1 − x) log (1 + x) + Li2 (−x) (1.8) = dx, 0 x (1 + x) where Lip (·) is the polylogarithm function. Proof. Consider ∞ X tn (1.9) V (j, t) = n + j n =1 n j ∞ ∞ X tn Γ(n)Γ(j + 1) X = = tn B (n, j + 1) , Γ(n + j + 1) n =1 n =1 AJMAA, Vol. 16, No. 1, Art. 14, pp. 1-14, 2019 AJMAA 4 ANTHONY SOFO where Γ(·) is the gamma function and B (·, ·) is the beta function. Now Z 1 j ∞ Z 1 j (1 − x) X n (1 − x) V (j, t) = (tx) dx = t dx, x 1 − tx 0 n =1 0 and differentiating p times with respect to j and then letting j → 0 with t = −1, results in Z 1 p (p) log (1 − x) p 1 V (0, −1) = dx = (−1) p!Lip+1 . 0 1 + x 2 For p = 3 1 (1.10) V (0, −1)(3) = −6Li 4 2 From (1.9) we also have ∞ n (p) X (−1) σn V (0, −1) = n n =1 where " !# dp n + j −1 σn = lim , j→0 djp j when p = 3, ∞ n+1 (3) X (−1) φ 1 (1.11) V (0, −1) = n = 6Li n 4 2 n =1 3 (2) (3) and φn = Hn + 3HnHn + 2Hn . From the paper [8], we have the results ∞ n+1 (3) X (−1) Hn 19 3 = ζ (4) − ζ (3) ln 2, n 16 4 n =1 ∞ n+1 (2) X (−1) HnHn 1 = 2Li − ζ (4) − 4ζ (2) ln2 2 n 4 2 n =1 1 7 + ln4 2 + ζ (3) ln 2, 4 8 and substituting into (1.11) we have the result (1.7). The representation of the integral (1.8) is obtained in the following way. From [11], we can express, for p ∈ N p Z 1 Z 1 p n−1 p p Hn Y Y (−1) p = ··· xj ln (1 − xj) dxj n 0 0 j=1 j=1 R 1 R 1 where 0 ··· 0 is a p- fold integration procedure, for p = 3, 3 Z 1 Z 1 Z 1 n Hn (x1x2x3) ln (1 − x1) ln (1 − x2) ln (1 − x3) − 3 = dx1dx2dx3. n 0 0 0 x1x2x3 (1.12) AJMAA, Vol. 16, No. 1, Art. 14, pp. 1-14, 2019 AJMAA CUBIC ALTERNATING HARMONIC NUMBER SUMS 5 Now, ∞ n+1 X (−1) H3 Z 1 Z 1 Z 1 n = − ln (1 − x) ln (1 − y) ln (1 − z) n n =1 0 0 0 ∞ X n−1 × n2 (−xyz) dxdydz n =1 Z 1 Z 1 Z 1 ln (1 − x) ln (1 − y) ln (1 − z) (1 − xyz) = − 2 dxdydz 0 0 0 (1 + xyz) Z 1 Z 1 ln (1 − x) ln (1 − y) (1 − ln (1 + xy)) = 2 dxdy 0 0 (1 + xy) Z 1 ln (1 − x) ln2 (1 + x) x = − Li2 dx 0 x (1 + x) 2 1 + x and applying Landen’s identity x ln2 (1 + x) Li = − − Li (−x) 2 1 + x 2 2 we obtain (1.8). By association we conclude Z 1 log (1 − x) log2 (1 + x) + Li (−x) 5 3 2 dx = ζ (4) + ζ (2) ln2 2 0 x (1 + x) 8 4 1 9 − ln4 2 − ζ (3) ln 2 4 8 ∞ n+1 X (−1) H3 = X (0) = n . n n =1 An alternate manipulation of (1.7) leads to the following result and adds to the results on cubic sums some of which are published in [10]. Lemma 1.3. X H3 5 67 3 3 2n = ζ (4) + ζ (3) + 2ζ (2) + ζ (2) ln2 2 − ζ (2) ln 2 2n (2n − 1) 8 16 4 2 n≥1 9 1 3 − ζ (3) ln 2 + ln 2 − ln4 2 + ln3 2 − ln2 2. 8 4 2 (1.13) Proof. By re-arrangement ∞ 1 3 3 ! X H2n − H X (0) = 2n − 2n 2n − 1 2n n =1 3 2 X H − 3H X 3H2n X 1 = 2n 2n + − . 2n (2n − 1) 4n2 (2n − 1) 8n3 (2n − 1) n≥1 n≥1 n≥1 AJMAA, Vol. 16, No. 1, Art. 14, pp. 1-14, 2019 AJMAA 6 ANTHONY SOFO Hence X H3 − 3H2 5 3 1 9 2n 2n = ζ (4) + ζ (2) ln2 2 − ln4 2 − ζ (3) ln 2 2n (2n − 1) 8 4 4 8 n≥1 33 3 3 + ln 2 + ζ (3) − ζ (2) + ln2 2 16 2 2 1 1 − ζ (3) − 3 ln 2 − ζ (2) .