Applied Mathematics E-Notes, 12(2012), 110-117 c ISSN 1607-2510 Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
Quadratic Harmonic Number Sums
Anthony Sofoy and Mehdi Hassaniz
Received 12 July 2012
Abstract 2 In this paper, we obtain some identities for the series n1=1 Hn=(n(n+k)) and 2 n+k n 1 n1=1 Hn= n k , where Hn = j=1 j and k is a positive integer. Then we obtain some series representations for the Apéry’sconstant,P (3). P P 1 Introduction
s The Riemann zeta function is de…ned for s C with (s) > 1 by (s) = j1=1 j . n s 2 < For integer n > 1 we let n(s) = j=1 j . We de…ne the n-th harmonic number by (r) P Hn = n(1), and the generalized nP-th harmonic number by Hn = n(r), for any real (r) number r. Moreover, we set H0 = 0. Identities for sums involving harmonic numbers, generalized harmonic numbers, and their powers are rare in number in the literature. A classical example is due to L. Euler [3], where for integers q > 3 he proved that
q 2 1 H 2 n = (q + 2) (q + 1) (m + 1) (q m) : nq n=1 m=1 X X 2 Some recently obtained identities are n1=1 Hn=n = 17(4)=4 due to D. Borwein and J. M. Borwein [2], the following one due to A. Sofo [7] which is valid for integers P k > 2 1 H2 k 2 n = (2) + H(2) ; n+k k 1 2 k 1 n=1 k (k 1) ! X n 2 2 3 and 3 j=1 Hj =j H j=j = Hn n(3) due to M. Hassani [5]. In this paper, we obtain some identities for the series P 1 H2 1 H2 S(m) := n ; and B(k) := n ; n(n + m) n+k n=1 n=1 n k X X where m and k are positive integers. More precisely, we show the following.
Mathematics Subject Classi…cations: 05A10, 11B65, 11M06, 33B15, 33D60, 33C20. yVictoria University College, Victoria University, P. O. Box: 14428, Melbourne City, VIC 8001, Australia zDepartment of Mathematics, University of Zanjan, University Blvd., P. O. Box: 45371-38791, Zanjan, Iran
110 A. Sofo and M. Hassani 111
THEOREM 1. Assume that m > 1 is an integer, and let
(2) (3) 3 (m) = Hm 1(2) + Hm 1Hm 1 Hm 1 + Hm 1: F Also, for j = m let 6
Hj 1 1 2 (2) (m; j) = 2 + Hj 1 + Hj 1 : A mj 2j(m j) Then, we have m 1 3(3) + (m) S(m) = F (m; j): (1) m A j=1 X
THEOREM 2. Assume that k > 1 is an integer, and let
(2) (3) 3 Hk 1Hk 1 Hk 1 Hk 1 (k) = (2)Hk 1 + + + : G 2 3 6 Also, for j = r let 6 (2) 2 r 1 2 (2) Hr 1 Hr 1 + Hr 1 H + H Hj 1 j 1 j 1 (r) = + : C r rj2 2j(r j) j=1 ! X Then, we have k k B(k) = 3(3) + (k) + ( 1)r+1r (r): (2) r G r=1 C X
Then we obtain some new series representations for (3), which is known as the Apéry’sconstant (see [4], pp 40–52). More precisely, applying (1) with m = 5, and (2) with k = 1 and k = 4, respectively, we get the following.
COROLLARY 1. We have
1 H2 3(3) 5(2) 8737 n = + + : n(n + 5) 5 12 8640 n=1 X
COROLLARY 2. We have
1 H2 1 H2 49 128587 n = 3(3) and n = 3(3) (2) + : n(n + 1) n+4 20 216000 n=1 n=1 n 4 X X 112 Quadratic Harmonic Number Sums
2 Auxiliary Lemmas
In this section we introduce two auxiliary lemmas, which are the base of proofs of our (1) results. In what follows below, we will use both of notations Hn and Hn for the n-th harmonic number, and we will apply the following known [2] integral representation
H 1 n+1 = xn ln(1 x)dx: (3) n + 1 Z0 j n Also, we recall the polylogarithm function de…ned by Lin : z 1 z =j for integral 7! j=1 n > 2 and z in the unit disk. The function Li2 is known as dilogarithm function. We P note that Lin(1) = (n). The identity (4) and its proof of the following Lemma is due to Furdui.
LEMMA 1. Assume that m > 1 is an integer, and let
(1) (1) 1 Hn H (m) = n+m : H n(n + m) n=1 X Then, we have
(1) 2(3) (2)H(1) 1 m H (m) = + m j + T (m); (4) H m m m j2 j=1 X where m 1 m 1 1 3 2 T (m) = ( 1)j : j (j + 1)3 j + 1 m j=0 X Also, we have
2 3 (1) H(1) H(1) 2(3) (2)Hm 1 m 1 m 1 (m) = + + + H m m 2m2 2m (2) (1) (2) m 1 (1) Hm 1 3Hm 1Hm 1 Hj 1 + + : (5) 2m2 2m mj2 j=1 X PROOF. For x (0; 1) we de…ne the function f by 2 1 1 x 1 2 ln(u 1) f(x) = ln(x) ln(1 x) ln (1 x) + Li2(x) + du: 2 u Z1
Since f 0(x) = 0 and limx 0+ f(x) = 0, we imply f(x) = 0. Thus, for x (0; 1) we obtain ! 2 1 1 x 1 2 ln(u 1) ln(x) ln(1 x) ln (1 x) + Li2(x) = du: (6) 2 u Z1 A. Sofo and M. Hassani 113
By using (3), we get 1 1 1 m 1 n 1 m (m) = x ln(1 x) ln(1 y) (xy) dydx = x ln(1 x) (x)dx; H I 0 0 n=1 0 Z Z X Z where 1 ln(1 y) (x) = dy: I 1 xy Z0 By letting 1 xy = t in (x) we obtain I 1 1 ln(x 1 + t) (x) = ln(x) ln(1 x) + dt : I x 1 x t Z Then, we substitute t = u (1 x), and we combine the result with (6) to get 1 1 2 (x) = ln (1 x) + Li2(x) : I x 2 Thus, we obtain 1 1 1 m 1 3 m 1 (m) = x ln (1 x)dx x ln(1 x)Li2(x)dx: (7) H 2 Z0 Z0 We have 1 m 1 m 1 3 j+1 m 1 6 x ln (1 x)dx = ( 1) ; (8) j (j + 1)4 0 j=0 Z X and similarly
1 m 1 m 1 2 j m 1 2 x ln (1 x)dx = ( 1) : (9) j (j + 1)3 0 j=0 Z X To evaluate the second part of the integral in (7) we use integration by parts by setting m 1 ln(1 x) u(x) = Li2(x) and v0(x) = x ln(1 x), from which we get u0(x) = x and i v(x) = 1 (xm 1) ln(1 x) m x . Hence, by considering (9) we obtain m i=1 i 1 1 m 1 P 1 m 1 2 x ln(1 x)Li2(x)dx = x ln (1 x)dx m Z0 Z0 m 1 Hi = 2(3) (2)Hm : (10) m i2 i=1 ! X Combining (7), (8) and (10) completes the proof of (4). To prove (5) we consider (4). In the interest of expressing T (m) in terms of harmonic numbers, we note that 3 2 (1) (1) Hm 3H(1)H(2) H(3) Hm H(2) T (m) = + m m + m m 2m 2m m m2 m2 2 3 (1) H(1) H(1) (2) (1) (2) (3) 1 Hm 1 m 1 m 1 Hm 1 3Hm 1Hm 1 Hm 1 = + + + + + + : m4 m3 2m2 2m 2m2 2m m 114 Quadratic Harmonic Number Sums
Hence, we get
(1) m 1 (1) 2 (3) (2) (1) 1 Hm 1 1 Hj 1 (m) = + Hm 1 + 3 2 H m m m m m j j=1 X 2 3 (1) H(1) H(1) (2) (1) (2) Hm 1 m 1 m 1 Hm 1 3Hm 1Hm 1 + + + + + ; m3 2m2 2m 2m2 2m and consequently, we obtain (5). This completes the proof.
Our next lemma, gives identities for n1=1 Hn=(n(n+m)(n+r)), where m and r are positive integers. We distinguish two cases r = m and r = m, which the last case results in identities involving (3). During the proof,P 6 we need a continuous version of harmonic numbers (to di¤erentiate). Such continuous versions are available by considering the relation of harmonic numbers with digamma (psi) function and polygamma functions of order m, which are de…ned by (x) := d(log (x))=dx, and (m)(x) := dm (x)=dxm, 1 t x 1 respectively. Note that (x) = 0 e t dt is the Euler gamma function. Since (x + 1) = x (x), we have Hn = (1) + (n + 1). On the other hand, it is known that (1) = (see [1], page 258),R where is the Euler–Mascheroni constant (see [4], pp 28–40). Thus Hn = + (n + 1): (11) Similar relation for generalized harmonic numbers asserts that ( 1)r H(r+1) = (r + 1) + (r)(n + 1); (12) n r! where r > 1 is an integer (see [1], page 260). LEMMA 2. Assume that m and r are positive integers, and let
(1) 1 H (m; r) = n : n(n + m)(n + r) J n=1 X Then, we have
(1) (2) (3) (2) Hm 1 (m; m) = J m2 m m 2 (2) (2) (3) (Hm 1) Hm 1 Hm 1Hm 1 Hm 1 + + + + : (13) 2m2 2m2 m m Also, for r = m we have 6 2 2 H(1) (2) H(1) (2) (2) m 1 Hm 1 r 1 Hr 1 (m; r) = + + : (14) J mr 2m (m r) 2m (m r) 2r (m r) 2r (m r) PROOF. To prove (13) we start from the known (see [6]) identity
1 (n) 2m = 2(m + 1) 2 + (2) (1)(m + 1) := ` (m); (15) n(n + m) 1 n=1 X A. Sofo and M. Hassani 115 say. Di¤erentiating both sides of (15) with respect to m gives us
1 (n) ` (m) ` (m) = 1 2 ; n(n + m)2 2m2 2m n=1 X where d (1) (2) `2(m) = `1(m) = 2 (m + 1) (m + 1) (m + 1): dm By using (11), and considering the known property (x + 1) = (x) + 1=x (see [1]), we obtain 1 1 1 ` (m) ` (m) (m; m) = + + 1 2 : n(n + m)2 n2(n + m)2 2m2 2m J n=1 n=1 X X Following the method used in the paper by A. Sofo [8], we obtain
1 1 1 1 = n(n + m)2 m n(n + m) (n + m)2 n=1 n=1 X X = (m + 1) + + (2) m (1)(m + 1) ; (16) m2 and
1 1 1 1 1 2 1 = + n2(n + m)2 m2 n2 n(n + m) (n + m)2 n=1 n=1 X X 1 = m(2) + m (1)(m + 1) 2 (m + 1) 2 : (17) m3 (p) (p) p Combining (12), (16), (17) and Hm = Hm 1 + 1=m for p = 1; 2; 3, we get (13). To prove (14), we assume that r = m, and we apply (12) and (15) in 6 (1) 1 H 1 1 (m; r) = n : n (m r) n + r n + m J n=1 X This gives the identity (14) and completes the proof. Finally, by using the results and techniques from Wang [9] the following can be shown.
LEMMA 3. Assume that k > 1 is an integer. Then, we have
k r k (1) (1) ( 1) Hr 1 = Hk 1; (18) r r=1 X and k 3 (1) (2) (3) r k (3) (Hk 1) Hk 1Hk 1 Hk 1 ( 1) Hr 1 = + + : (19) r 6 2 3 r=1 X 116 Quadratic Harmonic Number Sums
3 Proof of Theorems
We now prove our Theorems. PROOF of Theorem 1. We consider the following identity
(1) (1) m 1 m 1 Hn Hn + j=1 n+j (m) = = S(m) + (m; j): H n(n + m) J n=1 P j=1 X X m Thus S(m) = (m) j=1 (m; j). By using (5), (13) and (14) in this identity, we obtain (1). H J P
PROOF of Theorem 2. We consider the following expansion
2 2 (1) (1) k 1 k! Hn 1 k! Hn A B(k) = = r ; k n n + r n=1 n=1 r=1 X n (n + r) X X r=1 Q where
(n + r) r+1 r k Ar = lim 0 1 = ( 1) : n r k k! r ! B (n + r)C B r=1 C @ A Thus, we obtain Q k k B(k) = ( 1)r+1r S(r): r r=1 X Now, we use (1), and then we apply (18) and (19) to get (2). This completes the proof. Acknowledgment. We thank O. Furdui for the statement of relation (4) and its proof.
References
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