Some Observations on Alternating Power Sums

Laala Khaldi LIMPAF Laboratory Department of Mathematics University of Bouira 10000 Bouira Algeria

                     l.khaldi@univ-bouira.dz

Rachid Boumahdi National Higher School of Mathematics Sidi Abdellah, Algiers Algeria

                        r boumehdi@esi.dz

Abstract

Let x ≥ 1 be a real number and T_{n}(m) = −1^{m} + 2^{m} − ⋯ + (−1)^{n}n^{m}, where n and m are nonnegative integers with n ≥ 1. In this note we obtain an explicit formula for T_{⌊x⌋}(m), where ⌊x⌋ is the greatest integer less than or equal to x, and we establish a new expression for alternating power sums T_{n}(m) in terms of Stirling numbers of the second kind. Moreover, we give a congruence involving alternating sums of falling factorial.

1 Introduction and statement of results

Although they are important in many branches of mathematics and science, such as combinatorics, statistics and number theory, alternating power sums of integers

∑_{k=1}^{n}(−1)^{k}k^{m}

where n and m are nonnegative integers with n ≥ 1, have attracted less interest, of mathematicians, compared to the sums of powers of integers with positive signs

∑_{k=1}^{n} k^{m}, (m ≥ 0).

These last sums have been studied since antiquity to the Renaissance by Faulhaber, Fermat, Pascal and Bernoulli, to the modern era, where we find an abundant wealth of formulas in terms of Bernoulli, Stirling, Euler numbers and various other number sequences. One can

find the expression ∑_{k=1}^{n}(−1)^{k}k^{m} in a manuscript of Euler [4, p. 499] entitled ”Institutiones

Calculi Differentialis”, published in 1755, and over the years many formulas were also found in term of many polynomials and number sequences, among them the Euler polynomials.

Recall that Euler polynomials (E_{n}(x))_{n≥0} can be defined via generating series by

∑_{n=0}^{∞} E_{n}(x)\frac{t^{n}}{n!} = \frac{2e^{xt}}{e^{t} + 1}, (|t| < π)

and Euler numbers (see \underline{A122045} in the On-Line Encyclopedia of Integer Sequences (OEIS) [8]) can be defined by E_{n} = 2^{n}E_{n}(\frac{1}{2}). It is well-known that the alternating power sum T_{n}(m) can be written via Euler polynomials as follows [1, p. 804] and [6, Theorem 1]:

T_{n}(m) = \frac{(−1)^{n}E_{m}(n + 1) + E_{m}(0)}{2}, for n, m ≥ 1.

For example, the first four alternating power sums are

T_{n}(0) = ∑_{k=1}^{n}(−1)^{k} = \frac{−1 + (−1)^{n}}{2}

T_{n}(1) = ∑_{k=1}^{n}(−1)^{k}k = −\frac{1}{4} + \frac{(−1)^{n}}{4} (2n + 1)

T_{n}(2) = ∑_{k=1}^{n}(−1)^{k}k^{2} = \frac{(−1)^{n}}{2} n(n + 1)

T_{n}(3) = ∑_{k=1}^{n}(−1)^{k}k^{3} = \frac{1}{8} + \frac{(−1)^{n}}{8} (2n + 1)(2n^{2} + 2n − 1).

For nonnegative integers n, k, the binomial coefficient is defined by

\binom{n}{k} = \begin{cases}\frac{n!}{k!(n−k)!}, & if 0 ≤ k ≤ n; \\ 0, & otherwise. \end{cases}

Named after James Stirling, who defined and studied them in 1730, the Stirling numbers of the second kind S(n, k) (\underline{A008277}) may be defined as follows: S(n, k) is the number of partitions of the set {1, …, n} into k non-empty subsets. For more details on Stirling numbers and their properties, the reader is referred to [3, 5].

The following result expresses T_{⌊x⌋}(m) by means of Euler polynomials:

Theorem 1. Let x ∈ [1, ∞[. For a nonnegative integer m ≥ 1, we have

∑_{k=1}^{⌊x⌋}(−1)^{k}k^{m} = \frac{1}{2} (E_{m}(0) + (−1)^{⌊x⌋} ∑_{j=0}^{m}(−1)^{j}\binom{m}{j}E_{j}(\{x\})x^{m−j})

where ⌊x⌋ = max{k ∈ ℤ/ k ≤ x} and {x} = x − ⌊x⌋ is the fractional part of x.

For x a rational number we have the following corollary:

Corollary 2. Let p be an odd prime and m a nonnegative integer. Then

∑_{k=1}^{\frac{p−1}{2}}(−1)^{k}k^{2m} = \begin{cases}^{\frac{−1+(−1)^{⌊ \frac{p}{2} ⌋}}{2}} , & if m = 0; \\ ^{\frac{(−1)^{⌊ \frac{p}{2} ⌋}}{2^{2m+1}}} ∑_{j=0}^{2m}(−1)^{j}\binom{2m}{j}p^{2m−j}E_{j}, & if m ≥ 1, \end{cases}

and

∑_{k=1}^{\frac{p−1}{2}}(−1)^{k}k^{2m−1} = \frac{1 − 4^{m}}{2m} B_{2m} + \frac{(−1)^{⌊ \frac{p}{2} ⌋}}{2^{2m}} ∑_{j=0}^{2m−1}(−1)^{j}\binom{2m − 1}{j}p^{2m−1−j}E_{j},

where B_{n} is the nth Bernoulli number.

Note that by using the extended Boole summation formula, Schumacher [7, Section 4] obtained the following expression

∑_{k=1}^{⌊x⌋}(−1)^{k+1}k^{m} = η(−m) + \frac{(−1)^{x−⌊x⌋}}{2(m + 1)} ∑_{k=0}^{m}(−1)^{k+1}(k + 1)\binom{m + 1}{k + 1}E_{k}(\{x\})x^{m−k},

where η(s) := ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{k^{s}} is the Dirichlet eta function.

The following theorem provides an expression of ∑_{k=0}^{n−1}(−1)^{k}k^{m} in terms of Stirling numbers

of second kind.

Theorem 3. Let n and m be nonnegative integers with n ≥ 1. Then

∑_{k=0}^{n−1}(−1)^{k}k^{m} = (−1)^{m} ∑_{k=1}^{m+1} \frac{(−1)^{k−1}(k − 1)!}{2^{k}}S(m + 1, k)

+ (−1)^{n+1} ∑_{k=0}^{m} ∑_{ℓ=1}^{k+1} \binom{m}{k}\frac{(−1)^{k+ℓ−1}(ℓ − 1)!}{2^{ℓ}}n^{m−k}S(k + 1, ℓ),

where S(n, k) is the kth Stirling number of the second kind.

The following theorem gives us a congruence modulo p of the alternating sum of the falling factorials.

Recall that a p-integer is a rational number whose denominator is coprime with p. In the ring of p-integers, denoted by ℤ_{p}, we write \frac{a}{b} ≡ \frac{c}{d} (mod p) when ad − bc ≡ 0 (mod p) in ℤ.

Theorem 4. Let p ≥ 3 be a prime and m a nonnegative integer. Then

∑_{k=0}^{p−1}(−1)^{k}k\underline{m} ≡ (−1)^{m} \frac{m!}{2^{m}} (mod p).

Note that the falling factorial x\underline{m} (\underline{A068424}) is the polynomial in x defined by x\underline{0} := 1

and x\underline{m} : = ∏_{k=0}^{m−1}(x − k) for integers m ≥ 1.

2 Proofs

The proof of Theorem 1 is mainly based on the following two lemmas and some properties of Euler polynomials [1, p. 804].

Lemma 5. For integers n ≥ 1 and m ≥ 0, we have

T_{n}(m) = \frac{1}{2}(E_{m}(0) + (−1)^{n+m}E_{m}(−n)).

Proof. By using the relation E_{n}(x + 1) + E_{n}(x) = 2x^{n}, we have

∑_{k=0}^{n−1}(−1)^{k}k^{m} = \frac{1}{2} ∑_{k=0}^{n−1}(−1)^{k}(E_{m}(k + 1) + E_{m}(k))

= \frac{1}{2} (∑_{k=0}^{n−1}(−1)^{k}E_{m}(k) − ∑_{k=1}^{n}(−1)^{k}E_{m}(k))

= \frac{1}{2}(E_{m}(0) − (−1)^{n}E_{m}(n)).

Then we deduce that

∑_{k=1}^{n}(−1)^{k}k^{m} = \frac{1}{2}(E_{m}(0) + (−1)^{n}E_{m}(n + 1)).

As E_{m}(1 − x) = (−1)^{m}E_{m}(x), for x = −n we obtain E_{m}(1 + n) = (−1)^{m}E_{m}(−n), then it follows that

∑_{k=1}^{n}(−1)^{k}k^{m} = \frac{1}{2}(E_{m}(0) + (−1)^{n+m}E_{m}(−n)).

Lemma 6. Let m, j, k be non-negative integers. Then

\binom{m}{k}\binom{m − k}{j} = \binom{m}{j}\binom{m − j}{k}.

Proof. A direct calculation gives

\binom{m}{k}\binom{m − k}{j} = \frac{m!}{k!(m − k)!} \frac{(m − k)!}{j!(m − k − j)!} = \frac{m!}{k!j!(m − k − j)!}

= \frac{m!}{j!(m − j)!} \frac{(m − j)!}{k!(m − j − k)!} = \binom{m}{j}\binom{m − j}{k}.

Proof of Theorem 1. With the help of the relation

E_{n}(x) = ∑_{k=0}^{n} \binom{n}{k}\frac{E_{k}}{2^{k}} (x − \frac{1}{2})^{n−k}

and Lemma 5, we have

∑_{k=1}^{n}(−1)^{k}k^{m} = \frac{1}{2} (E_{m}(0) + (−1)^{n+m} ∑_{k=0}^{m} \binom{m}{k}\frac{E_{k}}{2^{k}} (−n − \frac{1}{2})^{m−k}) .  (1)

Taking n = ⌊x⌋ = x − {x} for x ≥ 1 in Eq. (1), we obtain the following:

∑_{k=1}^{⌊x⌋}(−1)^{k}k^{m} = \frac{1}{2}E_{m}(0) + \frac{(−1)^{⌊x⌋}}{2^{m+1}} ∑_{k=0}^{m} \binom{m}{k}(−1)^{k}E_{k}(2x + 1 − \{x\})^{m−k}.  (2)

Putting a = 2x, b = 1 − 2{x} and n = m − k in Newton’s binomial formula

(a + b)^{n} = ∑_{j=0}^{n} \binom{n}{j}a^{j}b^{n−j},

we obtain

(2x + 1 − 2\{x\})^{m−k} = ∑_{j=0}^{m−k} \binom{m − k}{j}(2x)^{j}(1 − 2\{x\})^{m−k−j}.  (3)

Then by Relations (2) and (3), we deduce

T_{⌊x⌋}(m) = \frac{1}{2}E_{m}(0) + \frac{(−1)^{⌊x⌋}}{2} ∑_{j=0}^{m}(−1)^{m−j} \frac{E_{k}}{2^{k}} x^{j} ∑_{k=0}^{m−j} \binom{m}{k}\binom{m − k}{j} (\{x\} − \frac{1}{2})^{m−j−k} .

Finally, an appeal to Lemma 6, gives

T_{⌊x⌋}(m) = \frac{1}{2}E_{m}(0) + \frac{(−1)^{⌊x⌋}}{2} ∑_{j=0}^{m} \binom{m}{j}(−1)^{m−j}x^{j} ∑_{k=0}^{m−j} \binom{m − j}{k}\frac{E_{k}}{2^{k}} (\{x\} − \frac{1}{2})^{m−j−k}

= \frac{1}{2}E_{m}(0) + \frac{(−1)^{⌊x⌋}}{2} ∑_{j=0}^{m} \binom{m}{j}(−1)^{m−j}E_{m−j}(\{x\})x^{j}.

This completes the proof.

Proof of Theorem 3. We consider the function Φ_{n}(x) = ∑_{k=0}^{n−1}(−1)^{k}e^{kx}. Since Φ_{n}(x) is the sum

of the geometric sequence of common ratio −e^{x}, then we have

Φ_{n}(x) = ∑_{k=0}^{n−1}(−1)^{k}e^{kx} = (−1)^{n+1} \frac{e^{nx}}{e^{x} + 1} + \frac{1}{e^{x} + 1}.  (4)

Differentiating the members of Eq. (4) m times, we obtain

Φ_{n}^{(m)}(x) = ∑_{k=0}^{n−1}(−1)^{k}k^{m}e^{kx} = (−1)^{n+1}φ(x) + ψ(x),  (5)

where φ(x) = (\frac{e^{nx}}{e^{x}+1})^{(m)} and ψ(x) = (\frac{1}{e^{x}+1})^{(m)}.

Note that g^{(m)} := \frac{d^{m}g}{dx^{m}} is the mth derivative of g with respect to x. By applying Leibniz’s formula of derivation, we can write φ(x) as follows:

φ(x) = (\frac{e^{nx}}{e^{x} + 1})^{(m)} = ∑_{k=0}^{m} \binom{m}{k}n^{m−k}e^{nx} (\frac{1}{e^{x} + 1})^{(k)} .

Taking λ = −1 and α = 1 in the Formula (3.1) of [9, Theorem 3.1]

(\frac{1}{1 − λe^{αx}} )^{(m)} = (−α)^{m} ∑_{k=1}^{m+1} \frac{(−1)^{k−1}(k − 1)!}{(1 − λe^{αx})^{k}}S(m + 1, k),  (6)

allows us to write φ(x) and ψ(x) as follows:

φ(x) = n^{m} ∑_{k=0}^{m} ∑_{ℓ=1}^{k+1} \binom{m}{k}\frac{1}{n^{k}} (−1)^{k+ℓ−1}(ℓ − 1)!S(k + 1, ℓ)\frac{e^{nx}}{(e^{x} + 1)^{ℓ}} ,  (7)

and

ψ(x) = (−1)^{m} ∑_{k=1}^{m+1} \frac{(−1)^{k−1}(k − 1)!}{(1 + e^{x})^{k}}S(m + 1, k).  (8)

Replacing x by 0 in Eq. (5), and using the expressions of φ(x) and ψ(x) given by (7) and (8) gives us the wanted result.

Corollary 7. Let n and m be nonnegative integers with n ≥ 1. Then

∑_{k=1}^{n}(−1)^{k}k^{m} = ∑_{k=1}^{m+1} \frac{(−1)^{k}(k − 1)!}{2^{k}}S(m + 1, k)

+ (−1)^{n} ∑_{k=0}^{m} ∑_{ℓ=1}^{k+1} \binom{m}{k}\frac{(−1)^{ℓ−1}(ℓ − 1)!}{2^{ℓ}}n^{m−k}S(k + 1, ℓ).

Proof. Here, using the function Ψ_{n}(x) = ∑_{k=1}^{n}(−1)^{k}e^{kx}, differentiating m times, and then

according to Eq. (6) for λ = −1 and α = −1, the result follows.

Proof of Theorem 4. In a similar way to the proof of Theorem 3, one considers the geometric sequence of common ratio −x, then the sum of the n first terms is

∑_{k=0}^{n−1}(−1)^{k}x^{k} = \frac{1}{1 + x} + (−1)^{n+1} \frac{x^{n}}{1 + x}.  (9)

Differentiating the members of Eq. (9) m times provides

∑_{k=0}^{n−1}(−1)^{k}k\underline{m}x^{k−m} = (−1)^{m} \frac{m!}{(1 + x)^{m+1}} + (−1)^{n+1}ϕ(x),  (10)

where ϕ(x) = (\frac{x^{n}}{x+1})^{(m)}. Applying the Leibniz derivation formula, ϕ(x) may be written as follows:

ϕ(x) = ∑_{k=0}^{m}(−1)^{m−k}\binom{m}{k}n(n − 1) ⋯ (n − k + 1)\frac{(m − k)!}{(1 + x)^{m−k+1}} x^{n−k}

= m! ∑_{k=0}^{m} \binom{n}{k}(−1)^{m−k} \frac{x^{m−k}}{(1 + x)^{m−k+1}} .  (11)

Replacing (11) in (10), we get

∑_{k=0}^{n−1}(−1)^{k}k\underline{m}x^{k−m} = (−1)^{m} \frac{m!}{(1 + x)^{m+1}} + (−1)^{n+1}m! ∑_{k=0}^{m} \binom{n}{k}(−1)^{m−k} \frac{x^{m−k}}{(1 + x)^{m−k+1}} . (12)

Now, replacing x by 1 in Eq. (12), we get

∑_{k=0}^{n−1}(−1)^{k}k\underline{m} = (−1)^{m} \frac{m!}{2^{m+1}} + (−1)^{n+1}m! ∑_{k=0}^{m} \binom{n}{k}(−1)^{m−k} \frac{1}{2^{m−k+1}}

= (−1)^{m} \frac{m!}{2^{m+1}} (1 + (−1)^{n+1} ∑_{k=0}^{m} \binom{n}{k}(−2)^{k}) .  (13)

Finally, taking n = p in Eq. (13) and by the trivial congruence \binom{p}{k} ≡ 0 (mod p) for k = 1, …, p − 1, we have

∑_{k=0}^{p−1}(−1)^{k}k\underline{m} ≡ (−1)^{m} \frac{m!}{2^{m}} (mod p).

This evidently completes the proof.

Note that Eq. (13) has been proven by Bazsó and Mező [2] by using Cauchy’s residue theorem.

Remark 8. Taking m = p − 1 in Theorem 4 and using Fermat little theorem for a = 2 with Wilson theorem, we obtain the following congruence:

∑_{k=0}^{p−1}(−1)^{k}k\underline{p−1} ≡ −1 (mod p).

3 Acknowledgments

The authors are grateful to the editor-in-chief for many useful comments and suggestions. Thanks are also due to the anonymous referees for the careful reading of the manuscript and bringing our attention to the papers [6] and [7], which helped us to improve the presentation.

References

[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, 1964.

[2] A. Bazsó and I. Mező, Some notes on alternating power sums of arithmetic progressions, J. Integer Sequences 21 (2018), Article 18.7.8.

[3] C. A. Charalambides, Enumerative Combinatorics, CRC Press, 2002.

[4] L. Euler, Institutiones Calculi Differentialis, Academie Imperialis Scientiareum Petropolitanae, 1755.

[5] H. W. Gould, Combinatorial Identities for Stirling Numbers: The Unpublished Notes of H W Gould, World Scientific, 2015.

[6] T. Kim, A note on the alternating sums of powers of consecutive integers, preprint,

    2005. Available at http://arxiv.org/abs/math/0508233.

[7] R. Schumacher, Extension of summation formulas involving Stirling series, preprint,

    2016. Available at http://arxiv.org/abs/1605.09204.

[8] N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, 2022. Available

    at https://oeis.org.

[9] A.-M. Xu and Z.-D. Cen, Some identities involving exponential functions and Stirling numbers and applications, J. Comput. Appl. Math. 260 (2014), 201–207.

2020 Mathematics Subject Classification: Primary 11B68; Secondary 11B73, 11B65, 11A07. Keywords: power sum, Euler polynomial, Stirling number of the second kind, congruence.

(Concerned with sequences \underline{A008277}, \underline{A068424}, and \underline{A122045}.)

Received March 8 2022; revised versions received June 1 2022; June 7 2022. Published in Journal of Integer Sequences, June 10 2022.

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