Formulas for Sums of Powers of Integers and Their Reciprocals

Home , Harmonic number

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n+1 n 1 1 n +1 H(p) = = B(p). (5) n+1 kp n! j +1 j k j=0 X=1 X (p) Here, Bk is the kth poly-Bernoulli number deﬁned by [5, Eq. (1)] ∞ tk Li (1 − e−t) B(p) = p , (6) k k! 1 − e−t k X=1

2 where Lip (z) is the polylogarithm and has the generating function

∞ zk Li (z)= . p kp k X=1 The poly-Bernoulli numbers are a generalization of the classical Bernoulli (1) k number with Bk = (−1) Bk. They have interesting combinatorial interpre- tations, and also appear in special values of certain zeta functions. Now, we present formulas for the sum of powers of integers, ﬁrst of which is as a consequence of Theorem 1.

Corollary 2 For all non-negative integers n and p, we have

n p p +1 n kp = j! . j +1 j +1 k j=0 X=1 X Theorem 3 For all positive integers n and p, we have

n p p +1 n + j +1 kp = (−1)p+j j! . j +1 j +1 k j=0 X=1 X

n Here, k is the Stirling numbers of the second kind, count the number of ways to partition a set of n objects into k non-empty subsets.

2 Proofs

2.1 Proof of Theorem 1 Substituting t → 1 − e−t in the generating function of the generalized harmonic numbers ∞ Li (t) H(p)tk = p , |t| < 1, (7) k 1 − t k X=0 one can easily obtain that

−t ∞ Li (1 − e ) n p = (−1)n H(p) e−t − 1 e−t. 1 − e−t n+1 n=0 X We now utilize the following generating function of the second kind Stirling numbers [4, p. 351]

∞ k +1 zk (ez − 1)n = ez, n +1 k! n! k n X=

3 and deduce that

∞ k Li (1 − e−t) k +1 tk p = (−1)k−n n!H(p) . 1 − e−t n +1 n+1 k! k n=0 ! X=0 X Considering (6) and equating coefﬁcients of tn give

n n +1 B(p) = (−1)n−k k!H(p) . n k +1 k+1 k X=0 (p) (p) Finally, taking an = Bn and bk = k!Hk+1 in the well-known Stirling trans- form [4, p. 310]

n n n +1 n +1 a = (−1)n−k b if and only if b = a n k +1 k n k +1 k k k X=0 X=0 give the desired result.

2.2 Proof of Corollary 2

(−p) n p Since Hn = k=1 k , take −p (p > 0) in (5) and then utilize the following identity of poly-Bernoulli numbers [1, Theorem 2] P min{k,p} p +1 k +1 B(−p) = (j!)2 . k j +1 j +1 j=0 X We ﬁnd that

n+1 n n 1 p +1 n +1 k +1 kp = (j!)2 . n! j +1 k +1 j +1 k j=0 k j X=1 X X= Hence, thanks to the identity [9, Theorems 3.7 and 3.11]

n n +1 k +1 n! n +1 = , k +1 j +1 j! j +1 k j X= the proof is completed.

2.3 Proof of Theorem 3

(−p) n p With the aid of Hn = k=1 k ,p> 0, we may consider (7) as

n P∞ Li (t) kp tn = −p (1 − t) n=0 k ! X X=1

4 and use [10]

p p +1 −1 k+1 Li (t) = (−1)p+1 k! , p ≥ 1 −p k +1 1 − t k X=0 to conclude that

∞ n ∞ p p +1 k +1+ n kp tn = (−1)p+k k! tn k +1 k +1 n=0 k ! n=0 k ! X X=1 X X=0 which is what we wanted to prove.

References

[1] Arakawa T., Kaneko M. (1999). On poly-Bernoulli numbers. Comment. Math. Univ. St. Pauli 48(2): 159–167. [2] Cheon G.-S., El-Mikkawy M. E. A. (2008). Generalized harmonic numbers with Riordan arrays. J. Number Theory 128(2): 413–425. [3] Gould H. W. (1978). Evaluation of sums of convolved powers using Stir- ling and Eulerian numbers. Fibonacci Quart. 16: 488–497. [4] Graham R. L., Knuth D. E., Patashnik O. (1994). Concrete Mathematics. New York: Addison-Wesley. [5] Kaneko M. (1997). Poly-Bernoulli numbers. J. Theor. Nr. Bordx. 9: 221– 228. [6] Knuth D. E. (1993). Johann Faulhaber and sums of powers. Math. Com- put. 61(203): 277–294. [7] Conway J. H., Guy R. K. (1996). The Book of Numbers. New York: Springer-Verlag. [8] Merca M. (2015). An alternative to Faulhaber’s formula. Amer. Math. Monthly. 122(6): 599-601. [9] Nyul G., Racz´ G. (2015). The r-Lah numbers. Discrete Math. 338(10): 1660–1666 [10] Wood D. C. (1992). The computation of polylogarithms. Technical Report 15-92. Canterbury, UK: University of Kent Computing Laboratory.