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\begin{center}
\vskip 1cm{\LARGE\bf Harmonic Sums via Euler's Transform: Complementing the Approach of Boyadzhiev  \\ 
\vskip .11in }
\vskip 1cm
{\large  Robert Frontczak\footnote{Disclaimer: Statements and conclusions made in this article are 
entirely those of the author. They do not necessarily reflect the views of LBBW.} \\
Landesbank Baden-W{\"u}rttemberg (LBBW) \\ 
Am Hauptbahnhof 2 \\ 
70173 Stuttgart \\ 
Germany \\
\href{mailto:robert.frontczak@lbbw.de}{\tt robert.frontczak@lbbw.de}
}
\end{center}

\vskip .2 in

\begin{abstract}
We prove a new expression for binomial sums with harmonic numbers. Our
derivation is based on an alternative argument for the Euler transform
of these sums. The findings complement a result of Boyadzhiev.
To demonstrate the usefulness of our alternative approach, several
examples are discussed.  We rediscover some known identities for
harmonic numbers and present some new ones.  In particular, we derive some new
identities involving harmonic numbers, and Fibonacci and Lucas numbers.
\end{abstract}

\section{Motivation}

Harmonic numbers $(H_n)_{n\geq 0}$ are defined by
\begin{displaymath}
H_0 = 0 \quad \mbox{and for $n\geq 1:$} \qquad H_n = \sum_{k=1}^n \frac{1}{k} = H_{n-1} + \frac{1}{n}.
\end{displaymath}
They have the following integral form
\begin{displaymath}
H_n = \int_0^1 \frac{1-x^n}{1-x} dx.
\end{displaymath}
Harmonic numbers and generalized harmonic numbers have been studied recently by many mathematicians 
and a considerable amount of research results has been produced (see \cite{2} - \cite{13}, to name a few articles). 
In 2009, Boyadzhiev \cite{2} studied binomial sums with harmonic numbers using the Euler transform. 
His main result is the following identity valid for $n\geq 1$
\begin{equation} \label{main_boy}
\sum_{k=1}^n \binom{n}{k} a^k b^{n-k} H_k = (a+b)^n H_n - \Big (b(a+b)^{n-1} + \frac{b^2}{2}(a+b)^{n-2} + \cdots + \frac{b^n}{n}\Big ),
\end{equation}
where $a$ and $b$ are arbitrary complex numbers. His proof is based on the Euler transform for power series and the existence of a power series near zero of the form
\begin{displaymath}
\frac{\ln(1-cz)}{1-dz} = -\sum_{n=1}^\infty \Big ( cd^{n-1} + \frac{1}{2}c^2d^{n-2}+ \cdots + \frac{1}{n}c^n\Big ) z^n.
\end{displaymath}

In this article, we show how Boyadzhiev's arguments may be modified to derive an alternative expression for 
the binomial sums on the left-hand side of \eqref{main_boy}. To demonstrate the usefulness of our alternative approach, 
several examples will be discussed. We will rediscover some known identities and present some new. 
In particular, we will derive some new identities involving harmonic numbers, and Fibonacci and Lucas numbers.

\section{Results}

Let $A(z)$ be the ordinary generating function for the harmonic numbers, i.e.,
\begin{displaymath}
A(z) = \sum_{n=0}^\infty H_n z^n = - \frac{\ln(1-z)}{1-z}.
\end{displaymath}
For $a,b\in\mathbb{C}$, let further $S_n(a,b)$ be defined as
\begin{equation}
S_n(a,b) = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} H_k. 
\end{equation}

Then we have the following theorem.

\begin{theorem} \label{thm1}
For all $n\geq 1$, we have the identity
\begin{equation} \label{main1}
S_n(a,b) = \Big ( (a+b)^n - b^n\Big ) H_n - a \sum_{k=0}^{n-1} (a+b)^k b^{n-1-k} H_{n-1-k}.
\end{equation}
\end{theorem}

\begin{proof}
Let $S(z)$ be the ordinary generating function for the sum $S_n(a,b)$. Then, by Euler's transform
\begin{eqnarray*}
S(z) = \sum_{n=0}^\infty S_n(a,b) z^n & = &  \frac{1}{1-bz} A\Big (\frac{az}{1-bz}\Big ) \\
& = & -\frac{\ln(1-(a+b)z)}{1-(a+b)z} + \frac{\ln(1-bz)}{1-(a+b)z}.
\end{eqnarray*}
Now, instead of searching for a power series for the second summand, we observe that
\begin{displaymath}
\frac{\ln(1-bz)}{1-(a+b)z} = \frac{az}{1-(a+b)z} \frac{\ln(1-bz)}{1-bz} + \frac{\ln(1-bz)}{1-bz}.
\end{displaymath}
Hence,
\begin{eqnarray*}
S(z) & = & -\frac{\ln(1-(a+b)z)}{1-(a+b)z} - \Big (- \frac{\ln(1-bz)}{1-bz}\Big ) - \frac{az}{1-(a+b)z}  \Big (- \frac{\ln(1-bz)}{1-bz}\Big ) \\
& = & \sum_{n=0}^\infty \Big ( (a+b)^n - b^n\Big ) H_n z^n - a \Big ( \sum_{n=0}^\infty (a+b)^n z^{n+1}\Big )\Big (\sum_{n=0}^\infty b^n H_n z^n\Big ).
\end{eqnarray*}
Using Cauchy's product rule for power series and comparing the coefficients of $z^n$ gives the result.
\end{proof}

For $(a;b)=(-1;1)$ we get as a special case
\begin{equation}
S_n(-1,1) = \sum_{k=0}^n \binom{n}{k} (-1)^{k} H_k = - H_n + H_{n-1} = -\frac{1}{n},
\end{equation}
which is an old result and has reappeared as Identity 20 in Spivey's paper \cite{12}. 
In addition, using the integral form for $H_n$ it is not difficult to show that
\begin{equation}
\sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{k} = H_n.
\end{equation}
This shows that the sequences $(H_n)_{n\geq 1}$ and $(1/n)_{n\geq 1}$ are connected by the binomial transform. 
More information about the binomial transform can be found in the book \cite{1}.

\begin{corollary}
For $n\geq 1$, the harmonic numbers allow the representation  
\begin{equation}
H_n = \sum_{k=1}^n \Big ( \frac{1}{k} 2^{n-k} - 2^{k-1} H_{n-k}\Big ).
\end{equation}
\end{corollary}
\begin{proof}
Setting $(a;b)=(a;a)$ in \eqref{main1} yields
\begin{displaymath}
\sum_{k=0}^n \binom{n}{k} H_k = (2^n - 1)H_n - \sum_{k=0}^{n-1} 2^k H_{n-1-k}.
\end{displaymath}
Now, compare with the well-known identity (\cite[Equation (20)]{2}, or \cite[Identity 14]{12})
\begin{displaymath}
\sum_{k=0}^n \binom{n}{k} H_k = 2^n\Big ( H_n - \sum_{k=1}^{n} \frac{1}{k2^k}\Big ).
\end{displaymath}
\end{proof}

\begin{corollary}
For $n\geq 1$, it is true that 
\begin{equation}
\sum_{k=0}^n \binom{n}{k} 2^k H_k = (3^n - 1)H_n - 2 \sum_{k=0}^{n-1} 3^k H_{n-1-k}.
\end{equation}
\end{corollary}
\begin{proof}
Set $(a;b)=(2;1)$ in \eqref{main1}.
\end{proof}

\begin{corollary}
For $n\geq 1$, we have the identity
\begin{equation}
\sum_{k=0}^n \binom{n}{k}(-1)^{n-k} 2^{k-1} H_k = \left.
  \begin{cases}
    \sum_{k=0}^{n-1} (-1)^k H_{n-1-k}, & \mbox{if $n$ is even}; \\
    H_n - \sum_{k=0}^{n-1} (-1)^k H_{n-1-k}, & \mbox{if $n$ is odd.}
  \end{cases}
  \right.
\end{equation}
\end{corollary}
\begin{proof}
Setting $(a;b)=(2;-1)$ in \eqref{main1} yields 
\begin{displaymath}
\sum_{k=0}^n \binom{n}{k}(-1)^{n-k} 2^k H_k = (1-(-1)^n) H_n - 2 \sum_{k=0}^{n-1} (-1)^{n-1-k} H_{n-1-k},
\end{displaymath}
from which the result is deduced easily.
\end{proof}

Theorem \ref{thm1} also allows to establish some identities involving harmonic numbers and Fibonacci (Lucas) numbers.
Recall, that Fibonacci numbers $(F_n)_{n\geq 0}$ and Lucas numbers $(L_n)_{n\geq}$ are defined by 
$F_0=0, F_1=1, L_0=2, L_1=1$, and for $n\geq 2$ we have the recurrences $F_{n} = F_{n-1}+F_{n-2}$ and $L_{n} = L_{n-1}+L_{n-2}$.

\begin{corollary}
Let $F_n$ and $L_n$ be the Fibonacci and Lucas numbers, respectively. Then, we have the relations
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} F_{k} H_{k} = F_{2n} H_n - \sum_{k=0}^{n-1} F_{2k+1} H_{n-1-k},
\end{equation}
and
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} L_{k} H_{k} = (L_{2n}-2) H_n - \sum_{k=0}^{n-1} L_{2k+1} H_{n-1-k}.
\end{equation}
\end{corollary} 
\begin{proof}
Evaluate \eqref{main1} at $(a;b)=(\alpha;1)$ and $(a;b)=(\beta;1)$, respectively, 
where $\alpha=(1+\sqrt{5})/2$ and $\beta=-1/\alpha$. This gives
\begin{displaymath}
S_n(\alpha,1) = (\alpha^{2n}-1)H_n - \sum_{k=0}^{n-1} \alpha^{2k+1}H_{n-1-k}
\end{displaymath}
and
\begin{displaymath}
S_n(\beta,1) = (\beta^{2n}-1)H_n - \sum_{k=0}^{n-1} \beta^{2k+1}H_{n-1-k},
\end{displaymath}
where we have used the relations $\alpha^2=\alpha+1$ and $\beta^2=\beta+1$. Now, calculate $S_n(\alpha,1)\pm S_n(\beta,1)$ and
use the Binet forms for $F_n$ and $L_n$, respectively.
\end{proof}

\begin{corollary}
Let $F_n$ and $L_n$ be the Fibonacci and Lucas numbers, respectively. Then, the following identities hold
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} F_{2k} H_{k} = F_{n} H_n - \sum_{k=0}^{n-1} (-1)^{n-1-k} F_{k+2} H_{n-1-k},
\end{equation}
and
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} L_{2k} H_{k} = (L_{n}-2(-1)^n) H_n - \sum_{k=0}^{n-1} (-1)^{n-1-k} L_{k+2} H_{n-1-k}.
\end{equation}
\end{corollary} 
\begin{proof}
Evaluate \eqref{main1} at $(a;b)=(\alpha^2;-1)$ and $(a;b)=(\beta^2;-1)$, respectively. 
Combine the results as in the previous proof. 
\end{proof}

\begin{corollary}
Let $F_n$ and $L_n$ be the Fibonacci and Lucas numbers, respectively. Then
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 2^k (-1)^{n-k} F_{n-k} H_{k} = -(F_{2n}+(-1)^n F_n) H_n + \sum_{k=0}^{n-1} F_{3k+1-n} H_{n-1-k},
\end{equation}
and
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} 2^k (-1)^{n-k} L_{n-k} H_{k} = (L_{2n}-(-1)^n L_n) H_n - \sum_{k=0}^{n-1} L_{3k+1-n} H_{n-1-k}.
\end{equation}
\end{corollary} 
\begin{proof}
Evaluate \eqref{main1} at $(a;b)=(2;-\alpha)$ and $(a;b)=(2;-\beta)$, respectively. 
Simplify using $2-\alpha=\alpha^{-2}$ and $2-\beta=\alpha^2$. Finally, calculate $S_n(2,-\alpha)\pm S_n(2,-\beta)$ 
and keep in mind that $F_{-n}=(-1)^{n+1}F_n$ and $L_{-n}=(-1)^{n}L_n$, respectively.
\end{proof}

\section{Harmonic sums with integer powers}

Boyadzhiev \cite[Proposition 10]{2} obtained a representation for the combinatorial sum
\begin{equation}
S_n(a,1,m) = S_n(a,m) = \sum_{k=0}^{n} \binom{n}{k} k^m a^k H_{k}, \quad m\geq 1,
\end{equation}
in terms of Stirling numbers of the second kind $S(m,k)$. 
The theorem below contains an alternative expression for the sum.

\begin{theorem} \label{thm2}
For all $n\geq 1$ we have
\begin{equation} \label{main2}
S_n(a,m) = \Big ( \sum_{k=0}^{n} \binom{n}{k} k^m a^k \Big ) H_n - \sum_{j=0}^{n-1} \sum_{k=j}^{n-1} \binom{k}{j} (j+1)^m a^{j+1} H_{n-1-k}.
\end{equation}
\end{theorem}
\begin{proof}
Apply the differential operator $(a\frac{d}{da})^m$ to both sides of \eqref{main1} with $S_n(a,b)=S_n(a,1)$. 
Note that
\begin{displaymath}
\Big (a\frac{d}{da}\Big )^m (a+1)^n = \sum_{k=0}^{n} \binom{n}{k} k^m a^k.
\end{displaymath}
\end{proof}

\begin{remark}
The Stirling numbers of the second kind $S(n,k)$ are defined by
\begin{displaymath}
x^n = \sum_{k=0}^n S(n,k) (x)_k,
\end{displaymath}
with $x_0=1$ and $(x)_n=x(x-1)\cdots (x-n+1), n\geq 1,$ being the falling factorial. From the equation
\begin{displaymath}
(x+1)_n = (x)_n + n (x)_{n-1},
\end{displaymath} 
it becomes clear, that we can restate the above result using Stirling numbers.
\end{remark}

\begin{corollary}
For $n\geq 1$, we have the following expression 
\begin{equation}
\sum_{k=0}^n \binom{n}{k} k H_k = n 2^{n-1} H_n - \sum_{k=0}^{n-1} (k+2) 2^{k-1} H_{n-1-k}.
\end{equation}
\end{corollary}
\begin{proof}
Use $(a;m)=(1;1)$ in \eqref{main2} as well as the obvious identities
\begin{displaymath}
\sum_{k=0}^n \binom{n}{k} = 2^n \qquad \mbox{and} \qquad \sum_{k=0}^n \binom{n}{k} k = n 2^{n-1}.
\end{displaymath}
\end{proof}

\begin{corollary}
For $n\geq 1$, we have the identity
\begin{equation}
\sum_{k=0}^n \binom{n}{k}(-1)^{k} k H_k = \left.
  \begin{cases}
    -1, & \mbox{if $n=1$}; \\
    \frac{1}{n-1}, & \mbox{if $n\geq 2$}.
  \end{cases}
  \right.
\end{equation}
\end{corollary}
\begin{proof}
Evaluate \eqref{main2} at $(a;m)=(-1;1)$ and simplify.
\end{proof}

\begin{corollary}
For $n\geq 1$, we have the identity
\begin{equation}
\sum_{k=j}^{n-1} \binom{k}{j} H_{n-1-k} = \binom{n}{j+1} ( H_{n} - H_{j+1}).
\end{equation}
\end{corollary}
\begin{proof}
We have
\begin{eqnarray*}
S_n(a,m) & = & \Big ( \sum_{k=0}^{n} \binom{n}{k} k^m a^k \Big ) H_n - \sum_{j=0}^{n} \binom{n}{j} j^m a^{j} ( H_{n} - H_{j}) \\
& = & \Big ( \sum_{k=0}^{n} \binom{n}{k} k^m a^k \Big ) H_n - \sum_{j=1}^{n} \binom{n}{j} j^m a^{j} ( H_{n} - H_{j}) \\
& = & \Big ( \sum_{k=0}^{n} \binom{n}{k} k^m a^k \Big ) H_n - \sum_{j=0}^{n-1} \binom{n}{j+1} (j+1)^m a^{j+1} ( H_{n} - H_{j+1}).
\end{eqnarray*}
Comparing with \eqref{main2} gives the result.
\end{proof}

\begin{corollary}
Let $F_n$ and $L_n$ be the Fibonacci and Lucas numbers, respectively. Then
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} k F_{k} H_{k} = n F_{2n-1} H_n - \sum_{k=0}^{n-1} (k F_{2k}+F_{2k+1}) H_{n-1-k},
\end{equation}
and
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} k L_{k} H_{k} = n L_{2n-1} H_n - \sum_{k=0}^{n-1} (k L_{2k}+L_{2k+1}) H_{n-1-k}.
\end{equation}
\end{corollary} 
\begin{proof}
Evaluate \eqref{main2} at $(a;m)=(\alpha;1)$ and $(a;m)=(\beta;1)$, respectively, using
\begin{displaymath}
\sum_{k=1}^n \binom{n}{k} k x^k = n x (1+x)^{n-1},
\end{displaymath}
in combination with 
\begin{displaymath}
\sum_{k=0}^n \binom{n}{k} \alpha^k = \alpha^{2n}, \quad \mbox{and} \quad \sum_{k=0}^n \binom{n}{k} \beta^k = \beta^{2n}.
\end{displaymath}
Combine the sums as in the previous proofs. 
\end{proof}

Using similar elementary arguments we can prove the following identities for $m=2$:

\begin{corollary}
For $n\geq 1$, it is true that 
\begin{equation}
\sum_{k=0}^n \binom{n}{k} k^2 H_k = n (n+1) 2^{n-2} H_n - \sum_{k=0}^{n-1} (k+1)(k+4) 2^{k-2} H_{n-1-k}.
\end{equation}
\end{corollary}

\begin{corollary}
For $n\geq 1$, we have the identity
\begin{equation}
\sum_{k=0}^n \binom{n}{k} (-1)^{k} k^2 H_k = \left.
  \begin{cases}
    -1, & \mbox{if $n=1$}; \\
		4, & \mbox{if $n=2$}; \\
    -\frac{n}{(n-2)(n-1)}, & \mbox{if $n\geq 3$.}
  \end{cases}
  \right.
\end{equation}
\end{corollary}

\begin{corollary}
For $n\geq 1$, the following relations hold:
\begin{eqnarray}
\sum_{k=0}^n \binom{n}{k} k^2 F_k H_k & = & (n F_{2n-3} + n^2 F_{2n-2})H_n \nonumber \\
&& - \sum_{k=0}^{n-1} \Big ( k F_{2k-2} + k^2 F_{2k-1} + 2k F_{2k} + F_{2k+1}\Big ) H_{n-1-k},
\end{eqnarray}
and
\begin{eqnarray}
\sum_{k=0}^n \binom{n}{k} k^2 L_k H_k & = & (n L_{2n-3} + n^2 L_{2n-2})H_n \nonumber \\
&& - \sum_{k=0}^{n-1} \Big ( k L_{2k-2} + k^2 L_{2k-1} + 2k L_{2k} + L_{2k+1}\Big ) H_{n-1-k}.
\end{eqnarray}
\end{corollary}

\section{Concluding comments}

Using the main results of this paper, it is possible to derive more identities involving harmonic numbers and Fibonacci (Lucas) numbers.
In addition, we mention that from Theorems \ref{thm1} and \ref{thm2} relations between harmonic numbers and other important number sequences, such as Mersenne numbers or Pell numbers, are deducible.

\section{Acknowledgments}

The author is thankful to K. N. Boyadzhiev and T. Goy for their remarks on the first draft of the manuscript.
He is also grateful to the anonymous referee and the editor-in-chief for valuable suggestions which 
improved the quality of the presentation.

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\end{thebibliography}

\bigskip
\hrule
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B37; Secondary 11B39, 11B65, 05A15.

\noindent \emph{Keywords: }
harmonic number, binomial sum, Fibonacci number, generating function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A001008}, and
\seqnum{A002805}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received October 20 2019;
revised version received January 21 2020. 
Published in {\it Journal of Integer Sequences}, February 23 2020.

\bigskip
\hrule
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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