Defining Sums of Products of Power Sums

Jitender Singh Department of Mathematics Guru Nanak Dev University Amritsar-143005 Punjab India

                      sonumaths@gmail.com

Abstract We study the sums of products of power sums of positive integers and their generalizations, using the multiple products of their exponential generating functions. The generalizations include a closed form expression for the sums of products of infinite series of the form ∑_{n=0}^{∞} α^{n}n^{k}, 0 < |α| < 1, k ∈ ℕ_{0} and the related Abel sum, which define, in a unified way, the sums of products of the power sums for all integers k and connect them with the zeta function.

1 Background

The discrete power sum of a positive integer n is defined as S_{k}(n) = 1^{k}+2^{k}+⋯+n^{k}, k ∈ ℕ_{0}. The power sums are not new and have been studied since Pascal (1654) who obtained the following expression for S_{k}(n) in terms of S_{k−1}(n), …, S_{0}(n)

(n + 1)^{k+1} − 1 = ∑_{m=0}^{k} \binom{k + 1}{m}S_{m}(n),  (1)

which shows that S_{k}(n) is a polynomial of degree k + 1 in n. Properties of these polynomials were observed by Faulhaber (1631). Since then, the power sums have remained fascinating even today in one or the other way. So, the literature is vast in context of different approaches and generalizations of the power sums. However, for a quick review of power sums, the reader may refer to a relatively new work in [1, 2, 3, 4, 5, 6].

As an extension of the discrete power-sum, the polynomial S_{k}(x) for k ∈ ℕ_{0} and complex number x is defined via its exponential generating function (hereafter referred to as egf) [8]

\frac{e^{(1+x)t} − e^{t}}{e^{t} − 1} = ∑_{k=0}^{∞} S_{k}(x)\frac{t^{k}}{k!}, |t| < 2π.  (2)

For fixed k, S_{k}(x) is given by the classical Faulhaber formula

S_{k}(x) = \frac{1}{k + 1} ∑_{m=0}^{k} \binom{k + 1}{m}(−1)^{m}B_{m}x^{k+1−m},  (3)

where the rational sequence {B_{k}}_{k∈ℕ_{0}} is generated by the egf

\frac{t}{e^{t} − 1}, |t| < 2π,  (4)

such that B_{k} is the k-th Bernoulli number.

For N ∈ ℕ, the sum of products of Bernoulli numbers denoted B_{k}^{N} is generated by the multiple product of the egf (4) in the ring of egfs over ℚ. The study has attracted the attention of researchers for several number theoretical aspects such as in evaluation of the sums of products of Riemann zeta functions. Srivastava and Todorov [9] have obtained a closed form expression for B_{k}^{N}.

Dilcher [10] has also evaluated B_{k}^{N} in closed form as the following expression:

B_{k}^{N} = N\binom{k}{N} ∑_{i=0}^{N}(−1)^{N−1−i}s(N, N − j)\frac{B_{k−i}}{k − i}, k > N,  (5)

where s(N, N − j) is the Stirling number of first kind.

In a short paper, Petojević and Srivastava [11] have computed the Dilcher sums of products of Bernoulli numbers in an elegant way. More generally, Kamano [12] has investigated sums of products of hypergeometric Bernoulli numbers and used them to study multiple hypergeometric zeta function.

On the other hand, Kim [13] has obtained sums of products of Bernoulli numbers, using an analytic continuation of the multiple hypergeometric zeta function. The work has been used by Kim and Hu [14] to describe the sums of products of Apostol-Bernoulli polynomials.

In the review of the sums of products, the author did not find any work on the sums of products of the power sums and their connection with the Riemann zeta function except for some work on the sum of products of the power sums of φ(n) integers in Singh [7], where φ denotes the Euler’s phi function. One may argue that the power sum is defined via its Bernoulli polynomial up to a constant, so all the properties of S_{k}(x) and its sums of products are contained in the study of the Bernoulli polynomial B_{k}(x). However, the power sums are fundamental, and the study of the multiple products of their egfs is important in its own right. The present work is an attempt to consider multiple products of egfs of the power sums and their generalizations in a very natural way via a considerably simple theory.

The paper is outlined as follows. The sums of products of power sums via the multiple product of their egfs are discussed in Section 2. Section 3 and 4 are devoted to explore the properties of the higher order α-Euler numbers [8] (a variant of the Apostol-Bernoulli numbers), their connection with the higher order Bernoulli numbers, and the sums of products of the alternating power sums. The related sums of products of the Abel sum of alternating infinite series are obtained in Section 4. In Section 5, the sums of products of the power sums are generalized to negative integer values, and their connection with the sum of products of zeta function is established.

2 Sums of products

Definition 1. In the ring of formal power series over a field 𝔽, where 𝔽 is one of the fields, ℚ, ℝ, and ℂ; an element f is said to be an exponential generating function (egf) if f is equal to the formal power series ∑_{k=0}^{∞} a_{k} \frac{t^{k}}{k!}, where each a_{k} ∈ 𝔽. The set of all egfs is also a ring with the component-wise addition (+) and multiplication (·), such that for f = ∑_{k=0}^{∞} a_{k} \frac{t^{k}}{k!} and g = ∑_{k=0}^{∞} b_{k} \frac{t^{k}}{k!}, a_{k}, b_{k} ∈ 𝔽,

f + g = ∑_{k=0}^{∞}(a_{k} + b_{k})\frac{t^{k}}{k!}; f · g = ∑_{k=0}^{∞} ∑_{m=0}^{k} \binom{k}{m}a_{m}b_{k−m} \frac{t^{k}}{k!},  (6)

where the sequence defined by c_{k} = \binom{k}{m}a_{m}b_{k−m}, k ∈ ℕ_{0} in 𝔽 is the Cauchy type product [15] of the sequences (a_{k})_{k∈ℕ_{0}} and (b_{k})_{k∈ℕ_{0}}.

For a detailed introduction to the ring of formal power series and the ring of egfs, the reader may refer to the review by Wilf [16].

Definition 2. Let N ∈ ℕ, and let f_{i} is the egf of the sequence (a_{ik})_{k∈ℕ_{0}}, for each i = 1, 2, …, N, where a_{ik} ∈ 𝔽. In the multiple product of egfs, i.e., f_{1} ⋯ f_{N} = ∑_{k=0}^{∞} c_{k} \frac{t^{k}}{k!}, the sequence defined by c_{k} = ∑_{k_{1}+···+k_{N} =k} \binom{k}{k_{1},...,k_{N}}a_{1k_{1}} ⋯ a_{Nk_{N}} in 𝔽 is called the sum of products of the sequences (a_{1k})_{k∈ℕ_{0}}, …, (a_{Nk})_{k∈ℕ_{0}}.

Definition 3. For x ∈ ℂ, let T_{k}^{N}(x) denotes the sum of products of the power sums, which is generated by the egf

H(x, t, N) = (\frac{e^{(1+x)t} − e^{t}}{e^{t} − 1} )^{N} = ∑_{k=0}^{∞} T_{k}^{N}(x)\frac{t^{k}}{k!}, |t| < 2π, N ∈ ℕ_{0}.  (7)

Remark 4. Observe from (7) that T_{0}^{0}(x) = 1; T_{k}^{0}(x) = 0 for all k ∈ ℕ; T_{k}^{N}(1) = N^{k} and T_{k}^{1}(x) = S_{k}(x). Remark 5. Since for each k ∈ ℕ_{0}, S_{k}(x) is a polynomial in x of degree k + 1 over ℚ with coefficient of the leading term (k + 1)^{−1} (see (3)), it follows that for each fixed N ∈ ℕ and partition {k_{1}, …, k_{N}} of k, the product S_{k_{1}}(x) ⋯ S_{k_{N}} (x) is a polynomial in x over ℚ of degree ∑_{i=1}^{N}(k_{i} + 1) = k + N with leading coefficient ∏_{i=1}^{N}(k_{i} + 1)^{−1} > 0. Also, as S_{k_{i}}(0) = 0

for all i = 1, …, k_{N}, x^{N} is always a factor of the polynomial S_{k_{1}}(x) ⋯ S_{k_{N}} (x). Thus, T_{k}^{N}(x) is a polynomial in x of degree k + N over ℚ since coefficient of x^{k+N} in T_{k}^{N}(x) is equal to the N-fold product of the egf of (k + 1)^{−1}. Therefore, T_{k}^{N}(x) is a linear combination of {x^{N}, x^{N+1}, …, x^{N+k}} over ℚ. In particular, the coefficient of x^{i}, i = 0, 1, …, N − 1, N > 0, in the expansion of T_{k}^{N}(x) in powers of x, is always zero. Also, note that T_{k}^{N}(−1) = 0 for all k ∈ ℕ. Thus, x^{N}(x + 1) is always a factor of T_{k}^{N}(x).

Proposition 6. For N ∈ ℕ_{0} and x ∈ ℝ,

(1) T_{0}^{N}(x) = x^{N}; T_{k}^{N}(x) = Nk ∫_{0}^{x} T_{k−1}^{N}(t) dt + N ∑_{m=0}^{k} \binom{k}{m}(−1)^{m}B_{m} ∫_{0}^{x} T_{k−m}^{N−1}(x), k ∈ ℕ.

(2) T_{k}^{N}(x) = (−1)^{k+N} ∑_{n=0}^{N} \binom{N}{n}T_{k}^{n}(−1 − x) for all k ∈ ℕ_{0}.

Proof. (1) Observe that T_{0}^{N}(x) = lim_{t→0} H(x, t, N) = x^{N}. The recurrence follows from the following

\frac{∂H}{∂x} = NtH(x, t, N) + NH(x, t, N − 1)\frac{te^{t}}{e^{t} − 1}.  (8)

(2) Follows from the following symmetry of H

(−1)^{N}H(x, t, N) = (\frac{e^{(1−1−x)(−t)} − e^{−t}}{e^{−t} − 1} + 1)^{N} = ∑_{n=0}^{N} \binom{N}{n}H(−1 − x, −t, n)  (9)

Definition 7. Higher order Bernoulli polynomial B_{k}^{N}(x), x ∈ ℂ is defined by the following egf:

\frac{t^{N}e^{xt}}{(e^{t} − 1)^{N}} = ∑_{k=0}^{∞} B_{k}^{N}(x)\frac{t^{k}}{k!}, N ∈ ℕ_{0}, |t| < 2π.  (10)

Definition 8. Higher order Bernoulli number is defined by B_{k}^{N} = B_{k}^{N}(0).

Remark 9. Note that B_{k}^{1}(x) = B_{k}(x) is the k-th Bernoulli polynomial and B_{k}^{1}(0) = B_{k}. Now the following recurrences are immediate: B_{0}^{0}(x) = 1, B_{k}^{0}(x) = x^{k} for k ∈ ℕ, and B_{k}^{N}(x) = ∑_{m=0}^{k} \binom{k}{m}B_{m}(x)B_{k−m}^{N−1}, which, for N = 1, reduces to the usual identity of the Bernoulli polynomials. Also,

B_{0}^{N+1} = 1; B_{k}^{N+1} = (1 − \frac{k}{N} )B_{k}^{N} − kB_{k−1}^{N}, k ∈ ℕ,  (11)

which can be used to evaluate B_{k}^{N} for N = 2, 3, . . ..

Remark 10. It is not hard to see that T_{k}^{N}(x) can be expressed in terms of B_{k}^{N}(x) by

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

Proposition 11. Let N, k ∈ ℕ_{0}, and let x ∈ ℂ. Then

T_{k}^{N}(x) = \frac{k!}{(k + N)!} ∑_{m=0}^{k} \binom{k + N}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}(N + nx)^{k+N−m}.  (13)

The case N = 1 recovers the classical Faulhaber formula. Proof. Consider the following computational steps:

\begin{aligned}∑_{k=0}^{∞} T_{k}^{N}(x)\frac{t^{k}}{k!} & = (\frac{e^{(1+x)t} − e^{t}}{e^{t} − 1} )^{N} \\ & = \frac{1}{t^{N}} (\frac{t}{e^{t} − 1})^{N}(e^{(1+x)t} − e^{t})^{N} \\ & = ∑_{m=0}^{∞} B_{m}^{N} \frac{t^{m−N}}{m!} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}e^{(N+nx)t} \\ & = ∑_{m=0}^{∞} B_{m}^{N} \frac{t^{m−N}}{m!} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n} ∑_{s=0}^{∞}(N + nx)^{s} \frac{t^{s}}{s!} \\ & = ∑_{m=0}^{∞} ∑_{s=0}^{∞} B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}(N + nx)^{s} \frac{t^{s+m−N}}{s!m!} \\ & = ∑_{k=0}^{∞} ∑_{m=0}^{k} \binom{k}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}(N + nx)^{k−m} \frac{t^{k−N}}{k!} , \end{aligned}  (14)

which on comparing the like powers of t gives

∑_{m=0}^{k} \binom{k}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}(N + nx)^{k−m} = 0,  (15)

for each k = 0, 1, …, N − 1, and

T_{k}^{N}(x) = \frac{k!}{(k + N)!} ∑_{m=0}^{k+N} \binom{k + N}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}(N + nx)^{k+N−m}.  (16)

As remarked earlier, T_{k}^{N}(x) is a linear combination of x^{N}, …, x^{N+k}, and for N ∈ ℕ, the coefficient of x^{i} for i = 0, 1, …, N − 1, k + N + 1, … vanishes in (16). So, we have

∑_{m=k+1}^{k+N} \binom{k + N}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}(N + nx)^{k+N−m} = 0.  (17)

Therefore, (16) reduces through (17) to (13).

Corollary 12. For k and N ∈ ℕ_{0},

T_{k}^{N}(x) = ∑_{m=0}^{k} \binom{k}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n} ∑_{s=0}^{k−m} \binom{k − m}{s}\frac{s!(nx)^{N+s}N^{k−m−s}}{(N + s)!}  (18)

Proof. For N = 0, (13) gives T_{k}^{0}(x) = ∑_{m=0}^{k} \binom{k}{m}B_{m}^{0}(1+x)^{k−m}, which takes value 1 for k = 0 and vanishes otherwise. Thus the case N = 0 follows. Now let N ∈ ℕ, and observe that for each i = 0, 1, …, N − 1,

∑_{m=0}^{k} \binom{k + N}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}\binom{k + N − m}{i}(nx)^{i}N^{k+N−m−i} = 0.  (19)

Using (19) in (13), we get

T_{k}^{N}(x) = ∑_{m=0}^{k} \binom{k + N}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n} ∑_{i=N}^{k+N−m} \binom{k + N − m}{i}\frac{k!(nx)^{i}N^{k+N−m−i}}{(k + N)!}

= ∑_{m=0}^{k} \binom{k + N}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n} ∑_{s=0}^{k−m} \binom{k + N − m}{N + s}\frac{k!(nx)^{N+s}N^{k−m−s}}{(k + N)!}

= ∑_{m=0}^{k} \binom{k}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n} ∑_{s=0}^{k−m} \binom{k + N − m}{N + s}\frac{(k − m)!(nx)^{N+s}N^{k−m−s}}{(k + N − m)!}

= ∑_{m=0}^{k} \binom{k}{m}B_{m}^{N} ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n} ∑_{s=0}^{k−m} \binom{k − m}{s}\frac{s!(nx)^{N+s}N^{k−m−s}}{(N + s)!}

as required.

3 Apostol-Bernoulli numbers

Definition 13. For α ∈ ℂ, the Apostol-Bernoulli number β_{k}(α) is defined by the following egf:

\frac{t}{αe^{t} − 1} = ∑_{k=0}^{∞} β_{k}(α)\frac{t^{k}}{k!}, k ∈ ℕ_{0}, α ≠ 1, |t| < | log α|.  (20)

Remark 14. Singh [8] introduced the number E_{k}(α) via the following egf:

\frac{α}{αe^{t} − 1} = ∑_{k=0}^{∞} E_{k}(α)\frac{t^{k}}{k!}, α ≠ 1, |t| < | log α|; E_{k}(1) = \frac{E_{k}(−1)}{1 − 2^{k+1}} , k ∈ ℕ_{0},  (21)

which is related to the Apostol-Bernoulli number by

\frac{E_{k}(α)}{α} = \frac{β_{k+1}(α)}{k + 1},  (22)

and is known to satisfy the following beautiful identities

α^{−1}E_{k+1}(α) = α\frac{∂}{∂α}(α^{−1}E_{k}(α)), α^{−1}E_{k}(α) = (−1)^{k+1}αE_{k}(α^{−1}).  (23)

for all α ≠ 0, 1.

Remark 15. The Apostol-Bernoulli numbers and hence E_{k}(α)’s are used to express the power sums of the type

S_{k}(α, x) = α + α^{2}2^{k} + ⋯ + α^{x}x^{k}, x ∈ ℕ_{0},  (24)

and their generalizations. In fact, we have the following

lim_{x→∞} S_{k}(α) = −α^{−1}E_{k}(α) for 0 < |α| < 1.  (25)

Definition 16. Consider for h ∈ ℝ and f : ℝ → ℝ, the operators defined by E_{h}(f)(x) = f(x + h), Δ_{h}(f)(x) = (E_{h} − I)(f)(x) = f(x + h) − f(x), where I = E_{0} is the identity operator. The map E_{h} is called the shift operator and Δ_{h} is called the difference operator. For a nonnegative integer n, the n-th difference for f at x is defined as Δ_{h}^{n}(f(x)) = (E_{h} − I)^{n}(f(x)) = ∑_{m=0}^{n} \binom{n}{m}(−1)^{n−m}E_{h}^{m}(f(x)).

Note that E_{h}^{m}(f)(x) = f(x + mh). In particular, if we take f(x) = x^{k} then

Δ_{h}^{n}(x^{k}) = ∑_{m=0}^{n} \binom{n}{m}(−1)^{n−m}(x + mh)^{k}.  (26)

The closed form expressions for E_{k}(α) and E_{k}^{N}(α) are given by the next result.

Proposition 17. For α ≠ 0, 1 and k ∈ ℕ,

E_{k}(α) = ∑_{n=1}^{k}(\frac{α}{α − 1})^{n+1} ∑_{m=1}^{n} \binom{n}{m}(−1)^{m}m^{k},  (27)

E_{k}^{N+1}(α) = −\frac{α^{2}}{N} \frac{∂}{∂α}E_{k}^{N}(α), N ∈ ℕ,  (28)

and

E_{k}^{N+1}(α) = ∑_{n=1}^{k} \binom{N + n}{N}(\frac{α}{α − 1})^{n+N+1} ∑_{m=1}^{n} \binom{n}{m}(−1)^{m}m^{k}, N ∈ ℕ_{0}.  (29)

Proof. Consider the geometric power series G(α, t) = \frac{α}{αe^{t} − 1} = − ∑_{n=0}^{∞} α^{n+1}e^{nt} which is valid for all 0 < |t| < | log α|. Then for each k ∈ ℕ, we have

E_{k}(α) = lim_{t→0^{−}} \frac{∂^{k}}{∂t^{k}} (G(α, t)) = −α lim_{t→0^{−}} ∑_{n=0}^{∞} α^{n}n^{k}e^{nt}

= −α lim_{t→0^{−}}(∑_{n=0}^{∞} α^{n}E_{1}^{n})t^{k}

= α lim_{t→0^{−}}(αE_{1} − I)^{−1}t^{k}

= α lim_{t→0^{−}}(α(I + Δ_{1}) − I)^{−1}t^{k}

= \frac{α}{α − 1} lim_{t→0^{−}}(1 + \frac{α}{α − 1}Δ_{1})^{−1}t^{k}

= \frac{α}{α − 1} ∑_{i=1}^{k}(\frac{α}{1 − α})^{i} lim_{t→0} Δ_{1}^{i}(t^{k})

= \frac{α}{α − 1} ∑_{i=1}^{k}(\frac{α}{α − 1})^{i} ∑_{j=1}^{i} \binom{i}{j}(−1)^{j}j^{k}

where we have used the fact that Δ_{1}^{i}(t^{k}) = 0 for i > k and (26).

Equation (28) follows from the following for α and N as before

\frac{∂}{∂α}(G(α, t))^{N} = −\frac{N}{α^{2}} (G(α, t)^{N+1}),  (30)

and (29) can be proved using induction on N.

Another way of representing E_{k}^{N}(α) is the following.

Theorem 18. For each N ∈ ℕ, k ∈ ℕ_{0}, and α ≠ 1,

E_{0}^{N}(α) = \frac{α^{N}}{(α − 1)^{N}}; E_{k}^{N}(α) = \frac{(−α)^{N−1}}{(N − 1)!} (Δ_{1} + 2I)_{N−3}E_{k}(α), k ∈ ℕ,  (31)

where Δ_{1}(E_{k}(α)) = E_{k+1}(α) − E_{k}(α) and (Δ_{1} + 2I)_{N−3} = (Δ_{1} + 2I)(Δ_{1} + 3I) ⋯ (Δ_{1} + NI).

Proof. Observe that E_{0}^{N}(α) = lim_{t→0}(\frac{α}{αe^{t} − 1})^{N} = \frac{α^{N}}{(α − 1)^{N}}. Also,

\frac{∂}{∂t}(\frac{α}{αe^{t} − 1})^{N} = −N(\frac{α}{αe^{t} − 1})^{N} − \frac{N}{α} (\frac{α}{αe^{t} − 1})^{N+1},  (32)

which gives

E_{k}^{N+1}(α) = −\frac{α}{N} (E_{k+1}^{N}(α) + NE_{k}^{N}(α)) ; = −\frac{α}{N} (E_{k+1}^{N}(α) − E_{k}^{N}(α) + (N + 1)E_{k}^{N}(α)) ; = −\frac{α}{N} (Δ_{1} + (N + 1)I)E_{k}^{N}(α) ; = \frac{(−α)^{2}}{N(N − 1)}(Δ_{1} + (N + 1)I)(Δ_{1} + NI)E_{k}^{N−1}(α) ; ⋮ ; = \frac{(−α)^{N}}{N!} (Δ_{1} + (N + 1)I) ⋯ (Δ_{1} + 2I)E_{k}(α).  (33)

The last expression in (33) is valid for N → 0 in addition to N = 1, 2, . . .. The result follows on changing N to N − 1 in (33).

Lemma 19. For k, α, N as before,

E_{k}^{N}(α) = \frac{α^{N−1}}{(N − 1)!} ∑_{i=0}^{N}(−1)^{i−1}s(N, i)E_{k+i−1}(α)  (34)

where s(n, m) is the Stirling number of first kind.

Proof. For n, m ∈ ℕ_{0}, the Stirling number of first kind s(n, m) is defined by x(x − 1) ⋯ (x − n + 1) = ∑_{m=0}^{n} s(n, m)x^{m} which on changing x to −x gives x(x + 1) ⋯ (x + n − 1) = ∑_{m=0}^{n}(−1)^{n−m}s(n, m)x^{m}. Replacing the symbol x in the generating function for s(n, m) above by the shift operator E_{1} = Δ_{1} + I, we have

\begin{aligned}(Δ_{1} + 2I) ⋯ (Δ_{1} + (N − 1)I)E_{k}(α) & = E_{1}(E_{1} + I) ⋯ (E_{1} + (N − 1)I)E_{k−1}(α) \\ & = ∑_{i=0}^{N}(−1)^{N−i}s(N, i)E_{1}^{i}(E_{k−1})(α) \\ & = ∑_{i=0}^{N}(−1)^{N−i}s(N, i)E_{k+i−1}(α). \end{aligned}  (35)

Substituting for (Δ_{1} + 2I) ⋯ (Δ_{1} + (N − 1)I)E_{k}(α) from (35) in (31), we get (34).

Remark 20. The expression (34) is special in a sense that, it connects B_{k+N}^{N} with E_{k}^{N}(1). More precisely, we have the next result.

Theorem 21. For all k, N ∈ ℕ,

B_{k+N}^{N} = \frac{(k + N)!}{k!}E_{k}^{N}(1).  (36)

Proof. Using E_{k}(1) = \frac{B_{k+1}}{k + 1} for α = 1 in (34), we get

\frac{(k + N)}{k!}E_{k}^{N}(1) = \frac{(k + N)}{k!} \frac{1}{(N − 1)!} ∑_{i=0}^{N}(−1)^{i−1}s(N, i)\frac{B_{k+i}}{k + i}

= N\binom{k + N}{N} ∑_{i=0}^{N}(−1)^{N−i−1}s(N, N − i)\frac{B_{k+N−i}}{k + N − i}

= B_{k+N}^{N}.

where the last step has been obtained using (5).

4 Alternating power sums

Singh [8] defined for any α ∈ ℂ, k-th α-power sum S_{k}(α, x) by egf

\frac{α^{x+1}e^{(x+1)t} − αe^{t}}{αe^{t} − 1} = ∑_{k=0}^{∞} S_{k}(α, x)\frac{t^{k}}{k!}, α ≠ 1, |t| < | log α|.  (37)

Remark 22. The following closed form expression for S_{k}(α, x) is obtainable:

S_{k}(α, x) = \begin{cases}∑_{m=0}^{k} \binom{k}{m}E_{m}(α)\{α^{x}(1 + x)^{k−m} − 1\}, & if α ≠ 1; \\ S_{k}(x), & if α = 1. \end{cases}  (38)

Remark 23. Note that lim_{α→1} S_{k}(α, x) ≠ S_{k}(x) even when x is a positive integer. However, for x ∈ ℕ, S_{k}(α, x) = α + α^{2}2^{k} + ⋯ α^{x}x^{k}. Further, if 0 < |α| < 1 then lim_{x→∞} S_{k}(α, x) exists and is equal to −α^{−1}E_{k}(α) for all k ∈ ℕ, where for k = 0, the limit is −E_{0}(α). In particular, the Abel sum for the divergent series η(−k) = ∑_{n=0}^{∞}(−1)^{n}n^{k} can be obtained in closed form as the one sided limit at α = −1, i.e.,

η(−k) = lim_{α→−1^{+}}( lim_{x→∞} S_{k}(α, x)) = −E_{k}(−1) = − ∑_{i=1}^{k} \frac{1}{2^{i+1}} ∑_{j=1}^{i} \binom{i}{j}(−1)^{j}j^{k}.  (39)

We now consider sums of products of α-power sums and the related Abel sums.

Definition 24. For N ∈ ℕ_{0} and complex number x, sum of products of α-power sums T_{k}^{N}(α, x) is defined via the following egf:

G(α, x, N, t) = (\frac{α^{x+1}e^{(x+1)t} − αe^{t}}{αe^{t} − 1} )^{N} = ∑_{k=0}^{∞} T_{k}^{N}(α, x)\frac{t^{k}}{k!}, α ≠ 1, |t| < | log α|.  (40)

It is immediate to see that T_{k}^{N}(α, 0) = 0 and T_{k}^{1}(α, x) = S_{k}(α, x).

Remark 25. For 0 < |α| < 1, α^{x} → 0 as x → ∞. If this is the case then from (38), we have

lim_{x→∞} T_{k}^{N}(α, x) = (−1)^{N} ∑_{m=0}^{k} \binom{k}{m}E_{m}^{N}(α)N^{k−m}, 0 < |α| < 1,  (41)

which in particular gives the Abel sum of products

η^{N}(−k) = (−1)^{N} ∑_{m=0}^{k} \binom{k}{m}E_{m}^{N}(−1)N^{k−m}.  (42)

Theorem 26. For N ∈ ℕ, k ∈ ℕ_{0}, x ∈ ℂ, and α ≠ 0, 1,

T_{k}^{N}(α, x) = ∑_{n=0}^{k} \binom{k}{m}E_{m}^{N}(α) ∑_{n=0}^{N} \binom{N}{n}(−1)^{N−n}α^{nx}(N + nx)^{k−m}.  (43)

Proof. Follows from following computational steps:

∑_{k=0}^{∞} T_{k}^{N}(α, x)\frac{t^{k}}{k!} = (\frac{α}{αe^{t} − 1})^{N}(α^{x}e^{xt} − e^{t})^{N}

= ∑_{m=0}^{∞} E_{m}(α)\frac{t^{m}}{m!} ∑_{i=0}^{N} \binom{N}{i}(−1)^{N−i}α^{ix}e^{(N+ix)t}

= ∑_{m=0}^{∞} E_{m}(α)\frac{t^{m}}{m!} ∑_{i=0}^{N} \binom{N}{i}(−1)^{N−i}α^{ix} ∑_{n=0}^{∞}(N + ix)^{n} \frac{t^{n}}{n!}

= ∑_{m=0}^{∞} ∑_{n=0}^{∞} E_{m}(α) ∑_{i=0}^{N} \binom{N}{i}(−1)^{N−i}α^{ix}(N + ix)^{n} \frac{t^{m+n}}{m!n!}

= ∑_{k=0}^{∞} ∑_{m=0}^{k} \binom{k}{m}E_{m}(α) ∑_{i=0}^{N} \binom{N}{i}(−1)^{N−i}α^{ix}(N + ix)^{k−m} \frac{t^{k}}{k!}

from which the result can be obtained on comparing the like powers of t.

5 Further extensions

The sequence of functions (T_{k}^{N}(α, x))_{k∈ℕ_{0}}, x, α ∈ ℂ, α ≠ 0, 1 satisfies simple recurrence relation from which its terms can be obtained successively. Here is the special one, which we state as the following. Proposition 27. For k, N ∈ ℕ_{0}, x ∈ ℂ, and α ≠ 0, 1,

T_{0}^{N}(α, x) = α^{N}(\frac{α^{x} − 1}{α − 1})^{N}, T_{k}^{N}(α, x) = α\frac{∂}{∂α}T_{k−1}^{N}(α, x), k ∈ ℕ.  (44)

Proof. Expression for T_{0}^{N} is trivially true. It is also easy to see that (44) holds for N = 0. For N ∈ ℕ, the proof follows from \frac{∂}{∂t}G(α, x, N, t) = α\frac{∂}{∂α}G(α, x, N, t), where G(α, x, N, t) is as in (40).

The recurrence (44) can be used to extend the definition of the sum of products of the power sums to negative integer values of k.

Definition 28. The sum of products T_{k}^{N}(α, x) for a complex number N and real α ≠ 0, 1 is defined successively by

T_{0}^{N}(α, x) = α^{N}(\frac{α^{x} − 1}{α − 1})^{N}; T_{−k}^{N}(α, x) = ∫_{0}^{α} \frac{T_{−k+1}^{N}(θ, x)}{θ}dθ, k ∈ ℕ, α ≠ 0, 1.  (45)

Definition 29. For 0 < |α| < 1, define

T_{k}^{N}(α, ∞) = lim_{x→∞} T_{k}^{N}(α, x).  (46)

From (45), we have

T_{−1}^{N}(α, ∞) = ∫_{0}^{α} θ^{N−1}(1 − θ)^{−N}dθ.  (47)

For N = 1, Singh [8] proved the following result:

T_{−i}^{1}(α, ∞) = \begin{cases}− log |1 − α|, & if i = 1; \\ \frac{(−1)^{i−1}}{(i − 1)!} ∫_{0}^{α} \frac{(log(t))^{i−1}}{1 − t} dt − ∑_{β=1}^{i−1} \frac{(− log |α|)^{β}}{β!} T_{−i+β}(α, ∞), & if i ≥ 2. \end{cases} (48)

Observe from (48) that

lim_{α→1^{−}} T_{−k}^{1}(α, ∞) = ζ(k), lim_{α→−1^{+}} T_{−k}^{1}(α, ∞) = −η(k),  (49)

for all k ∈ ℕ.

In the next result, we generalize the formula (48) for the underlying sums of products, i.e., for all positive integers N ≠ 1. So, we have the following main result.

Theorem 30. For all k ∈ ℕ_{0}, N = 2, 3, …, and 0 < |α| < 1,

\begin{aligned}T_{−k}^{N}(α, ∞) & = ∑_{i_{1}=1}^{N−1} ∑_{i_{2}=1}^{i_{1}−1} ⋯ ∑_{i_{k}=1}^{i_{k−1}−1}(\frac{α}{1 − α})^{i_{k}} \frac{(−1)^{N−k−i_{k}}}{i_{1} ⋯ i_{k}} \\ & + ∑_{i=1}^{k−1} \frac{s(N, k + 1 − i)}{(N − 1)!}T_{−i}^{1}(α, ∞) + (−1)^{N−1}T_{−k}^{1}(α, ∞). \end{aligned}  (50)

Proof. From (45) and (47), we have

T_{−1}^{N}(α, ∞) = ∑_{i=1}^{N−1}(−1)^{N−1−i} \frac{T_{0}^{i}(α, ∞)}{i} + (−1)^{N−1}T_{−1}^{1}(α, ∞),  (51)

for N = 2, 3, . . .. Similarly,

T_{−2}^{N}(α, ∞) = ∑_{i=1}^{N−1} ∑_{j=1}^{i−1}(−1)^{N−2−j} \frac{T_{0}^{j}(α, ∞)}{ij} + (−1)^{N−1}T_{−1}^{1} ∑_{i=1}^{N−1} \frac{1}{i} + (−1)^{N−1}T_{−2}^{1},  (52)

which inductively leads to the following:

\begin{aligned}T_{−k}^{N}(α, ∞) & = ∑_{i_{1}=1}^{N−1} ∑_{i_{2}=1}^{i_{1}−1} ⋯ ∑_{i_{k}=1}^{i_{k−1}−1}(\frac{α}{1 − α})^{i_{k}} \frac{(−1)^{N−k−i_{k}}}{i_{1} ⋯ i_{k}} \\ & + ∑_{i_{1}=1}^{N−1} ∑_{i_{2}=1}^{i_{1}−1} ⋯ ∑_{i_{k−1}=1}^{i_{k−2}−1} \frac{(−1)^{N−k}}{i_{1} ⋯ i_{k−1}} T_{−1}^{1}(α, ∞) \\ & + ⋯ \\ & + ∑_{i_{1}=1}^{N} \frac{(−1)^{N−1}}{i_{1}} T_{−k+1}^{1}(α, ∞) + (−1)^{N−1}T_{−k}^{1}(α, ∞), \end{aligned}  (53)

where we have used (48) and the fact that T_{0}^{i}(α, ∞) = (\frac{α}{1−α})^{i} for all i ∈ ℕ.

On substituting for the following expression of the Stirling numbers of the first kind [18] in (53)

s(n + 1, m + 1) = (−1)^{n+m}n! ∑_{ℓ=1}^{n} ∑_{ℓ_{2}=1}^{ℓ_{1}−1} ⋯ ∑_{ℓ_{m}=1}^{ℓ_{m−1}−1} \frac{1}{ℓ_{1}ℓ_{2} ⋯ ℓ_{m}} ,  (54)

we obtain (50).

Corollary 31. Let k, N ∈ ℕ, such that N = 2, 3, . . .. Then

lim_{α→1^{−}} T_{−k}^{N}(α, ∞) = \frac{1}{(N − 1)!} ∑_{i=2}^{k−1} s(N, k + 1 − i)ζ(i) + (−1)^{N−1}ζ(k) for N < k,  (55)

and

\begin{aligned}lim_{α→−1^{+}} T_{−k}^{N}(α, ∞) & = ∑_{i_{1}=1}^{N−1} ∑_{i_{2}=1}^{i_{1}−1} ⋯ ∑_{i_{k}=1}^{i_{k}−1}(\frac{1}{2})^{i_{k}} \frac{(−1)^{N−k−i_{k}}}{i_{1} ⋯ i_{k}} \\ & + ∑_{i=1}^{k−1} \frac{s(N, k + 1 − i)η(i)}{(N − 1)!} + (−1)^{N−1}η(k). \end{aligned}  (56)

Remark 32. For a positive integer N and 0 < |α| < 1, we have \frac{(−α)^{N}e^{Nt}}{(αe^{t} − 1)^{N}} = ∑_{k=0}^{∞} T_{k}^{N}(α, ∞)\frac{t^{k}}{k!}, from which, it follows that

lim_{α→1^{−}} T_{−k}^{N}(α, ∞) = (−1)^{N} lim_{t_{1}→0}(∫_{−∞}^{t_{1}} ⋯ ∫_{−∞}^{t_{k}} \frac{e^{Nt_{k+1}}}{(e^{t_{k+1}} − 1)^{N}} dt_{k+1} ⋯ dt_{2})

= \frac{1}{(N − 1)!} ∑_{i=k+1−N}^{k} s(N, k + 1 − i)ζ(i) if N < k.

(57)

Example 33. Table 1 below shows the values of lim_{α→1^{−}} T_{−k}^{N}(α, ∞) for few values of k and N. The computations were done in Mathematica with reference to (55).

Table 1: lim_{α→1^{−}} T_{−k}^{N}(α, ∞) for some values of k and N
k \ N  2  3  4
3  \frac{π^{2}}{6} − ζ(3)
4  −\frac{π^{4}}{90} + ζ(3) \frac{π^{2}}{12} + \frac{π^{4}}{90} − 3\frac{ζ(3)}{2}
5  \frac{π^{4}}{90} − ζ(5) −\frac{π^{4}}{60} + \frac{ζ(3)}{2} + ζ(5) \frac{π^{2}}{36} + \frac{11π^{4}}{540} − ζ(3) − ζ(5)
6 −\frac{π^{6}}{945} + ζ(5) \frac{π^{4}}{180} + \frac{π^{6}}{945} − \frac{3ζ(5)}{2} −\frac{π^{4}}{90} − \frac{π^{6}}{945} + \frac{ζ(3)}{6} + \frac{11ζ(5)}{6}

Remark 34. The preceding Example 33, together with (57), establishes that

lim_{α→1^{−}} T_{−k}^{N}(α, ∞) = k ; ∑ ; ∑_{i=1}^{N} k_{i}=k, 0≤k_{i}≤k, k_{i}≠1 \binom{k}{k_{1}, …, k_{N}}ζ(k_{1}) ⋯ ζ(k_{N}),  (58)

for N < k, where the right hand side is the sum of product of Riemann zeta functions. The classical Euler’s formula is given by

∫0^{\frac{1}{2}} \frac{log(1 − t)}{t}dt = −T_{−2}^{1}(\frac{1}{2}, ∞) = \frac{(log(2))^{2}}{2} − \frac{ζ(2)}{2}.  (59)

The next result generalize the formula (59) to the sums of products.

Theorem 35. For N = 2, 3, …, the generalized Euler’s formula is given by

(−1)^{N}T_{−2}^{N}(\frac{1}{2}, ∞) = ∑_{i=1}^{N−1} ∑_{j=1}^{i−1} \frac{(−1)^{j}}{ij} + log(2) ∑_{i=1}^{N−1} \frac{1}{i} + \frac{(log(2))^{2}}{2} − \frac{ζ(2)}{2}.  (60)

Proof. First observe that for α = \frac{1}{2}, T_{−2}^{N}(α, ∞) = T_{−2}^{N}(1 − α, ∞). Along with this, if we use the following identity due to Singh [8]

− ∫_{0}^{1−α} \frac{log(1 − t)}{t}dt = − log(α) log(1 − α) + ∫_{0}^{1} \frac{log(t)}{1 − t}dt + ∫_{0}^{α} \frac{log(1 − t)}{t}dt,  (61)

and the fact that T_{0}^{N}(\frac{1}{2}) = 1 for all N, we obtain (60) by taking α = \frac{1}{2} in (52), where

∫_{0}^{1} \frac{log(t)}{1 − t}dt = ζ(2).

6 Acknowledgments

The author is thankful to the referee for the valuable suggestions regarding improvement in the presentation of the paper.

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2010 Mathematics Subject Classification: Primary 11A25; Secondary 11B68, 05A10, 11B65. Keywords: power sum of integers, Bernoulli number, Bernoulli polynomial, Riemann zeta function, exponential generating function, Apostol-Bernoulli number.

(Concerned with sequences \underline{A0xxxxx}.)

Received May 14 2015; revised versions received September 21 2015; September 28 2015; November 26 2015. Published in Journal of Integer Sequences, December 17 2015.

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