International Journal of Combinatorics
Volume 2011 (2011), Article ID 208260, 14 pages
Research Article

Harmonic Numbers and Cubed Binomial Coefficients

Victoria University College, Victoria University, P.O. Box 14428, Melbourne City, VIC 8001, Australia

Received 18 January 2011; Accepted 3 April 2011

Academic Editor: Toufik Mansour

Copyright © 2011 Anthony Sofo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Euler related results on the sum of the ratio of harmonic numbers and cubed binomial coefficients are investigated in this paper. Integral and closed-form representation of sums are developed in terms of zeta and polygamma functions. The given representations are new.

1. Introduction

The well-known Riemann zeta function is defined as 𝜁(𝑧)=𝑟=11𝑟𝑧,Re(𝑧)>1.(1.1) The generalized harmonic numbers of order 𝛼 are given by 𝐻𝑛(𝛼)=𝑛𝑟=11𝑟𝛼for(𝛼,𝑛)×,={1,2,3,},(1.2) and for 𝛼=1,𝐻𝑛(1)=𝐻𝑛=101𝑡𝑛1𝑡𝑑𝑡=𝑛𝑟=11𝑟=𝛾+𝜓(𝑛+1),(1.3) where 𝛾 denotes the Euler-Mascheroni constant defined by 𝛾=lim𝑛𝑛𝑟=11𝑟log(𝑛)=𝜓(1)0.5772156649,(1.4) and where 𝜓(𝑧) denotes the Psi, or digamma function defined by 𝑑𝜓(𝑧)=Γ𝑑𝑧logΓ(𝑧)=(𝑧)=Γ(𝑧)𝑛=011𝑛+1𝑛+𝑧𝛾,(1.5) and the Gamma function Γ(𝑧)=0𝑢𝑧1𝑒𝑢𝑑𝑢, for (𝑧)>0.

Variant Euler sums of the form 𝑛=1𝐻(1)2𝑛(1/2)𝐻𝑛(1)𝑥2𝑛𝑛𝑝for𝑝=1and𝑝=2(1.6) have been considered by Chen [1], and recently Boyadzhiev [2] evaluated various binomial identities involving power sums with harmonic numbers 𝐻𝑛(1). Other remarkable harmonic number identities known to Euler are𝑛=1𝐻𝑛(1)(𝑛+1)2=𝜁(3),𝑛=1𝐻𝑛(1)𝑛3=54𝜁(4),(1.7) there is also a recurrence formula (2𝑛+1)𝜁(2𝑛)=2𝑛1𝑟=1𝜁(2𝑟)𝜁(2𝑛2𝑟),(1.8) which shows that in particular, for 𝑛=2,5𝜁(4)=2(𝜁(2))2 and more generally that 𝜁(2𝑛) is a rational multiple of (𝜁(𝑛))2. Another elegant recursion known to Euler [3] is 2𝑛=1𝐻𝑛(1)𝑛𝑞=(𝑞+2)𝜁(𝑞+1)𝑞2𝑟=1𝜁(𝑟+1)𝜁(𝑞𝑟).(1.9) Further work in the summation of harmonic numbers and binomial coefficients has also been done by Flajolet and Salvy [4] and Basu [5]. In this paper it is intended to add, in a small way, some results related to (1.7) and to extend the result of Cloitre, as reported in [6], 𝑛=1(𝐻𝑛(1)/𝑘𝑛+𝑘)=(𝑘/(𝑘1)2) for 𝑘>1. Specifically, we investigate integral representations and closed form representations for sums of harmonic numbers and cubed binomial coefficients. The works of [713] also investigate various representations of binomial sums and zeta functions in simpler form by the use of the Beta function and other techniques. Some of the material in this paper was inspired by the work of Mansour, [8], where he used, in part, the Beta function to obtain very general results for finite binomial sums.

2. Integral Representations and Identities

The following Lemma, given by Sofo [11], is stated without proof and deals with the derivative of a reciprocal binomial coefficient.

Lemma 2.1. Let 𝑎 be a positive real number, 𝑧0,𝑛 is a positive integer and let 𝑄(𝑎𝑛,𝑧)=𝑧𝑎𝑛+𝑧1 be an analytic function of 𝑧. Then, 𝑄(𝑎𝑛,𝑧)=𝑑𝑄=[]𝑑𝑧𝑄(𝑎𝑛,𝑧)𝜓(𝑧+1+𝑎𝑛)𝜓(𝑧+1)for𝑧>0,𝐻𝑛(1),for𝑧=0and𝑎=1.(2.1)

Theorem 2.2. Let 𝑎,𝑏,𝑐,𝑑0 be real positive numbers, |𝑡|1,𝑝0 and let 𝑗,𝑘,𝑙,𝑚0 be real positive numbers. Then 𝑛1𝑡𝑛𝑛+𝑝1𝑛1[𝜓](𝑗+1+𝑎𝑛)𝜓(𝑗+1)𝑛𝑗𝑎𝑛+𝑗𝑘𝑏𝑛+𝑘𝑙𝑐𝑛+𝑙𝑚𝑑𝑛+𝑚=𝑎𝑘𝑙𝑚𝑡10(1𝑥)𝑗(1𝑦)𝑘1(1𝑧)𝑙1(1𝑤)𝑚11𝑡𝑥𝑎𝑦𝑏𝑧𝑐𝑤𝑑𝑝+1×ln(1𝑥)𝑥𝑎1𝑦𝑏𝑧𝑐𝑤𝑑𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤.(2.2)

Proof. Expand 𝑛1𝑡𝑛𝑛+𝑝1𝑛1𝑛𝑗𝑎𝑛+𝑗𝑘𝑏𝑛+𝑘𝑙𝑐𝑛+𝑙𝑚𝑑𝑛+𝑚=𝑛1𝑡𝑛𝑛+𝑝1𝑛1𝑎𝑛Γ(𝑎𝑛)Γ(𝑗+1)Γ(𝑏𝑛+1)𝑘Γ(𝑘)×𝑛Γ(𝑎𝑛+𝑗+1)Γ(𝑏𝑛+𝑘+1)Γ(𝑐𝑛+1)𝑙Γ(𝑙)Γ(𝑑𝑛+1)𝑚Γ(𝑚)Γ(𝑐𝑛+𝑙+1)Γ(𝑑𝑛+𝑚+1)=𝑎𝑘𝑙𝑚10(1𝑥)𝑗𝑥𝑛1𝑥𝑎𝑛𝑡𝑛𝑛+𝑝1𝑛1×𝐵(𝑏𝑛+1,𝑘)𝐵(𝑐𝑛+1,𝑙)𝐵(𝑑𝑛+1,𝑚)𝑑𝑥,(2.3) where 𝐵(𝛼,𝛽)=Γ(𝛼)Γ(𝛽)=Γ(𝛼+𝛽)10(1𝑦)𝛼1𝑦𝛽1=𝑑𝑦10(1𝑦)𝛽1𝑦𝛼1𝑑𝑦,for𝛼>0and𝛽>0(2.4) is the classical Beta function. Differentiating with respect to the parameter 𝑗, and utilizing Lemma 2.1 implies the resulting equation is as follows: 𝑛1𝑡𝑛𝑛+𝑝1𝑛1(𝜓(𝑗+1+𝑎𝑛)𝜓(𝑗+1))𝑛𝑗𝑎𝑛+𝑗𝑘𝑏𝑛+𝑘𝑙𝑐𝑛+𝑙𝑚𝑑𝑛+𝑚=𝑎𝑘𝑙𝑚10(1𝑥)𝑗𝑥ln(1𝑥)𝑛1𝑥𝑎𝑛𝑡𝑛𝑛+𝑝1𝑛1×𝐵(𝑏𝑛+1,𝑘)𝐵(𝑐𝑛+1,𝑙)𝐵(𝑑𝑛+1,𝑚)𝑑𝑥,=𝑎𝑘𝑙𝑚10(1𝑥)𝑗ln(1𝑥)𝑥(1𝑦)𝑘1(1𝑧)𝑙1(1𝑤)𝑚1×𝑛1𝑛+𝑝1𝑛1𝑡𝑥𝑎𝑦𝑏𝑧𝑐𝑤𝑑𝑛𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤=𝑎𝑘𝑙𝑚𝑡10(1𝑥)𝑗ln(1𝑥)(1𝑦)𝑘1(1𝑧)𝑙1(1𝑤)𝑚11𝑡𝑥𝑎𝑦𝑏𝑧𝑐𝑤𝑑𝑝+1×𝑥𝑎1𝑦𝑏𝑧𝑐𝑤𝑑𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤(2.5) for |𝑡𝑥𝑎𝑦𝑏𝑧𝑐𝑤𝑑|<1.

In the following three corollaries we encounter harmonic numbers at possible rational values of the argument, of the form 𝐻(𝛼)(𝑟/𝑏)1 where 𝑟=1,2,3,,𝑘,𝛼=1,2,3,, and 𝑘. The polygamma function 𝜓(𝛼)(𝑧) is defined as 𝜓(𝛼)𝑑(𝑧)=𝛼+1𝑑𝑧𝛼+1=𝑑logΓ(𝑧)𝛼𝑑𝑧𝛼[]𝜓(𝑧),𝑧{0,1,2,3,}.(2.6) To evaluate 𝐻(𝛼)(𝑟/𝑏)1 we have available a relation in terms of the polygamma function 𝜓(𝛼)(𝑧), for rational arguments 𝑧,𝐻(𝛼+1)(𝑟/𝑏)1=𝜁(𝛼+1)+(1)𝛼𝜓𝛼!(𝛼)𝑟𝑏,(2.7) where 𝜁(𝑧) is the Riemann zeta function. We also define𝐻(1)(𝑟/𝑏)1𝑟=𝛾+𝜓𝑏,𝐻0(𝛼)=0.(2.8) The evaluation of the polygamma function 𝜓(𝛼)(𝑟/𝑏) at rational values of the argument can be explicitly done via a formula as given by Kölbig [14] (see also [15]) or Choi and Cvijović [16] in terms of the polylogarithmic or other special functions. Some specific values are given as 𝜓(𝑛)12=(1)𝑛2𝑛!𝑛+1𝐻1𝜁(𝑛+1)(5)(2/3)1=𝜁(5)𝜓24(4)13=2𝜋539120𝜁(5),𝐻(2)1/4=168𝐺5𝜁(2),(2.9) and can be confirmed on a mathematical computer package, such as Mathematica [17].

Corollary 2.3. Let 𝑎=1,𝑑=𝑐=𝑏>0,𝑡=1,𝑝=0,𝑗=0 and let 𝑙=𝑚=𝑘1 be a positive integer. Then 𝑛1𝐻𝑛(1)𝑛𝑘𝑏𝑛+𝑘3=𝑘310[](1𝑦)(1𝑧)(1𝑤)𝑘11𝑥(𝑦𝑧𝑤)𝑏ln(1𝑥)(𝑦𝑧𝑤)𝑏=𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤(2.10)𝑘𝑟=1(1)𝑟+1𝑘𝑟3×𝑟24𝑏2𝑟𝜁(4)2𝑏2𝐻(1)(𝑟/𝑏)1+𝑟𝑏+𝑟2𝑏+𝑟𝑋𝑅(𝑘)𝜁(3)1𝑏𝐻(1)(𝑟/𝑏)1𝑟2𝑏2𝐻(2)(𝑟/𝑏)1𝑟+𝑟1𝑏𝐻(1)(𝑟/𝑏)1𝑋𝑅(𝑘)+𝑟2+1𝑌𝑅(𝑘)𝜁(2)2𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+𝑟𝑏𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1+𝑟22𝑏2𝐻(2)(𝑟/𝑏)12+2𝐻(1)(𝑟/𝑏)1𝐻(3)(𝑟/𝑏)1+3𝐻(4)(𝑟/𝑏)1+𝑟2𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+𝑟2𝑏𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1+𝑟𝑋𝑅(𝑘)22𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1,𝑌𝑅(𝑘)(2.11) where 𝐻𝑋𝑅(𝑘)=3(1)𝑘𝑟𝐻(1)𝑟13,(2.12)𝑌𝑅(𝑘)=2𝑋𝑅2(𝑘)3+𝐻(2)𝑘𝑟+𝐻(2)𝑟1.(2.13)

Proof. Let 𝑛1𝐻𝑛(1)𝑛𝑘𝑏𝑛+𝑘3=𝑛1𝐻𝑛(1)(𝑘!)3𝑛𝑘𝑟=1(𝑏𝑛+𝑟)3=𝑛1𝐻𝑛(1)(𝑘!)3𝑛𝑘𝑟=1𝐴𝑟+𝐵𝑏𝑛+𝑟𝑟(𝑏𝑛+𝑟)2+𝐶𝑟(𝑏𝑛+𝑟)3,(2.14) where 𝐶𝑟=lim𝑛(𝑟/𝑏)(𝑏𝑛+𝑟)3𝑘𝑟=1(𝑏𝑛+𝑟)3=(1)𝑟+1𝑟𝑘𝑟𝑘!3,𝐵𝑟=lim𝑛(𝑟/𝑏)𝑑𝑑𝑛(𝑏𝑛+𝑟)3𝑘𝑟=1(𝑏𝑛+𝑟)3=3(1)𝑟𝑟𝑘𝑟𝑘!3𝐻(1)𝑘𝑟𝐻(1)𝑟1,𝐴𝑟=12lim𝑛(𝑟/𝑏)𝑑2𝑑𝑛2(𝑏𝑛+𝑟)3𝑘𝑟=1(𝑏𝑛+𝑟)3=32(1)𝑟+1𝑟𝑘𝑟𝑘!33𝐻(1)𝑘𝑟𝐻(1)𝑟12+𝐻(2)𝑘𝑟+𝐻(2)𝑟1.(2.15) Now, by interchanging sums, we have 𝑛1𝐻𝑛(1)𝑛𝑘𝑏𝑛+𝑘3=𝑘𝑟=1𝐴𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛(𝑏𝑛+𝑟)+𝐵𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛(𝑏𝑛+𝑟)2+𝐶𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛(𝑏𝑛+𝑟)3.(2.16) We can evaluate 𝑛1𝐻𝑛(1)=𝑛(𝑏𝑛+𝑟)𝑛1𝜓(𝑛)+(1/𝑛)+𝛾𝑛(𝑏𝑛+𝑟)=𝑛(𝑏𝑛+𝑟)𝑛1𝜓(𝑛)+1𝑛(𝑏𝑛+𝑟)𝑛2+𝛾(𝑏𝑛+𝑟)=𝑛(𝑏𝑛+𝑟)𝑛1𝜓(𝑛)+1𝑛(𝑏𝑛+𝑟)𝑟𝑛2+(𝛾(𝑏/𝑟))𝑟𝑛011𝑛+1+1𝑛+(𝑟/𝑏)1𝑛+(𝑟/𝑏)𝑛+1+(𝑟/𝑏)=𝑏𝛾𝑟2𝛾2+2𝑟3𝜁(2)+2𝑟(𝜓(𝑟/𝑏))2𝜓2𝑟(𝑟/𝑏)+𝛾2𝑟𝑟𝜓𝑟𝑏,+𝛾+𝑏(2.17) here we have used the result from [18] 𝑛1𝜓(𝑛)=1𝑛(𝑏𝑛+𝑟)𝜓𝑟2𝑟𝑏+12𝛾2+𝜁(2)𝜓𝑟𝑏.+1(2.18)
Now using (2.7) and (2.8), we may write 𝑛1𝐻𝑛(1)=𝑛(𝑏𝑛+𝑟)𝜁(2)𝑟+1𝐻2𝑟(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1.(2.19) Similarly 𝑛1𝐻𝑛(1)𝑛(𝑏𝑛+𝑟)2=1𝑟𝑛1𝐻𝑛(1)𝑏𝑛(𝑏𝑛+𝑟)𝑟𝑛1𝐻𝑛(1)(𝑏𝑛+𝑟)2=𝜁(3)+𝑏𝑟𝜁(2)𝑟2𝜁(2)𝐻(1)(𝑟/𝑏)1+1𝑏𝑟2𝑟2𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+1𝐻𝑏𝑟(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1,𝑛1𝐻𝑛(1)𝑛(𝑏𝑛+𝑟)3=𝑛1𝐻𝑛(1)1𝑟2𝑏𝑛(𝑏𝑛+𝑟)𝑟2(𝑏𝑛+𝑟)2𝑏𝑟(𝑏𝑛+𝑟)3=𝜁(4)4𝑏2𝑟𝜁(3)𝑏𝑟2𝜁(3)𝐻(1)(𝑟/𝑏)1𝑏2𝑟+𝜁(2)𝑟3𝜁(2)𝐻(1)(𝑟/𝑏)1𝑏𝑟2𝜁(2)𝐻(2)(𝑟/𝑏)1𝑏2𝑟+12𝑟3𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+1𝑏𝑟2𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1+12𝑏2𝑟𝐻(2)(𝑟/𝑏)12+2𝐻(1)(𝑟/𝑏)1𝐻(3)(𝑟/𝑏)1+3𝐻(4)(𝑟/𝑏)1.(2.20) Substituting (2.19), (2.20) into (2.16) where 𝑋𝑅(𝑘) and 𝑌𝑅(𝑘) are given by (2.12) and (2.13), respectively, on simplifying the identity (2.11) is realized.

For 𝑘=1 and 𝑏=1 the following identity is valid: 𝑛1𝐻𝑛(1)𝑛(𝑛+1)3=𝜁(2)𝜁(3)𝜁(4)4.(2.21)

Theorem 2.4. 𝑛1𝑡𝑛𝑛+𝑝1𝑛1(𝜓(𝑗+1+𝑎𝑛)𝜓(𝑗+1))𝑛2𝑗𝑎𝑛+𝑗𝑘𝑏𝑛+𝑘𝑙𝑐𝑛+𝑙𝑚𝑑𝑛+𝑚=𝑎𝑏𝑙𝑚𝑡10(1𝑥)𝑗(1𝑦)𝑘(1𝑧)𝑙1(1𝑤)𝑚11𝑡𝑥𝑎𝑦𝑏𝑧𝑐𝑤𝑑𝑝+1×ln(1𝑥)𝑥𝑎1𝑦𝑏1𝑧𝑐𝑤𝑑𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤.(2.22)

Proof. The proof of this theorem is very similar to that of Theorem 2.2 and will not be given here.

Corollary 2.5. Let 𝑎=1,𝑑=𝑐=𝑏>0,𝑡=1,𝑝=0,𝑗=0, and let 𝑙=𝑚=𝑘1 be a positive integer. Then 𝑛1𝐻𝑛(1)𝑛2𝑘𝑏𝑛+𝑘3=𝑏𝑘210(1𝑦)𝑘[](1𝑧)(1𝑤)𝑘11𝑥(𝑦𝑧𝑤)𝑏ln(1𝑥)𝑦𝑏1(𝑧𝑤)𝑏=𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤(2.23)𝑘𝑟=1(1)𝑟+1𝑘𝑟3×𝑟𝑟4𝑏𝜁(4)+4+𝑏𝐻(1)(𝑟/𝑏)1+3𝑟𝑋𝑅(𝑘)+2𝑟2+𝑌𝑅(𝑘)𝜁(3)3𝑏𝑟+2𝐻(1)(𝑟/𝑏)1+𝑟𝑏𝐻(2)(𝑟/𝑏)1+𝑟𝐻(1)(𝑟/𝑏)12𝑏𝑋𝑅(𝑘)𝑏𝑟𝑌𝑅(𝑘)𝜁(2)3𝑏𝐻2𝑟(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1𝐻2(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1𝑟𝐻2𝑏(2)(𝑟/𝑏)12+2𝐻(1)(𝑟/𝑏)1𝐻(3)(𝑟/𝑏)1+3𝐻(4)(𝑟/𝑏)1𝑏𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1𝐻+𝑟(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1𝑋𝑅(𝑘)𝑏𝑟2𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1,𝑌𝑅(𝑘)(2.24) where 𝑋𝑅(𝑘) is given by (2.12) and 𝑌𝑅(𝑘) is given by (2.13).

Proof. Following similar steps to Corollary 2.3, we may write 𝑛1𝐻𝑛(1)𝑛2𝑘𝑏𝑛+𝑘3=𝑘𝑟=1𝐴𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛2(𝑏𝑛+𝑟)+𝐵𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛2(𝑏𝑛+𝑟)2+𝐶𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛2(𝑏𝑛+𝑟)3,(2.25) and evaluate 𝑛1𝐻𝑛(1)𝑛2=(𝑏𝑛+𝑟)𝑛1𝐻𝑛(1)1𝑟𝑛2𝑏=𝑟𝑛(𝑏𝑛+𝑟)2𝜁(3)𝑟𝑏𝜁(2)𝑟2𝑏2𝑟2𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1,𝑛1𝐻𝑛(1)𝑛2(𝑏𝑛+𝑟)2=3𝜁(3)𝑟22𝑏𝜁(2)𝑟3+𝜁(2)𝐻(1)(𝑟/𝑏)1𝑟2𝑏𝑟3𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)11𝑟2𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1,𝑛1𝐻𝑛(1)𝑛2(𝑏𝑛+𝑟)3=𝜁(4)4𝑏𝑟23𝑏𝜁(2)𝑟4+2𝜁(2)𝐻(1)(𝑟/𝑏)1𝑟3+𝜁(2)𝐻(2)(𝑟/𝑏)1𝑏𝑟2+4𝜁(3)𝑟3+𝜁(3)𝐻(1)(𝑟/𝑏)1𝑏𝑟23𝑏2𝑟4𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)12𝑟3𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)112𝑏𝑟2𝐻(2)(𝑟/𝑏)12+2𝐻(1)(𝑟/𝑏)1𝐻(3)(𝑟/𝑏)1+3𝐻(4)(𝑟/𝑏)1.(2.26) By substituting (2.26) into (2.25) and collecting zeta functions, the identity (2.24) is obtained.

For 𝑘=1 and 𝑏=1 the following identity is valid: 𝑛1𝐻𝑛(1)𝑛2(𝑛+1)3=𝜁(4)4+4𝜁(3)3𝜁(2).(2.27)

Theorem 2.6. 𝑛1𝑡𝑛𝑛+𝑝1𝑛1(𝜓(𝑗+1+𝑎𝑛)𝜓(𝑗+1))𝑛3𝑗𝑎𝑛+𝑗𝑘𝑏𝑛+𝑘𝑙𝑐𝑛+𝑙𝑚𝑑𝑛+𝑚=𝑎𝑏𝑐𝑚𝑡10(1𝑥)𝑗(1𝑦)𝑘(1𝑧)𝑙(1𝑤)𝑚11𝑡𝑥𝑎𝑦𝑏𝑧𝑐𝑤𝑑𝑝+1×ln(1𝑥)𝑥𝑎1𝑦𝑏1𝑧𝑐1𝑤𝑑𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤.(2.28)

Proof. The proof of this theorem is very similar to that of Theorem 2.2 and will not be given here.

Corollary 2.7. Let 𝑎=1,𝑑=𝑐=𝑏>0,𝑡=1,𝑗=0, and let 𝑙=𝑚=𝑘1 be a positive integer. Then 𝑛1𝐻𝑛(1)𝑛3𝑘𝑏𝑛+𝑘3=𝑏2𝑘10((1𝑦)(1𝑧))𝑘(1𝑤)𝑘11𝑥(𝑦𝑧𝑤)𝑏ln(1𝑥)(𝑦𝑧)𝑏1𝑤𝑏=𝑑𝑥𝑑𝑦𝑑𝑧𝑑𝑤(2.29)𝑘𝑟=1(1)𝑟+1𝑘𝑟3×51+45𝑋𝑅(𝑘)+4𝑟2𝑌𝑅(𝑘)𝜁(4)9𝑏𝑟+𝐻(1)(𝑟/𝑏)1++5𝑏𝑋𝑅(𝑘)+2𝑏𝑟𝑌𝑅(𝑘)𝜁(3)6𝑏2𝑟23𝑏𝐻(1)(𝑟/𝑏)1𝑟𝐻(2)(𝑟/𝑏)13𝑏2𝑟+𝑏𝐻(1)(𝑟/𝑏)1𝑋𝑅(𝑘)+𝑏2+𝑌𝑅(𝑘)𝜁(2)3𝑏2𝑟2𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+3𝑏𝑟𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1+12𝐻(2)(𝑟/𝑏)12+2𝐻(1)(𝑟/𝑏)1𝐻(3)(𝑟/𝑏)1+3𝐻(4)(𝑟/𝑏)1+3𝑏2𝐻2𝑟(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1𝐻+𝑏(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1+𝑏𝑋𝑅(𝑘)22𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1,𝑌𝑅(𝑘)(2.30) where 𝑋𝑅(𝑘) is given by (2.12) and 𝑌𝑅(𝑘) is given by (2.13).

Proof. We follow similar steps as the previous corollary so that 𝑛1𝐻𝑛(1)𝑛3𝑘𝑏𝑛+𝑘3=𝑘𝑟=1𝐴𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛3(𝑏𝑛+𝑟)+𝐵𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛3(𝑏𝑛+𝑟)2+𝐶𝑟(𝑘!)3𝑛1𝐻𝑛(1)𝑛3(𝑏𝑛+𝑟)3.(2.31) After much algebraic simplification, the following identity is obtained: 𝑛1𝐻𝑛(1)𝑛3(𝑏𝑛+𝑟)3=𝑛1𝐻𝑛(1)1𝑟3𝑛33𝑏𝑟4𝑛2𝑏3𝑟3(𝑏𝑛+𝑟)33𝑏3𝑟4(𝑏𝑛+𝑟)2+6𝑏2𝑟4𝑛=(𝑏𝑛+𝑟)𝜁(4)𝑟39𝑏𝜁(3)𝑟4𝜁(3)𝐻(1)(𝑟/𝑏)1𝑟3+6𝑏2𝜁(2)𝑟53𝑏𝜁(2)𝐻(1)(𝑟/𝑏)1𝑟4𝜁(2)𝐻(2)(𝑟/𝑏)1𝑟3+3𝑏2𝑟5𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+3𝑏𝑟4𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1+12𝑟3𝐻(2)(𝑟/𝑏)12+2𝐻(1)(𝑟/𝑏)1𝐻(3)(𝑟/𝑏)1+3𝐻(4)(𝑟/𝑏)1,𝑛1𝐻𝑛(1)𝑛3(𝑏𝑛+𝑟)2=5𝜁(4)4𝑟25𝑏𝜁(3)𝑟3+3𝑏2𝜁(2)𝑟4𝑏𝜁(2)𝐻(1)(𝑟/𝑏)1𝑟3+3𝑏22𝑟4𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1+𝑏𝑟3𝐻(1)(𝑟/𝑏)1𝐻(2)(𝑟/𝑏)1+𝐻(3)(𝑟/𝑏)1,𝑛1𝐻𝑛(1)𝑛3=(𝑏𝑛+𝑟)5𝜁(4)+𝑏4𝑟2𝜁(2)𝑟32𝑏𝜁(3)𝑟2+𝑏22𝑟3𝐻(1)(𝑟/𝑏)12+𝐻(2)(𝑟/𝑏)1.(2.32) Now we can substitute (2.32) into (2.31), collecting zeta functions and using (2.12) and (2.13) for 𝑋𝑅(𝑘) and 𝑌𝑅(𝑘), respectively, the identity (2.30) is obtained.

Some specific examples of Corollary 2.7 are as follows.

For 𝑘=1 and 𝑏=1 the following identity is valid, 𝑛1𝐻𝑛(1)𝑛3(𝑛+1)31=𝜁(4)9𝜁(3)+6𝜁(2),𝑛1𝐻𝑛(1)𝑛384𝑛+83=1272907300027264101740083211025ln2+4403800186883675(ln2)216604790784315𝐺+11272192𝐺2+157674700835𝐺ln21390679293952𝜋33075172233728𝜋31052039927759892450𝜁(2)4069970159210𝜁(3)+28974848ln2𝜁(3)622467372𝜁(4),(2.33) where 𝐺 is Catalan's constant, defined by 1𝐺=210𝐾(𝑠)𝑑𝑠=𝑟=1(1)𝑟(2𝑟+1)20.915965,(2.34) and 𝐾(𝑠) is the complete elliptic integral of the first kind. The degenerate case 𝑘=0, gives the well-known result 𝑛1𝐻𝑛(1)𝑛3=54𝜁(4).(2.35)

Remark 2.8. Corollaries 2.3, 2.5, and 2.7 are important and can be evaluated as demonstrated independently of their integral representations. Similarly the proofs of Corollaries 2.3, 2.5, and 2.7 are not obvious therefore their explicit representations is desired.

Remark 2.9. Theoretically it should be possible to obtain an integral representation for the general sum 𝑛1𝑡𝑛𝑛+𝑝1𝑛1(𝜓(𝑗+1+𝑎𝑛)𝜓(𝑗+1))𝑛𝑞𝑗𝑎𝑛+𝑗𝑘𝑏𝑛+𝑘𝑙𝑐𝑛+𝑙𝑚𝑑𝑛+𝑚,for𝑞=1,2,3,,(2.36) with its associated corollaries. This work will be investigated in a forthcoming paper.


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