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\begin{center}
\vskip 1cm{\LARGE\bf 
A Unified Treatment of Certain Classes \\
\vskip .12in
of Combinatorial Identities
}
\vskip 1cm
\large
Necdet Bat{\i}r and Anthony Sofo \\
Department of Mathematics\\
Nev{\c{s}}ehir Hac{\i} Bekta{\c{s}} Veli University \\
50300 Nev{\c{s}}ehir \\
Turkey \\
and\\
College of Engineering and Science \\
Victoria University \\
VIC 8001 \\
Australia\\
\href{mailto:nbatir@hotmail.com}{\tt nbatir@hotmail.com} \\
\href{mailto:anthony.sofo@vu.edu.au}{\tt anthony.sofo@vu.edu.au}\\
\end{center}

\vskip .2 in

\begin{abstract} We propose and prove some general combinatorics formulas. Applying these
formulas, we obtain some new binomial harmonic and harmonic Fibonacci and
Lucas number identities. We also recover some known identities included in
the works of Frontczak and Boyadzhiev.
\end{abstract}

\section{Introduction}

For $s\in\mathbb{C}$ and $n\in\mathbb{N}$, a generalized harmonic number $H_n^{(s)}$ of order $s$ is defined by
\begin{equation*}
H_n^{(s)}=\sum\limits_{k=1}^{n}\frac{1}{k^s}, \quad \mbox{and} \quad
H_n^{(1)}=H_n,\quad H_0^{(s)}=0,
\end{equation*}
where $H_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ are the usual
harmonic numbers; see \cite{5,9,SoSri}. Alternating or skew-harmonic
numbers $H_n^{-}$ are defined by
\begin{equation*}
H_n^{-}=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k},
\end{equation*}
which is the partial sum of $\log2=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}$; see \cite{7,14,16}.

In 2009, Boyadzhiev \cite{6} studied some binomial sums involving the
harmonic numbers using Euler's transform and obtained the following
identity:
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}a^{k}b^{n-k}H_{k}=(a+b)^{n}H_{n}-\sum_{k=1}^{n}\frac{b^{k}}{k}(a+b)^{n-k},
\end{equation*}
which can be equivalently written, by setting $a/b=x$, as follows:
\begin{equation}
\sum_{k=1}^{n}\binom{n}{k}H_{k}x^{k}=(x+1)^{n}\bigg(H_{n}-\sum_{k=1}^{n}\frac{1}{k(x+1)^k}\bigg).  \label{e:1}
\end{equation}
Batir, in a recent paper \cite{2}, offers many interesting finite sum
identities with harmonic numbers, including (\ref{e:1}). Frontczak \cite{11,12} obtained a complement of  the identity given in (\ref{e:1})  and derived
\begin{equation}
\sum_{k=1}^{n}\binom{n}{k}H_{k}x^k=(x+1)^nH_n-\sum_{k=0}^{n-1}H_{n-k}(x+1)^k,
\label{e:2}
\end{equation}
which can be regarded as a complement of (\ref{e:1}). Frontczak \cite{12} 
also derived an analogue formula for the skew-harmonic numbers. Throughout this paper, we take
\begin{equation*}
\alpha =\frac{1+\sqrt{5}}{2}\quad \mbox{and}\quad \beta =\frac{1-\sqrt{5}}{2}.
\end{equation*}
The Fibonacci sequence $(F_{n})_{n\geq 0}$ is given by $F_{0}=0$, $F_{1}=1$,
and for $n\geq 2$ it satisfies the recursion relation $F_{n}=F_{n-1}+F_{n-2}$. The Lucas sequence $(L_{n})_{n\geq 0}$ satisfies the same recursion
relation, that is, $L_{n}=L_{n-1}+L_{n-2}$ with the initial values $L_{0}=2$
and $L_{1}=1$, where $n\in\mathbb{N}$. The Binet formulas for the Fibonacci
numbers $F_{n}$ and Lucas numbers $L_{n}$ state that
\begin{equation*}
F_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\sqrt{5}}\quad \mbox{and}\quad
L_{n}=\alpha ^{n}+\beta ^{n}.
\end{equation*}
In this article we develop a general formula which contains (\ref{e:1}) and many others as special cases. Our results allow to
establish many interesting new identities involving the harmonic numbers, Fibonacci numbers, and Lucas numbers. More
precisely, for any sequence $(a_{n})_{n\geq 1}$ in $\mathbb{C}$ and $x\in\mathbb{C}$ we consider the following general binomial sums:
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}\sum_{j=1}^{k}a_{j}x^{k}.
\end{equation*}
To demonstrate the usefulness of our formulas we give many applications. The
following simple, but useful, identities will be required in our ongoing
analysis:
\begin{equation}
\binom{n+1}{k}=\frac{n+1}{k}\binom{n}{k-1},  \label{e:3}
\end{equation}
\begin{equation}
\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}.  \label{e:4}
\end{equation}
We continue with the following two lemmas.
\begin{lemma}
\bigskip\label{LemmaA} Let $(a_n)_{n\geq 1}$ be any real or complex sequence. Then we have
\begin{equation*}
\sum\limits_{k=1}^n\frac{1}{k}\sum\limits_{j=1}^k\binom{k}{j}a_j=\sum\limits_{k=1}^n\binom{n}{k}\frac{a_k}{k}.
\end{equation*}
\end{lemma}
This lemma is not new, and a generalization of it can be found in Boyadzhiev \cite[Eq. 3]{B1}. As stated there, Boyadzhiev attributed this lemma  to 't Woord \cite{19}. Also see equation (5.6) in \cite{7}. It is  worthy to note that  Bat{\i}r \cite{1} rediscovered this lemma in 2017, and established many interesting combinatorial identities and series involving the harmonic numbers.
\begin{lemma}
\bigskip\label{LemmaB} Let $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ be any two sequences of real or
complex numbers. Then we have
\begin{equation*}
\sum\limits_{k=1}^{n}a_k\sum\limits_{j=1}^{k}b_j=\sum\limits_{p=0}^{n-1}\sum\limits_{k=1}^{n-p}b_ka_{p+k}.
\end{equation*}
\end{lemma}

This lemma was proved in \cite{3} and helps us to reduce some double sums to a single sum.

\section{Main Results}

In this section we collect our main results.

\begin{theorem}
\bigskip\label{TheoremA} Let $(a_n)_{n\geq 1}$ be any sequence in $\mathbb{C}$. Then we have
\begin{equation*}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\sum_{j=1}^{k}a_j=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}a_{k+1}.
\end{equation*}
\end{theorem}

\begin{proof}
By (\ref{e:4}), we obtain
\begin{align}  \label{e:5}
&\sum_{k=1}^{n}(-1)^k\binom{n}{k}\sum_{j=1}^{k}a_j=\sum_{k=1}^{n}(-1)^k\bigg(\binom{n-1}{k}+\binom{n-1}{k-1}\bigg)\sum_{j=1}^{k}a_j  \notag \\
&=\sum_{k=1}^{n-1}(-1)^k\binom{n-1}{k}\sum_{j=1}^{k}a_j+\sum_{k=1}^{n}(-1)^k\binom{n-1}{k-1}\sum_{j=1}^{k}a_j,
\end{align}
where we have used $\binom{n}{n+1}=0$ and $\binom{n-1}{-1}=0$. On the other hand,
we have
\begin{align}  \label{e:6}
&\sum_{k=1}^{n}(-1)^k\binom{n-1}{k-1}\sum_{j=1}^{k}a_j=\sum_{k=0}^{n-1}(-1)^{k-1}\binom{n-1}{k}\sum_{j=1}^{k+1}a_j  \notag \\
&=\sum_{k=0}^{n-1}(-1)^{k-1}\binom{n-1}{k}\bigg(a_{k+1}+\sum_{j=1}^{k}a_j\bigg)  \notag \\
&=\sum_{k=0}^{n-1}(-1)^{k-1}\binom{n-1}{k}a_{k+1}-\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\sum_{j=1}^{k}a_j.
\end{align}
Replacing (\ref{e:6}) in (\ref{e:5}), we complete the proof.
\end{proof}

\begin{theorem}
\bigskip\label{TheoremB} Let $(a_n)_{n\geq 1}$ be any sequence in $\mathbb{C}$, $x\in\mathbb{C}$, and $A_n$ be the $n$th partial sum of $(a_n)_{n\geq 1}$, that is, $A_n=a_1+a_2+\cdots+a_n$. Then
\begin{equation}  \label{e:7}
\sum_{k=1}^{n}\binom{n}{k}A_k x^k=(1+x)^{n-1}\sum_{k=0}^{n-1}\bigg(\sum_{j=0}^{k}\binom{k}{j}a_{j+1}x^{j+1}\bigg)\frac{1}{(1+x)^k}.
\end{equation}
Here the sum of $0$ terms is taken $0$, that is, $A_0 = 0$.
\end{theorem}

\begin{proof}
Let us define
\begin{equation*}
g_n(x)=\sum_{k=1}^{n}(-1)^k\binom{n}{k}A_kx^k.
\end{equation*}
Clearly, we have by Theorem \ref{TheoremA} that
\begin{align*}
g_n(x)&=\sum_{k=1}^{n}(-1)^k\binom{n}{k}A_jx^j=\sum_{k=0}^{n}(-1)^k\binom{n}{k}\sum_{j=1}^{k}\left(A_jx^j-A_{j-1}x^{j-1}\right) \notag \\
&=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}\left(A_{k+1}x^{k+1}-A_kx^k\right).
\end{align*}
Using $A_{k+1}=A_k+a_{k+1}$, we obtain
\begin{align}\label{e:8}
g_n(x)&=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}\left((A_{k}+a_{k+1})x^{k+1}-A_kx^k\right)
\notag \\
&=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}A_{k}x^{k+1}+\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}a_{k+1}x^{k+1}  \notag \\
&-\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}A_{k}x^{k}  \notag \\
&=(1-x)g_{n-1}(x)+\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}a_{k+1}x^{k+1}.
\end{align}
Now we shall show the following expression by mathematical induction.
\begin{align}  \label{e:9}
g_n(x)=(1-x)^{n-1}\sum_{k=0}^{n-1}\bigg(\sum_{j=0}^{k}(-1)^{j+1}\binom{k}{j}a_{j+1}x^{j+1}\bigg)\frac{1}{(1-x)^k}.
\end{align}
It is easy to see that (\ref{e:9}) is true for $n=1$. We assume that (\ref{e:9}) is true for $n$. Then by (\ref{e:8}) and the induction hypothesis,
we have
\begin{align*}
g_{n+1}(x)&=(1-x)^n\sum_{k=0}^{n-1}\bigg(\sum_{j=0}^{k}(-1)^{j+1}\binom{k}{j}a_{j+1}x^{j+1}\bigg)\frac{1}{(1-x)^k} \\
&+\sum_{k=0}^{n}(-1)^{k+1}\binom{n}{k}a_{k+1}x^{k+1} \\
&=(1-x)^n\sum_{k=0}^{n}\bigg(\sum_{j=0}^{k}(-1)^{j+1}\binom{k}{j}a_{j+1}x^{j+1}\bigg)\frac{1}{(1-x)^k},
\end{align*}
which shows that (\ref{e:9}) is valid for $n+1$, and the proof of (\ref{e:9}) is complete. Replacing $x$ by $-x$ in (\ref{e:9}) we prove (\ref{e:7}).
Setting $x=1$ in (\ref{e:7}), we obtain the following conclusion:
\end{proof}
\begin{corollary}
Let $(a_n)_{n\geq 1}$ be any sequence of complex numbers. Then we have
\begin{equation}  \label{e:10}
\sum_{k=1}^{n}\binom{n}{k}A_k=2^{n-1}\sum_{k=0}^{n-1}\frac{1}{2^k}\sum_{j=0}^{k}\binom{k}{j}a_{j+1}.
\end{equation}
\end{corollary}

\section{Applications}

In this section we present many applications of our main theorems. Our first
identity recovers (\ref{e:1}).

\begin{identity}
\bigskip\label{IDEN1}
 Let $n\in\mathbb{N}$ and $x\in\mathbb{C}$. Then
\begin{equation}  \label{e:11}
\sum_{k=1}^{n}\binom{n}{k}H_kx^k=(1+x)^n\bigg(H_n-\sum_{k=1}^{n}\frac{1}{k(1+x)^{k}}\bigg).
\end{equation}
\end{identity}


\begin{proof}
Putting $a_k=\frac{1}{k}$ in Theorem \ref{TheoremB}, we get
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}H_kx^k=(1+x)^{n-1}\sum_{k=0}^{n-1}\bigg(\sum_{j=0}^{k}\binom{k}{j}\frac{x^{j+1}}{j+1}\bigg)\frac{1}{(1+x)^{k}}.
\end{align*}
From (\ref{e:3}), it is very easy to see that $\sum_{j=0}^{k}\binom{k}{j}\frac{x^{j+1}}{j+1}=\frac{(1+x)^{k+1}-1}{k+1}$, so that
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}H_kx^k=(1+x)^{n}\sum_{k=1}^{n}\frac{(1+x)^{k}-1}{k(1+x)^{k}},
\end{align*}
which is equivalent to (\ref{e:11}).
\end{proof}

In the particular cases of $x=-1$ and $x=1$ in (\ref{e:11}), we get
\begin{align}  \label{e:12}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k=-\frac{1}{n}\quad \mbox{and}\quad
\sum_{k=1}^{n}\binom{n}{k}H_k=2^n\sum_{k=1}^{n}\frac{2^k-1}{k2^k}.
\end{align}

\begin{remark}
The first identity here is very old and due to Euler \cite{10}. The second
identity is also well known and recently, it has been rediscovered by many
authors; (see \cite{2,6, 8,13,17}).
\end{remark}

\begin{identity}
\bigskip\label{IDEN2}  Let $n\in\mathbb{N}$ and $x\in\mathbb{C}$. Then we have
\begin{align}  \label{e:13}
\sum_{k=1}^{n}\binom{n}{k}H_k^2x^k=(1+x)^{n}\bigg(H_n^2-\sum_{k=1}^{n}\frac{H_n-2H_k+H_{n-k}}{k(1+x)^{k}}-2\sum_{k=1}^{n}\frac{1}{k^2(1+x)^{k}}\bigg).
\end{align}
\end{identity}


\begin{proof}
Letting $a_{k}=(H_{k})^{2}-(H_{k-1})^{2}$ in (\ref{e:7}) and using the
relation $H_{k+1}=H_{k}+\frac{1}{k+1}$, we get
\begin{align}
& \sum_{k=1}^{n}\binom{n}{k}H_{k}^{2}x^{k}=\sum_{k=1}^{n}\binom{n}{k}\sum_{j=1}^{k}(H_{j}^{2}-H_{j-1}^{2})x^{k}  \notag  \label{e:14} \\
& =(1+x)^{n-1}\sum_{k=0}^{n-1}\bigg(\sum_{j=0}^{k}\binom{k}{j}\frac{H_{j+1}+H_{j}}{j+1}x^{j+1}\bigg)(1+x)^{-k}.
\end{align}
On the other hand, by (\ref{e:3}) we have
\begin{align}
& \sum_{j=0}^{k}\binom{k}{j}\frac{H_{j+1}+H_{j}}{j+1}x^{j+1}=\frac{1}{k+1}\sum_{j=0}^{k}\binom{k+1}{j+1}\bigg(2H_{j+1}-\frac{1}{j+1}\bigg)x^{j+1}
\notag  \label{e:15} \\
& =\frac{1}{k+1}\sum_{j=1}^{k+1}\binom{k+1}{j}\bigg(2H_{j}-\frac{1}{j}\bigg)x^{j}  \notag \\
& =\frac{2}{k+1}\sum_{j=1}^{k+1}\binom{k+1}{j}H_{j}x^{j}-\frac{1}{k+1}\sum_{j=1}^{k+1}\binom{k+1}{j}\frac{x^{j}}{j}.
\end{align}
By (\ref{e:11}), it follows that
\begin{equation*}
\sum_{j=1}^{k+1}\binom{k+1}{j}H_{j}x^{j}=(1+x)^{k+1}\sum_{j=1}^{k+1}\frac{(1+x)^{j}-1}{j(1+x)^{j}}.
\end{equation*}
Using this identity, we conclude from (\ref{e:14}) and (\ref{e:15}), and Lemma \ref{LemmaA}
\begin{align}
\sum_{k=1}^{n}\binom{n}{k}H_{k}^{2}x^{k}&=2(1+x)^{n}\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{(1+x)^{j}-1}{j(1+x)^{j}}  \notag  \\
&-(1+x)^{n}\sum_{k=1}^{n}\frac{1}{k(1+x)^{k}}\sum_{j=1}^{k}\binom{k}{j}\frac{x^{j}}{j}\label{e:17}.
\end{align}Applying Lemma \ref{LemmaA}, one easily gets
\begin{equation}
\sum_{j=1}^{k}\binom{k}{j}\frac{x^{j}}{j}=\sum_{j=1}^{k}\frac{(1+x)^{j}-1}{j}. \label{e:18}
\end{equation}Substituting (\ref{e:18}) in (\ref{e:17}), we get, after some simplifications, that
\begin{align}  \label{e:19}
&\sum_{k=1}^{n}\binom{n}{k}H_k^2x^k=2(1+x)^{n}\sum_{k=1}^{n}\frac{H_k}{k}-2(1+x)^{n}\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{1}{j(1+x)^{j}}
\notag \\
&-(1+x)^{n}\sum_{k=1}^{n}\frac{1}{k(1+x)^{k}}\sum_{j=1}^{k}\frac{(1+x)^j}{j}+(1+x)^n\sum_{k=1}^{n}\frac{H_k}{k(1+x)^{k}}.
\end{align}
Setting
\begin{equation*}
a_k=\frac{(1+x)^{-k}}{k}\quad \mbox{and} \quad b_k=\frac{(1+x)^{k}}{k}
\end{equation*}
in Lemma \ref{LemmaB}, we obtain
\begin{equation}  \label{e:20}
\sum_{k=1}^{n}\frac{1}{k(1+x)^{k}}\sum_{j=1}^{k}\frac{(1+x)^j}{j}=H_n^{(2)}+\sum_{k=1}^{n}\frac{H_k+H_{n-k}-H_n}{k(1+x)^{k}}.
\end{equation}
Hassani and Rahimpour \cite{15} proved that for any double sequence $(a_{ij})_{i,j\geq 1}$ the following summation formula is valid.
\begin{equation*}
\sum_{j,k=1}^{n}a_{jk}=\sum_{k=1}^n\sum_{j=1}^{k}(a_{jk}+a_{kj})-\sum_{k=1}^{n}a_{kk}.
\end{equation*}
Setting $a_{ij}=\frac{(1+x)^{-j}}{ij}$ here, it follows that
\begin{align}  \label{e:21}
\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{1}{j(1+x)^{j}}=\sum_{k=1}^{n}\frac{H_n-H_k}{k(1+x)^{k}}+\sum_{k=1}^{n}\frac{1}{k^2(1+x)^{k}}.
\end{align}
Substituting (\ref{e:20}) and (\ref{e:21}) in (\ref{e:19}) the proof is
completed.
\end{proof}

\begin{identity}
\bigskip\label{IDEN3}
Putting $x=-1$ and $x=1$ in (\ref{e:13}), we obtain
\begin{equation}  \label{e:22}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^2=\frac{H_n}{n}-\frac{2}{n^2},
\end{equation}
and
\begin{align}  \label{e:23}
\sum_{k=1}^{n}\binom{n}{k}H_k^2=2^{n}\bigg(H_n^2-\sum_{k=1}^{n}\frac{H_n-2H_k+H_{n-k}}{k2^{k}}-2\sum_{k=1}^{n}\frac{1}{k^22^k}\bigg).
\end{align}
\end{identity}

\begin{remark}
The first identity here has previously been obtained by Wang \cite{18} by the
method of Riordan arrays, and rediscovered by Boyadzhiev \cite{B1}.
\end{remark}

\begin{identity}
\bigskip\label{IDEN4}
For $n\in\mathbb{N}$ we have
\begin{align*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^3=\frac{3H_n}{n^2}-\frac{H_n^2}{2n}+\frac{5H_n^{(2)}}{2n}-\frac{6}{n^3}.
\end{align*}
\end{identity}


\begin{proof}
Using $H_k=H_{k+1}-\frac{1}{k+1}$, we see that
\begin{equation}  \label{e:24}
H_{k+1}^{3}-H_{k}^{3}=\frac{3H_{k+1}^{2}}{k+1}-\frac{3H_{k+1}}{(k+1)^2}+\frac{1}{(k+1)^3}.
\end{equation}
Applying Theorem \ref{TheoremA} with $a_k=H_{k}^3-H_{k-1}^3$ and using (\ref{e:24}) we
obtain
\begin{align*}
&\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^3=\sum_{k=1}^{n}(-1)^k\binom{n}{k}\sum_{j=1}^{k}\left(H_j^3-H_{j-1}^3\right)  \notag \\
&=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}\left(H_{k+1}^3-H_k^3\right)
\notag \\
&=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}\left(\frac{3H_{k+1}^{2}}{k+1}-\frac{3H_{k+1}}{(k+1)^2}+\frac{1}{(k+1)^3}\right).
\end{align*}
Letting $k+1=k^\prime$ and then dropping the prime, we get by (\ref{e:3})
\begin{align}  \label{e:25}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^3&=\frac{3}{n}\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}H_{k}^2\notag\\
&-\frac{3}{n}\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}\frac{H_{k}
}{k}+\frac{1}{n}\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}\frac{1}{k^2}.
\end{align}
By Lemma \ref{LemmaA} and the first identity in (\ref{e:12}) we have
\begin{equation}  \label{e:26}
\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}\frac{H_{k}}{k}=\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}(-1)^j\binom{k}{j}H_j=-H_n^{(2)}.
\end{equation}
We also have from \cite[p.~5]{B1} or  \cite[Eq.\ (9.6)]{7} that
\begin{equation}  \label{e:27}
\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}\frac{1}{k^2}=-\frac{H_n^2+H_n^{(2)}}{2}.
\end{equation}
Replacing (\ref{e:26}) and (\ref{e:27}) in (\ref{e:25}) and using (\ref{e:22}), we obtain the desired result.
\end{proof}

\begin{remark}
Using Theorem \ref{TheoremA} it is possible to evaluate all the sums
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}\left(H_k\right)^m,
\end{equation*} for $m\in\mathbb{N}$,
but it requires lengthy calculations when $m$ is large.
\end{remark}

Identity \ref{IDEN1} and Identity \ref{IDEN2} enable us to establish new identities involving the product of the
harmonic numbers and Fibonacci numbers, and harmonic numbers and Lucas numbers.

\begin{identity}
\bigskip\label{IDEN5}
Let $n\in\mathbb{N}$. Then
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_kF_k=H_nF_{2n}-\sum_{k=1}^{n}\frac{F_{2n-2k}}{k},
\end{equation*}
\begin{equation*}
\mbox{and} \quad  \sum_{k=1}^{n}\binom{n}{k}H_kL_k=H_nL_{2n}-\sum_{k=1}^{n}\frac{L_{2n-2k}}{k}.
\end{equation*}
\end{identity}
\begin{proof}
Let $P_{n}(x)$ denote the left-hand side of (\ref{e:11}). Evaluate $
P_{n}(x)$ at $x=\alpha $ and $x=\beta $, where $\alpha =\frac{1+\sqrt{5}}{2}$
and $\beta =\frac{1-\sqrt{5}}{2}$. This gives
\begin{equation*}
P_{n}(\alpha )=\alpha ^{2n}H_{n}-\sum_{k=1}^{n}\frac{\alpha ^{2n-2k}}{k}\quad \mbox{and} \quad  P_{n}(\beta )=\beta ^{2n}H_{n}-\sum_{k=1}^{n}\frac{\beta ^{2n-2k}}{k},
\end{equation*}
where we have used $\alpha ^{2}=\alpha +1$ and $\beta ^{2}=\beta +1$. Now
calculate $P_{n}(\alpha )\mp P_{n}(\beta )$, and then use the Binet formulas
for $F_{n}$ and $L_{n}$.
\end{proof}
\begin{remark} Identity \ref{IDEN5} can  be easily  derived from \cite[Eq.~10]{B2} by letting $c_k=(-1)^{k-1}F_k$; see also \cite[Thm.\ 6.1]{7}. Furthermore, it  can be compared with equations (9) and (10) in \cite{11}. If in the first part of Identity \ref{IDEN5} we put
\[
\sum\limits_{k=1}^{n}\frac{F_{2n-2k}}{k}=\sum\limits_{k=1}^{n}F_{2k-1}H_{n-k},
\]
then we obtain the result in \cite{11}. Similarly, if in the second part of
Identity \ref{IDEN5} we put
\[
\sum\limits_{k=1}^{n}\frac{L_{2n-2k}}{k}=2H_{n}+\sum\limits_{k=1}^{n}L_{2k-1}H_{n-k},
\]
then we obtain the result in \cite{11}.
\end{remark}

\begin{identity}
\bigskip\label{IDEN6}
For $n\in\mathbb{N}$ we have
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_kF_{3k}=(-1)^n2^n\bigg(H_nF_n-\sum_{k=1}^{n}\frac{(-1)^kF_{n-k}}{k2^k}\bigg),
\end{equation*}
and
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_kL_{3k}=(-1)^n2^n\bigg(H_nL_n-\sum_{k=1}^{n}\frac{(-1)^kL_{n-k}}{k2^k}\bigg).
\end{equation*}
\end{identity}


\begin{proof}
Letting $x=-\alpha ^{3}$ and $x=-\beta ^{3}$ in (\ref{e:11}), and using $1-\alpha ^{3}=-2\alpha $ and $1-\beta ^{3}=-\beta $, respectively, we get
\begin{equation*}
\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}H_{k}\alpha ^{3}=(-1)^{n}2^{n}\alpha
^{n}H_{n}-2^{n}\sum_{k=1}^{n}\frac{(-1)^{n-k}\alpha ^{n-k}}{k2^{k}},
\end{equation*}
and
\begin{equation*}
\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}H_{k}\beta ^{3}=(-1)^{n}2^{n}\beta
^{n}H_{n}-2^{n}\sum_{k=1}^{n}\frac{(-1)^{n-k}\beta ^{n-k}}{k2^{k}}.
\end{equation*}
The combination of these two identities according to the Binet formula
yields the desired results.
\end{proof}

\begin{identity}
\bigskip\label{IDEN7}
For $n\in\mathbb{N}$ we have
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_kF_{3k}=2^n\bigg(H_nF_{2n}-\sum_{k=1}^{n}\frac{F_{2n-2k}}{k2^k}\bigg),
\end{equation*}
\begin{equation*}
\mbox{and} \quad  \sum_{k=1}^{n}\binom{n}{k}H_kL_{3k}=2^n\bigg(H_nL_{2n}-\sum_{k=1}^{n}\frac{L_{2n-2k}}{k2^k}\bigg).
\end{equation*}
\end{identity}


\begin{proof}
First we note that $1+\alpha ^{3}=2\alpha ^{2}$ and $1+\beta ^{3}=2\beta
^{2} $. Computing (\ref{e:11}) at $x=\alpha ^{3}$ and $x=\beta ^{3}$ yields
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_{k}\alpha ^{3k}=2^{n}\alpha
^{2n}H_{n}-\sum_{k=1}^{n}\frac{2^{n-k}\alpha ^{2n-2k}}{k}
\end{equation*}
\begin{equation*}
\mbox{and} \quad  \sum_{k=1}^{n}\binom{n}{k}H_{k}\beta ^{3k}=2^{n}\beta
^{2n}H_{n}-\sum_{k=1}^{n}\frac{2^{n-k}\beta ^{2n-2k}}{k}.
\end{equation*}
Now the proof follows from combining these two sums according to the Binet
formula for $F_{n}$ and $L_{n}$.
\end{proof}

By the same method used in the proof of Identity \ref{IDEN6} and Identity \ref{IDEN7} we can prove the
next two results by the help of Identity \ref{IDEN2}.

\begin{identity}
\bigskip\label{IDEN8}
For $n\in\mathbb{N}$ we have
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}H_k^2F_k=F_{2n}H_n^2-\sum_{k=1}^{n}\frac{(H_n-2H_k+H_{n-k})F_{2n-2k}}{k}-2\sum_{k=1}^{n}\frac{F_{2n-2k}}{k^2}
\end{align*}
and
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}H_k^2L_k=L_{2n}H_n^2-\sum_{k=1}^{n}\frac{(H_n-2H_k+H_{n-k})L_{2n-2k}}{k}-2\sum_{k=1}^{n}\frac{L_{2n-2k}}{k^2},
\end{align*}
where we have used here $\sum_{k=1}^{n}\frac{H_k}{k}=\frac{H_n^2+H_n^{(2)}}{2}$; see \cite{A} and \cite{1}.
\end{identity}

\begin{identity}
\bigskip\label{IDEN9}
Putting $x=\alpha^3$ in (\ref{e:13}), and using $1+\alpha^3=2\alpha^2$, $1-\alpha^3=-2\alpha$ and $1+\beta^3=2\beta^2$, $1-\beta^3=-2\beta$, the
evaluation of (\ref{e:13}) yields for $n\in\mathbb{N}$
\begin{align*}
&\sum_{k=1}^{n}\binom{n}{k}H_k^2F_{3k}\\
&=2^nF_{2n}H_n^2-2^n\sum_{k=1}^{n}\frac{(H_n-2H_k+H_{n-k})F_{2n-2k}}{k2^k}-2^{n+1}\sum_{k=1}^{n}\frac{F_{2n-2k}}{k^22^k}
\end{align*}
and
\begin{align*}
&\sum_{k=1}^{n}\binom{n}{k}H_k^2L_{3k}\\
&=2^nL_{2n}H_n^2-2^n\sum_{k=1}^{n}\frac{(H_n-2H_k+H_{n-k})L_{2n-2k}}{k2^k}-2^{n+1}\sum_{k=1}^{n}\frac{L_{2n-2k}}{k^22^k}.
\end{align*}
\end{identity}

The proof of the following identity can be done similar to the proof of
Identity \ref{IDEN6} by letting $x=-\alpha^3$ and $x=-\beta^3$ in Identity \ref{IDEN2}, respectively.

\begin{identity}
\bigskip\label{IDEN10}  For $n\in\mathbb{N}$ we have
\begin{align*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^2F_{3k}&=(-1)^n2^n\bigg(F_nH_n^2-2\sum_{k=1}^{n}\frac{(-1)^kF_{n-k}}{k^22^k} \\
&-\sum_{k=1}^{n}\frac{(-1)^k(H_n-2H_k+H_{n-k})F_{n-k}}{k2^k}\bigg),
\end{align*}
and
\begin{align*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^2L_{3k}&=(-1)^n2^n\bigg(L_nH_n^2-2\sum_{k=1}^{n}\frac{(-1)^kL_{n-k}}{k^22^k} \\
&-\sum_{k=1}^{n}\frac{(-1)^k(H_n-2H_k+H_{n-k})L_{n-k}}{k2^k}\bigg).
\end{align*}
\end{identity}

\begin{identity}
\bigskip\label{IDEN11}
For $n\in\mathbb{N}$ we have
\begin{equation}  \label{e:28}
\sum_{k=1}^{n}\binom{n}{k}H_{k}^{-}x^k=(1+x)^n\sum_{k=1}^{n}\frac{1-(1-x)^k}{k(1+x)^k}.
\end{equation}
\end{identity}
\begin{proof}
The proof follows from Theorem \ref{TheoremB} by setting $a_k=\frac{(-1)^{k+1}}{k}$.
\end{proof}
\begin{remark} Identity \ref{IDEN11} can be easily obtained from \cite[Cor.\ 1]{B2} by setting $a_k=(-1)^kH_k^{-}$; see also \cite[Thm.\ 1.6]{7}. Letting  $x=-1$ and $x=1$ in (\ref{e:28}), respectively, we get
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_{k}^{-}=\frac{1-2^n}{n}\quad \mbox{and}
\quad \sum_{k=1}^{n}\binom{n}{k}H_{k}^{-}=2^n\sum_{k=1}^{n}\frac{1}{k2^k}.
\end{equation*}
The first identity  here can be found in  \cite[Eq.\ (9.20)] {7}, and \cite[Eq.\ (14)]{B1}. The second identity recovers (\ref{e:2}).
\end{remark}
The next identity is not new and can be found in entry (9.43) in \cite {7}.
\begin{identity}
\bigskip\label{IDEN12}
For integers $n\geq 1$ we have
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_k^{-}F_{3k}=2^n\sum_{k=1}^{n}\frac{F_{2n-2k}}{2^kk}-2^n\sum_{k=1}^{n}\frac{(-1)^kF_{2n-k}}{k}
\end{equation*}
\begin{equation*}
\mbox{and}\quad \sum_{k=1}^{n}\binom{n}{k}H_k^{-}L_{3k}=2^n\sum_{k=1}^{n}\frac{L_{2n-2k}}{2^kk}-2^n\sum_{k=1}^{n}\frac{(-1)^kL_{2n-k}}{k}.
\end{equation*}
\end{identity}


\begin{proof}
Clearly, we have $1+\alpha^3=2\alpha^2$, $1-\alpha^3=-2\alpha$ and $1+\beta^3=2\beta^2$, $1-\beta^3=-2\beta$. Thus, evaluating (\ref{e:28}) at $x=\alpha^3$ yields
\begin{equation*}
Q_n(\alpha):=\sum_{k=1}^{n}\binom{n}{k}H_k^{-}\alpha^{3k}=2^n\sum_{k=1}^{n}\frac{\alpha^{2n-2k}}{k2^k}-\sum_{k=1}^{n}\frac{(-1)^k\alpha^{2n-k}}{k},
\end{equation*}
and similarly
\begin{equation*}
Q_n(\beta):=\sum_{k=1}^{n}\binom{n}{k}H_k^{-}\beta^{3k}=2^n\sum_{k=1}^{n}\frac{\beta^{2n-2k}}{k2^k}-2^n\sum_{k=1}^{n}\frac{(-1)^k\beta^{2n-k}}{k}.
\end{equation*}
The evaluation of $Q_n(\alpha)\mp Q_n(\beta)$ leads to the desired results.
\end{proof}

The proof of the following result is similar to that of Identity \ref{IDEN6}.

\begin{identity}
\bigskip\label{IDEN13}
For $n\in\mathbb{N}$ we have
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^{-}F_{3k}=(-1)^n2^n\bigg(\sum_{k=1}^{n}\frac{(-1)^kF_{n-k}}{2^kk}-\sum_{k=1}^{n}\frac{(-1)^kF_{n+k}}{k}\bigg),
\end{equation*}
and
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^{-}L_{3k}=(-1)^n2^n\bigg(\sum_{k=1}^{n}\frac{(-1)^kL_{n-k}}{2^kk}-\sum_{k=1}^{n}\frac{(-1)^kL_{n+k}}{k}\bigg).
\end{equation*}
\end{identity}
The proof of the next two identities can be done by proceeding as in the
proof of Identity \ref{IDEN2} by setting $a_k=\frac{1}{k^2}$ and $a_k=\left(H_k^{-}\right)^2-\left(H_{k-1}^{-}\right)^2$ in Theorem \ref{TheoremB}. We omit the proofs.
\begin{identity}
\bigskip\label{IDEN14}
Letting $a_k=\frac{1}{k^2}$ in (\ref{e:7}), we get after some simple
computations
\begin{align}  \label{e:29}
\sum_{k=1}^{n}\binom{n}{k}H_k^{(2)}x^k&=(1+x)^n\bigg(H_n^{(2)}-\sum_{k=1}^{n}\frac{H_n -H_{n-k}}{k(1+x)^{k}}\bigg).
\end{align}
Considering the two values of $x=1$ and $x=-1$, we have
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_k^{(2)}=2^n\left(H_n^{(2)}-\sum_{k=1}^{n}\frac{H_n -H_{n-k}}{k2^k}\right)
\end{equation*}
\begin{equation*}
\mbox{and\quad} \sum_{k=1}^{n}(-1)^k\binom{n}{k}H_k^{(2)}=-\frac{H_n}{n}.
\end{equation*}
\end{identity}

\begin{identity}
\bigskip\label{IDEN15}
Letting $a_k=\frac{(-1)^{k+1}}{k}$ in Theorem \ref{TheoremB},
we get for  $n\in\mathbb{N}$ and $x\in\mathbb{C}$
\begin{align}  \label{e:30}
&\sum_{k=1}^{n}\binom{n}{k}\left(H_k^{-}\right)^2x^k=(1+x)^n\bigg(2\sum_{k=1}^{n}\frac{H_k+H_{n-k}-H_n}{k}\bigg(\frac{1-x}{1+x}\bigg)^k  \notag
\\
&+H_n^{(2)}+\sum_{k=1}^{n}\frac{H_n-H_{n-k}}{k(1+x)^{k}}-2\sum_{k=1}^{n}\frac{1}{k}\bigg(\frac{1-x}{1+x}\bigg)^k\sum_{j=1}^{k}\frac{(1-x)^{-j}}{j}
\bigg).
\end{align}
Considering the two values of $x=1$ and $x=-1$, we get
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}\left(H_k^{-}\right)^2=2^n\bigg(H_n^{(2)}+\sum_{k=1}^{n}\frac{H_n-H_{n-k}}{2^kk}-2\sum_{k=1}^{n}\frac{1}{2^kk^2}\bigg),
\end{align*}
\begin{align*}
\mbox{and} \quad \sum_{k=1}^{n}(-1)^k\binom{n}{k}\left(H_k^{-}\right)^2=\frac{H_n}{n}-\frac{2^{n+1}}{n}\sum_{k=1}^{n}\frac{1}{k2^k}.
\end{align*}
\end{identity}

\begin{identity}
\bigskip\label{IDEN16}
Letting $x=\alpha $ and $x=\beta $ in (\ref{e:29}) and combining the resulting expressions according
to the Binet formula one gets
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_{k}^{(2)}F_{k}=F_{2n}H_{n}^{(2)}-\sum_{k=1}^{n}\frac{(H_{n}-H_{n-k})F_{2n-2k}}{k},
\end{equation*}
\begin{equation*}
\mbox{and}\quad \sum_{k=1}^{n}\binom{n}{k}H_{k}^{(2)}L_{k}=L_{2n}H_{n}^{(2)}-\sum_{k=1}^{n}\frac{(H_{n}-H_{n-k})L_{2n-2k}}{k}.
\end{equation*}
\end{identity}

\begin{identity}
\bigskip\label{IDEN17}
Letting $x=\alpha ^{3}$ and $x=\beta ^{3}$ in (\ref{e:29}) and combining the
results according to the Binet formula one gets
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}H_{k}^{(2)}F_{3k}=2^{n}H_{n}^{(2)}F_{2n}-2^{n}\sum_{k=1}^{n}\frac{(H_{n}-H_{n-k})F_{2n-2k}}{k2^{k}}
\end{equation*}
\begin{equation*}
\mbox{and}\quad \sum_{k=1}^{n}\binom{n}{k}H_{k}^{(2)}L_{3k}=2^{n}H_{n}^{(2)}L_{2n}-2^{n}\sum_{k=1}^{n}\frac{(H_{n}-H_{n-k})L_{2n-2k}}{k2^{k}}.
\end{equation*}
\end{identity}

\begin{identity}
\bigskip\label{IDEN18}
Setting $x=-\alpha ^{3}$ and $x=-\beta ^{3}$ in (\ref{e:29}), respectively,
and then combining the resulting identities according to the Binet formula,
we get
\begin{align*}
\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}H_{k}^{(2)}F_{3k}=(-1)^{n}2^{n}\bigg(H_{n}^{(2)}F_{n}-\sum_{k=1}^{n}\frac{(-1)^{k}(H_{n}-H_{n-k})F_{n-k}}{k}\bigg),
\end{align*}
and
\begin{align*}
\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}H_{k}^{(2)}L_{3k} =(-1)^{n}2^{n}\bigg(H_{n}^{(2)}L_{n}-\sum_{k=1}^{n}\frac{(-1)^{k}(H_{n}-H_{n-k})L_{n-k}}{k}\bigg).
\end{align*}
\end{identity}

\begin{identity}
\bigskip\label{IDEN19}
Setting $x=\alpha ^{3}$ and $x=\beta ^{3}$ in (\ref{e:30}), respectively,
and then combining the resulting identities according to the Binet formula,
we get
\begin{align*}
& \sum_{k=1}^{n}\binom{n}{k}\left( H_{k}^{-}\right)
^{2}F_{3k}=2^{n}F_{2n}H_{n}^{(2)}-2^{n+1}\sum_{k=1}^{n}\frac{(-1)^{k}}{k}\sum_{j=1}^{k}\frac{(-1)^{j}F_{2n+k-j}}{j} \\
& +2^{n+1}\sum_{k=1}^{n}\frac{(-1)^{k}(H_{k}+H_{n-k}-H_{n})F_{2n-k}}{k2^{k}}+2^{n}\sum_{k=1}^{n}\frac{(H_{n}-H_{k})F_{2n-k}}{k2^{k}}.
\end{align*}
and
\begin{align*}
& \sum_{k=1}^{n}\binom{n}{k}\left( H_{k}^{-}\right)
^{2}L_{3k}=2^{n}L_{2n}H_{n}^{(2)}-2^{n+1}\sum_{k=1}^{n}\frac{(-1)^{k}}{k}\sum_{j=1}^{k}\frac{(-1)^{j}L_{2n+k-j}}{j} \\
& +2^{n+1}\sum_{k=1}^{n}\frac{(-1)^{k}(H_{k}+H_{n-k}-H_{n})L_{2n-k}}{k2^{k}}+2^{n}\sum_{k=1}^{n}\frac{(H_{n}-H_{k})L_{2n-k}}{k2^{k}}.
\end{align*}
\end{identity}


The following identity has been posed as a problem by Ohtsuka in \cite{Oht}.

\begin{identity}
\bigskip\label{IDEN20}
For all positive integers $n$ the following identity holds:
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}\sum_{1\leq i\leq j\leq k}\frac{1}{ij}=\sum_{1\leq
i\leq j\leq n}\frac{2^n-2^{n-i}}{ij}.
\end{equation*}
\end{identity}

\begin{proof}
Since
\begin{align*}
\sum_{1\leq i\leq j\leq k}\frac{1}{ij}&=1+\frac{1}{2}\left(1+\frac{1}{2}\right)+\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)+\cdots \\
&+\frac{1}{k}\left(1+\frac{1}{2}+\cdots+\frac{1}{k}\right)=\sum_{k=1}^{n}\frac{H_k}{k}
\end{align*}
and similarly
\begin{equation*}
\sum_{1\leq i\leq j\leq n}\frac{2^n-2^{n-i}}{ij}=\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{2^n-2^{k-j}}{j},
\end{equation*}
it suffices to show
\begin{equation*}
\sum_{k=1}^{n}\binom{n}{k}\sum_{k=1}^n\frac{H_k}{k}=\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{2^n-2^{k-j}}{j}.
\end{equation*}
Substituting $a_k=\frac{H_k}{k}$ in (\ref{e:10}), we get by the help of (\ref{e:3})
\begin{align*}
\sum_{k=1}^{n}\binom{n}{k}\sum_{j=1}^{k}\frac{H_j}{j}&=2^{n-1}+2^{n-1}\sum_{k=1}^{n-1}\frac{1}{2^k}\sum_{j=1}^{k}\binom{k}{j}\frac{H_{j+1}}{j+1}
\notag \\
&=2^{n-1}+2^{n}\sum_{k=2}^{n}\frac{1}{k2^{k}}\sum_{j=1}^{k}\binom{k}{j}H_j.
\end{align*}
Using the second identity in (\ref{e:12}), we find
\begin{align*}
&\sum_{k=1}^{n}\binom{n}{k}\sum_{j=1}^{k}\frac{H_j}{j}=2^{n-1}+2^n\sum_{k=2}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{2^j-1}{j2^{j}}  \notag \\
&=2^{n}\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{2^j-1}{j2^{j}}=\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\frac{2^n-2^{n-j}}{j},
\end{align*}
as claimed.
\end{proof}


As usual let $S(n,k)$ denote the  Stirling numbers of the second kind, that is
\begin{equation*}
S(n,k)=\frac{(-1)^k}{k}\sum_{j=0}^{k}(-1)^j\binom{k}{j}j^n.
\end{equation*}
The following identity provides a new proof, based on Theorem \ref{TheoremA}, of a
well-known formula for the sum of powers of consecutive positive integers;
see, for example, \cite{4}.

\begin{identity}
\bigskip\label{IDEN21}
For $m,n\in\mathbb{N}$ the following identity holds:
\begin{equation*}
\sum_{k=1}^{n}k^m=\sum_{k=1}^{n}\binom{n}{k}(k-1)!S(m+1,k).
\end{equation*}
\end{identity}

\begin{proof}
Putting $a_k=k^m$ in Theorem \ref{TheoremA}, we get
\begin{align*}
&\sum_{k=1}^{n}(-1)^k\binom{n}{k}\sum_{k=1}^{n}k^m=\sum_{k=0}^{n-1}(-1)^{k+1}\binom{n-1}{k}(k+1)^m \\
&=\frac{1}{n}\sum_{k=1}^{n}(-1)^k\binom{n}{k}k^{m+1}=(-1)^n(n-1)!S(m+1,n).
\end{align*}
Employing the well-known binomial inversion formula the proof is completed.
\end{proof}

\section{Acknowledgments}
We wish to thank the anonymous referee for useful comments and bringing our attention to the papers \cite{B1} and \cite{B2}.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 05A19.

\noindent \emph{Keywords: }
combinatorial identity, binomial coefficient, harmonic number,
Fibonacci sequence, Lucas sequence, harmonic number sum.

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\noindent (Concerned with sequences
\seqnum{A000032} and 
\seqnum{A000045}.)

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\vspace*{+.1in}
\noindent
Received November 16 2020; revised versions
received November 28 2020; November 30 2020; December 30 2020; 
January 6 2021; February 2 2021.
Published in {\it Journal of Integer Sequences}, February 2 2021.

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