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\begin{center}
\vskip 1cm{\LARGE\bf
Harmonic Number Identities  \\
\vskip .1in Via Euler's Transform} \vskip 1cm \large
Khristo N. Boyadzhiev\\
Department of Mathematics\\
Ohio Northern University\\
Ada, Ohio 45810 \\
USA \\
\href{mailto:k-boyadzhiev@onu.edu}{\tt k-boyadzhiev@onu.edu}
\end{center}

\vskip .2in

\begin{abstract}
We evaluate several binomial transforms by using Euler's transform
for power series. In this way we obtain various binomial identities
involving power sums with harmonic numbers.
\end{abstract}

\section{Introduction and prerequisites}

Given a sequence $\{ a_{k} \} $, its \textit{binomial transform}
$\{ b_{k} \} $ is the sequence defined by

  $$ b_{n} =\sum _{k=0}^{n}\binom{n}{k}  a_{k}, \ \mbox{with inversion}  \ a_{n} =\sum _{k= 0}^{n} \binom{n}{k}(-1)^{n-k}  b_{k}, $$

\noindent or,  in the symmetric version

$$b_{n} =\sum _{k=0}^{n}\binom{n}{k}(-1)^{k+
1}   a_{k} \ \mbox{with inversion} \ a_{n} =\sum
_{k=0}^{n}\binom{n}{k}(-1)^{k+ 1}
b_{k} $$

\noindent (see \cite{gould1,prodinger,spivey}). The
binomial transform is related to the \textit{Euler transform} of
series defined in the following lemma. Euler's transform is used
sometimes for improving the convergence of certain series 
\cite{amore,guillera,prodinger,sondow}.

\begin{lemma} \label{Lemma 1} Given a function analytical on the unit disk

\begin{equation} \label{GrindEQ__1_}
f(t)=\sum _{n= 0}^{\infty }  a_{n}  t^{n} ,
\end{equation}
then the following representation is true

\begin{equation} \label{GrindEQ__2_}
\frac{1}{1-t} \ f \!   \left ( \frac{t}{1-t}   \right )=\sum _{n= 0}^{\infty } t^{n}   \left (\sum
_{k=0}^{n}\binom{n}{k}a_{k}  \right )  .
\end{equation}
\end{lemma}
\noindent (Proof can be found in the Appendix.)

 If we have a convergent series

\begin{equation} \label{GrindEQ__3_}
s=\sum _{n= 0}^{\infty }  a _{n},
\end{equation}
we can define the function

\begin{equation} \label{GrindEQ__4_}
f(t)=\sum _{n=  0}^{\infty }  a_{n}  t^{n},  \quad |t|&lt;1.
\end{equation}
Then, with $t={\frac{1}{2}} $ in \eqref{GrindEQ__2_} we obtain

\begin{equation} \label{GrindEQ__5_}
s=\sum _{n= 0}^{\infty }
\left ( \sum _{k=0}^{n}\binom{n}{k}a_{k}  \right )
\frac{1}{2^{n+1} }.
\end{equation}
This formula is a classical version of Euler's series
transformation. Sometimes the new series converges faster,
sometimes not -- see the examples in \cite{knopp}.

We shall use Euler's transform for the evaluation of several
interesting binomial transformations, thus obtaining binomial
identities of combinatorial and analytical character. Evaluating a
binomial transform is reduced to finding the Taylor
coefficients of the function on the left hand side of
\eqref{GrindEQ__2_}. In Section 2 we obtain several identities
with harmonic numbers. In Section 3 we prove Dilcher's formula via
Euler's transform.

This paper is close in spirit to the classical article \cite{gould1} of
Henry Gould.

  \begin{rema}  \label{Remark 1} The representation \eqref{GrindEQ__2_} can be put in a more flexible equivalent form

\begin{equation} \label{GrindEQ__6_}
\frac{1}{1-\lambda t} \ f\! \left ( \frac{\mu
t}{1-\lambda  t} \right  )=\sum _{n= 0}^{\infty
} t^{n}  \left (  \sum
_{k=0}^{n}\binom{n}{k}\mu ^{k} \lambda ^{n-k} a_{k}
\right ) ,
\end{equation}
where $\lambda ,\mu $ are appropriate parameters.

 To show the equivalence of \eqref{GrindEQ__2_} and \eqref{GrindEQ__6_} we first write

\begin{equation} \label{GrindEQ__7_}
f\! \left (\frac{\mu  t}{\lambda } \right )=\sum _{n=
0}^{\infty }  a_{n}  \left ( \frac{\mu }{\lambda
}   \right )^{n}  t^{n},  \quad
\end{equation}
and then apply \eqref{GrindEQ__2_} to the function
$g(t)=f(\frac{\mu }{\lambda } t)$. This provides

\begin{equation} \label{GrindEQ__8_}
\frac{1}{1-t} \ f \! \left (  \frac{\mu }{\lambda }
\frac{t}{1-t} \right )=\sum _{n= 0}^{\infty }
t^{n}   \left ( \sum _{k=0}^{n}\binom{n}{k}\left(\frac{\mu }{\lambda } \right)^{k} a_{k}  \right ).
\end{equation}
Replacing here $t$ by $\lambda  t$ yields
\eqref{GrindEQ__6_}.

 With $\lambda =1$ and $\mu =-1$ we have

\begin{equation} \label{GrindEQ__9_}
\frac{1}{t-1} \  f\! \left ( \frac{t}{t-1} \right )=\sum _{n=
0}^{\infty } t^{n}   \left ( \sum
_{k=0}^{n}\binom{n}{k}(-1)^{k+1} a_{k}  \right ) ,
\end{equation}
corresponding to the symmetrical binomial transform.
\end{rema}
\begin{lemma} \label{Lemma 2} Given a formal power series
\begin{equation} \label{GrindEQ__10_}
g(t)=\sum _{n=  0}^{\infty }  b_{n}  t^{n} , 
\end{equation}
we have
\begin{equation} \label{GrindEQ__11_}
\frac{g(t)}{1-t} =\sum _{n=  0}^{\infty }
\left ( \sum _{k=0}^{n}b_{k}  \right )  t^{n}.
\end{equation}
\end{lemma}
This is a well-known property. To prove it we just need to
multiply both sides of \eqref{GrindEQ__11_} by $1-t$ and simplify
the right hand side.

\section{Identities with harmonic numbers }

\noindent \begin{proposition} \label{Proposition 1}The following expansion holds in
a neighborhood of zero

\begin{equation} \label{GrindEQ__12_}
\frac{\log  (1-\alpha
t)}{1-\beta   t} =-\sum _{n=1}^{\infty
}\left (\alpha  \beta ^{n-1}  +\frac{1}{2} \alpha ^{2} \beta
^{n-2} +\cdots +\frac{1}{n} \alpha ^{n} \right ) t^{n}
\end{equation}
where $\alpha ,\beta $ are appropriate parameters.
\end{proposition}
\begin{proof} It is sufficient to prove \eqref{GrindEQ__12_} when $\beta =1$ and then rescale the variable $t$, i.e. we only need
\begin{equation} \label{GrindEQ__13_}
\frac{\log  (1-\alpha  t)}{1- t}
=-\sum _{n=1}^{\infty }\left (\alpha   +\frac{1}{2} \alpha
^{2} +\cdots+\frac{1}{n} \alpha ^{n} \right ) t^{n} .
\end{equation}
This follows immediately from  Lemma \ref{Lemma 2}.
\end{proof}

 \begin{corollary} \label{Corollary 1} With $\alpha =1$ in \eqref{GrindEQ__13_} we obtain the generating function of the harmonic numbers

\begin{equation} \label{GrindEQ__14_}
- \frac{\log   (1- t)}{1- t}
=\sum _{n=0}^{\infty }H_{n}   t^{n}, \qquad  H_{n} =1+\frac{1}{2}
+\cdots +\frac{1}{n}  .
\end{equation}
\end{corollary}
The next proposition is one of our main results

\begin{proposition} \label{Proposition 2} For every positive integer $n$ and every
two complex numbers$\lambda ,\mu $,
\begin{equation} \label{GrindEQ__15_}
\sum _{k=1}^{n}\binom{n}{k}  H_{k}  \lambda^{n-k} \mu^{k} =H_{n} (\lambda +\mu )^{n} -\left ( \lambda
 (\lambda +\mu )^{n-1} +\frac{\lambda^{2} }{2} (\lambda +\mu
)^{n-2} +\cdots+\frac{\lambda^{n} }{n}  \right ) .
\end{equation}
\end{proposition}
\begin{proof} We apply \eqref{GrindEQ__6_} to the function
\begin{equation} \label{GrindEQ__16_}
f(t)= - \frac{\log   (1- t)}{1- t} =\sum _{n=0}^{\infty }H_{n}   t^{n}  .
\end{equation}
On the left hand side we obtain

\begin{equation} \label{GrindEQ__17_}
\frac{-1}{1-\lambda t} \frac{\log   (1-\frac{\mu
t}{1-\lambda  t} )}{1-\frac{\mu  t}{1-\lambda  t} } =-\frac{\log  (1-(\lambda +\mu ) t)}{1-(\lambda +\mu
) t} +\frac{\log  (1-\lambda  t)}{1-(\lambda +\mu
) t}  ,
\end{equation}
which equals, according to Corollary \ref{Corollary 1} and Proposition \ref{Proposition 1},
\begin{equation} \label{GrindEQ__18_}
\sum _{n=1}^{\infty }H_{n}  (\lambda +\mu )^{n} t^{n} -\sum
_{n=1}^{\infty } \left (  \lambda  (\lambda +\mu )^{n-1}
+\frac{\lambda ^{2} }{2} (\lambda +\mu )^{n-2} +\cdots+\frac{\lambda
^{n} }{n}  \right ) t^{n} .
\end{equation}
At the same time, by Euler's transform the right hand side is

\begin{equation} \label{GrindEQ__19_}
\sum _{n=1}^{\infty }t^{n}  \left( \sum _{k=1}^{n}\binom{n}{k} H_{n}   \lambda ^{n-k} \mu^{k} \right )  .
\end{equation}
Comparing coefficients in \eqref{GrindEQ__18_} and
\eqref{GrindEQ__19_} we obtain the desired result.
\end{proof}
\begin{corollary} \label{Corollary 2} Setting $\lambda =\mu =1$ in
\eqref{GrindEQ__15_} yields the well-known identity (see, for
instance, \cite{gould,spivey}):

\begin{equation} \label{GrindEQ__20_}
\sum _{k=1}^{n}\binom{n}{k}  H_{k}   =
 2^{n}  \left ( H_{n} -\sum _{k=1}^{n} \frac{1}{k 2^{k}
}   \right ).
\end{equation}
\end{corollary}
\begin{corollary} \label{Corollary 3} Setting $\lambda =1$ in \eqref{GrindEQ__15_}
reduces it to

\begin{equation} \label{GrindEQ__21_}
\sum _{k=1}^{n}\binom{n}{k}  H_{k} \mu
^{k} =H_{n} (1+\mu )^{n} -  \left ( (1+\mu )^{n-1} +\frac{(1+\mu
)^{n-2} }{2} +\cdots+\frac{1+\mu }{n-1} +\frac{1}{n}  \right ).
\end{equation}
\end{corollary}
We shall use this last identity to obtain a representation for the
combinatorial sum

\begin{equation} \label{GrindEQ__22_}
\sum _{k=1}^{n}\binom{n}{k}  H_{k}  k^{m} \mu  ^{k} ,
\end{equation}
by applying the operator $(\mu \frac{d}{d\mu }  )^{m} $ to both
sides in \eqref{GrindEQ__21_}. First, however, we need the
following lemma.

 \begin{lemma} \label{Lemma 3}  For every positive integer $m$ define the quantities

\begin{equation} \label{GrindEQ__23_}
a(m,n,\mu )=\left (\mu \frac{d}{d\mu }  \right )^{m} (1+\mu )^{n} =\sum
_{k=0}^{n}\binom{n}{k}  k ^{m} \mu   ^{k} .
\end{equation}
Then

\begin{equation} a(m,  n,\mu )=\sum _{k=0}^{n}\binom{n}{k}  k !  S (m,k)\mu   ^{k}  (1+\mu )^{n-k}     . \end{equation}
\end{lemma}
This is a known identity that can be found, for example, in \cite{gould}.

 From Lemma \ref{Lemma 3} we obtain another of our main results.

\begin{proposition} \label{Proposition 3} For every two positive integers $m$ and
$n$,
\begin{equation} \label{GrindEQ__25_}
\sum _{k=1}^{n}\binom{n}{k}  H_{k}  k^{m} \mu^{k} = a  (m,  n,\mu ) H_{n}
-\sum _{p=1}^{n-1}\frac{1}{n-p}   a (m,  p,\mu ).
\end{equation}
\end{proposition}
 \begin{proof} Apply $(\mu \frac{d}{d \mu }  )^{m} $ to both sides of \eqref{GrindEQ__21_} and note that $(\mu \frac{d}{d\mu }  )^{m} \mu^{k} =k^{m} \mu^{k} $.
\end{proof}
 The sums \eqref{GrindEQ__22_} were recently studied by M. Coffey \cite{coffey} by using a different method (a recursive formula) and a representation was given in terms of the hypergeometric function


\section{Stirling functions of a negative
argument. Dilcher's formula}

Some time ago Karl Dilcher obtained the nice identity

\begin{equation} \label{GrindEQ__26_}
\sum _{k=1}^{n}\binom{n}{k}  \frac{(-1)^{k-
1} }{k^{ m} } =  \sum \frac{1}{j_{1} j_{2} \cdots j_{m} }  
,\quad 1\le j_{1} \le j_{2} \le \cdots \le j_{m} \le n,
\end{equation}
as a corollary from a certain multiple series representation 
\cite[Corollary 3]{dilcher}; see also a similar result in \cite{flajolet}.
As this is one
binomial transform, it is good to have a direct proof by Euler's
transform method. Before giving such a proof, however, we want to
point out one interesting interpretation of the sum on the left
hand side in \eqref{GrindEQ__26_}.

Let $S(m,n)$ be the Stirling numbers of the second kind \cite{graham}.
Butzer et al. \cite{butzer} defined an extension $ S(\alpha ,n)$ for
any complex number $\alpha \ne 0$. The functions $S(\alpha ,n)$ of the
complex variable $\alpha $ are called Stirling functions of the second
kind. The extension is given by the formula

\begin{equation} \label{GrindEQ__27_}
S(\alpha ,n)=\frac{1}{n!} \sum _{k=1}^{n}\binom{n}{k}
(-1)^{n-k} k^{\alpha },
\end{equation}
with $S(\alpha ,0)=0$. Thus, for $m , n\ge 1$,

\begin{equation} \label{GrindEQ__28_}
(-1)^{n-1} n!S(-m  ,n)=\sum _{k=1}^{n}\binom{n}{k}  \frac{(-1)^{k-1} }{k^{ m} }.
\end{equation}
 For the next proposition we shall need the polylogarithmic function \cite{lewin}

\begin{equation} \label{GrindEQ__29_}
{\rm Li}_{m} (t)=\sum _{n= 1}^{\infty } \frac{t^{n} }{n^{m} }.
\end{equation}

\begin{proposition} \label{Proposition 4} For any integer $m\ge 1$ we have
\begin{equation} \label{GrindEQ__30_}
(-1)^{n-1} n!S(-m ,n)=  \sum \frac{1}{j_{1} j_{2} \cdots j_{m} }
 ,\quad 1\le j_{1} \le j_{2} \le \cdots \le j_{m} \le n .
\end{equation} \end{proposition}
\begin{proof}
The proof is based on the representation
\begin{equation} \label{GrindEQ__31_}
{\rm Li}_{m}\! \left (\frac{-t}{1-  t} \right )=- \sum \frac{t^{j_{m} }
}{j_{1} j_{2} \cdots j_{m} }    ,\quad 1\le j_{1} \le j_{2} \le
\cdots \le j_{m} ,
\end{equation}
(see \cite{ulanskii}) from which, in view of Lemma 2,

\begin{equation} \label{GrindEQ__32_}
\frac{-1}{1-t} \  {\rm Li}_{m} \! \left (\frac{-t}{1-t} \right )=\sum _{n=1}^{\infty
}A _{n}   t^{n}  ,
\end{equation}
with coefficients

\begin{equation} \label{GrindEQ__33_}
A_{n} =\sum \frac{1}{j_{1} j_{2} \cdots j_{m} }   ,\quad 1\le j_{1}
\le j_{2} \le \cdots \le j_{m} \le n .
\end{equation}
The assertion now follows from \eqref{GrindEQ__9_}.
\end{proof}
In conclusion, many thanks to the referee for a correction and for
some interesting comments.



\section{Appendix}

 We prove Euler's transform representation \eqref{GrindEQ__2_} by using Cauchy's integral formula, both for the Taylor coefficients of a holomorphic function and for the function itself. Thus, given a holomorphic function $f$ as in \eqref{GrindEQ__1_}, we have

\begin{equation} \label{GrindEQ__34_}
a_{k} =\frac{1}{2\pi i} \oint_{L} \frac{1}{\lambda^{k} }
\frac{f(\lambda )}{\lambda }   d\lambda,
\end{equation}
for an appropriate closed curve $L$ around the origin. Multiplying
both sides by $\binom{n}{k}$ and summing for $k$ we find

\begin{equation} \label{GrindEQ__35_}
\sum _{k=0}^{n}\binom{n}{k}a_{k}  =\frac{1}{2\pi  i} \oint_{L} \left ( \sum_{k=0}^{n}\binom{n}{k}
\frac{1}{\lambda^{k} } \right )  \frac{f(\lambda )}{\lambda }
  d\lambda =\frac{1}{2\pi  i} \oint _{L}
\left(1+\frac{1}{\lambda } \right)^{n}   \frac{f(\lambda
)}{\lambda }   d\lambda.
\end{equation}
Multiplying this by $t^{n} $ (with $t$ small enough) and summing
for $n$ we arrive at the desired representation
\eqref{GrindEQ__2_}, because

\begin{equation} \label{GrindEQ__36_}
\sum _{n=0}^{\infty }t^{n} \left ( 1+\frac{1}{\lambda }  \right )^{n}
=\frac{1}{1-t (1+\frac{1}{\lambda } )} =\frac{1}{1-t}  
\frac{\lambda }{\lambda -\frac{t}{1-t} },
\end{equation}
and therefore,
\begin{equation} \label{GrindEQ__37_}
\sum _{n= 0}^{\infty } t^{n}
\left ( \sum _{k=0}^{n}\binom{n}{k}a_{k}  \right )
=\frac{1}{1-t}   \frac{1}{2\pi i} \oint_{L} \frac{f(\lambda
)}{\lambda -\frac{t}{1-t} }   d\lambda =\frac{1}{1-t} \
f\! \left (\frac{t}{1-t}\right ).
\end{equation}



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\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
05A20; Secondary 11B73.


\noindent \emph{Keywords: } Harmonic numbers, binomial transform,
Euler transform, Stirling number of the second kind, Stirling
function, combinatorial identity, polylogarithm, Dilcher's
formula.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 12 2009;
revised version received  July 25 2009.
Published in {\it Journal of Integer Sequences}, August 30 2009.

\bigskip
\hrule
\bigskip

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


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