aa r X i v : . [ m a t h . N T ] M a r Summation formulas of q -hyperharmonic numbers Takao Komatsu
Department of Mathematical Sciences, School of ScienceZhejiang Sci-Tech UniversityHangzhou 310018 China [email protected]
Rusen Li
School of MathematicsShandong UniversityJinan 250100 China [email protected]
Abstract
In this paper, several weighted summation formulas of q -hyperharmonicnumbers are derived. As special cases, several formulas of hyperharmonicnumbers of type P nℓ =1 ℓ p H ( r ) ℓ and P nℓ =0 ℓ p H ( r ) n − ℓ are obtained. Keywords:
Hyperharmonic numbers, Stirling numbers, q -generalizations Spieß[15] gives some identities including the types of P nℓ =1 ℓ k H ℓ , P nℓ =1 ℓ k H n − ℓ and P nℓ =1 ℓ k H ℓ H n − ℓ . In particular, explicit forms for r = 0 , , , P nℓ =1 ℓ k H ( r ) ℓ and P nℓ =1 ℓ k H ( r ) n − ℓ are shownas special cases of more general results, where H ( r ) ℓ denotes hyperharmonicnumbers defined in (4). When r = 1, H n = H (1) n is the original harmonicnumber defined by H n = P nj =1 /j . This paper is also motivated from thesummation P nℓ =1 ℓ k , which is related to Bernoulli numbers. In [1], Stirlingnumbers are represented via harmonic numbers and hypergeometric functionsrelated to Euler sums. In this paper, the sums involving harmonic numbersand their q -generalizations are expressed by using Stirling numbers and their q -generalizations.There are many generalizations of harmonic numbers. Furthermore, some q -generalizations of hyperharmonic numbers have been proposed. In this paper,based upon a certain type of q -harmonic numbers H ( r ) n ( q ) defined in (3), sev-eral formulas of q -hyperharmonic numbers are also derived as q -generalizations.1hese results are also motivated from the q -analogues of the sums of consecutiveintegers ([9, 14, 16]).In order to consider the weighted summations, we are motivated by the factthat the sum of powers of consecutive integers 1 k +2 k + · · · + n k can be explicitlyexpressed in terms of Bernoulli numbers or Bernoulli polynomials. After seeingthe sums of powers for small k : n X ℓ =1 ℓ = n ( n + 1)2 , n X ℓ =1 ℓ = n ( n + 1)(2 n + 1)6 , n X ℓ =1 ℓ = (cid:18) n ( n + 1)2 (cid:19) , . . . , the formula can be written as n X ℓ =1 ℓ k = 1 k + 1 k X j =0 (cid:18) k + 1 j (cid:19) B j n k +1 − j (1)= 1 k + 1 ( B k +1 ( n + 1) − B k +1 (1)) [6] , (2)where Bernoulli numbers B n are determined by the recurrence formula k X j =0 (cid:18) k + 1 j (cid:19) B j = k + 1 ( k ≥ t − e − t = ∞ X n =0 B n t n n ! , and Bernoulli polynomials B n ( x ) are defined by the following generating func-tion te xt e t − ∞ X n =0 B n ( x ) t n n ! . If Bernoulli numbers B n are defined by te t − ∞ X n =0 B n t n n ! , we can see that B n = ( − n B n . Then n X ℓ =1 ℓ k = 1 k + 1 k X j =0 (cid:18) k + 1 j (cid:19) ( − j B j n k +1 − j . We recall the well-known Abel’s identity, which is frequently used in the presentpaper. 2 emma 1. (Abel’s identity) For any positive integer n , n X ℓ =1 a ℓ b ℓ = s n b n + n − X ℓ =1 s ℓ ( b ℓ − b ℓ +1 ) . where s n = n X ℓ =1 a ℓ . In the weight of harmonic numbers H n , we have the following formulas. Proposition 1.
For n, k ≥ , n X ℓ =1 ℓ k H ℓ = H n k + 1 k X j =0 (cid:18) k + 1 j (cid:19) B j n k +1 − j − n − X ℓ =1 ( H n − H ℓ ) ℓ k . Proof.
Set a ℓ = ℓ k and b ℓ = H ℓ in Lemma 1. With n − X ℓ =1 s ℓ ( H ℓ − H ℓ +1 ) = s ( H − H ) + · · · + s n − ( H n − − H n )= 1 k H + · · · + ( n − k H n − − s n − H n = − n − X ℓ =1 ( H n − H ℓ ) ℓ k , formula (1) gives the result. Proposition 2.
For n, k ≥ , n X ℓ =1 ℓ k H ℓ = H n k + 1 ( B k +1 ( n + 1) − B k +1 (1)) − n − X ℓ =1 B k +1 ( ℓ + 1) − B k +1 (1)( k + 1)( ℓ + 1) . Proof.
Set a ℓ = ℓ k and b ℓ = H ℓ in Lemma 1. Formula (2) gives the result. q -hyperharmonic num-bers Many types of q -generalizations have been studied for harmonic numbers (e.g.,[11,17]). In this paper, a q -hyperharmonic number H ( r ) n ( q ) (see [12]) is defined by H ( r ) n ( q ) = n X j =1 q j H ( r − j ( q ) ( r, n ≥
1) (3)with H (0) n ( q ) = 1 q [ n ] q n ] q = 1 − q n − q . Note that lim q → [ n ] q = n . In this q -generalization, H n ( q ) = H (1) n ( q ) = n X j =1 q j − [ j ] q is a q -harmonic number. When q → H n = lim q → H n ( q ) is the originalharmonic number and H ( r ) n = lim q → H ( r ) n ( q ) is the r -th order hyperharmonicnumber, defined by H ( r ) n = n X ℓ =1 H ( r − ℓ with H (1) n = H n . (4)Mansour and Shattuck [12, Identity 4.1, Proposition 3.1] give the followingidentities H ( r ) n ( q ) = (cid:18) n + r − r − (cid:19) q (cid:0) H n + r − ( q ) − H r − ( q ) (cid:1) (5)= n X j =1 (cid:18) n + r − j − r − (cid:19) q q rj − [ j ] q , (6)where (cid:18) nk (cid:19) q = [ n ] q ![ k ] q ![ n − k ] q !is a q -binomial coefficient with q -factorials [ n ] q ! = [ n ] q [ n − q · · · [1] q . Notethat the identities (5) and (6) are q -generalization of the identities (7) and (8),respectively. H ( r ) n = (cid:18) n + r − r − (cid:19) ( H n + r − − H r − ) [7] (7)= n X j =1 (cid:18) n + r − j − r − (cid:19) j [2] . (8)So, we can see the recurrence relation for r ≥ H ( r +1) n = n + rr H ( r ) n − r (cid:18) n + r − r (cid:19) . The generating function of this type of q -hyperharmonic numbers is givenby ∞ X n =1 H ( r ) n ( q ) z n = − log q (1 − q r z ) q ( z ; q ) r ( r ≥
0) (9)4[12, Theorem 3.2]), where − log q (1 − t ) = ∞ X m =1 t m [ m ] q is the q -logarithm function and( z ; q ) k := k − Y j =0 (1 − zq j )is the q -Pochhammer symbol. When q →
1, (9) is reduced to the generatingfunction of hyperharmonic numbers: ∞ X n =1 H ( r ) n z n = − log(1 − z )(1 − z ) r ( r ≥ . In fact, the same form is given by Knuth [10] as ∞ X n = r − (cid:18) nr − (cid:19) ( H n − H r − ) z n − r +1 = − log(1 − z )(1 − z ) r ( r ≥ . By (5), we have H ( r +1) n ( q ) − [ n + r ] q [ r ] q H ( r ) n ( q ) = − q r − [ r ] q (cid:18) n + r − r (cid:19) q . Hence, H ( r +1) n ( q ) = [ n + r ] q [ r ] q H ( r ) n ( q ) − q r − [ r ] q (cid:18) n + r − r (cid:19) q . (10)By replacing n by n + 1 and r by r − n + r ] q H ( r ) n ( q )= [ n + 1] q H ( r ) n +1 ( q ) − q n + r − (cid:18) n + r − r − (cid:19) q . (11)Mansour and Shattuck [12, Theorem 3.3] also give the following formula, H ( r ) n ( q ) = n X j =1 q j ( r − m ) (cid:18) n + r − m − j − r − m − (cid:19) H ( m ) j ( q ) . (12)When q →
1, (12) is reduced to H ( r ) n = n X j =1 (cid:18) n + r − m − j − r − m − (cid:19) H ( m ) j (see also [2],[3, 2.4.Theorem]). When m = 0, (12) is reduced to (6).We prove a more general result of (5).5 heorem 1. For nonnegative integers n and k and a positive integer r , we have (cid:18) k + r − k (cid:19) q H ( k + r ) n ( q ) = (cid:18) n + kn (cid:19) q H ( r ) n + k ( q ) − (cid:18) n + k + r − n (cid:19) q H ( r ) k ( q ) . Remark. If r = 1 and k is replaced by r − q → Proof of Theorem 1.
The proof is done by induction on k . When k = 0, theidentity is clear since both sides are equal to H ( r ) n ( q ). Assume, then, thatthe identity has been proved for 0 , , · · · , k . We give some explanations forthe following calculation. Firstly, by replacing r by k + r in (10), we get thefirst identity. Secondly, by using the inductive assumption, we get the secondidentity. Thirdly, by replacing n by n + k and n by k respectively in (11), weget the third identity. Then, we have (cid:18) k + rk + 1 (cid:19) q H ( k + r +1) n ( q )= (cid:18) k + rk + 1 (cid:19) q [ n + k + r ] q [ k + r ] q H ( k + r ) n ( q ) − (cid:18) k + rk + 1 (cid:19) q q k + r − [ k + r ] q (cid:18) n + k + r − k + r (cid:19) q = [ n + k + r ] q [ k + 1] q (cid:18) n + kn (cid:19) q H ( r ) n + k ( q ) − [ n + k + r ] q [ k + 1] q (cid:18) n + k + r − n (cid:19) q H ( r ) k ( q ) − q k + r − [ k + r ] q (cid:18) k + rk + 1 (cid:19) q (cid:18) n + k + r − k + r (cid:19) q = [ n + k + 1] q [ k + 1] q (cid:18) n + kn (cid:19) q H ( r ) n + k +1 ( q ) − q n + k + r − [ k + 1] q (cid:18) n + kn (cid:19) q (cid:18) n + k + r − r − (cid:19) q − [ n + k + r ] q [ k + 1] q (cid:18) n + k + r − n (cid:19) q [ k + 1] q [ k + r ] q H ( r ) k +1 ( q )+ [ n + k + r ] q [ k + 1] q (cid:18) n + k + r − n (cid:19) q q k + r − [ k + r ] q (cid:18) k + r − r − (cid:19) q − q k + r − [ k + r ] q (cid:18) k + rk + 1 (cid:19) q (cid:18) n + k + r − k + r (cid:19) q = (cid:18) n + k + 1 n (cid:19) q H ( r ) n + k +1 ( q ) − (cid:18) n + k + rn (cid:19) q H ( r ) k +1 ( q ) . We used the relation [ n + k + r ] q − q n [ k + r ] q = [ n ] q in the final part.Cereceda [5] gives the following formula,lim n →∞ H ( n +1) n +1 H ( n ) n = 4 . q -hyperharmonic numbers of type H ( n ) n ( q ) has a differentphenomenon. Proposition 3.
For | q | < , we have lim n →∞ H ( n +1) n +1 ( q ) H ( n ) n ( q ) = q . Proof.
Since (1 − q n +1 )(1 − q n )(1 − q n +1 )(1 − q n ) → | q | < , n → ∞ )and H n +1 ( q ) − H n ( q ) H n − ( q ) − H n − ( q ) = q (cid:18) n + 1] q + q [ n + 2] q + · · · + q n [2 n + 1] q (cid:19) n ] q + q [ n + 1] q + · · · + q n − [2 n − q → q ( n → ∞ ) , from (5), H ( n +1) n +1 ( q ) H ( n ) n ( q ) = (cid:0) n +1 n (cid:1) q (cid:0) H n +1 ( q ) − H n ( q ) (cid:1)(cid:0) n − n − (cid:1) q (cid:0) H n − ( q ) − H n − ( q ) (cid:1) → · q = q . Theorem 2.
For positive integers n and r , n X ℓ =1 q ℓ − [ ℓ ] q H ( r ) ℓ ( q ) = [ n ] q [ n + r ] q [ r + 1] q H ( r ) n ( q ) − q r [ n − q [ n ] q ([ r + 1] q ) (cid:18) n + r − r − (cid:19) q = [ n ] q [ r ] q [ r + 1] q H ( r +1) n ( q ) + q r − [ r + 1] q (cid:18) n + rr + 1 (cid:19) q . (13) Proof.
Set a ℓ = q ℓ − (cid:0) ℓ + r − r (cid:1) q and b ℓ = H ℓ + r − ( q ). By using Lemma 1, we have n X ℓ =1 q ℓ − (cid:18) ℓ + r − r (cid:19) q H ℓ + r − ( q )= n X ℓ =1 q ℓ − (cid:18) ℓ + r − r (cid:19) q H n + r − ( q ) − n − X ℓ =1 q ℓ + r − [ ℓ + r ] q (cid:18) ℓ + rr + 1 (cid:19) q (cid:18) n + rr + 1 (cid:19) q H n + r − ( q ) − q r [ r + 1] q (cid:18) n + r − r + 1 (cid:19) q . (14)Hence, n X ℓ =1 q ℓ − [ ℓ ] q H ( r ) ℓ ( q )= n X ℓ =1 q ℓ − [ ℓ ] q (cid:18) ℓ + r − r − (cid:19) q ( H ℓ + r − ( q ) − H r − ( q ))= [ r ] q n X ℓ =1 q ℓ − (cid:18) ℓ + r − r (cid:19) q ( H ℓ + r − ( q ) − H r − ( q ))= [ r ] q n X ℓ =1 (cid:18) ℓ + r − r (cid:19) q H ℓ + r − ( q ) − [ r ] q H r − ( q ) (cid:18) n + rr + 1 (cid:19) q . (15)With the help of (5), (14) and (15), we get the desired result.When q →
1, Theorem 2 is reduced to the following.
Corollary 1.
For n, r ≥ , n X ℓ =1 ℓH ( r ) ℓ = n ( n + r ) r + 1 H ( r ) n − ( n − ( r +1) ( r − r + 1) = nrr + 1 H ( r +1) n + 1 r + 1 (cid:18) n + rr + 1 (cid:19) , where ( x ) ( n ) = x ( x + 1) · · · ( x + n − ( n ≥ ) denotes the rising factorial with ( x ) (0) = 1 . In order to establish similarly structured theorems of q -hyperharmonic num-bers, we recall the q -Stirling numbers of the second kind, denoted by S q ( n, m ),defined by Carlitz (see e.g. [4]) as([ x ] q ) n = n X m =0 q ( m ) S q ( n, m )([ x ] q ) ( m ) , ( n ∈ N ) , (16)where ([ x ] q ) ( m ) = [ x ] q [ x − q · · · [ x − m + 1] q denotes the q -falling factorial with([ x ] q ) = 1. The q -Stirling numbers of the second kind S q ( n, m ) satisfy therecurrence relation S q ( n + 1 , m ) = S q ( n, m −
1) + [ m ] q · S q ( n, m )with boundary values S q ( n,
0) = S q (0 , n ) = δ n , ( n ≥ q -version of the relation by Spieß[15], which is essential in theproof of the following structured theorem of q -hyperharmonic numbers of type P nℓ =0 q ℓ − ([ ℓ ] q ) p H ( r ) ℓ ( q ). Lemma 2.
Given summation formulas P nℓ =0 q ℓ − (cid:0) ℓj (cid:1) q [ c ℓ ] q = F q ( n, j ) for n, j ∈ N , one has n X ℓ =0 q ℓ − ([ ℓ ] q ) p [ c ℓ ] q = p X ℓ =0 q ( ℓ ) S q ( p, ℓ ) · [ ℓ ] q ! · F q ( n, ℓ ) . where S q ( p, ℓ ) denote the q -Stirling numbers of the second kind.Proof. Using (16), we have n X ℓ =0 q ℓ − ([ ℓ ] q ) p [ c ℓ ] q = n X ℓ =0 q ℓ − [ c ℓ ] q p X j =0 q ( j ) S q ( p, j ) · ([ ℓ ] q ) ( j ) = p X j =0 q ( j ) S q ( p, j )[ j ] q ! n X ℓ =0 q ℓ − (cid:18) ℓj (cid:19) q [ c ℓ ] q = p X j =0 q ( j ) S q ( p, j ) · [ j ] q ! · F q ( n, j ) . We introduce some notations. For n, r, p ∈ N , set n X ℓ =0 q ℓ − [ ℓ ] qp H ( r ) ℓ ( q ) = A q ( p, r, n ) H ( r ) n ( q ) − B q ( p, r, n ) . From (10), for p = 0, A q (0 , r, n ) = [ n + r ] q [ r ] q , B q (0 , r, n ) = q r − [ r ] q (cid:0) n + r − r (cid:1) q . FromTheorem 2, for p = 1, we know that A q (1 , r, n ) = [ n ] q [ n + r ] q [ r + 1] q ,B q (1 , r, n ) = q r [ n − q [ n ] q ([ r + 1] q ) (cid:18) n + r − r − (cid:19) q . Theorem 3.
For n, r, p ≥ , n X ℓ =0 q ℓ − [ ℓ ] qp H ( r ) ℓ ( q ) = A q ( p, r, n ) H ( r ) n ( q ) − B q ( p, r, n ) , where A q ( p, r, n ) 9 p X ℓ =0 q ( ℓ ) + p − S q ( p, ℓ )[ ℓ ] q ! (cid:18) n + r − r − (cid:19) − q (cid:18) r + ℓ − ℓ (cid:19) q (cid:18) r + nr + ℓ (cid:19) q ,B q ( p, r, n ) = p X ℓ =0 q ( ℓ ) + r +2 p − [ r + ℓ ] q S q ( p, ℓ )[ ℓ ] q ! (cid:18) r + ℓ − ℓ (cid:19) q (cid:18) r + n − r + ℓ (cid:19) q . Proof.
Set [ c ℓ ] q = H ( r ) ℓ ( q ) in Lemma 2. Then by using Lemma 1, we have F q ( n, p ) = n X ℓ =0 q ℓ − (cid:18) ℓp (cid:19) q H ( r ) ℓ ( q )= n X ℓ =1 q ℓ − (cid:18) ℓp (cid:19) q (cid:18) ℓ + r − r − (cid:19) q ( H ℓ + r − ( q ) − H r − ( q ))= n X ℓ =1 q ℓ − (cid:18) r + p − p (cid:19) q (cid:18) ℓ + r − r + p − (cid:19) q ( H ℓ + r − ( q ) − H r − ( q ))= q p − (cid:18) r + p − p (cid:19) q (cid:18) r + nr + p (cid:19) q H n + r − ( q ) − q p + r − (cid:18) r + p − p (cid:19) q n − X ℓ =1 q ℓ − [ ℓ + r ] q (cid:18) r + ℓr + p (cid:19) q − (cid:18) r + p − p (cid:19)(cid:18) r + nr + p (cid:19) H r − = q p − (cid:18) r + p − p (cid:19) q (cid:18) r + nr + p (cid:19) q ( H n + r − ( q ) − H r − ( q )) − q r +2 p − [ r + p ] q (cid:18) r + p − p (cid:19) q (cid:18) r + n − r + p (cid:19) q . (17)With the help of (5) and (17), Lemma 2 gives the result.When q →
1, Theorem 3 is reduced to the following.
Corollary 2.
For n, r, p ≥ , n X ℓ =0 ℓ p H ( r ) ℓ = A ( p, r, n ) H ( r ) n − B ( p, r, n ) , where A ( p, r, n ) = p X ℓ =0 S ( p, ℓ ) ℓ ! (cid:18) n + r − r − (cid:19) − (cid:18) r + ℓ − ℓ (cid:19)(cid:18) r + nr + ℓ (cid:19) ,B ( p, r, n ) = p X ℓ =0 r + ℓ S ( p, ℓ ) ℓ ! (cid:18) r + ℓ − ℓ (cid:19)(cid:18) r + n − r + ℓ (cid:19) . xample 1. p = 2 gives n X ℓ =1 q ℓ − ([ ℓ ] q ) H ( r ) ℓ ( q )= [ n ] q [ n + r ] q (1 + q [ r + 1] q [ n ] q )[ r + 1] q [ r + 2] q H ( r ) n ( q ) − q r [ n − q [ n ] q (cid:18) n + r − r − (cid:19) q q [ r + 1] q [ n ] q − q [ r ] q + [2] q [ r + 1] q [ r + 2] q . (18)Note that [ ℓ + 1] q = 1 + q · [ ℓ ] q and [ ℓ + 2] q = [2] q + q · [ ℓ ] q . With the helpof Theorem 3 and identities (13) and (18), we have the following identities. Forpositive integers n and r , n X ℓ =1 q ℓ − [ ℓ ] q [ ℓ + 1] q H ( r ) ℓ ( q )= [ n ] q [ n + r ] q ([2] q [ n + 2] q + q [ r − q [ n + 1] q )[ r + 1] q [ r + 2] q H ( r ) n ( q ) − q r [ n − q [ n ] q (cid:18) n + r − r − (cid:19) q [2] q [ r + 2] q + q [ r + 1] q [ n − q [ r + 1] q [ r + 2] q . n X ℓ =1 q ℓ − [ ℓ ] q [ ℓ + 1] q [ ℓ + 2] q H ( r ) ℓ ( q )= [ n ] q [ n + r ] q (cid:0) ( r + 1)( r + 2) n + 3( r + 1)( r + 4) n + 2( r + 6 r + 11) (cid:1) [ r + 1] q [ r + 2] q [ r + 3] q H ( r ) n ( q ) − q r [ n − q [ n ] q (cid:16) n + r − r − (cid:17) q ( r + 1) ( r + 2) n + ( r + 1) ( r + 16 r + 34) n + 12(3 r + 12 r + 11)([ r + 1] q ) ([ r + 2] q ) ([ r + 3] q ) . To give a more general result, we need the q -unsigned Stirling numbers ofthe first kind s uq ( n, k ) defined by[ ℓ ] ( n ) q = [ ℓ ] q [ ℓ + 1] q · · · [ ℓ + n − q = n X k =0 s uq ( n, k )([ ℓ ] q ) k , ( n ∈ N ) . The q -unsigned Stirling numbers of the first kind s uq ( n, k ) are well defined since[ ℓ + m ] q = [ m ] q + q m · [ ℓ ] q . Theorem 4.
For positive integers n, p and r , n X ℓ =1 q ℓ − [ ℓ ] ( p ) q H ( r ) ℓ ( q ) = A q ( p, r, n ) H ( r ) n − B q ( p, r, n ) , where A q ( p, r, n ) = p X m =0 s uq ( p.m ) A q ( m, r, n ) , q ( p, r, n ) = p X m =0 s uq ( p.m ) B q ( m, r, n ) . Proof. n X ℓ =1 q ℓ − [ ℓ ] ( p ) q H ( r ) ℓ ( q )= n X ℓ =1 q ℓ − p X m =0 s uq ( p, m )[ ℓ ] qm H ( r ) ℓ ( q )= p X m =0 s uq ( p, m ) n X ℓ =1 q ℓ − [ ℓ ] qm H ( r ) ℓ ( q )= p X m =0 s uq ( p, m )( A q ( m, r, n ) H ( r ) n ( q ) − B q ( m, r, n ))= p X m =0 s uq ( p, m ) A q ( m, r, n ) ! H ( r ) n ( q ) − p X m =0 s uq ( p, m ) B q ( m, r, n ) ! . When q →
1, Theorem 4 is reduced to the following.
Corollary 3.
For positive integers n, p and r , n X ℓ =1 ( ℓ ) ( p ) H ( r ) ℓ = A ( p, r, n ) H ( r ) n − B ( p, r, n ) , where A ( p, r, n ) = p X m =0 ( − p + m s ( p.m ) A ( m, r, n ) ,B ( p, r, n ) = p X m =0 ( − p + m s ( p.m ) B ( m, r, n ) . Now we consider backward summations of q -hyperharmonic numbers. Theorem 5.
For positive integers n and r , X ℓ =1 q n − ℓ [ ℓ ] q H ( r ) n − ℓ ( q )= [ n ] q [ n + r ] q [ r ] q [ r + 1] q H ( r ) n ( q ) − (cid:18) n + rr + 1 (cid:19) q (cid:18) q r − [ r ] q + q r [ r + 1] q − q n + r − [ n + r ] q (cid:19) . Proof.
Set a ℓ = q n − ℓ H ( r ) n − ℓ ( q ), and b ℓ = q n − ℓ [ ℓ ] q . By using Lemma 1 and[ ℓ + 1] q − q [ ℓ ] q = 1, we have n X ℓ =1 q n − ℓ [ ℓ ] q H ( r ) n − ℓ ( q )= [ n ] q · H ( r +1) n − ( q ) + n − X ℓ =1 ( H ( r +1) n − ( q ) − H ( r +1) n − ℓ − ( q ))( q n − ℓ [ ℓ ] q − q n − ℓ − [ ℓ + 1] q )= [ n ] q · H ( r +1) n − ( q ) + n − X ℓ =1 H ( r +1) n − ( q )( q n − ℓ [ ℓ ] q − q n − ℓ − [ ℓ + 1] q )+ n − X ℓ =1 H ( r +1) n − ℓ − ( q )( − q n − ℓ [ ℓ ] q + q n − ℓ − [ ℓ + 1] q )= q n − H ( r +1) n − ( q ) + n − X ℓ =1 q n − ℓ − H ( r +1) n − ℓ − ( q )= H ( r +2) n − ( q ) . With the help of (5), we get the desired result.When q →
1, Theorem 5 is reduced to the following.
Corollary 4.
For positive integers n and r , n X ℓ =1 ℓH ( r ) n − ℓ = n ( n + r ) r ( r + 1) H ( r ) n − ( n ) ( r ) (cid:0) (2 r + 1) n + r (cid:1) ( r − r ( r + 1) . It is more complicated to get a summation formula for the backward sum-mations of higher power. In the case where q →
1, we have more relations,including the following.
Theorem 6.
For positive integers n, p and r , n X ℓ =0 ℓ p H ( r ) n − ℓ = A ( p, r, n ) H ( r ) n − B ( p, r, n ) . where A ( p, r, n ) and B ( p, r, n ) satisfy the following relations: A ( p, r, n ) = A (0 , r, n ) p − X j =0 (cid:18) pj (cid:19) A ( j, r + 1 , n − , ( p, r, n )= B (0 , r, n ) p − X j =0 (cid:18) pj (cid:19) A ( j, r + 1 , n − + p − X j =0 (cid:18) pj (cid:19) B ( j, r + 1 , n − , with the initial values A (0 , r, n ) = nr and B (0 , r, n ) = r (cid:0) n + r − r (cid:1) . Nevertheless, we can have a different backward summation formula withoutweights.
Theorem 7.
For positive integers n, p and r , n X ℓ =1 q p ( n − ℓ ) H ( r ) n − ℓ ( q ) = C q ( p, r, n ) H ( r ) n ( q ) − D q ( p, r, n ) , where C q ( p, r, n ) and D q ( p, r, n ) satisfy the following recurrence relation. C q ( p, r, n ) = [ n ] q [ r ] q (cid:16) q ( p − n − + (1 − q p − ) C q ( p − , r + 1 , n − (cid:17) D q ( p, r, n )= q r − [ n ] q ([ r ] q ) (cid:18) n + r − r (cid:19) q (cid:16) q ( p − n − + (1 − q p − ) C q ( p − , r + 1 , n − (cid:17) + (1 − q p − ) D q ( p − , r + 1 , n − . Proof.
Set a ℓ = q n − ℓ H ( r ) n − ℓ ( q ) and b ℓ = q ( p − n − ℓ ) . By using Lemma 1 and[ ℓ + 1] q − q [ ℓ ] q = 1, we have n X ℓ =1 q p ( n − ℓ ) H ( r ) n − ℓ ( q )= H ( r +1) n − ( q ) + n − X ℓ =1 ( H ( r +1) n − ( q ) − H ( r +1) n − ℓ − ( q ))( q ( p − n − ℓ ) − q ( p − n − ℓ − )= H ( r +1) n − ( q ) + n − X ℓ =1 H ( r +1) n − ( q )( q ( p − n − ℓ ) − q ( p − n − ℓ − )+ n − X ℓ =1 H ( r +1) n − ℓ − ( q )( − q ( p − n − ℓ ) + q ( p − n − ℓ − )= q ( p − n − H ( r +1) n − ( q ) + (1 − q p − ) n − X ℓ =1 q ( p − n − ℓ − H ( r +1) n − ℓ − ( q )= q ( p − n − H ( r +1) n − ( q )+ (1 − q p − ) (cid:16) C q ( p − , r + 1 , n − H ( r +1) n − ( q ) − D q ( p − , r + 1 , n − (cid:17) . With the help of (5), we get the desired result.14 cknowledgement
Authors are grateful to the anonymous referee for helpful comments.
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