On a recurrence relation for the sums of powers of integers

Abstract

Recently, Thomas and Namboothiri [1] derived a recurrence identity expressing an exponential power sum with negative powers in terms of another exponential power sum with positive powers. From this result, the authors obtained a corresponding recurrence relation for the ordinary power sums $S_k(n) = 1^k + 2^k + \cdots + n^k$. In this short note, we provide an alternative simple proof of the latter recurrence. Our proof is based on the following two ingredients: (i) an expression for $S_k(n+m)$, and (ii) the symmetry property of the power sum polynomials $S_k(n)$.

Full text

On a recurrence relation for the sums
of powers of integers
*
Jos´e Luis Cereceda
Collado Villalba, 28400 (Madrid), Spain
jl.cereceda@movistar.es
Abstract
Recently, Thomas and Namboothiri [1] derived a recurrence identity expressing an exponen-
tial power sum with negative powers in terms of another exponential power sum with positive
powers. From this result, the authors obtained a corresponding recurrence relation for the or-
dinary power sums Sk(n)=1k+ 2k+· · · +nk. In this short note, we provide an alternative
simple proof of the latter recurrence. Our proof is based on the following two ingredients: (i) an
expression for Sk(n+m), and (ii) the symmetry property of the power sum polynomials Sk(n).
1 Introduction
In a remarkable paper [1], Thomas and Namboothiri derived a recurrence identity expressing the
exponential power sum Pk−1
s=1 spe
−2πims
kin terms of another exponential power sum with positive
powers (see Proposition 2.1 of [1]). As a consequence of this result, the authors obtained a corre-
sponding recurrence relation for the ordinary power sums Sk(n)=1k+ 2k+· · · +nk. Specifically,
they showed that (see, equation (8) of [1])
k−1
X
s=1
sp=kp(k−1) +
p−1
X
a=0
(−1)p−ap
aka
k−1
X
s=1
sp−a.
It is easy to see that the above equation is equivalent to the recurrence relation given in Propo-
sition 1.1 below. In Section 2, we provide a simple novel proof of the recurrence in question. In
Section 3, we derive a variant of Proposition 1.1. We conclude in Section 4 with some final remarks.
Proposition 1.1. For integers k, n ≥1, let Sk(n)denote the power sum 1k+ 2k+· · · +nk. Then,
k−1
X
j=0
(−1)jk
jnk−jSj(n−1) = 2Sk(n−1),if k is odd;
0,if k is even,(1)
where S0(n) = n, and where it is assumed that Sj(0) = 0 for all j.
2 Proof of Proposition 1.1
Let us first observe that, as shown in [2], Sk(n+m) can be expressed as
Sk(n+m) = Sk(n) + Sk(m) +
k−1
X
j=0 k
jnk−jSj(m),(2)
*
This is a slightly corrected version of the paper published in the Far East Journal of Mathematical Education
24, 2023, pp. 45-50. The two corrections made here are: the inclusion of the factor (−1)kin the right-hand side of
equation (3), and the quotation of Theorem 10 of [3], instead of Theorem 3 of [3].
1

because
Sk(n+m)=1k+ 2k+· · · +nk+ (n+ 1)k+· · · + (n+m)k
=Sk(n) +
k
X
j=0 k
jnk−j1j+
k
X
j=0 k
jnk−j2j
+· · · +
k
X
j=0 k
jnk−jmj
=Sk(n) +
k
X
j=0 k
jnk−j1j+ 2j+· · · +mj.
Therefore, making the transformations n→ −nand m→n−1, it follows that
Sk(−1) = Sk(−n) + Sk(n−1) + (−1)k
k−1
X
j=0
(−1)jk
jnk−jSj(n−1).(3)
On the other hand, by invoking the symmetry property of the power sum polynomials Sk(n) (cf.,
e.g., Theorem 10 of [3])
Sk(−n−1) = −δk,0+ (−1)k+1Sk(n),(4)
we immediately deduce that, for all k≥1,
Sk(−n) = (−1)k+1Sk(n−1).(5)
In particular, from (5) we find that Sk(−1) = 0 provided k≥1. Hence, substituting (5) into (3)
and taking into account that Sk(−1) = 0 for k≥1 yields
1+(−1)k+1 Sk(n−1) =
k−1
X
j=0
(−1)jk
jnk−jSj(n−1), k ≥1,
which is just the recurrence relation (1).
Remark 2.1. By letting n= 2 in (1), we obtain the following identity for k≥1:
k−1
X
j=0 −1
2jk
j=1/2k−1,if k is odd;
0,if k is even.
3 A variant of Proposition 1.1
It is worthwhile to point out that, by making the transformation n→ −nand using the symmetry
property (4), we can express (1) in the alternate form
nk+
k−1
X
j=0
(−1)jk
jnk−jSj(n) = 2Sk(n),if kis odd ;
0,if kis even.(6)
2

Let us note, incidentally, that letting n= 1 in the preceding equation leads to the well-known
binomial identity
k
X
j=0
(−1)jk
j= 0,for all k≥1.
On the other hand, putting n=−1 in (2) and recalling that Sk(−1) = 0 for all k≥1, we obtain
mk= (−1)k−1
k−1
X
j=0
(−1)jk
jSj(m), k ≥1.
Thus, renaming mas nin the last equation and substituting the resulting expression for nkinto
(6) gives rise to the following variant of (1):
k−1
X
j=0
(−1)jk
jnk−j−(−1)kSj(n) = 2Sk(n),if kis odd ;
0,if kis even.(7)
For the case where k≥1 is odd, from (7) we have
Sk(n) = 1
2
k−1
X
j=0
(−1)jk
jnk−j+ 1Sj(n),odd k≥1,
which should be compared with the formula for Sk(n) that is obtained from (1) when k≥1 is odd,
namely
Sk(n) = 1
2
k−1
X
j=0
(−1)jk
j(n+ 1)k−jSj(n),odd k≥1.(8)
Likewise, for the case in which k≥2 is even, from (7) we obtain the identity
k−1
X
j=0
(−1)jk
jnk−j−1Sj(n)=0,even k≥2.(9)
Notice that (9) allows one to obtain the odd-indexed power sum Sk−1(n) in terms of S0(n),
S1(n), . . . , Sk−2(n).
4 Concluding remarks
We end this note with the following remarks.
Remark 4.1. The recurrence relation (1)was already proved by El-Mikkawy and Atlan in Theorem
3.1 (viii) of [4] by manipulating the exponential generating function of Sk(n).
Remark 4.2. In Proposition 8.6.1 of [5], Treeby obtained the following recursive formula
Sk(n) = 1
k−1
k−1
X
j=1
(−1)jk
j−1(n+ 1)k−jSj(n),(10)
which holds for any odd integer k≥3.
3

Remark 4.3. By combining (8)and (10), we obtain in turn the recursive formula
Sk(n) = 1
k+ 1
k−1
X
j=0
(−1)jk+ 1
j(n+ 1)k−jSj(n),odd k≥1,(11)
which applies to any odd integer k≥1. Note that, for each j= 0,1, . . . , k −1, the product
(n+ 1)k−jSj(n)is a polynomial in nof degree k+ 1.
Example 4.4. When k= 3, from (11)it follows that
S3(n) = 1
4(n+ 1)3S0(n)−4(n+ 1)2S1(n) + 6(n+ 1)S2(n)
=1
4n(n+ 1)3−2n(n+ 1)3+n(n+ 1)2(2n+ 1),
which, after simplifying, gives us the well-known result
S3(n) = 1
4n2(n+ 1)2=S1(n)2.
A beautiful proof without words of the identity S3(n)=(S1(n))2can be found in the recent
paper [6]. Finally, it should be noted that the formula (11) can also be obtained directly starting
from (9). We leave it as an easy exercise for the interested reader to check this fact.
References
[1] N. E. Thomas and K. V. Namboothiri, On an exponential power sum, preprint (2023), available
at https://arxiv.org/abs/2303.10853v1
[2] D. R. Snow, Some identities for sums of powers of integers, Proceedings of Utah Academy of
Sciences, Arts, and Letters, 52 Part 2, pp. 29–31 (1975).
[3] N. J. Newsome, M. S. Nogin, and A. H. Sabuwala, A proof of symmetry of the power sum
polynomials using a novel Bernoulli number identity, Journal of Integer Sequences 20, Article
17.6.6 (2017).
[4] M. El-Mikkawy and F. Atlan, Notes on the power sum 1k+ 2k+. . . +nk,Annals of Pure and
Applied Mathematics 9(2), 215–232 (2015).
[5] D. Treeby, Optimal block stacking and combinatorial identities via Archimedes’ method, PhD
thesis, Monash University, 2018.
[6] F. Mart´ınez de la Rosa, Proof without words: sums of cubes in puzzle form, Far East Journal
of Mathematical Education 23, 7 (2022).
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