Harmonic Number Sums in Closed Form 1. Introduction

MATHEMATICAL COMMUNICATIONS 335 Math. Commun. 16(2011), 335–345.

Harmonic number sums in closed form

Anthony Sofo1,∗ 1 Victoria University College, Victoria University, PO Box 14 428, Melbourne City 8001, VIC, Australia

Received March 1, 2010; accepted November 17, 2010

Abstract. 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 are developed in terms of Polygamma functions. The given representations are new. AMS subject classifications: Primary 11B65; Secondary 33C20 Key words: polylogarithms, harmonic numbers, integral representations, binomial coefficients

1. Introduction

The generalized harmonic numbers of order α are given by

Xn 1 H(α) = , for n = 1, 2, 3, . . . , α = 1, 2, 3,... n rα r=1 and Z 1 1 − tn Xn 1 H(1) = H = dt = = γ + ψ (n + 1) , n n 1 − t r t=0 r=1 where γ denotes the Euler-Mascheroni constant, deﬁned by Ã ! Xn 1 γ = lim − log n = −ψ (1) ≈ 0.57721566490153286 ... n→∞ r r=1 Recently Jung, Cho and Choi [10] evaluated Euler sums from integrals and obtained results like X∞ H(1) 1 1 n = ζ (3) − (ln 2)3 . (1) n 2 4 3 n=1 2 (n + 1) In this paper we expand (1) to obtain results of the form

X H(1) 7 1 n = ζ (6) − ζ2 (3) − 18ζ (5) + 6ζ (3) ζ (2) − 15ζ (4) 5 3 4 2 (2) n≥1 n (2n + 1) −440 ζ (3) + 56 ln 2 ζ (3) + 240 ln 2 ζ (2) + 480 (ln 2)2

∗Corresponding author. Email address: [email protected] (A. Sofo)

http://www.mathos.hr/mc °c 2011 Department of Mathematics, University of Osijek 336 A. Sofo to add, in a small way, some results related to (1) and (3) and to extend the result of Cloitre, as reported in [19]. A remarkable recursion known to Euler [8] is

q−2 X∞ H(1) X 2 n = (q + 2) ζ (q + 1) − ζ (r + 1) ζ (q − r) (3) nq n=1 r=1 and there is also a recurrence formula nX−1 (2n + 1) ζ (2n) = 2 ζ (2r) ζ (2n − 2r) r=1 which shows that in particular, for n = 2, 5ζ (4) = 2 (ζ (2))2 and more generally that ζ (4n) is a rational multiple of (ζ (2n))2 . The Riemann zeta function

X∞ 1 ζ (z) = , Re (z) > 1. rz r=1 Here we are also interested in evaluating closed form and integral representations of (α) Euler type sums containing both harmonic numbers, Hn and powers of binomial coefficients, two types of special numbers of enumerative combinatorics. There are many 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. Chu and Zheng [7] also obtained many other identities involving harmonic numbers and central binomial coefficients.

2. Identities

Theorem 1. Let a, b, c, d ≥ 0 be real positive numbers, |t| < 1 and j, k, l, m ≥ 0, then P X tn an 1 ³ ´ ³ r=1´ ³r+j ´ ³ ´ 5 an+j bn+k cn+l dn+m n≥1 n j k l m Z Z Z Z 1 1 1 1 (1 − x)j (1 − y)k (1 − z)l (1 − w)m (4) = −abcd 0 0 0 0 xyzw ¡ ¢ × ln (1 − x) ln 1 − txaybzcwd dx dy dz dw

Proof. Consider X tn ³ ´ ³ ´ ³ ´ ³ ´ 5 an+j bn+k cn+l dn+m n≥1 n j k l m X tnn4Γ(an)Γ(bn)Γ(cn)Γ(dn)Γ(j + 1) Γ (k + 1) Γ (l + 1) Γ (m + 1) = abcd n5Γ(an + j + 1) Γ (bn + k + 1) Γ (cn + l + 1) Γ (dn + m + 1) n≥1 Harmonic number sums in closed form 337

X tnΓ(bn)Γ(cn)Γ(dn)Γ(k + 1) Γ (l + 1) Γ (m + 1) = abcd B (an, j + 1) nΓ(bn + k + 1) Γ (cn + l + 1) Γ (dn + m + 1) n≥1 Z X tn 1 = abcd B (bn, k + 1) B (cn, l + 1) B (dn, m + 1) xan−1 (1 − x)j dx, n n≥1 0 where the Beta function Z 1 Γ(s)Γ(w) B (s, z) = ws−1 (1 − w)z−1 dw = 0 Γ(s + w) for Re (s) > 0, Re (z) > 0 and the Gamma function

Z ∞ Γ(z) = wz−1e−w dw for Re (z) > 0. 0 Now diﬀerentiate with respect to the parameter j, then P X tn an 1 ³ ´ ³ r=1´ ³r+j ´ ³ ´ 5 an+j bn+k cn+l dn+m n≥1 n j k l m Z 1 (1 − x)j = −abcd ln (1 − x) 0 x ·Z Z Z ¸ X xant 1 1 1 × ybn−1 (1 − y)k zcn−1 (1 − z)l wdn−1 (1 − w)m dw dz dy dx n n≥1 0 0 0 Z Z Z Z 1 1 1 1 (1 − x)j ln (1 − x) · (1 − y)k (1 − z)l (1 − w)m = −abcd 0 0 0 0 xyzw ¡ ¢n X txaybzcwd × dx dy dz dw n n≥1 by a subsequent allowable change of sum and integral, hence P X tn an 1 ³ ´ ³ r=1´ ³r+j ´ ³ ´ 5 an+j bn+k cn+l dn+m n≥1 n j k l m Z Z Z Z 1 1 1 1 (1 − x)j (1 − y)k (1 − z)l (1 − w)m = −abcd xyzw 0 0 ¡0 0 ¢ × ln (1 − x) ln 1 − txaybzcwd dx dy dz dw.

We now investigate three new corollaries that are a consequence of the Main Theorem 1. 338 A. Sofo

Corollary 1. For b = c = d = a > 0, t = 1, j = 0 = l = m and k ≥ 1 an integer, we have

X H(1) ³ n ´ 5 n+k n≥1 n k Z Z Z Z 1 1 1 1 (1 − y)k ln (1 − x) ln (1 − (xyzw)a) = −a4 dx dy dz dw xyzw 0 0 0 0 ½ ¾ 7 ζ2 (3) 5a2 ³ ´2 = ζ (6) − − 3aH(1)ζ (5) + aH(1)ζ (3) ζ (2) + H(1) + H(2) ζ (4) 4 2 k k 8 k k ½ ¾ a3 ³ ´3 − H(1) + 3H(1)H(2) + 2H(3) ζ (3) 3 k k k k ½ ¾ a4 ³ ´4 ³ ´2 ³ ´2 + H(1) + 6 H(1) H(2) + 3 H(2) + 8H(1)H(3) + 6H(4) ζ (2) 24 k k k k k k k µ ¶½ ¾ 4 Xk r+1 ³ ´2 a (−1) k (1) (2) + H r −1 + H r −1 . (5) 2 r4 r a a r=1 Proof. Expand

X H(1) X k!H(1) X k!H(1) Xk A ³ n ´ n n r = Qk = 5 , 5 an+k n5 (an + r) n an + r n≥1 n k n≥1 r=1 n≥1 r=1 where ( ) µ µ ¶¶ an + s s+1 s k As = lim Qk = (−1) , n→ − s k! s ( a ) s=1 (an + s) hence ³ ´ (1) k (−1)r+1 r k X Hn X r n5 an + r n≥1 r=1 µ ¶ Xk k X H(1) = (−1)r+1 r n r n5 (an + r) r=1 n≥1 µ ¶ · ¸ Xk k X 1 a a2 a3 a4 = (−1)r+1 r H(1) − + − + r n rn5 r2n4 r3n3 r4n2 r4n (an + r) r=1 n≥1 µ ¶ · Xk k 7ζ (6) ζ2 (3) 3aζ (5) aζ (3) ζ (2) = (−1)r+1 r − − + r 4r 2r r2 r2 r=1 ½ ¾¸ 2 3 4 4 ³ ´2 5a ζ (4) 2a ζ (3) a ζ (2) a (1) (2) + − + + H r −1 + H r −1 . (6) 4r3 r4 r5 2r5 a a

Simplifying (6), we obtain (5). Harmonic number sums in closed form 339

Remark 1. The degenerate case, k = 0 gives the known result

X H(1) 7 1 n = ζ (6) − ζ2 (3) . n5 4 2 n≥1

Similarly,

X H(1) 7 1 22 425 9200 ¡ n ¢ = ζ (6) − ζ2 (3) − 22ζ (5) + ζ (3) ζ (2) + ζ (4) − ζ (3) n5 4n+3 4 2 3 9 27 n≥1 3 15616 30976 30368 247808 − ζ (2) + π ln 2 + (ln 2)2 − G, 27 27 9 81 X H(1) 7 1 9 3 ³ n ´ 2 n = ζ (6) − ζ (3) − ζ (5) + ζ (3) ζ (2) 5 4 +2 4 2 8 8 n≥1 n 2 315 15 31 12601699 + ζ (4) − ζ (3) + ζ (2) + , 2305 256 4096 722534400 where G is Catalan’s constant, deﬁned by

X∞ (−1)r G = 2 ≈ 0.915965 ... r=0 (2r + 1)

Corollary 2. For d = c = b = a > 0, t = 1, j = 0 = m, l = k ≥ 1 integers, we have

(1) X Hn ³ ´2 (7) an+k n≥1 5 n k Z Z Z Z 1 1 1 1 [(1 − y) (1 − z)]k = −a4 ln (1 − x) ln (1 − (xyzw)a) dx dy dz dw 0 0 0 0 (1 − xyzw) µ ¶ · 7 1 Xk k 2 6a = ζ (6) − ζ2 (3) + − (1 − rX (k, r)) ζ (5) 4 2 r r r=1 2a 5a2 + (1 − rX (k, r)) ζ (3) ζ (2) + (3 − 2rX (k, r)) ζ (4) r 2r2 3 3 ³ ´ a a (1) − 3 (9 − 4rX (k, r)) ζ (3) + 4 5a − rH r −1 − 2raX (k, r) ζ (2) r · ½ r ¾ a 4 ³ ´2 ³ ´ 1 5a (1) (2) 3 (1) (2) (3) + H r −1 + H r −1 + ra H r −1H r −1 + H r −1 r4 2 a a a a a ½ ¾ ¸¸ ³ ´2 4 (1) (2) −a r H r + H r X (k, r) , (8) a −1 a −1 where

(1) (1) X (k, r) := Hk−r − Hr−1. (9) 340 A. Sofo

Proof. Expand

(1) (1) 2 X Hn X Hn (k!) ³ ´2 = ¡ ¢2 an+k 5 n≥1 5 n≥1 n (an + 1) n k k+1 " # X H(1) (k!)2 Xk A B = n r + r , n5 an + r 2 n≥1 r=1 (an + r) where ( ) µ µ ¶¶ (an + s)2 s k 2 Bs = lim Qk 2 = n→ − s k! s ( a ) s=1 (an + s) and ( ) µ µ ¶¶ d (an + s)2 s k 2 As = lim Qk 2 = −2 X (k, s) , n→ − s dn k! s ( a ) s=1 (an + s) where X (k, r) is given by (9). Now, by interchanging the order of summation, we have X (1) Xk X (1) Xk X (1) Hn 2 Hn 2 Hn ³ ´2 = (k!) Ar 5 + (k!) Br 2 , (10) an+k n (an + r) 5 n≥1 5 r=1 n≥1 r=1 n≥1 n (an + r) n k where

(1) X Hn 5 2 n≥1 n (an + r) " # X 1 2a 3a2 4a3 a5 5a4 = H(1) − + − − + n r2n5 r3n4 r4n3 r5n2 5 2 r5n (an + r) n≥1 r (an + r) 7ζ (6) ζ2 (3) 6aζ (5) 2aζ (3) ζ (2) = − − + 4r2 2r2 r3 r3 3 (1) 2 3 4 a ζ (2) H r 15a ζ (4) 9a ζ (3) 5a ζ (2) a −1 + 4 − 5 + 6 − 5 4r½ r ¾ r r 4 ³ ´2 3 n o 5a (1) (2) a (1) (2) (3) + H r −1 + H r −1 + H r −1H r −1 + H r −1 (11) 2r6 a a r5 a a a and from (11) and the inner part of (6) into (10), we obtain

(1) X Hn ³ ´2 an+k n≥1 5 n k µ µ ¶¶ · Xk k 2 7ζ (6) ζ2 (3) 6aζ (5) 2aζ (3) ζ (2) = r − − + r 4r2 2r2 r3 r3 r=1 Harmonic number sums in closed form 341

3 (1) 2 3 4 a ζ (2) H r 15a ζ (4) 9a ζ (3) 5a ζ (2) a −1 + 4 − 5 + 6 − 5 4r½ r ¾ r r ¸ 4 ³ ´2 3 n o 5a (1) (2) a (1) (2) (3) + H r −1 + H r −1 + H r −1H r −1 + H r −1 2r6 a a r5 a a a µ µ ¶¶ · Xk k 2 7ζ (6) 3aζ (5) 5a2ζ (4) ζ2 (3) −2 r − + − r 4r r2 4r3 2r r=1 ½ ¾¸ 3 4 4 ³ ´2 aζ (3) ζ (2) 2a ζ (3) a ζ (2) a (1) (2) + − + + H r −1 + H r −1 X (k, r) . r r4 r5 2r5 a a

Now we collect the zeta functions, and using the property that µ ¶ Xk k 2 [1 − 2rX (k, r)] = 1 r r=1 then (8) is conﬁrmed. Remark 2. Some examples are

X (1) Hn 7 1 2 ¡ ¢2 = ζ (6) − ζ (3) − 36ζ (5) + 12ζ (3) ζ (2) + 115ζ (4) − 8376ζ (3) 5 4n+2 4 2 n≥1 n 2 −128π3 − 2304ζ (2) + 4656ζ (2) ln 2 + 4608π ln 2 +14112 (ln 2)2 − 12288G + 1024πG + 6144G ln 2, X (1) Hn 7 1 2 11 11 ³ ´2 = ζ (6) − ζ (3) − ζ (5) + ζ (3) ζ (2) n +3 4 2 3 9 n≥1 5 3 n 3 970 305 2496877 1108486911367 − ζ (4) − ζ (3) − ζ (2) + . 216 216 7348320 648121824000 Corollary 3. For d = c = b = a > 0, t = 1, j = 0, m = l = k ≥ 1 an integer, then

(1) X Hn ³ ´3 an+k n≥1 5 n k Z Z Z Z 1 1 1 1 [(1 − w) (1 − z) (1 − y)]k = −a4 0 0 0 0 xyzw × ln (1 − x) ln (1 − (xyzw)a) dx dy dz dw (12) µ ¶ · µ ¶ 7 1 Xk k 3 9 = ζ (6) − ζ2 (3) + (−1)r+1 −a + 18X (k, r) + 3rY (k, r) ζ (5) 4 2 r r µ r=1 ¶ 3 +a − 6X (k, r) + rY (k, r) ζ (3) ζ (2) r µ ¶ 29 45X (k, r) 5 +a2 − + Y (k, r) ζ (4) 4r2 4r 4 342 A. Sofo   (1) 25a H r −1 27aX (k, r) 2a +a2 − − a + + Y (k, r) ζ (3) r3 r2 r2 r    (1) (2) (1) 2  2  15a 5aH r −1 H r −1 5a aH r −1 +a2  − a − a − 3 − a X (k, r) r4 r3 r2  r3 r2  ¶ ½ ¾ 2 4 ³ ´2 3 n o a 15a (1) (2) 5a (1) (2) (3) + Y (k, r) ζ (2) + H r −1 + H r −1 + H r −1H r −1 + H r −1 r2 2r4 a a r3 a a a ½ ¾ 2 ³ ´2 a (2) (1) (3) (4) + H r −1 + 2H r −1H r −1 + 3H r −1 2r2 a a a a µ ½ ¾ ¶ 4 ³ ´2 3 n o 5a (1) (2) a (1) (2) (3) −3 H r −1 + H r −1 + H r −1H r −1 + H r −1 X (k, r) 2r3 a a r2 a a a ½ ¾ ¸ 4 ³ ´2 a (1) (2) + H r −1 + H r −1 Y (k, r) , (13) 2r2 a a where X (k, r) is given by (9) and

3 n o Y (k, r) := 3X2 (k, r) + H(2) + H(2) . (14) 2 k−r r−1 Proof. Expand

(1) (1) 3 X Hn X Hn (k!) ³ ´3 = Qk 3 an+k 5 n≥1 5 n≥1 n r=1 (an + r) n k " # X H(1) (k!)3 Xk A B C = n r + r + r , n5 an + r 2 3 n≥1 r=1 (an + r) (an + r) where ( ) 3 µ µ ¶¶3 (an + s) s+1 s k Cs = lim Qk 3 = (−1) , n→(− s ) (an + s) k! s a (s=1 ) 3 µ µ ¶¶3 d (an + s) s s k Bs = lim Qk 3 = 3 (−1) X (k, s) n→ − s dn k! s ( a ) s=1 (an + s) and ( ) 1 d2 (an + s)3 As = lim 2 Qk 3 2 n→ − s dn ( a ) s=1 (an + s) µ µ ¶¶ 3 s k 3 h i = (−1)s+1 3X2 (k, s) + H(2) + H(2) , 2 k! s k−s s−1 where X (k, r) is given by (9). Harmonic number sums in closed form 343

Now we can write

X (1) Xk X (1) Xk X (1) Hn 3 Hn 3 Hn ³ ´3 = Ar (k!) 5 + Br (k!) 2 an+k n (an + r) 5 n≥1 n5 r=1 n≥1 r=1 n≥1 n (an + r) k (15) Xk X (1) 3 Hn + Cr (k!) , 5 3 r=1 n≥1 n (an + r) where

· X H(1) X 1 3a 6a2 10a3 n = H(1) − + − 5 3 n r3n5 r4n4 r5n3 r6n2 n≥1 n (an + r) n≥1 # a5 5a5 15a4 − − + r5 (an + r)3 r6 (an + r)2 r6n (an + r) 7ζ (6) ζ2 (3) 9aζ (5) 3aζ (3) ζ (2) 29a2ζ (4) = − − + + 4r3 2r3 r4 r4 4r5 (1) 3 2 4 25a ζ (3) a ζ (3) H r −1 15a ζ (2) − − a + r6 r5 r7 3 (1) 2 (2) ½ ¾ 5a ζ (2) H r a ζ (2) H r 4 ³ ´2 a −1 a −1 15a (1) (2) − − + H r −1 + H r −1 r6 r5 2r7 a a 3 n o 5a (1) (2) (3) + 6 H r −1H r −1 + H r −1 r ½ a a a ¾ 2 ³ ´2 a (2) (1) (3) (4) + H r −1 + 2H r −1H r −1 + 3H r −1 (16) 2r5 a a a a

Substituting (16), (11) and the inner part of (6) into (15) we get

X (1) Xk µ µ ¶¶3 h i Hn 3 r+1 r k 2 (2) (2) ³ ´3 = (−1) 3X (k, r) + Hk−r + Hr−1 an+k 2 k! r n≥1 5 r=1 n k · 7ζ (6) 3aζ (5) 5a2ζ (4) ζ2 (3) × − + − 4r r2 4r3 2r ½ ¾¸ 3 4 4 ³ ´2 aζ (3) ζ (2) 2a ζ (3) a ζ (2) a (1) (2) + − + + H r −1 + H r −1 r r4 r5 2r5 a a µ µ ¶¶ · Xk k 3 7ζ (6) ζ2 (3) 6aζ (5) + 3 (−1)r r X (k, r) − − r 4r2 2r2 r3 r=1 2aζ (3) ζ (2) 15a2ζ (4) 9a3ζ (3) 5a4ζ (2) + − − + r3 4r4 r5 r6 344 A. Sofo  (1) 3 4 ½³ ´ ¾ 3 n o a ζ (2) H r −1 5a 2 a a (1) (2) (1) (2) (3)  − + H r −1 + H r −1 + H r −1H r −1 + H r −1 r5 2r6 a a r5 a a a µ µ ¶¶ · Xk k 3 7ζ (6) ζ2 (3) 9aζ (5) 3aζ (3) ζ (2) + (−1)r+1 r − − + r 4r3 2r3 r4 r4 r=1 (1) (1) 2 3 2 4 3 29a ζ (4) 25a ζ (3) a ζ (3) H r −1 15a ζ (2) 5a ζ (2) H r −1 + − − a + − a 4r5 r6 r5 r7 r6 2 (2) ½ ¾ a ζ (2) H r 2 ³ ´2 a −1 a (2) (1) (3) (4) − + H r −1 + 2H r −1Hr−1 + 3H r −1 r5 2r5 a a a ½ ¾ ¸ 4 ³ ´2 3 n o 15a (1) (2) 5a (1) (2) (3) + H r −1 + H r −1 + H r −1H r −1 + H r −1 , 2r7 a a r6 a a a 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 used the fact that µ µ ¶¶ Xk k 3 £ ¤ (−1)r+1 r 1 − 3rX (k, r) + r2Y (k, r) = 1, r r=1 and Y (k, r) is given by (14). Remark 3. Some examples are X (1) Hn 7 1 2 75 25 145805 ¡ ¢3 = ζ (6) − ζ (3) − ζ (5) + ζ (3) ζ (2) − ζ (4) 5 2n+4 4 2 2 2 48 n≥1 n 4 68053 35840 ln 2 2767747 4096 ln 2 − ζ (3) + ζ (3) − ζ (2) − ζ (2) 216 9 144 9 311296 157696 391657 + (ln 2)2 − ln 2 + . 27 27 48 For a = 2, k = 1 we obtain (2). Remark 4. In Corollaries 1, 2 and 3 we encounter harmonic numbers at possible (α) rational values of the argument, of the form H r where r = 1, 2, 3, . . . , k and a −1 (α) k ∈ N. To evaluate H r we have available a relation in terms of the Polygamma a −1 function ψ(α) (z) , for rational arguments z, α ³ ´ (α+1) (−1) (α) r H r −1 = ζ (α + 1) + ψ , a α! a where ζ (z) is the Riemann Zeta function. We also define ³ ´ (1) r (α) H r −1 = γ + ψ , and H0 = 0. a a ¡ ¢ (α) r 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[12], (see also [11]), or Choi and Cvijović[5] in terms of the Polylogarithmic or other special functions. Harmonic number sums in closed form 345

In concluding we remark that the general results obtained by the evaluation of identities (5), (8) and (12) are an extension and generalization of the identities obtained by Sofo [18].

References

[1] H. Alzer, D. Karaannakis, H. M. Srivastava, Series representations of some mathematical constants, J. Math. Anal. Appl. 320(2006), 145–162. [2] A. Basu, A new method in the study of Euler sums, Ramanujan J. 16(2008), 7–24. [3] D. Borwein, J. Borwein, R. Girgensohn, Explicit evaluation of Euler sums, Proc. Edinburgh Math. Soc. 38(1995), 277–294. [4] X. Chen, W. Chu, Dixon’s 3F2 (1)-series and identities involving harmonic numbers and the Riemann Zeta function, Discrete Math. 310(2009), 83–91. [5] J. Choi, Dj. Cvijovic´, Values of the polygamma functions at rational arguments, J. Phys. A 40(2007), 15019–15028. [6] W. Chu, A. M. Fu, Dougall-Dixon formula and harmonic number identities, Ra- manujan J. 18(2009), 11–31. [7] W. Chu, D. Zheng, Infinite series with harmonic numbers and central binomial coefficients, Int. J. Number Theory 5(2009), 429–448. [8] L. Euler, Opera Omnia, Vol XV, Teubner, Berlin, 1917. [9] P. Flajolet, B. Salvy, Euler sums and contour integral representations, Exp. Math. 7(1998), 15–35. [10] M. Jung, Y. j. Cho, J. Choi, Euler sums evaluatable from integrals, Commun. Korean Math. Soc. 19(2008), 545–555. [11] K. Kolbig¨ , The polygamma function ψ (x) for x = 1/4 and x = 3/4, J. Comput. Appl. Math. 75(1996), 43–46. [12] K. Kolbig¨ , The polygamma function and the derivatives of the cotangent function for rational arguments, CERN-IT-Reports, CERN-CN-96-005, 1996. [13] C. Krattenthaler, K. S. Rao, Automatic generation of hypergeometric identities by the beta integral method, J. Comput. Math. Appl. 160(2003), 159–173. [14] A. Sofo, Integral forms of sums associated with Harmonic numbers, Appl. Math. Comput. 207(2009), 365–372. [15] A. Sofo, Computational Techniques for the Summation of Series, Kluwer Aca- demic/Plenum Publishers, New York, 2003. [16] A. Sofo, Sums of derivatives of binomial coefficients, Advances Appl. Math. 42(2009), 123–134. [17] A. Sofo, Harmonic numbers and double binomial coefficients, Integral Transforms Spec. Funct. 20(2009), 847–857. [18] A. Sofo, Closed form representations of Harmonic sums, Hacettepe J. Maths. Stats. 39(2010), 255–263. [19] J. Sondow, E. W. Weisstein, Harmonic number, From MathWorld-A Wolfram Web Rescources, available at http://mathworld.wolfram.com/HarmonicNumber.html. [20] Wolfram Research Inc., Mathematica, Wolfram Research Inc., Champaign, IL.