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\begin{center}
\vskip 1cm{\LARGE\bf Defining Sums of Products of Power Sums}
\vskip 1cm
\large
Jitender Singh\\
Department of Mathematics \\
Guru Nanak Dev University\\
Amritsar-143005\\
Punjab \\
India\\
\href{mailto:sonumaths@gmail.com}{\tt sonumaths@gmail.com}
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\begin{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 $\sum_{n=0}^{\infty}\alpha^n n^k$, $0&lt;|\alpha|&lt;1$, $k\in\A$ 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.
\end{abstract}

\section{Background}
The discrete power sum of a positive integer $n$ is defined as
    $S_k(n)=1^k+2^k+\cdots+n^k,~k\in\A$.
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 $S_{k-1}(n),\ldots,S_{0}(n)$
\begin{equation}
(n+1)^{k+1}-1=\sum_{m=0}^{k}{k+1\choose m}S_{m}(n),
\end{equation} 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 \cite{s1,s2,krish,beardon,s3,s4}.

As an extension of the discrete power-sum, the polynomial $S_k(x)$ for $k\in\A$ and complex number $x$ is defined via its exponential generating function (hereafter referred  to as egf)~\cite{JS9}
\begin{equation}\label{eq1}
\frac{e^{(1+x)t}-e^t}{e^t-1}=\sum_{k=0}^{\infty}S_k(x)\frac{t^k}{k!},~|t|&lt;2\pi.
\end{equation}
For fixed $k$, $S_k(x)$ is given by the classical Faulhaber formula
 \begin{equation}\label{eq2}
S_k(x)=\frac{1}{k+1}\sum_{m=0}^{k}{k+1\choose m} (-1)^m B_m x^{k+1-m},
\end{equation}
where  the rational sequence $\{B_k\}_{k\in\A}$ is generated by the egf
\begin{equation}\label{eq3}
    \displaystyle \frac{t}{e^t-1},~|t|&lt;2\pi,
\end{equation}
such that $B_k$ is the $k$-th Bernoulli number.

For $N\in\mathbb{N}$, the sum of products of Bernoulli numbers denoted $B_k^N$ is generated by the multiple product of the egf \eqref{eq3} in the ring of egfs over $\mathbb{Q}$.
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~\cite{p0} have obtained a closed form expression for $B_k^N$.

 Dilcher~\cite{karl} has also evaluated $B_k^N$ in closed form as the following expression:
\begin{equation}\label{eq001}
    B_k^N=N{k\choose N}\sum_{i=0}^{N}(-1)^{N-1-i}s(N,N-j)\frac{B_{k-i}}{k-i},~k&gt;N,
\end{equation}
where  $s(N,N-j)$ is the Stirling number of first kind.

In a short paper, Petojevi\'c and Srivastava~\cite{p1} have computed the Dilcher sums of products of Bernoulli numbers in an elegant way. More generally, Kamano~\cite{p2} has investigated sums of products of hypergeometric Bernoulli numbers and used them to study multiple hypergeometric zeta function.

On the other hand, Kim~\cite{p3} 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~\cite{kim} 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 $\varphi(n)$ integers in Singh~\cite{singh2014}, where $\varphi$ 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 \ref{Sec:3}. Section \ref{Sec:4} and \ref{Sec:5} are devoted to   explore the properties of the higher order $\alpha$-Euler numbers \cite{JS9} (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 \ref{Sec:5}. In Section \ref{Sec:6}, 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.
\section{Sums of products}\label{Sec:3}
\begin{definition}
In the ring of formal power series over a field $\mathbb{F}$, where $\mathbb{F}$ is one of the fields,  $\mathbb{Q},\mathbb{R}$, and $\mathbb{C}$; an element $f$ is said to be an exponential generating function (egf) if $f$ is equal to the formal power series $\sum_{k=0}^\infty a_k \frac{t^k}{k!}$, where each $a_k\in\mathbb{F}$. The set of all egfs is also a ring with the component-wise addition (+) and  multiplication ($\cdot$), such that for $f=\sum_{k=0}^\infty a_k \frac{t^k}{k!}$ and $g=\sum_{k=0}^\infty b_k \frac{t^k}{k!},~a_k,b_k\in \mathbb{F}$,
\begin{equation}
f+g=\sum_{k=0}^\infty (a_k+b_k) \frac{t^k}{k!};~f\cdot g=\sum_{k=0}^\infty\sum_{m=0}^k {k\choose m} a_m b_{k-m}\frac{t^k}{k!},
\end{equation}
where the sequence defined by $c_k={k\choose m} a_m b_{k-m},~k\in\A$ in $\mathbb{F}$ is the Cauchy type product \cite{singh14} of the sequences $(a_k)_{k\in\A}$ and $(b_k)_{k\in\A}$.
\end{definition}
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 \cite{generatingfunctionology}.
\begin{definition}
Let $N\in\mathbb{N}$, and let $f_i$ is the egf of the sequence $(a_{ik})_{k\in\A}$, for each $i=1,2,\ldots, N$,  where $a_{ik}\in\mathbb{F}$. In the multiple product of egfs, i.e., $f_1\cdots f_{N}=\sum_{k=0}^\infty c_k \frac{t^k}{k!}$, the sequence defined by $c_k=\sum_{k_1+\cdots+k_N=k}{k\choose k_1,\ldots,k_N}a_{1k_1}\cdots a_{N k_N}$ in $\mathbb{F}$ is called the sum of products of the sequences $(a_{1k})_{k\in\A},\ldots,(a_{Nk})_{k\in\A}$.
\end{definition}
\begin{definition}
For $x\in\mathbb{C}$, let $T_k^N(x)$  denotes the sum of products of the power sums, which is generated by the egf
    \begin{equation}\label{eq14}
H(x,t,N)=\Bigl(\frac{e^{(1+x)t}-e^t}{e^t-1}\Bigr)^N=\sum_{k=0}^{\infty} T_k^N(x)\frac{t^k}{k!},~|t|&lt;2\pi,~N\in\A.\end{equation}
\end{definition}
\begin{remark}
Observe from \eqref{eq14} that
$T_0^0(x)=1;~T_k^0(x)=0~\mbox{for all}~ k\in\mathbb{N}$; $~T_k^N(1)=N^k$ and $T_k^1(x)=S_k(x)$.
\end{remark}
\begin{remark}
Since for each $k\in\A$, $S_k(x)$ is a polynomial in $x$ of degree $k+1$ over $\mathbb{Q}$ with coefficient of the leading term $(k+1)^{-1}$ (see \eqref{eq2}), it follows that for each fixed $N\in\mathbb{N}$ and partition $\{k_1,\ldots,k_N\}$ of $k$,  the product $S_{k_1}(x)\cdots S_{k_N}(x)$ is a polynomial in $x$ over $\mathbb{Q}$ of degree $\sum_{i=1}^{N}(k_i+1)=k+N$ with leading coefficient $\displaystyle \prod_{i=1}^N (k_i+1)^{-1}&gt;0$.
Also, as $S_{k_i}(0)=0$ for all $i=1,\ldots,k_N$,  $x^N$ is always a factor of the polynomial $S_{k_1}(x)\cdots S_{k_N}(x)$. Thus, $\displaystyle T_k^N(x)$ is  a polynomial in $x$ of degree $k+N$ over $\mathbb{Q}$ 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 $\{x^N,x^{N+1},\ldots,x^{N+k}\}$ over $\mathbb{Q}$. In particular, the coefficient of $x^i,~i=0,1,\ldots,N-1,~N&gt;0$,  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 $k\in\mathbb{N}$.
 Thus, $x^N(x+1)$ is always a factor of $T_k^N(x)$.
\end{remark}
\begin{proposition}\label{th3} For  $N\in\A$ and $~x\in\mathbb{R}$,
\begin{enumerate}
\item[$(1)$] $T_0^N(x)=x^N$; $\displaystyle T_k^N(x)=Nk\int_{0}^{x}T_{k-1}^N(t) ~dt+N\sum_{m=0}^{k}{k\choose m}(-1)^mB_{m}\int_{0}^{x}T_{k-m}^{N-1}(x),
$  $k\in\mathbb{N}$.
\item[$(2)$] $T_k^N(x)=(-1)^{k+N} \sum_{n=0}^{N}{N\choose n}T_k^n(-1-x)$ for all $k\in\A$.
\end{enumerate}
\end{proposition}
\begin{proof}(1) Observe that $T_0^N(x)=\displaystyle \lim_{t\rightarrow 0} H(x,t,N)=x^N$. The recurrence follows from the following
\begin{equation}
\frac{\partial H}{\partial x}=NtH(x,t,N)+NH(x,t,N-1)\frac{te^t}{e^t-1}.
\end{equation}
(2) Follows from the following symmetry of $H$
\begin{equation}
(-1)^N H(x,t,N)=\Bigl(\frac{e^{(1-1-x)(-t)}-e^{-t}}{e^{-t}-1}
+1\Bigr)^N=\sum_{n=0}^{N}{N\choose n}
H(-1-x,-t,n)
\end{equation}
\end{proof}
\begin{definition}Higher order Bernoulli polynomial $B_k^N(x),~x\in\mathbb{C}$ is defined by the following egf:
\begin{equation}\label{eq16}
    \frac{t^N e^{xt}}{(e^{t}-1)^N}=\sum_{k=0}^{\infty}B_k^N(x)\frac{t^k}{k!},~N\in\A,|t|&lt;2\pi.
\end{equation}
\begin{definition}
Higher order Bernoulli number is defined by $B_k^N=B_k^N(0)$.
\end{definition}
\end{definition}
\begin{remark}
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:
$B_0^0(x)=1,~B_k^0(x)=x^k~\mbox{for}~k\in\mathbb{N}$, and $B_{k}^N(x)=\sum_{m=0}^k {k\choose m}B_m(x) B_{k-m}^{N-1}$,
                                             which, for $N=1$, reduces to the usual identity of the Bernoulli polynomials. Also,
\begin{equation}
  B_0^{N+1}=1;~B_k^{N+1}=\Bigl(1-\frac{k}{N}\Bigr)B_k^N-kB_{k-1}^N,~k\in\mathbb{N},
\end{equation}
 which  can be used to evaluate $B_k^N$ for $N=2,3,\ldots$.
\end{remark}
\begin{remark}
It is  not hard to see that $T_k^N(x)$  can be expressed in terms of $B_k^N(x)$ by
\begin{equation}\label{eq17}
    T_k^N(x)=\frac{k!}{(N+k)!}\sum_{n=0}^{N}{N\choose n}(-1)^{N-n} B_{N+k}^N(N+(1+x)n).
\end{equation}
\end{remark}
\begin{proposition}\label{th7} Let $N,~k\in\A$, and let $x\in\mathbb{C}$. Then
\begin{equation}\label{eq171}
    T_k^N(x)=\frac{k!}{(k+N)!}\sum_{m=0}^{k}{k+N\choose m}B_{m}^N\sum_{n=0}^{N}
      {N\choose n}(-1)^{N-n}(N+nx)^{k+N-m}.
\end{equation}
The case $N=1$ recovers the classical Faulhaber formula.
\end{proposition}
\begin{proof}Consider the following computational steps:
  \begin{equation}\label{eq172}
 \begin{split}
   \sum_{k=0}^{\infty} T_k^N(x)\frac{t^k}{k!}&amp;=\Bigl(\frac{e^{(1+x)t}-e^t}{e^t-1}\Bigr)^N\\
   &amp;=\frac{1}{t^N}\Bigl(\frac{t}{e^t-1}\Bigr)^N(e^{(1+x)t}-e^t)^N\\
   &amp;=\sum_{m=0}^{\infty}B_{m}^N\frac{t^{m-N}}{m!}
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}e^{(N+nx)t}\\
   &amp;=\sum_{m=0}^{\infty}B_{m}^N\frac{t^{m-N}}{m!}
     \sum_{n=0}^{N} {N\choose n}(-1)^{N-n}\sum_{s=0}^{\infty}(N+nx)^{s}\frac{t^s}{s!}\\
   &amp;=\sum_{m=0}^{\infty}\sum_{s=0}^{\infty}B_{m}^N
     \sum_{n=0}^{N} {N\choose n}(-1)^{N-n}(N+nx)^{s}\frac{t^{s+m-N}}{s!m!}\\
   &amp;=\sum_{k=0}^{\infty}\sum_{m=0}^{k}{k\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}(N+nx)^{k-m}\frac{t^{k-N}}{k!},
\end{split}
\end{equation}
which on comparing the like powers of $t$ gives
 \begin{equation}
 \sum_{m=0}^{k}{k\choose m}B_{m}^N
     \sum_{n=0}^{N} {N\choose n}(-1)^{N-n}(N+nx)^{k-m}=0,
     \end{equation}
for each $k=0,1,\ldots,N-1$, and
 \begin{equation}\label{eq46}
  T_k^N(x)=\frac{k!}{(k+N)!}\sum_{m=0}^{k+N}{k+N\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}(N+nx)^{k+N-m}.
 \end{equation}
 As remarked earlier, $T_k^N(x)$ is a linear combination of $x^N,\ldots,x^{N+k}$, and for $N\in\mathbb{N}$, the coefficient of $x^i$ for $i=0,1,\ldots,N-1,k+N+1,\ldots$ vanishes in \eqref{eq46}. So, we have
 \begin{equation}\label{eq47}
\sum_{m=k+1}^{k+N}{k+N\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}(N+nx)^{k+N-m}=0.
 \end{equation}
Therefore, \eqref{eq46} reduces through \eqref{eq47} to \eqref{eq171}.
\end{proof}
\begin{corollary}For $k$ and $N\in\A$,
    {\begin{equation}\label{eq50}
    \begin{split}
        T_k^N(x)= \sum_{m=0}^{k}{k\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n} \sum_{s=0}^{k-m}
                                            {k-m\choose s}\frac{s!(nx)^{N+s} N^{k-m-s}}{(N+s)!}
                                             \end{split}
    \end{equation}}
\end{corollary}
\begin{proof}For $N=0$, \eqref{eq171} gives $T_k^0(x)=\sum_{m=0}^k{k\choose m}B_{m}^0 (1+x)^{k-m}$, which takes value $1$ for $k=0$ and vanishes otherwise. Thus the case $N=0$ follows.
Now let $N\in\mathbb{N}$, and observe that for each $i=0,1,\ldots,N-1$,
 \begin{equation}\label{eq48}
\sum_{m=0}^{k}{k+N\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}{k+N-m\choose i}(nx)^i N^{k+N-m-i}=0.
 \end{equation}
Using \eqref{eq48} in  \eqref{eq171}, we get
{\begin{equation*}\label{eq49}
\begin{split}
T_k^N(x)&amp;=\sum_{m=0}^{k}{k+N\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}\sum_{i=N}^{k+N-m}{k+N-m\choose i}\frac{k!(nx)^i N^{k+N-m-i}}{(k+N)!}\\
 &amp;=\sum_{m=0}^{k}{k+N\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}\sum_{s=0}^{k-m}{k+N-m\choose N+s}\frac{k!(nx)^{N+s} N^{k-m-s}}{(k+N)!}\\
&amp;=\sum_{m=0}^{k}{k\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}\sum_{s=0}^{k-m}
                                            {k+N-m\choose N+s}\frac{(k-m)!(nx)^{N+s} N^{k-m-s}}{(k+N-m)!}\\                              &amp;=\sum_{m=0}^{k}{k\choose m}B_{m}^N
      \sum_{n=0}^{N}{N\choose n}(-1)^{N-n}\sum_{s=0}^{k-m}
                                             {k-m\choose s}\frac{s!(nx)^{N+s} N^{k-m-s}}{(N+s)!}                                        \end{split}
 \end{equation*}}
as required.
\end{proof}
\section{Apostol-Bernoulli numbers}\label{Sec:4}
\begin{definition}
For $\alpha\in\mathbb{C}$,  the  Apostol-Bernoulli number $\beta_k(\alpha)$ is defined by the following egf:
\begin{equation}\label{eq21111}
\frac{t}{\alpha e^{t}-1}=\sum_{k=0}^{\infty} \beta_k(\alpha)\frac{t^k}{k!},~k\in \A,~\alpha\neq 1,~|t|&lt;|\log \alpha|.
\end{equation}
\end{definition}
\begin{remark}
Singh \cite{JS9} introduced the number $E_k(\alpha)$ via the following egf:
\begin{equation}\label{eq21}
\frac{\alpha}{\alpha e^{t}-1}=\sum_{k=0}^{\infty} E_k(\alpha)\frac{t^k}{k!},~\alpha\neq 1,~|t|&lt;|\log\alpha|;~E_k(1)=\frac{E_k(-1)}{1-2^{k+1}},~k\in \A,
\end{equation} which is related to the Apostol-Bernoulli number by
\begin{equation}
    \frac{E_k(\alpha)}{\alpha}=\frac{\beta_{k+1}(\alpha)}{k+1},
\end{equation}
and  is known to satisfy the following beautiful identities
\begin{equation}\label{eq22}
    \alpha^{-1}E_{k+1}(\alpha)=\alpha\frac{\partial}{\partial\alpha}(\alpha^{-1}E_k(\alpha)),~\alpha^{-1}E_k(\alpha)=(-1)^{k+1}\alpha E_k(\alpha^{-1}).
\end{equation}
for all $\alpha\neq 0,1$.
\end{remark}
\begin{remark}
The Apostol-Bernoulli numbers and hence $E_k(\alpha)$'s are used to express the power sums of the type
\begin{equation}
  S_k(\alpha,x)=\alpha+\alpha^2 2^k+\cdots+\alpha^x x^k,~x\in\A,
\end{equation}
and their generalizations. In fact, we have the following
\begin{equation}
\lim_{x\rightarrow \infty}{S_k(\alpha)}=-\alpha^{-1}E_k(\alpha)~\text{for}~ 0&lt;|\alpha|&lt;1.
\end{equation}
\end{remark}
\begin{definition}
Consider for $h\in\mathbb{R}$ and $f:\mathbb{R}\rightarrow \mathbb{R}$,
the operators defined by $E_h(f)(x)=f(x+h),~\Delta_h (f)(x)=(E_h-I)(f)(x)=f(x+h)-f(x)$, where $I=E_0$ is the identity operator.  The map $E_h$ is called the shift operator and $\Delta_h$ is called the difference operator.
For a nonnegative integer $n$, the $n$-th difference for $f$ at $x$ is defined as
$\Delta_h^n(f(x))=(E_h-I)^n (f(x))=\sum_{m=0}^{n}{n\choose m}(-1)^{n-m}E_h^m(f(x))$.
\end{definition}
Note that $E_h^m(f)(x)=f(x+mh)$.
In particular, if we take $f(x)=x^k$  then
\begin{equation}\label{eq26}
\Delta_h^n(x^k)=\sum_{m=0}^{n}{n\choose m}(-1)^{n-m}(x+mh)^k.
\end{equation}
The closed form expressions for $E_k(\alpha)$ and $E_k^N(\alpha)$ are given by the next result.
\begin{proposition}\label{L1}
For $\alpha\neq 0,1$ and $k\in\mathbb{N}$,
\begin{equation}\label{eq23}
    E_k(\alpha)=\sum_{n=1}^{k}\Bigl(\frac{\alpha}{\alpha-1}\Bigr)^{n+1}\sum_{m=1}^{n}{n\choose m}(-1)^m  m^k,
\end{equation}
\begin{equation}\label{eq35}
    E_{k}^{N+1}(\alpha)= -\frac{\alpha^2}{N}\frac{\partial}{\partial\alpha}E_{k}^N(\alpha),~N\in\mathbb{N},
  \end{equation}
and
  \begin{equation}\label{eq77}
  E_{k}^{N+1}(\alpha)=\sum_{n=1}^{k}{N+n\choose N}
    \Bigl(\frac{\alpha}{\alpha-1}\Bigr)^{n+N+1}\sum_{m=1}^{n}{n\choose m}(-1)^m  m^k,~N\in\mathbb{N}_0.
  \end{equation}
\end{proposition}
\begin{proof}
Consider the geometric power series $\displaystyle G(\alpha,t)= \frac{\alpha}{\alpha e^t-1}=$ $-\sum_{n=0}^{\infty}\alpha^{n+1} e^{nt}$  which is valid for all $0&lt;|t|&lt;|\log\alpha|$.
Then for each $k\in\mathbb{N}$, we have
\begin{equation*}\label{eq24}
\begin{split}
E_k(\alpha)=\lim_{t\rightarrow 0^-}\frac{\partial^k}{\partial t^k}(G(\alpha,t))&amp;=-\alpha\lim_{t\rightarrow
0^-}\sum_{n=0}^{\infty}\alpha^{n} n^k e^{nt}\\
&amp;=-\alpha\lim_{t\rightarrow
0^-}\Bigl(\sum_{n=0}^{\infty}\alpha^n E_1^n\Bigr)t^k\\
&amp;=\alpha\lim_{t\rightarrow
0^-}(\alpha E_1 -I)^{-1}t^k\\
&amp;=\alpha\lim_{t\rightarrow
0^-}(\alpha (I+\Delta_1) -I)^{-1}t^k\\
&amp;=\frac{\alpha}{\alpha-1}\lim_{t\rightarrow
0^-}\Bigl(1+\frac{\alpha}{\alpha-1} \Delta_1\Bigr)^{-1}t^k\\
&amp;=\frac{\alpha}{\alpha-1}\sum_{i=1}^{k}\Bigl(\frac{\alpha}{1-\alpha}
\Bigr)^i\lim_{t\rightarrow 0}\Delta_{1}^i(t^k)\\
&amp;=\frac{\alpha}{\alpha-1}\sum_{i=1}^{k}\Bigl(\frac{\alpha}{\alpha-1}\Bigr)^{i}\sum_{j=1}^{i}{i\choose j}(-1)^j  j^k
\end{split}
\end{equation*}
where we have used the fact that $\Delta_1^i(t^k)=0$ for $i&gt;k$ and \eqref{eq26}.

Equation \eqref{eq35} follows from the following for $\alpha$ and $N$ as before
\begin{equation}
    \frac{\partial}{\partial \alpha}\Bigl(G(\alpha,t)\Bigr)^N=-\frac{N}{\alpha^2} \Bigl(G(\alpha,t)^{N+1}\Bigr),
\end{equation}
and \eqref{eq77} can be proved using induction on $N$.
\end{proof}
 Another way of representing $E_k^N(\alpha)$ is the following.
\begin{theorem}\label{th5} For each $N\in\mathbb{N}$, $k\in\A$, and $\alpha\neq 1$,
\begin{equation}\label{eq31}
    E_0^N(\alpha)=\frac{\alpha^N}{(\alpha-1)^N};~E^N_k(\alpha)=\frac{(-\alpha)^{N-1}}{(N-1)!}(\Delta_1+2I)_{N-3} E_k(\alpha),k\in\mathbb{N},
\end{equation}
where $\Delta_1 (E_k(\alpha))=E_{k+1}(\alpha)-E_k(\alpha)$ and $(\Delta_1+2I)_{N-3}=(\Delta_1+2I)(\Delta_1+3I)\cdots (\Delta_1+N I)$.
\end{theorem}
\begin{proof}
Observe that $\displaystyle E_0^N(\alpha)=\lim_{t\rightarrow 0}\Bigl(\frac{\alpha}{\alpha e^t-1}\Bigr)^N=\frac{\alpha^N}{(\alpha-1)^N}$.
Also,
\begin{equation}\label{eq32}
\frac{\partial}{\partial t}\Bigl(\frac{\alpha}{\alpha e^t-1}\Bigr)^{N}=-N\Bigl(\frac{\alpha}{\alpha e^t-1}\Bigr)^N-\frac{N}{\alpha}\Bigl(\frac{\alpha}{\alpha e^t-1}\Bigr)^{N+1},
\end{equation}
which gives
\begin{equation}\label{eq33}
\begin{split}
E_k^{N+1}(\alpha)&amp;=-\frac{\alpha}{N}(E_{k+1}^N(\alpha)+N E_k^N(\alpha))\\
&amp;=-\frac{\alpha}{N}(E_{k+1}^N(\alpha)-E_k^N(\alpha)+(N+1) E_k^N(\alpha))\\
&amp;=-\frac{\alpha}{N}(\Delta_1+(N+1)I) E_k^N(\alpha)\\
&amp;=\frac{(-\alpha)^2}{N(N-1)}(\Delta_1+(N+1)I) (\Delta_1+NI)E_k^{N-1}(\alpha)\\
&amp;~~\vdots\\
&amp;=\frac{(-\alpha)^{N}}{N!}(\Delta_1+(N+1)I) \cdots (\Delta_1+2 I) E_k(\alpha).
\end{split}
\end{equation}
The last expression in \eqref{eq33} is valid for $N\rightarrow 0$ in addition to $N=1,2,\ldots$.
The result follows on changing $N$ to $N-1$ in \eqref{eq33}.
\end{proof}
\begin{lemma}\label{L2} For $k,~\alpha,~N$ as before,
    \begin{equation}\label{eq40}
E_k^{N}(\alpha)=\frac{\alpha^{N-1}}{(N-1)!}\sum_{i=0}^{N}(-1)^{i-1} s(N,i)E_{k+i-1}(\alpha)
\end{equation}
where $s(n,m)$ is the Stirling number of first kind.
\end{lemma}
\begin{proof}
For $n,m\in\A$, the Stirling number of first kind $s(n,m)$ is defined
by $x(x-1)\cdots (x-n+1)=\sum_{m=0}^{n}s(n,m)x^m$ which on changing $x$ to $-x$ gives
$x(x+1)\cdots (x+n-1)=\sum_{m=0}^{n}(-1)^{n-m} s(n,m)x^m$.
Replacing the symbol $x$ in the generating function  for $s(n,m)$ above by the shift operator $E_1=\Delta_1+I$, we have
\begin{equation}\label{eq39}
 \begin{split}
(\Delta_1+2I)\cdots (\Delta_1+(N-1)I)E_k(\alpha)&amp;=E_1 (E_1+I)\cdots (E_1+(N-1)I)E_{k-1}(\alpha)\\
&amp;=\sum_{i=0}^{N}(-1)^{N-i} s(N,i)E_1^i (E_{k-1})(\alpha)\\
&amp;=\sum_{i=0}^{N}(-1)^{N-i} s(N,i)E_{k+i-1}(\alpha).
\end{split}
\end{equation}
Substituting for $(\Delta_1+2I)\cdots (\Delta_1+(N-1)I)E_k(\alpha)$ from \eqref{eq39} in \eqref{eq31}, we get \eqref{eq40}.
\end{proof}
\begin{remark}
The expression \eqref{eq40} is special in a sense that, it connects $B_{k+N}^N$ with $E_k^N(1)$. More precisely, we have the next result.
\end{remark}
\begin{theorem}\label{th6}  For all $k$, $N\in\mathbb{N}$,
\begin{equation}\label{eq42}
B_{k+N}^N=\frac{(k+N)!}{k!}E_k^N(1).
\end{equation}
\end{theorem}
\begin{proof}
Using $\displaystyle {E_{k}}{(1)}=\frac{B_{k+1}}{k+1}$ for $\alpha=1$ in \eqref{eq40}, we get
 \begin{equation*}\label{eq52}
 \begin{split}
   \frac{(k+N)}{k!}E_k^N(1)&amp;=
 \frac{(k+N)}{k!}\frac{1}{(N-1)!}\sum_{i=0}^{N}(-1)^{i-1}s(N,i)\frac{B_{k+i}}{k+i}\\
 &amp;=N{k+N\choose N}\sum_{i=0}^{N}(-1)^{N-i-1}s(N,N-i)\frac{B_{k+N-i}}{k+N-i}\\&amp;=B_{k+N}^N.
\end{split}
 \end{equation*}
where the last step has been obtained using \eqref{eq001}.
\end{proof}
\section{Alternating power sums}\label{Sec:5}
Singh \cite{JS9} defined for any $\alpha\in \mathbb{C}$,  $k$-th $\alpha$-power sum $S_k(\alpha,x)$ by egf
\begin{equation}
    \frac{\alpha^{x+1}e^{(x+1)t}-\alpha e^t}{\alpha e^t-1}=\sum_{k=0}^{\infty}S_k(\alpha,x)\frac{t^k}{k!},~\alpha\neq 1,~|t|&lt;|\log\alpha|.
\end{equation}
\begin{remark}
The following closed form expression for $S_k(\alpha,x)$ is obtainable:
\begin{equation}\label{eq41}
    S_k(\alpha,x)=\begin{cases}\sum_{m=0}^{k}{k \choose m}E_m(\alpha)\{\alpha^x(1+x)^{k-m}-1\},&amp;~\text{if}~\alpha\neq 1;\\
                                S_k(x), &amp;~\text{if}~ \alpha=1.
                                \end{cases}
\end{equation}
\end{remark}
\begin{remark}
Note that $\lim_{\alpha\rightarrow 1}S_k(\alpha,x)\neq S_{k}(x)$ even when $x$ is a positive integer.  However, for $x\in\mathbb{N}$,  $S_k(\alpha,x)=\alpha+\alpha^2 2^k +\cdots \alpha^x x^k$. Further, if $0&lt;|\alpha|&lt;1$ then $\lim_{x\rightarrow \infty} S_k(\alpha,x)$ exists and is equal to $-\alpha^{-1}E_k(\alpha)$ for all $k\in\mathbb{N}$, where for $k=0$, the limit is $-E_0(\alpha)$. In particular, the Abel sum for the divergent series $\eta(-k)=\sum_{n=0}^\infty (-1)^n n^k$ can be obtained in closed form as the one sided limit at $\alpha=-1$, i.e.,
\begin{equation}
\displaystyle \eta(-k)=\lim_{\alpha\rightarrow -1^{+}}(\lim_{x\rightarrow \infty}S_{k}(\alpha,x))=-E_{k}(-1)=-\sum_{i=1}^{k}\frac{1}{2^{i+1}}\sum_{j=1}^{i}{i\choose j}(-1)^j  j^k.
\end{equation}
\end{remark}
We now consider sums of products of $\alpha$-power sums and the related Abel sums.
\begin{definition}
   For $N\in\A$ and complex number $x$, sum of products of $\alpha$-power sums $T_k^N(\alpha,x)$ is defined via the following egf:
   \begin{equation}\label{eq411}
   G(\alpha,x,N,t)=\Bigl( \frac{\alpha^{x+1}e^{(x+1)t}-\alpha e^t}{\alpha e^t-1}\Bigr)^N=\sum_{k=0}^{\infty}T_k^N(\alpha,x)\frac{t^k}{k!},~\alpha\neq 1,~|t|&lt;|\log\alpha|.
\end{equation}
\end{definition}
 It is immediate to see that $T_k^N(\alpha,0)=0$ and $T_k^1(\alpha,x)=S_k(\alpha,x)$.
 \begin{remark}
   For $0&lt;|\alpha|&lt;1$, $\alpha^x\rightarrow 0$ as $x\rightarrow \infty$. If this is the case then from \eqref{eq41}, we have
\begin{equation}
\lim_{x\rightarrow\infty}T_k^N(\alpha,x)=(-1)^N\sum_{m=0}^{k}{k\choose m}E_m^N(\alpha)N^{k-m},~0&lt;|\alpha|&lt;1,
\end{equation}
which in particular gives the Abel sum of products
\begin{equation}\label{0001}
    \eta^N(-k)=(-1)^N\sum_{m=0}^{k}{k\choose m}E_m^N(-1)N^{k-m}.
\end{equation}
 \end{remark}
\begin{theorem}\label{th60}  For $N\in \mathbb{N}$, $k\in\A,~x\in\mathbb{C}$, and $\alpha\neq 0,1$,
\begin{equation}\label{eq43}
    T_k^N(\alpha,x)=\sum_{n=0}^{k}{k\choose m}E_m^N(\alpha)\sum_{n=0}^{N}{N\choose n}(-1)^{N-n}\alpha^{nx}(N+nx)^{k-m}.
\end{equation}
\end{theorem}
\begin{proof} Follows from following computational steps:
\begin{equation*}\label{eq45}
\begin{split}
   \sum_{k=0}^{\infty}T_k^N(\alpha,x)\frac{t^k}{k!}&amp;=\Bigl( \frac{\alpha}{\alpha e^t-1}\Bigr)^N(\alpha^x e^{xt}-e^t)^N\\
   &amp;=\sum_{m=0}^{\infty}E_m(\alpha)\frac{t^m}{m!}\sum_{i=0}^{N}{N\choose i}(-1)^{N-i}\alpha^{ix} e^{(N+ix)t}\\
   &amp;=\sum_{m=0}^{\infty}E_m(\alpha)\frac{t^m}{m!}\sum_{i=0}^{N}{N\choose i}(-1)^{N-i}\alpha^{ix} \sum_{n=0}^{\infty}(N+ix)^n\frac{t^n}{n!}\\
   &amp;=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}E_m(\alpha)\sum_{i=0}^{N}{N\choose i}(-1)^{N-i}\alpha^{ix} (N+ix)^n\frac{t^{m+n}}{m!n!}\\
&amp;=\sum_{k=0}^{\infty}\sum_{m=0}^{k}{k\choose m}E_m(\alpha)\sum_{i=0}^{N}{N\choose i}(-1)^{N-i}\alpha^{ix} (N+ix)^{k-m}\frac{t^{k}}{k!}\\
\end{split}
\end{equation*}
from which the result can be obtained on comparing the like powers of $t$.
 \end{proof}
\section{Further extensions}\label{Sec:6}
The sequence of functions $\bigl(T_k^N(\alpha,x)\bigr)_{k\in\mathbb{N}_0},~x,~\alpha\in\mathbb{C},~\alpha\neq 0,~1$ satisfies simple recurrence relation from which its terms can be obtained successively. Here is the special one, which we state as the following.
\begin{proposition} For $k,N\in\A,~x\in\mathbb{C}$, and $\alpha\neq 0,~1$,
    \begin{equation}\label{eq54}
    T_0^N(\alpha,x)=\alpha^N\Bigl(\frac{\alpha^x-1}{\alpha-1}\Bigr)^N,~T_{k}^{N}(\alpha,x)=
    \alpha\frac{\partial}{\partial\alpha}T_{k-1}^N(\alpha,x),~k\in\mathbb{N}.
    \end{equation}
\end{proposition}
\begin{proof}Expression for $T_0^N$ is trivially true. It is also easy to see that \eqref{eq54} holds for $N=0$. For $N\in\mathbb{N}$, the proof follows from
  $\frac{\partial}{\partial t}G(\alpha,x,N,t)=\alpha\frac{\partial}{\partial \alpha}G(\alpha,x,N,t)$,
where $G(\alpha,x,N,t)$ is as in \eqref{eq411}.
\end{proof}
The recurrence \eqref{eq54} can be used to extend the definition of the sum of products of the power sums to negative integer values of $k$.
\begin{definition}\label{def1}
The sum of products $T_k^N(\alpha,x)$ for a complex number $N$  and real $\alpha\neq 0,~1$ is defined successively by
\begin{equation}\label{eq56}
    T_0^N(\alpha,x)=\alpha^N\Bigl(\frac{\alpha^x-1}{\alpha-1}\Bigr)^N;
    ~T_{-k}^N(\alpha,x)=\int_{0}^{\alpha
    }\frac{T^N_{-k+1}(\theta,x)}{\theta}d\theta,~k\in\mathbb{N},~\alpha\neq 0,~1.
\end{equation}
\end{definition}
\begin{definition}
For $0&lt;|\alpha|&lt;1$, define \begin{equation}\label{zz}
T_k^N(\alpha,\infty)=\lim_{x\rightarrow \infty}T_{k}^N(\alpha,x).
\end{equation}
\end{definition}
From \eqref{eq56}, we have
\begin{equation}\label{eq57}
\begin{split}
\displaystyle T_{-1}^N(\alpha,\infty)&amp;=\int_{0}^{\alpha}{\theta^{N-1}}(1-\theta)^{-N} d\theta.
\end{split}
\end{equation}
For $N=1$, Singh \cite{JS9} proved the following result:
\begin{equation}\label{eq67}
   T_{-i}^{1}(\alpha,\infty)=\begin{cases}-\log|1-\alpha|,&amp;~\text{if}~i=1;\\ \displaystyle \frac{(-1)^{i-1}}{(i-1)!}\int_{0}^{\alpha}\frac{(\log(t))^{i-1}}{1-t}dt-
\sum_{\beta=1}^{i-1}\frac{(-\log|\alpha|)^\beta}{\beta!}T_{-i+\beta}(\alpha,\infty),&amp;~\text{if}~i\geq 2.
\end{cases}
\end{equation}
Observe from \eqref{eq67} that
\begin{equation}
\lim_{\alpha\rightarrow 1^{-}}T_{-k}^1(\alpha,\infty)=\zeta(k),~\lim_{\alpha\rightarrow -1^{+}}T_{-k}^1(\alpha,\infty)=-\eta(k),
 \end{equation}
for all $k\in\mathbb{N}$.

In the next result, we generalize the formula \eqref{eq67} for the underlying sums of products, i.e., for all positive integers $N\neq 1$. So, we have the following main result.
\begin{theorem} For all $k\in\A$, $N=2,3,\ldots$, and $0&lt;|\alpha|&lt;1$,
\begin{equation}\label{eq64}
\begin{split}
    T_{-k}^N(\alpha,\infty)&amp;=\sum_{i_1=1}^{N-1}\sum_{i_2=1}^{i_1-1}\cdots
\sum_{i_k=1}^{i_{k-1}-1}\Bigl(\frac{\alpha}{1-\alpha}\Bigr)^{i_k}\frac{(-1)^{N-k-i_k}}{i_1\cdots i_k} \\&amp;+ \sum_{i=1}^{k-1}\frac{{s(N,k+1-i)}}{(N-1)!}T_{-i}^1(\alpha,\infty)+(-1)^{N-1} T_{-k}^1(\alpha,\infty).
\end{split}
\end{equation}
\end{theorem}
\begin{proof}From \eqref{eq56} and \eqref{eq57}, we have
\begin{equation}\label{eq61}
\displaystyle T_{-1}^N(\alpha,\infty)=
\sum_{i=1}^{N-1}(-1)^{N-1-i}\frac{T_0^{i}(\alpha,\infty)}{i}+
(-1)^{N-1}T_{-1}^{1}(\alpha,\infty),
\end{equation}
for $N=2,3,\ldots$. Similarly,
\begin{equation}\label{eq62}
\displaystyle T_{-2}^N(\alpha,\infty)=\sum_{i=1}^{N-1}\sum_{j=1}^{i-1}(-1)^{N-2-j}\frac{T_0^{j}(\alpha,\infty)}{i j}+(-1)^{N-1}T_{-1}^1\sum_{i=1}^{N-1}\frac{1}{i}+(-1)^{N-1}T_{-2}^1,
\end{equation}
which inductively leads to the following:
\begin{equation}\label{eq65}
\begin{split}
    T_{-k}^N(\alpha,\infty)&amp;=\sum_{i_1=1}^{N-1}\sum_{i_2=1}^{i_1-1}\cdots
\sum_{i_k=1}^{i_{k-1}-1}\Bigl(\frac{\alpha}{1-\alpha}\Bigr)^{i_k}\frac{(-1)^{N-k-i_k}}{i_1\cdots i_k}\\&amp;+\sum_{i_1=1}^{N-1}\sum_{i_2=1}^{i_1-1}\cdots
\sum_{i_{k-1}=1}^{i_{k-2}-1}\frac{(-1)^{N-k}}{i_1\cdots i_{k-1}}T_{-1}^1(\alpha,\infty)\\&amp;+\cdots\\
&amp;+\sum_{i_1=1}^{N}\frac{(-1)^{N-1}}{i_1}T_{-k+1}^1(\alpha,\infty)+(-1)^{N-1} T_{-k}^1(\alpha,\infty),
\end{split}
\end{equation}
where we have used \eqref{eq67} and the fact that $T_0^i(\alpha,\infty)=\bigl(\frac{\alpha}{1-\alpha}\bigr)^{i}$ for all $i\in\mathbb{N}$.

On substituting for  the following expression of
the Stirling numbers of the first kind \cite{Q} in \eqref{eq65}
\begin{equation}\label{eq70}
   \displaystyle s(n+1,m+1)=(-1)^{n+m}n!\sum_{\ell=1}^{n}\sum_{\ell_2=1}^{\ell_1-1}\cdots \sum_{\ell_{m}=1}^{\ell_{m-1}-1}\frac{1}{\ell_1\ell_2\cdots \ell_{m}},
\end{equation}
we obtain   \eqref{eq64}.
\end{proof}
\begin{corollary} Let  $k,~N\in\mathbb{N}$, such that $N=2,3,\ldots$. Then
\begin{equation}\label{eq71}
    \lim_{\alpha\rightarrow 1^-}{T_{-k}^N(\alpha,\infty)}=\frac{1}{(N-1)!}\sum_{i=2}^{k-1}s(N,k+1-i)\zeta(i)+(-1)^{N-1}\zeta(k)~\mbox{for}~N&lt;k,
\end{equation}
and
\begin{equation}\label{eq72}
\begin{split}
    \lim_{\alpha\rightarrow -1^+}{T_{-k}^N(\alpha,\infty)}&amp;=\sum_{i_1=1}^{N-1}\sum_{i_2=1}^{i_1-1}\cdots
\sum_{i_k=1}^{i_k-1}\Bigl(\frac{1}{2}\Bigr)^{i_k}\frac{(-1)^{N-k-i_k}}{i_1\cdots i_k}\\&amp;+\sum_{i=1}^{k-1}\frac{s(N,k+1-i)\eta(i)}{(N-1)!}+(-1)^{N-1}\eta(k).
\end{split}
\end{equation}
\end{corollary}
\begin{remark}
For a positive integer $N$ and $0&lt;|\alpha|&lt;1$, we have $\displaystyle \frac{(-\alpha)^N e^{Nt}}{(\alpha e^t-1)^N}=\sum_{k=0}^{\infty}T_{k}^N(\alpha,\infty)\frac{t^k}{k!}$, from which, it follows that
\begin{equation}\label{eq81}
\begin{split}
    \lim_{\alpha\rightarrow 1^-}T_{-k}^N(\alpha,\infty)&amp;=(-1)^N\lim_{t_1\rightarrow 0}\Bigl(\int_{-\infty}^{t_1} \cdots \int_{-\infty}^{t_{k}} \frac{e^{Nt_{k+1}}}{(e^{t_{k+1}}-1)^N}d{t_{k+1}}\cdots d{t_2}\Bigr)\\&amp;=\frac{1}{(N-1)!}\sum_{i=k+1-N}^{k}s(N,k+1-i)\zeta(i)~\mbox{if}~N&lt;k.
\end{split}
\end{equation}
\end{remark}
\begin{example}\label{ex1}
Table \ref{table1} below shows the values of $\displaystyle\lim_{\alpha\rightarrow 1^-}T_{-k}^N(\alpha,\infty)$ for few values of $k$ and $N$. The computations were done in {\tt Mathematica} with reference to \eqref{eq71}.
\begin{table}[h!]
\caption{$\displaystyle\lim_{\alpha\rightarrow 1^-}T_{-k}^N(\alpha,\infty)$ for some values of $k$ and $N$}\label{table1}
\begin{center}
\begin{tabular}{cccc}
  \hline
  $k \setminus N$ &amp;2&amp;3&amp;4\\
  \hline
 3&amp; $\displaystyle\frac{\pi^2}{6}-\zeta(3)$ &amp;
 &amp; \vspace{.3cm} \\
 4&amp; $\displaystyle-\frac{\pi^4}{90}+\zeta(3)$ &amp; $\displaystyle\frac{\pi^2}{12}+\frac{\pi^4}{90}-
 3\frac{\zeta(3)}{2}$ &amp; \vspace{.3cm}
 \\
 5&amp;$\displaystyle\frac{\pi^4}{90}-\zeta(5)$  &amp; $\displaystyle-\frac{\pi^4}{60}+\frac{\zeta(3)}{2}+\zeta(5)$ &amp; $\displaystyle\frac{\pi^2}{36}+\frac{11\pi^4}{540}-
 {\zeta(3)}-\zeta(5)$ \vspace{.3cm}
 \\
 6 &amp;$\displaystyle-\frac{\pi^6}{945}+\zeta(5)$  &amp; $\displaystyle\frac{\pi^4}{180}+
 \frac{\pi^6}{945}-\frac{3\zeta(5)}{2}$ &amp;$\displaystyle-\frac{\pi^4}{90}-\frac{\pi^6}{945}+
 \frac{\zeta(3)}{6}+\frac{11\zeta(5)}{6}$   \\
   &amp;  &amp;  &amp;      \\
\hline
\end{tabular}
\end{center}
\end{table}
\end{example}
\begin{remark}
The preceding Example \ref{ex1}, together with
 \eqref{eq81}, establishes that
 \begin{equation}\label{abb}
   \lim_{\alpha\rightarrow 1^-}T_{-k}^N(\alpha,\infty)=\sum_{\sum_{i=1}^N k_i=k,~0\leq k_i\leq k,~k_i\neq 1}^{k} {k\choose k_1,\ldots,k_N}\zeta(k_1)\cdots \zeta(k_N),
 \end{equation}
 for $N&lt;k$, where the right hand side is the sum of product of Riemann zeta functions.
\end{remark}
The classical Euler's  formula is given by
\begin{equation}\label{abc}
  \int_{0}^{\frac{1}{2}}\frac{\log(1-t)}{t}dt=-T_{-2}^1\Bigl(\frac{1}{2},\infty\Bigr)=\frac{(\log(2))^2}{2}-\frac{\zeta(2)}{2}.
\end{equation}The next result generalize the formula \eqref{abc} to the sums of products.
\begin{theorem}\label{th44} For $N=2,3,\ldots$,  the generalized Euler's formula is given by
   \begin{equation}\label{eq63}
\begin{split}
\displaystyle (-1)^NT_{-2}^N\Bigl(\frac{1}{2},\infty\Bigr)=\sum_{i=1}^{N-1}\sum_{j=1}^{i-1}\frac{(-1)^{j}}{i j}+\log(2)\sum_{i=1}^{N-1}\frac{1}{i}+\frac{(\log(2))^2}{2}-\frac{\zeta(2)}{2}.
\end{split}
\end{equation}
\end{theorem}
\begin{proof}
First observe that for  $\alpha=\frac{1}{2}$, $T_{-2}^{N}(\alpha,\infty)=T_{-2}^{N}(1-\alpha,\infty)$.
Along with this, if we use the following identity due to Singh \cite{JS9}
\begin{equation}
-\int_{0}^{1-\alpha}\frac{\log(1-t)}{t}dt=-\log(\alpha)\log(1-\alpha)+
\int_{0}^{1}\frac{\log(t)}{1-t}dt+\int_{0}^\alpha\frac{\log(1-t)}{t}dt,
\end{equation}
and the fact that $T_0^N(\frac{1}{2})=1$ for all $N$, we obtain \eqref{eq63} by taking $\alpha=\frac{1}{2}$ in \eqref{eq62}, where $\displaystyle \int_{0}^{1}\frac{\log(t)}{1-t}dt=\zeta(2)$.
\end{proof}

\section{Acknowledgments}
The author is thankful to the referee for the valuable suggestions regarding improvement in the presentation of the paper.

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\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11B68, 05A10, 11B65.

\noindent \emph{Keywords: } 
power sum of integers, Bernoulli number, Bernoulli polynomial, Riemann
zeta function, exponential generating function, Apostol-Bernoulli
number.


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\vspace*{+.1in}
\noindent
Received  May 14 2015;
revised versions received  September 21 2015; September 28 2015;
November 26 2015.
Published in {\it Journal of Integer Sequences}, December 17 2015.

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