We extend some results of Euler related sums. Integral and closed form representation of sums with products of harmonic numbers and cubed binomial coefficients are developed in terms of Polygamma functions. The given representations are new

Anthony Sofo

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Sofo, Anthony

Victoria University Institutional Repository (VUIR)

Harmonic Number Sums in Closed Form

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Harmonic number sums in closed form

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MATHEMATICAL COMMUNICATIONS 335Math. Commun. 16(2011), 335–345.Harmonic number sums in closed formAnthony Sofo1,∗1 Victoria University College, Victoria University, PO Box 14 428, Melbourne City 8001,VIC, AustraliaReceived March 1, 2010; accepted November 17, 2010Abstract. We extend some results of Euler related sums. Integral and closed form repre-sentation of sums with products of harmonic numbers and cubed binomial coefficients aredeveloped in terms of Polygamma functions. The given representations are new.AMS subject classifications: Primary 11B65; Secondary 33C20Key words: polylogarithms, harmonic numbers, integral representations, binomial coeffi-cients1. IntroductionThe generalized harmonic numbers of order α are given byH(α)n =n∑r=11rα, for n = 1, 2, 3, . . . , α = 1, 2, 3, . . .andH(1)n = Hn =∫ 1t=01− tn1− t dt =n∑r=11r= γ + ψ (n + 1) ,where γ denotes the Euler-Mascheroni constant, defined byγ = limn→∞(n∑r=11r− log n)= −ψ (1) ≈ 0.57721566490153286 . . .Recently Jung, Cho and Choi [10] evaluated Euler sums from integrals and obtainedresults like ∞∑n=1H(1)n2n (n + 1)2=14ζ (3)− 13(ln 2)3 . (1)In this paper we expand (1) to obtain results of the form∑n≥1H(1)nn5 (2n + 1)3=74ζ (6)− 12ζ2 (3)− 18ζ (5) + 6ζ (3) ζ (2)− 15ζ (4)−440 ζ (3) + 56 ln 2 ζ (3) + 240 ln 2 ζ (2) + 480 (ln 2)2(2)∗Corresponding author. Email address: Anthony.Sofo@vu.edu.au (A. Sofo)http://www.mathos.hr/mc c©2011 Department of Mathematics, University of Osijek336 A. Sofoto add, in a small way, some results related to (1) and (3) and to extend the resultof Cloitre, as reported in [19]. A remarkable recursion known to Euler [8] is2∞∑n=1H(1)nnq= (q + 2) ζ (q + 1)−q−2∑r=1ζ (r + 1) ζ (q − r) (3)and there is also a recurrence formula(2n + 1) ζ (2n) = 2n−1∑r=1ζ (2r) ζ (2n− 2r)which shows that in particular, for n = 2, 5ζ (4) = 2 (ζ (2))2 and more generallythat ζ (4n) is a rational multiple of (ζ (2n))2 . The Riemann zeta functionζ (z) =∞∑r=11rz, Re (z) > 1.Here we are also interested in evaluating closed form and integral representations ofEuler type sums containing both harmonic numbers, H(α)n and powers of binomialcoefficients, two types of special numbers of enumerative combinatorics. There aremany works investigating sums of both harmonic numbers and binomial coefficients,see for example [1, 2, 3, 4, 6, 9, 13, 14, 15, 16, 17], and references therein. Chuand Zheng [7] also obtained many other identities involving harmonic numbers andcentral binomial coefficients.2. IdentitiesTheorem 1. Let a, b, c, d ≥ 0 be real positive numbers, |t| < 1 and j, k, l,m ≥ 0,then∑n≥1tn∑anr=11r+jn5(an+jj)(bn+kk)(cn+ll) (dn+mm)= −abcd∫ 10∫ 10∫ 10∫ 10(1− x)j (1− y)k (1− z)l (1− w)mxyzw× ln (1− x) ln (1− txaybzcwd)dxdy dz dw(4)Proof. Consider∑n≥1tnn5(an+jj)(bn+kk)(cn+ll)(dn+mm)= abcd∑n≥1tnn4Γ (an) Γ (bn) Γ (cn) Γ (dn) Γ (j + 1) Γ (k + 1)Γ (l + 1) Γ (m + 1)n5Γ (an + j + 1)Γ (bn + k + 1)Γ (cn + l + 1) Γ (dn + m + 1)Harmonic number sums in closed form 337= abcd∑n≥1tnΓ (bn) Γ (cn) Γ (dn) Γ (k + 1) Γ (l + 1)Γ (m + 1)nΓ (bn + k + 1) Γ (cn + l + 1) Γ (dn + m + 1)B (an, j + 1)= abcd∑n≥1tnnB (bn, k + 1) B (cn, l + 1) B (dn,m + 1)∫ 10xan−1 (1− x)j dx,where the Beta functionB (s, z) =∫ 10ws−1 (1− w)z−1 dw = Γ (s) Γ (w)Γ (s + w)for Re (s) > 0, Re (z) > 0 and the Gamma functionΓ (z) =∫ ∞0wz−1e−w dw for Re (z) > 0.Now differentiate with respect to the parameter j, then∑n≥1tn∑anr=11r+jn5(an+jj)(bn+kk)(cn+ll) (dn+mm)= −abcd∫ 10(1− x)jxln (1− x)×∑n≥1xantn[∫ 10ybn−1 (1− y)k∫ 10zcn−1 (1− z)l∫ 10wdn−1 (1− w)m dw dz dy]dx= −abcd∫ 10∫ 10∫ 10∫ 10(1− x)j ln (1− x) · (1− y)k (1− z)l (1− w)mxyzw×∑n≥1(txaybzcwd)nndx dy dz dwby a subsequent allowable change of sum and integral, hence∑n≥1tn∑anr=11r+jn5(an+jj)(bn+kk)(cn+ll)(dn+mm)= −abcd∫ 10∫ 10∫ 10∫ 10(1− x)j (1− y)k (1− z)l (1− w)mxyzw× ln (1− x) ln (1− txaybzcwd) dxdy dz dw.We now investigate three new corollaries that are a consequence of the MainTheorem 1.338 A. SofoCorollary 1. For b = c = d = a > 0, t = 1, j = 0 = l = m and k ≥ 1 an integer,we have∑n≥1H(1)nn5(n+kk)= −a4∫ 10∫ 10∫ 10∫ 10(1− y)k ln (1− x) ln (1− (xyzw)a)xyzwdxdy dz dw=74ζ (6)− ζ2 (3)2− 3aH(1)k ζ (5) + aH(1)k ζ (3) ζ (2) +5a28{(H(1)k)2+ H(2)k}ζ (4)−a33{(H(1)k)3+ 3H(1)k H(2)k + 2H(3)k}ζ (3)+a424{(H(1)k)4+ 6(H(1)k)2H(2)k + 3(H(2)k)2+ 8H(1)k H(3)k + 6H(4)k}ζ (2)+a42k∑r=1(−1)r+1r4(kr){(H(1)ra−1)2+ H(2)ra−1}. (5)Proof. Expand∑n≥1H(1)nn5(an+kk) =∑n≥1k!H(1)nn5∏kr=1 (an + r)=∑n≥1k!H(1)nn5k∑r=1Aran + r,whereAs = limn→(− sa ){an + s∏ks=1 (an + s)}= (−1)s+1(sk!(ks)),hence∑n≥1H(1)nn5k∑r=1(−1)r+1 r(kr)an + r=k∑r=1(−1)r+1 r(kr) ∑n≥1H(1)nn5 (an + r)=k∑r=1(−1)r+1 r(kr) ∑n≥1H(1)n[1rn5− ar2n4+a2r3n3− a3r4n2+a4r4n (an + r)]=k∑r=1(−1)r+1 r(kr)[7ζ (6)4r− ζ2 (3)2r− 3aζ (5)r2+aζ (3) ζ (2)r2+5a2ζ (4)4r3− 2a3ζ (3)r4+a4ζ (2)r5+a42r5{(H(1)ra−1)2+ H(2)ra−1}]. (6)Simplifying (6), we obtain (5).Harmonic number sums in closed form 339Remark 1. The degenerate case, k = 0 gives the known result∑n≥1H(1)nn5=74ζ (6)− 12ζ2 (3) .Similarly,∑n≥1H(1)nn5(4n+33) = 74ζ (6)− 12ζ2 (3)− 22ζ (5) + 223ζ (3) ζ (2) +4259ζ (4)− 920027ζ (3)−1561627ζ (2) +3097627π ln 2 +303689(ln 2)2 − 24780881G,∑n≥1H(1)nn5(n4 +22) = 74ζ (6)− 12ζ2 (3)− 98ζ (5) +38ζ (3) ζ (2)+3152305ζ (4)− 15256ζ (3) +314096ζ (2) +12601699722534400,where G is Catalan’s constant, defined byG =∞∑r=0(−1)r(2r + 1)2≈ 0.915965 . . .Corollary 2. For d = c = b = a > 0, t = 1, j = 0 = m, l = k ≥ 1 integers, we have∑n≥1H(1)nn5(an+kk)2 (7)= −a4∫ 10∫ 10∫ 10∫ 10[(1− y) (1− z)]k(1− xyzw) ln (1− x) ln (1− (xyzw)a) dxdy dz dw=74ζ (6)− 12ζ2 (3) +k∑r=1(kr)2 [−6ar(1− rX (k, r)) ζ (5)+2ar(1− rX (k, r)) ζ (3) ζ (2) + 5a22r2(3− 2rX (k, r)) ζ (4)−a3r3(9− 4rX (k, r)) ζ (3) + a3r4(5a− rH(1)ra−1 − 2raX (k, r))ζ (2)+1r4[5a42{(H(1)ra−1)2+ H(2)ra−1}+ ra3(H(1)ra−1H(2)ra−1 + H(3)ra−1)−a4r{(H(1)ra−1)2+ H(2)ra−1}X (k, r)]], (8)whereX (k, r) := H(1)k−r −H(1)r−1. (9)340 A. SofoProof. Expand∑n≥1H(1)nn5(an+kk)2 =∑n≥1H(1)n (k!)2n5((an + 1)k+1)2=∑n≥1H(1)n (k!)2n5k∑r=1[Aran + r+Br(an + r)2],whereBs = limn→(− sa ){(an + s)2∏ks=1 (an + s)2}=(sk!(ks))2andAs = limn→(− sa )ddn{(an + s)2∏ks=1 (an + s)2}= −2(sk!(ks))2X (k, s) ,where X (k, r) is given by (9). Now, by interchanging the order of summation, wehave∑n≥1H(1)nn5(an+kk)2 =k∑r=1(k!)2Ar∑n≥1H(1)nn5 (an + r)+k∑r=1(k!)2Br∑n≥1H(1)nn5 (an + r)2, (10)where∑n≥1H(1)nn5 (an + r)2=∑n≥1H(1)n[1r2n5− 2ar3n4+3a2r4n3− 4a3r5n2− a5r5 (an + r)2+5a4r5n (an + r)]=7ζ (6)4r2− ζ2 (3)2r2− 6aζ (5)r3+2aζ (3) ζ (2)r3+15a2ζ (4)4r4− 9a3ζ (3)r5+5a4ζ (2)r6−a3ζ (2) H(1)ra−1r5+5a42r6{(H(1)ra−1)2+ H(2)ra−1}+a3r5{H(1)ra−1H(2)ra−1 + H(3)ra−1}(11)and from (11) and the inner part of (6) into (10), we obtain∑n≥1H(1)nn5(an+kk)2=k∑r=1(r(kr))2 [7ζ (6)4r2− ζ2 (3)2r2− 6aζ (5)r3+2aζ (3) ζ (2)r3Harmonic number sums in closed form 341+15a2ζ (4)4r4− 9a3ζ (3)r5+5a4ζ (2)r6−a3ζ (2)H(1)ra−1r5+5a42r6{(H(1)ra−1)2+ H(2)ra−1}+a3r5{H(1)ra−1H(2)ra−1 + H(3)ra−1}]−2k∑r=1(r(kr))2 [7ζ (6)4r− 3aζ (5)r2+5a2ζ (4)4r3− ζ2 (3)2r+aζ (3) ζ (2)r− 2a3ζ (3)r4+a4ζ (2)r5+a42r5{(H(1)ra−1)2+ H(2)ra−1}]X (k, r) .Now we collect the zeta functions, and using the property thatk∑r=1(kr)2[1− 2rX (k, r)] = 1then (8) is confirmed.Remark 2. Some examples are∑n≥1H(1)nn5(4n+22)2 =74ζ (6)− 12ζ2 (3)− 36ζ (5) + 12ζ (3) ζ (2) + 115ζ (4)− 8376ζ (3)−128π3 − 2304ζ (2) + 4656ζ (2) ln 2 + 4608π ln 2+14112 (ln 2)2 − 12288G + 1024πG + 6144G ln 2,∑n≥1H(1)nn5(n3 +33)2 =74ζ (6)− 12ζ2 (3)− 113ζ (5) +119ζ (3) ζ (2)−970216ζ (4)− 305216ζ (3)− 24968777348320ζ (2) +1108486911367648121824000.Corollary 3. For d = c = b = a > 0, t = 1, j = 0, m = l = k ≥ 1 an integer, then∑n≥1H(1)nn5(an+kk)3= −a4∫ 10∫ 10∫ 10∫ 10[(1− w) (1− z) (1− y)]kxyzw× ln (1− x) ln (1− (xyzw)a) dxdy dz dw (12)=74ζ (6)− 12ζ2 (3) +k∑r=1(−1)r+1(kr)3 [−a(9r+ 18X (k, r) + 3rY (k, r))ζ (5)+a(3r− 6X (k, r) + rY (k, r))ζ (3) ζ (2)+a2(294r2− 45X (k, r)4r+54Y (k, r))ζ (4)342 A. Sofo+a2−25ar3−H(1)ra−1r2+27aX (k, r)r2+2arY (k, r) ζ (3)+a215a2r4−5aH(1)ra−1r3−H(2)ra−1r2− 35a2r3−aH(1)ra−1r2 X (k, r)+a2r2Y (k, r))ζ (2) +15a42r4{(H(1)ra−1)2+ H(2)ra−1}+5a3r3{H(1)ra−1H(2)ra−1 + H(3)ra−1}+a22r2{(H(2)ra−1)2+ 2H(1)ra−1H(3)ra−1 + 3H(4)ra−1}−3(5a42r3{(H(1)ra−1)2+ H(2)ra−1}+a3r2{H(1)ra−1H(2)ra−1 + H(3)ra−1})X (k, r)+a42r2{(H(1)ra−1)2+ H(2)ra−1}Y (k, r)], (13)where X (k, r) is given by (9) andY (k, r) :=32{3X2 (k, r) + H(2)k−r + H(2)r−1}. (14)Proof. Expand∑n≥1H(1)nn5(an+kk)3 =∑n≥1H(1)n (k!)3n5∏kr=1 (an + r)3=∑n≥1H(1)n (k!)3n5k∑r=1[Aran + r+Br(an + r)2+Cr(an + r)3],whereCs = limn→(− sa ){(an + s)3∏ks=1 (an + s)3}= (−1)s+1(sk!(ks))3,Bs = limn→(− sa )ddn{(an + s)3∏ks=1 (an + s)3}= 3 (−1)s(sk!(ks))3X (k, s)andAs =12limn→(− sa )d2dn2{(an + s)3∏ks=1 (an + s)3}=32(−1)s+1(sk!(ks))3 [3X2 (k, s) + H(2)k−s + H(2)s−1],where X (k, r) is given by (9).Harmonic number sums in closed form 343Now we can write∑n≥1H(1)nn5(an+kk)3 =k∑r=1Ar (k!)3∑n≥1H(1)nn5 (an + r)+k∑r=1Br (k!)3∑n≥1H(1)nn5 (an + r)2+k∑r=1Cr (k!)3∑n≥1H(1)nn5 (an + r)3,(15)where∑n≥1H(1)nn5 (an + r)3=∑n≥1H(1)n[1r3n5− 3ar4n4+6a2r5n3− 10a3r6n2− a5r5 (an + r)3− 5a5r6 (an + r)2+15a4r6n (an + r)]=7ζ (6)4r3− ζ2 (3)2r3− 9aζ (5)r4+3aζ (3) ζ (2)r4+29a2ζ (4)4r5−25a3ζ (3)r6−a2ζ (3) H(1)ra−1r5+15a4ζ (2)r7−5a3ζ (2)H(1)ra−1r6−a2ζ (2) H(2)ra−1r5+15a42r7{(H(1)ra−1)2+ H(2)ra−1}+5a3r6{H(1)ra−1H(2)ra−1 + H(3)ra−1}+a22r5{(H(2)ra−1)2+ 2H(1)ra−1H(3)ra−1 + 3H(4)ra−1}(16)Substituting (16), (11) and the inner part of (6) into (15) we get∑n≥1H(1)nn5(an+kk)3 =k∑r=132(−1)r+1(rk!(kr))3 [3X2 (k, r) + H(2)k−r + H(2)r−1]×[7ζ (6)4r− 3aζ (5)r2+5a2ζ (4)4r3− ζ2 (3)2r+aζ (3) ζ (2)r− 2a3ζ (3)r4+a4ζ (2)r5+a42r5{(H(1)ra−1)2+ H(2)ra−1}]+k∑r=13 (−1)r(r(kr))3X (k, r)[7ζ (6)4r2− ζ2 (3)2r2− 6aζ (5)r3+2aζ (3) ζ (2)r3− 15a2ζ (4)4r4− 9a3ζ (3)r5+5a4ζ (2)r6344 A. Sofo−a3ζ (2)H(1)ra−1r5+5a42r6{(H(1)ra−1)2+ H(2)ra−1}+a3r5{H(1)ra−1H(2)ra−1 + H(3)ra−1}+k∑r=1(−1)r+1(r(kr))3 [7ζ (6)4r3− ζ2 (3)2r3− 9aζ (5)r4+3aζ (3) ζ (2)r4+29a2ζ (4)4r5− 25a3ζ (3)r6−a2ζ (3) H(1)ra−1r5+15a4ζ (2)r7−5a3ζ (2) H(1)ra−1r6−a2ζ (2) H(2)ra−1r5+a22r5{(H(2)ra−1)2+ 2H(1)ra−1H(3)r−1 + 3H(4)ra−1}+15a42r7{(H(1)ra−1)2+ H(2)ra−1}+5a3r6{H(1)ra−1H(2)ra−1 + H(3)ra−1}],and after much algebraic manipulation with the aid of say, Mathematica [20], sim-plification and collection of the zeta functions, we arrive at (13) where we have usedthe fact thatk∑r=1(−1)r+1(r(kr))3 [1− 3rX (k, r) + r2Y (k, r)] = 1,and Y (k, r) is given by (14).Remark 3. Some examples are∑n≥1H(1)nn5(2n+44)3 =74ζ (6)− 12ζ2 (3)− 752ζ (5) +252ζ (3) ζ (2)− 14580548ζ (4)−68053216ζ (3) +35840 ln 29ζ (3)− 2767747144ζ (2)− 4096 ln 29ζ (2)+31129627(ln 2)2 − 15769627ln 2 +39165748.For a = 2, k = 1 we obtain (2).Remark 4. In Corollaries 1, 2 and 3 we encounter harmonic numbers at possiblerational values of the argument, of the form H(α)ra−1 where r = 1, 2, 3, . . . , k andk ∈ N. To evaluate H(α)ra−1 we have available a relation in terms of the Polygammafunction ψ(α) (z) , for rational arguments z,H(α+1)ra−1 = ζ (α + 1) +(−1)αα!ψ(α)( ra),where ζ (z) is the Riemann Zeta function. We also defineH(1)ra−1 = γ + ψ( ra), and H(α)0 = 0.The evaluation of the Polygamma function ψ(α)(ra)at rational values of the argu-ment can be explicitly done via a formula as given by Kölbig [12], (see also [11]), orChoi and Cvijović [5] in terms of the Polylogarithmic or other special functions.Harmonic number sums in closed form 345In concluding we remark that the general results obtained by the evaluationof identities (5), (8) and (12) are an extension and generalization of the identitiesobtained by Sofo [18].References[1] H.Alzer, D.Karaannakis, H.M. Srivastava, Series representations of somemathematical constants, J. 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We extend some results of Euler related sums. Integral and closed form representation of sums with products of harmonic numbers and cubed binomial coefficients are developed in terms of Polygamma functions. The given representations are new.\ud
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Harmonic number sums in closed form

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