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\begin{center}
\vskip 1cm{\LARGE\bf Alternating Sums of the Reciprocals \\
\vskip .1in of Binomial Coefficients } 
\vskip 1cm 
{\large 
Hac\`{e}ne Belbachir and Mourad Rahmani\\
University of Sciences and Technology Houari Boumediene\\ 
Faculty of Mathematics\\
P. O. Box 32 \\
El Alia, Bab-Ezzouar 16111\\
Algiers\\
Algeria\\
\href{mailto:hacenebelbachir@gmail.com}{\tt hacenebelbachir@gmail.com} \\
\href{mailto:mrahmani@usthb.dz}{\tt mrahmani@usthb.dz}\\
\ \\
B. Sury\\
Statistics \&amp; Mathematics Unit\\
Indian Statistical Institute\\
8th Mile Mysore Road\\
Bangalore 560059\\
India\\
\href{mailto:sury@isibang.ac.in}{\tt sury@isibang.ac.in}\\
}
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\vskip .2 in

\begin{abstract}
We investigate summations of the form
$\sum\nolimits_{0\leq k\leq n}\left(  -1\right)
^{k}k^{m}\binom{n}{k}^{-1}$. We give closed formulae in terms of
the Akiyama-Tanigawa matrix. Recurrence formulae, ordinary generating
functions and some other results are also given.
\end{abstract}

\section{Introduction and notation}
Binomial coefficients play an important role in many areas of
mathematics, such as combinatorics, number theory and special
functions. The inverse binomial coefficients have an integral
representation in terms of beta function
\begin{equation}
\binom{n}{k}^{-1}=\left(  n+1\right)  {\displaystyle\int\limits_{0}^{1}}
x^{k}\left(  1-x\right)  ^{n-k}dx.\label{betaa}%
\end{equation}
and one can obtain (see \cite{sury93})
\[%
{\displaystyle\sum\limits_{k=0}^{n}}
\binom{n}{k}^{-1}=%
{\displaystyle\frac{n+1}{2^{n+1}}%
{\displaystyle\sum\limits_{k=1}^{n+1}}
\frac{2^{k}}{k}.},
\]
There are many papers dealing with sums involving inverses of binomial
coefficients, see for instance
\cite{belrahsur,juan,rockett,sofo,sprungoli,staver,sury93,yang06,zhao}. For nonnegative integers
$n,m $ and $p$ we consider the sums
\begin{equation}
T_{n}^{\left(  m,p\right)  }:={\displaystyle\sum\limits_{k=0}^{n}}\left(
-1\right)  ^{k}k^{m}\binom{p+n}{p+k}^{-1}.\label{base}%
\end{equation}


These sums have been studied by many authors. Trif \cite{trif}, using
(\ref{betaa}) proved for $m=0$ that
\begin{equation}
T_{n}^{\left(  0,p\right)  }=\left(  \left(  -1\right)  ^{n}+\binom{p+n+1}%
{p}^{-1}\right)  \frac{p+n+1}{p+n+2}.\label{tri}%
\end{equation}
Sury, Wang and Zhao \cite{sury04}, studied (\ref{base}) for $m=1$ and $m=2,$
they obtain%

\begin{equation}
T_{n}^{\left(  1,p\right)  } =\frac{p+n+1}{p+n+3}\bigg(\frac{(-1)^{n}%
(n+1)(p+n+3)}{p+n+2}-{\binom{p+n+2}{p+1}}^{-1} -(-1)^{n}\bigg),
\end{equation}
and
\begin{align}
T_{n}^{\left(  2,p\right)  } &amp; =(p+n+1)\bigg(\frac{(-1)^{n}(n+1)^{2}}{p+n+2}-
\frac{(-1)^{n}(2n+3)}{p+n+3}\nonumber\\
&amp; +\frac{2}{p+n+4}\bigg({\binom{p+n+3}{p+2}}^{-1}+(-1)^{n}\bigg)\ -\frac{1}{p+n+3}{\binom{p+n+2}{p+1}}^{-1}\bigg).
\end{align}


Our aim is to give a closed form and recurrence relation for the sums
(\ref{base}). In order to investigate the summation of the form $S_{n}%
^{\left(  m\right)  }:=T_{n}^{\left(  m,0\right)  }$ and $T_{n}^{\left(
m,p\right)  }$, we shall use the following tools \cite{Abra,knuth}:

\begin{itemize}
\item The Stirling numbers of the first kind
$%
s(n,k)%
$ (see A008275 in \cite{sloane}), are defined by the generating function
\[
x\left(  x-1\right)  \cdots\left(  x-n+1\right)  ={\displaystyle\sum
\limits_{k\geq0}}%
s(n,k)%
x^{k},
\]
and satisfy the recurrence relation%
\[%
s(n+1,k)%
=%
s(n,k-1)%
-n%
s(n,k)%
,\text{ }\ \left(  1\leq k\leq n\right)  ,
\]
with $%
s(n,n)%
=1,$ $%
s(n,0)%
=0$ for $n\geq1$ and $%
s(n,k)%
=0$ for $k&lt;0$\ or $k&gt;n.$

\item The Stirling numbers of the second kind $%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
$ (see A008277 in \cite{sloane}), are defined by the generating function
\[
{\displaystyle\prod\limits_{j=1}^{k}}\frac{x}{1-jx}={\displaystyle\sum
\limits_{n\geq k}}%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
x^{n},
\]
and satisfy the recurrence relation%
\[%
\genfrac{\{}{\}}{0pt}{}{n+1}{k}%
=%
\genfrac{\{}{\}}{0pt}{}{n}{k-1}%
+k%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
,
\]
with $%
\genfrac{\{}{\}}{0pt}{}{n}{1}%
=%
\genfrac{\{}{\}}{0pt}{}{n}{n}%
$ $=1.$

They also verify the following important identity
\begin{equation}
x^{n}={\displaystyle\sum\limits_{k=0}^{n}}\left(  -1\right)  ^{n+k}%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
x\left(  x+1\right)  \cdots\left(  x+k-1\right)  .\label{stir}%
\end{equation}


\item The Eulerian numbers $%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
$ (see A008292 in \cite{sloane}) are defined by
\[%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
={\displaystyle\sum\limits_{i=0}^{k}}\left(  -1\right)  ^{i}\left(
k-i\right)  ^{n}\binom{n+1}{i},\text{ \ }\left(  1\leq k\leq n\right)  ,
\]
and satisfy the recursive identity%
\[%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
=k%
\genfrac{\langle}{\rangle}{0pt}{}{n-1}{k}%
+\left(  n-k+1\right)
\genfrac{\langle}{\rangle}{0pt}{}{n-1}{k-1}%
,
\]
with $%
\genfrac{\langle}{\rangle}{0pt}{}{1}{1}%
=1.$

\item The Worpitzky numbers $W_{n,k}$ (see A028246 in \cite{sloane}), are
defined by
\[
W_{n,k}={\displaystyle\sum\limits_{i=0}^{k}}\left(  -1\right)  ^{i+k}\left(
i+1\right)  ^{n}\binom{k}{i}.
\]
They can also be expressed through the Stirling numbers of the second kind as
follows
\begin{equation}
W_{n,k}=k!%
\genfrac{\{}{\}}{0pt}{}{n+1}{k+1}%
.\label{Ho5}%
\end{equation}
The Worpitzky numbers satisfy the recursive relation
\begin{equation}
W_{n,k}=\left(  k+1\right)  W_{n-1,k}+kW_{n-1,k-1}\text{ \ }\left(
n\geq1,k\geq1\right)  .\label{Ho4}%
\end{equation}


Some simple properties are given%

\begin{equation}
{\displaystyle\sum\limits_{k=0}^{n}}%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
x^{k}={\displaystyle\sum\limits_{k=0}^{n}}\left(  x-1\right)  ^{n-k}%
kW_{n-1,k-1},\label{Ho3}%
\end{equation}%
\begin{equation}
{\displaystyle\sum\limits_{k=0}^{n}}\binom{n}{k}%
\genfrac{\{}{\}}{0pt}{}{k}{t}%
=%
\genfrac{\{}{\}}{0pt}{}{n+1}{t+1}%
,\label{Ho1}%
\end{equation}
and
\begin{equation}
{\displaystyle\sum\limits_{k=0}^{n}}%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
\binom{k+1}{t}=W_{n,n-t}.\label{Ho2}%
\end{equation}


\item The Bernoulli numbers $B_{n}$ are defined by the exponential generating
function%
\[
\frac{x}{1-e^{-x}}=%
{\displaystyle\sum\limits_{n\geq0}}
B_{n}\frac{x^{n}}{n!}.
\]


The recursive relation is
\begin{align*}
B_{0} &amp;  =1,\\
B_{n} &amp;  =1-{\displaystyle\sum\limits_{k=0}^{n-1}}\binom{n}{k}\frac{B_{k}%
}{n-k+1},\text{ }\left(  n\geq1\right)  .
\end{align*}
Thus we have $B_{1}=\frac{1}{2},B_{2}=\frac{1}{6},B_{3}=0,$ and so on, they
can also be expressed through the Worpitzky numbers%
\[
B_{n}={\displaystyle\sum\limits_{k=0}^{n}}\left(  -1\right)  ^{k}\frac
{W_{n,k}}{k+1}.
\]


\item The Akiyama-Tanigawa matrix $\left(  A_{n,k}\right)  _{n,k\geq0}$
associated with initial sequence $A_{0,k}=\frac{1}{k+1}$ is defined by (see
\cite{akitan,chen,kaneko,spr})
\[
A_{n,k}=\left(  k+1\right)  \left(  A_{n-1,k}-A_{n-1,k+1}\right)  ,
\]
or equivalently by \cite{Inaba}
\begin{align}
A_{n,k} &amp;  =\frac{1}{k!}{\displaystyle\sum\limits_{i=0}^{k}}\left(  -1\right)
^{k}%
s(k+1,i+1)%
B_{n+i},\nonumber\\
&amp;  ={\displaystyle\sum\limits_{i=1}^{n}}\left(  -1\right)  ^{i-1}\binom
{k+i+1}{k+1}^{-1}W_{n,i}.\label{Ber}%
\end{align}
The Akiyama-Tanigawa matrix $A_{n,k}$ is then%
\[
A_{n,k}=%
\begin{pmatrix}
1 &amp; \frac{1}{2} &amp; \frac{1}{3} &amp; \frac{1}{4} &amp; \frac{1}{5} &amp; \cdots\\
&amp;  &amp;  &amp;  &amp;  &amp; \\
\frac{1}{2} &amp; \frac{1}{3} &amp; \frac{1}{4} &amp; \frac{1}{5} &amp; \frac{1}{6} &amp; \cdots\\
&amp;  &amp;  &amp;  &amp;  &amp; \\
\frac{1}{6} &amp; \frac{1}{6} &amp; \frac{3}{20} &amp; \frac{2}{15} &amp; \frac{5}{42} &amp;
\cdots\\
&amp;  &amp;  &amp;  &amp;  &amp; \\
0 &amp; \frac{1}{30} &amp; \frac{1}{20} &amp; \frac{2}{35} &amp; \frac{5}{84} &amp; \cdots\\
&amp;  &amp;  &amp;  &amp;  &amp; \\
-\frac{1}{30} &amp; -\frac{1}{30} &amp; -\frac{3}{140} &amp; -\frac{1}{105} &amp; 0 &amp; \cdots\\
&amp;  &amp;  &amp;  &amp;  &amp; \\
0 &amp; -\frac{1}{42} &amp; -\frac{1}{28} &amp; -\frac{4}{105} &amp; -\frac{1}{28} &amp; \cdots\\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\end{pmatrix}
.
\]

\end{itemize}

\section{Explicit formula for S$_{n}^{\left(  m\right)  }$}
For any nonnegative integer $m$, we consider the sums
\begin{equation}
S_{n}^{\left(  m\right)  }:={\displaystyle\sum\limits_{k=0}^{n}}\left(
-1\right)  ^{k}k^{m}\binom{n}{k}^{-1}.\label{base0}%
\end{equation}
Note that $S_n^{(0)} = (1 + (-1)^n) \frac{n+1}{n+2}$ (identity (14)
of \cite{sury04} or identity (\ref{tri}) here for $p=0$). The
following expression for $S_n^{(m)}$ holds and, in terms of
computational complexity, it is better than the sum defining
$S_n^{(m)}$ for $n \geq m$ as the expression in the theorem involves
$O(m)$ operations.

\begin{theorem}
For any nonnegative integers $n$ and $m$, we have%
\begin{equation}
S_{n}^{\left(  m\right)  }=\left(  n+1\right)  {\displaystyle\sum
\limits_{j=0}^{m}} \frac{\left(  -1\right)  ^{m+j}}{n+j+2}\left(  1+\left(
-1\right)  ^{n}\dbinom{n+j+1}{j}\right)  W_{m,j}.\label{exppp}%
\end{equation}

\end{theorem}



\begin{proof}
We can write $S_{n}^{\left(  m\right)  }$ as follows%
\begin{align*}
S_{n}^{\left(  m\right)  }  &amp;  ={\displaystyle\sum\limits_{k=0}^{n}}
(-1)^{k}\dbinom{n}{k}^{-1}\left(  \left(  k+1\right)  -1\right)  ^{m}\\
&amp;  ={\displaystyle\sum\limits_{k=0}^{n}} (-1)^{k}\dbinom{n}{k}^{-1}%
{\displaystyle\sum\limits_{i=0}^{m}} \left(  -1\right)  ^{m-i}\dbinom{m}%
{i}\left(  k+1\right)  ^{i},
\end{align*}
and with (\ref{stir}), we obtain
\begin{align*}
S_{n}^{\left(  m\right)  }  &amp;  ={\displaystyle\sum\limits_{k=0}^{n}}
(-1)^{k}\dbinom{n}{k}^{-1}{\displaystyle\sum\limits_{i=0}^{m}} \left(
-1\right)  ^{m-i}\dbinom{m}{i}{\displaystyle\sum\limits_{j=0}^{i}} \left(
-1\right)  ^{i+j}%
\genfrac{\{}{\}}{0pt}{0}{i}{j}%
\left(  k+1\right)  \cdots\left(  k+j\right) \\
&amp;  ={\displaystyle\sum\limits_{k=0}^{n}} {\displaystyle\sum\limits_{i=0}^{m}}
{\displaystyle\sum\limits_{j=0}^{i}} \frac{(-1)^{k}}{n!}\left(  -1\right)
^{m+j}\dbinom{m}{i}%
\genfrac{\{}{\}}{0pt}{0}{i}{j}%
k!\left(  k+1\right)  \cdots\left(  k+j\right)  (n-k)!\\
&amp;  ={\displaystyle\sum\limits_{i=0}^{m}} {\displaystyle\sum\limits_{j=0}^{i}}
\left(  -1\right)  ^{m+j}\dbinom{m}{i}%
\genfrac{\{}{\}}{0pt}{0}{i}{j}%
\left(  n+1\right)  \cdots\left(  n+j\right)  {\displaystyle\sum
\limits_{k=0}^{n}} (-1)^{k}\dbinom{n+j}{k+j}^{-1}.
\end{align*}


Now, from (\ref{tri}) and after some rearrangement, we get%
\[
S_{n}^{\left(  m\right)  }=\left(  n+1\right)  {\displaystyle\sum
\limits_{j=0}^{m}} \frac{\left(  -1\right)  ^{m+j}}{n+j+2}\left(  1+\left(
-1\right)  ^{n}\dbinom{n+j+1}{j}\right)  j!{\displaystyle\sum\limits_{i=0}%
^{m}} \dbinom{m}{i}%
\genfrac{\{}{\}}{0pt}{0}{i}{j}%
.
\]


From (\ref{Ho5}) and (\ref{Ho1}), the result holds.

\end{proof}


\section{Recurrence relation for $S_{n}^{\left(  m\right)  }$}

\begin{theorem}
\label{thh01} For any nonnegative integers $m$ and $n$, we have%
\begin{equation}
S_{n+1}^{\left(  m\right)  }=S_{n}^{\left(  m\right)  }-\frac{1}{n+1}%
S_{n}^{\left(  m+1\right)  }-\left(  -1\right)  ^{n}\left(  n+1\right)  ^{m}.\label{rec1}%
\end{equation}
\end{theorem}

\begin{proof}
The proof is based on the Sprugnoli \cite{sprugnoli1} observation $\binom{n+1}{k}%
^{-1}=\binom{n}{k}^{-1}-\frac{k}{n+1}\binom{n}{k}^{-1}.$
\end{proof}

The recurrence relation for $S_{n}^{\left(  m\right)  }$ is given in the following

\begin{theorem}
For any nonnegative integers $m$ and $n$, we have
\begin{align}
S_{n+1}^{\left(  m\right)  }=S_{n}^{\left(  m\right)  }-\left(  -1\right)
^{n}\left(  n+1\right)  ^{m}+{\displaystyle\sum\limits_{j=0}^{m+1}}
\frac{\left(  -1\right)  ^{m+j}}{n+j+2}\left(  1+\left(  -1\right)  ^{n}%
\binom{n+j+1}{j}\right)  W_{m+1,j},\label{rec2}
\end{align}
with the initial condition $S_{0}^{\left(  m\right)  }=\delta_{0m}$, where $\delta_{ij}$ is the Kronecker symbol.
\end{theorem}

\begin{proof}
This follows immediately from (\ref{exppp}) and (\ref{rec1}).
\end{proof}

Setting $m=1$ in (\ref{rec2}), we have the following

\begin{corollary}
If $n$ is nonnegative integer, then%
\[
S_{n+1}^{\left(  1\right)  }=S_{n}^{\left(  1\right)  }%
+\frac{n-\left(  -1\right)  ^{n}\left(  2n^{4}+17n^{3}+49n^{2}+57n+24\right)
}{\left(  n+2\right)  \left(  n+3\right)  \left(
n+4\right)  }.
\]

\end{corollary}

Our next goal is to calculate the ordinary generating functions of
$S_{n}^{(m)}$.

\section{Ordinary generating functions of $S_{n}^{(m)}$}

In $2002$, Mansour \cite{mansour}, generalized the idea of Sury \cite{sury93}
and gave an approach based on calculus to obtain the generating function for
some combinatorial identities.

\begin{theorem}
[Mansour \cite{mansour}]\label{Ma} Let $r,n\geq k$ be any nonnegative integer
numbers, and let $f_{r}\left(  n,k\right)  $ be given by
\begin{equation}
f_{r}\left(  n,k\right)  =\frac{\left(  n+r\right)  !}{n!}%
{\displaystyle\int\limits_{u_{1}}^{u_{2}}}
p^{k}\left(  t\right)  q^{n-k}\left(  t\right)  dt,\label{beta}%
\end{equation}
where $p\left(  t\right)  $ and $q\left(  t\right)  $ are two functions
defined on $\left[  u_{1},u_{2}\right]  .$ Let $\left\{  a_{n}\right\}
_{n\geq0}$ and $\left\{  b_{n}\right\}  _{n\geq0}$ be any two sequences, and
let $A\left(  x\right)  ,B\left(  x\right)  $ be the corresponding ordinary
generating functions. Then
\begin{equation}
{\displaystyle\sum\limits_{n=0}^{\infty}} \left[  {\displaystyle\sum
\limits_{k=0}^{n}} f_{r}\left(  n,k\right)  a_{k}b_{n-k}\right]  x^{n}=\frac
{d^{r}}{dx^{r}}\left[  x^{r}{\displaystyle\int\limits_{u_{1}}^{u_{2}}}
A\left(  xp\left(  t\right)  \right)  B\left(  xq\left(  t\right)  \right)
dt\right]  .\label{2}%
\end{equation}

\end{theorem}

We apply Theorem \ref{Ma}, for $a_{n}=\left(  -1\right)  ^{n}n^{m}$ $(m\geq1)$
and $b_{n}=1$, we have
\begin{align*}
A\left(  x\right)   &amp;  =\frac{1}{\left(  1+x\right)  ^{m+1}}\sum
\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -x\right)  ^{k+1}\\
&amp;  =\sum\limits_{k=0}^{m}\frac{(-1)^{m+k}}{\left(  1+x\right)  ^{k+1}}%
W_{m,k},\\
B\left(  x\right)   &amp;  =%
{\displaystyle\sum\limits_{n\geq0}}
x^{n}=\frac{1}{1-x}.
\end{align*}
For $r=1$, formula (\ref{2}) gives:
\begin{equation}%
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  m\right)  }x^{n}=\frac{d}{dx}\left[  x%
{\displaystyle\int\limits_{0}^{1}}
\frac{\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -xt\right)  ^{k+1}}{\left(  1+xt\right)  ^{m+1}\left(  1-x+xt\right)
}dt\right]  .\label{g1}%
\end{equation}
Making the substitution $xt=y$ in the right-hand side of (\ref{g1}), we
obtain
\[%
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  m\right)  }x^{n}=\frac{d}{dx}\left[
{\displaystyle\int\limits_{0}^{x}}
\frac{\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -y\right)  ^{k+1}}{\left(  1+y\right)  ^{m+1}\left(  1-x+y\right)
}dy\right]  ,
\]
Since the degree of the denominator is at least one higher than that of the
numerator, this fraction decomposes into partial fractions of the form%
\begin{equation}
\frac{\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -y\right)  ^{k+1}}{\left(  1+y\right)  ^{m+1}\left(  1-x+y\right)
}=\frac{\alpha^{\left(  m\right)  }\left(  x\right)  }{1-x+y}+\sum
\limits_{s=0}^{m}\frac{\alpha_{s}^{\left(  m\right)  }\left(  x\right)
}{\left(  1+y\right)  ^{m-s+1}},\label{g2}%
\end{equation}


We note in passing that (\ref{g2}) is equivalent to
\begin{align}
\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -y\right)  ^{k+1} &amp;  =\left(  1+y\right)  ^{m+1}\alpha^{\left(
m\right)  }\left(  x\right)  +\left(  1-x+y\right)  \sum\limits_{s=0}%
^{m}\left(  1+y\right)  ^{s}\alpha_{s}^{\left(  m\right)  }\left(  x\right)
\label{g3}\\
&amp;  =\sum\limits_{k=0}^{m}\left(  -1\right)  ^{m+k+1}y\left(  1+y\right)
^{m-k}W_{m-1,k-1}.\nonumber
\end{align}
For $y=-1$ and using the fact that $W_{p,p}=p!$ for $p\geq0,$ we immediately
obtain the well-known identity%
\[
\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
=m!.
\]
Next, if we set $y=0$ in (\ref{g3}) then we obtain a relation between
$\alpha^{\left(  m\right)  }\left(  x\right)  $ and $\alpha_{s}^{\left(
m\right)  }\left(  x\right)  $
\begin{equation}
\sum\limits_{s=0}^{m}\alpha_{s}^{\left(  m\right)  }\left(  x\right)
=\frac{\alpha^{\left(  m\right)  }\left(  x\right)  }{x-1}.\label{g5}%
\end{equation}


\begin{proposition}
\label{gl1}\bigskip For $m\geq1,$ we have%
\begin{align}
\alpha_{s}^{\left(  m\right)  }\left(  x\right)   &amp;  =\sum\limits_{i=0}%
^{s}\frac{\left(  -1\right)  ^{i+s+1}}{x^{i+1}}\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\dbinom{k+1}{s-i}\label{alpha}\\
&amp;  =\sum\limits_{j=m-s}^{m}\frac{\left(  -1\right)  ^{m+j+1}}{x^{s-m+1+j}%
}W_{m,j},\nonumber
\end{align}
and
\begin{align}
\alpha^{\left(  m\right)  }\left(  x\right)   &amp;  =\frac{1}{x^{m+1}}%
\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  1-x\right)  ^{k+1}\label{alphas}\\
&amp;  =\sum\limits_{j=0}^{m}\frac{\left(  -1\right)  ^{m+j}}{x^{j+1}}%
W_{m,j},\nonumber\\
&amp;  =-\alpha_{m}^{\left(  m\right)  }\left(  x\right)  .\nonumber
\end{align}

\end{proposition}

\begin{proof}
We verify that (\ref{alpha}) and (\ref{alphas}) satisfy (\ref{g3}). Denote the
right-hand side of (\ref{g3}) by $R^{\left(  m\right)  }(y)$, after some
rearrangement, we get%
\begin{multline*}
R^{\left(  m\right)  }(y)=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}
\left[  \frac{\left(  1+y\right)  ^{m+1}}{x^{m+1}}\left(  1-x\right)
^{k+1}\right. \\
+\left.  \left(  1-x+y\right)  \sum\limits_{s=0}^{m}\left(  1+y\right)
^{s}\sum\limits_{j=0}^{s}\frac{\left(  -1\right)  ^{j+1}}{x^{s-j+1}}%
\dbinom{k+1}{j}\right]  ,
\end{multline*}
using binomial formula and for $k\leq m,$ we obtain%
\begin{multline*}
R^{\left(  m\right)  }(y)=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}
\left[  \sum\limits_{s=m+1}^{m+1}\frac{\left(  1+y\right)  ^{s}}{x^{s}}%
\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)  ^{j}x^{j}\right. \\
-\left.  \frac{\left(  1-x+y\right)  }{x}\sum\limits_{s=0}^{m}\frac{\left(
1+y\right)  ^{s}}{x^{s}}\sum\limits_{j=0}^{s}\left(  -1\right)  ^{j}%
x^{j}\dbinom{k+1}{j}\right]
\end{multline*}%
\begin{multline*}
=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left[  \sum\limits_{s=m+1}^{m+1}\frac{\left(  1+y\right)  ^{s}}{x^{s}}%
\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)  ^{j}x^{j}\right. \\
+\sum\limits_{s=0}^{m}\frac{\left(  1+y\right)  ^{s}}{x^{s}}\sum
\limits_{j=0}^{s}\left(  -1\right)  ^{j}x^{j}\dbinom{k+1}{j}\\
-\left.  \sum\limits_{s=0}^{m}\frac{\left(  1+y\right)  ^{s+1}}{x^{s+1}}%
\sum\limits_{j=0}^{s}\left(  -1\right)  ^{j}x^{j}\dbinom{k+1}{j}\right]
\end{multline*}%
\begin{multline*}
=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left[  \sum\limits_{s=0}^{m+1}\frac{\left(  1+y\right)  ^{s}}{x^{s}}%
\sum\limits_{j=0}^{s}\left(  -1\right)  ^{j}x^{j}\dbinom{k+1}{j}\right. \\
-\left.  \sum\limits_{s=0}^{m}\frac{\left(  1+y\right)  ^{s+1}}{x^{s+1}}%
\sum\limits_{j=0}^{s}\left(  -1\right)  ^{j}x^{j}\dbinom{k+1}{j}\right]
\end{multline*}%
\begin{multline*}
=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left[  \sum\limits_{s=0}^{m+1}\frac{\left(  1+y\right)  ^{s}}{x^{s}}\left(
\sum\limits_{j=0}^{s}\left(  -1\right)  ^{j}x^{j}\dbinom{k+1}{j}\right.
\right. \\
\left.  -\left.  \sum\limits_{j=0}^{s-1}\left(  -1\right)  ^{j}x^{j}%
\dbinom{k+1}{j}\right)  \right]  .
\end{multline*}


Finally,%
\begin{align*}
R^{\left(  m\right)  }(y)  &amp;  =\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left[  \sum\limits_{s=0}^{k+1}\left(  1+y\right)  ^{s}\left(  \left(
-1\right)  ^{s}\dbinom{k+1}{s}\right)  \right] \\
&amp;  =\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -y\right)  ^{k+1}.
\end{align*}


According to (\ref{Ho5}) and (\ref{Ho2}), we have%
\begin{align*}
\alpha_{s}^{\left(  m\right)  }\left(  x\right)   &amp;  =\sum\limits_{i=0}%
^{s}\frac{\left(  -1\right)  ^{i+s+1}}{x^{i+1}}\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\dbinom{k+1}{s-i}\\
&amp;  =\sum\limits_{i=0}^{s}\frac{\left(  -1\right)  ^{i+s+1}(m-s+i)!}{x^{i+1}}%
\genfrac{\{}{\}}{0pt}{}{m+1}{m-s+i+1}%
\\
&amp;  =\sum\limits_{i=0}^{s}\frac{\left(  -1\right)  ^{i+s+1}}{\left(  x\right)
^{i+1}}W_{m,m-s+i}\\
&amp;  =\sum\limits_{j=m-s}^{m}\frac{\left(  -1\right)  ^{m+j+1}}{x^{s-m+1+j}%
}W_{m,j}.
\end{align*}


It follows from (\ref{Ho3}) that
\begin{align*}
\alpha^{\left(  m\right)  }\left(  x\right)   &amp;  =\frac{1}{x^{m+1}}%
\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  1-x\right)  ^{k+1}\\
&amp;  =\frac{1-x}{x^{m+1}}\sum\limits_{k=1}^{m}\left(  -1\right)  ^{m+k}%
x^{m-k}kW_{m-1,k-1}\\
&amp;  =\left(  1-x\right)  \sum\limits_{k=0}^{m}\left(  -1\right)  ^{m+k}\frac
{k}{x^{k+1}}W_{m-1,k-1}\\
&amp;  =\sum\limits_{k=0}^{m}\left(  -1\right)  ^{m+k}\frac{k}{x^{k+1}}%
W_{m-1,k-1}-\sum\limits_{k=0}^{m}\left(  -1\right)  ^{m+k}\frac{k}{x^{k}%
}W_{m-1,k-1}\\
&amp;  =\sum\limits_{k=0}^{m}\left(  -1\right)  ^{m+k}\frac{k}{x^{k+1}}%
W_{m-1,k-1}+\sum\limits_{k=0}^{m-1}\left(  -1\right)  ^{m+k}\frac{k+1}%
{x^{k+1}}W_{m-1,k}.
\end{align*}
Using (\ref{Ho4}), we get $\alpha^{\left(  m\right)  }\left(  x\right)  $ as
desired. This completes the proof.
\end{proof}

Now, integrating the right-hand side of (\ref{g2}) over $y$, we obtain%
\begin{equation}\label{g44}
{\displaystyle\int\limits_{0}^{x}}\frac{\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  -y\right)  ^{k+1}}{\left(  1+y\right)  ^{m+1}\left(  1-x+y\right)
}dy=\alpha_{m}^{\left(  m\right)  }\left(  x\right)  \ln(1-x^{2}%
)+\sum\limits_{s=0}^{m-1}\frac{\alpha_{s}^{\left(  m\right)  }\left(
x\right)  }{m-s}\left[  1-\left(  1+x\right)  ^{s-m}\right]  .
\end{equation}
By differentiating (\ref{g44}) we get the ordinary generating function of
$S_{n}^{\left(  m\right)  }$%
\begin{align}
{\displaystyle\sum\limits_{n\geq0}}S_{n}^{\left(  m\right)  }x^{n} &amp;
=\ln(1-x^{2})\frac{d}{dx}\alpha_{m}^{\left(  m\right)  }\left(  x\right)
+\sum\limits_{s=0}^{m-1}\frac{\frac{d}{dx}\alpha_{s}^{\left(  m\right)
}\left(  x\right)  }{m-s}\left(  \left[  1-\left(  1+x\right)  ^{s-m}\right]
\right) \nonumber\\
&amp;  +\sum\limits_{s=0}^{m-1}\left(  \left(  1+x\right)  ^{s-m-1}\alpha
_{s}^{\left(  m\right)  }\left(  x\right)  \right)  -\frac{2x}{1-x^{2}}%
\alpha_{m}^{\left(  m\right)  }\left(  x\right)  ,\label{generating}%
\end{align}
with
\[
\frac{d}{dx}\alpha_{s}^{\left(  m\right)  }\left(  x\right)  =\sum
\limits_{j=m-s}^{m}\frac{\left(  s-m+1+j\right)  \left(  -1\right)  ^{m+j}%
}{x^{s-m+2+j}}W_{m,j}.
\]


With Proposition \ref{gl1}, we can now rewrite (\ref{generating}) as follows

\begin{theorem}
For any real numbers $x$ and for all
nonnegative integer $m$, we have%

\begin{multline}
{\displaystyle\sum\limits_{n\geq0}}S_{n}^{\left(  m\right)  }x^{n}=\left(
\sum\limits_{j=0}^{m}\frac{\left(  1+j\right)  \left(  -1\right)  ^{m+j}%
}{x^{2+j}}W_{m,j}\right)  \ln(1-x^{2})\\
+\sum\limits_{0\leq j\leq s\leq m-1}\frac{\left(  -1\right)  ^{j}}{x^{s-j+2}%
}W_{m,m-j}\left(  \frac{s-j+1}{m-s}\left(  1-\left(  1+x\right)
^{s-m}\right)  -x\left(  1+x\right)  ^{s-m-1}\right) \\
+\frac{2}{1-x^{2}}\sum\limits_{j=0}^{m}\frac{\left(  -1\right)  ^{m+j}}{x^{j}%
}W_{m,j}.
\end{multline}

\end{theorem}

In particular for $m=0$ and $m=1$, we get%
\[
{\displaystyle\sum\limits_{n\geq0}}S_{n}^{\left(  0\right)  }x^{n}=\frac
{2}{1-x^{2}}+\frac{\ln\left(  1-x^{2}\right)  }{x^{2}},
\]
and
\[
{\displaystyle\sum\limits_{n\geq0}}S_{n}^{\left(  1\right)  }x^{n}=\frac
{2+3x}{x\left(  1+x\right)  ^{2}}+\frac{2-x}{x^{3}}\ln\left(  1-x^{2}\right)
.
\]

\section{The asymptotic expansion }

In the previous sections, $S_{n}^{\left(  m\right)  }$ becomes more complex
when, $m$ grows, so it is important to have asymptotic expansion of
$S_{n}^{\left(  m\right)  }.$

\begin{theorem}
For $m&gt;0$, we\bigskip\ have
\[
S_{2n}^{\left(  m\right)  }\sim\left(  2n\right)  ^{m}\text{ and }%
S_{2n+1}^{\left(  m\right)  }\sim-\left(  2n+1\right)  ^{m}.\text{ }%
\]

\end{theorem}

\begin{proof}
Write
\[
k^{m}=c_{0}+c_{1}\left(  k+1\right)  +c_{2}\left(  k+1\right)  \left(
k+2\right)  \cdots+c_{m}\left(  k+1\right)  \cdots\left(  k+m\right)  ,
\]
where $c_{i}$'s depend on $m$ $\left(  c_{m}=1\right)  $. we immediately have%
\[
S_{n}^{\left(  m\right)  }=c_{0}%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  -1\right)  ^{k}\binom{n}{k}^{-1}+c_{1}%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  -1\right)  ^{k}\left(  k+1\right)  \binom{n}{k}^{-1}+\cdots+%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  -1\right)  ^{k}\left(  k+1\right)  \cdots\left(  k+m\right)  \binom
{n}{k}^{-1}.
\]
After some rearrangement and using the fact that $\left(  k+1\right)  \binom{n}{k}^{-1}=\left(  n+1\right)  \binom{n+1}%
{k+1}^{-1}$, we have
\[
S_{n}^{\left(  m\right)  }=c_{0}T_{n}^{\left(  0,0\right)  }+c_{1}\left(
n+1\right)  T_{n}^{\left(  0,1\right)  }+c_{2}\left(  n+1\right)  \left(
n+2\right)  T_{n}^{\left(  0,2\right)  }+\cdots+\left(  n+1\right)
\cdots\left(  n+m\right)  T_{n}^{\left(  0,m\right)  }.
\]
Since $T_{2n}^{\left(  0,p\right)  }\rightarrow1$ and $T_{2n+1}^{\left(
0,p\right)  }\rightarrow-1,$ the result holds.
\end{proof}


\section{A connection to the Akiyama-Tanigawa matrix}

In this section we consider $T_{n}^{\left(  m,p\right)  }.$ The following
lemma will be useful in the proof of the main theorem of this section.

\begin{lemma}
For $m\geq1,$ we have%
\begin{align*}%
{\displaystyle\sum\limits_{k=0}^{n}}
k^{m}z^{k}  &amp;  =%
{\displaystyle\sum\limits_{k=0}^{m}}
W_{m,k}\left(  \frac{z}{1-z}\right)  ^{k+1}\\
&amp;  -z^{n+1}\text{ }%
{\displaystyle\sum\limits_{s=0}^{m}}
\binom{m}{s}\left(  n+1\right)  ^{m-s}%
{\displaystyle\sum\limits_{k=0}^{s}}
\left(  -1\right)  ^{s+k}W_{s,k}(1-z)^{-k-1}.
\end{align*}

\end{lemma}

\begin{proof}
Recall that, for $m\geq1$
\[%
{\displaystyle\sum\limits_{k=0}^{\infty}}
k^{m}z^{k}=%
{\displaystyle\sum\limits_{k=0}^{m}}
W_{m,k}\left(  \frac{z}{1-z}\right)  ^{k+1}=%
{\displaystyle\sum\limits_{k=0}^{m}}
\left(  -1\right)  ^{m+k}W_{m,k}(1-z)^{-k-1},
\]
we have%
\begin{align*}%
{\displaystyle\sum\limits_{k=0}^{n}}
k^{m}z^{k}  &amp;  =%
{\displaystyle\sum\limits_{k=0}^{\infty}}
k^{m}z^{k}-%
{\displaystyle\sum\limits_{k=n+1}^{\infty}}
k^{m}z^{k}\\
&amp;  =%
{\displaystyle\sum\limits_{k=0}^{m}}
W_{m,k}\left(  \frac{z}{1-z}\right)  ^{k+1}-z^{n+1}%
{\displaystyle\sum\limits_{i=0}^{\infty}}
\left(  i+n+1\right)  ^{m}z^{i}\\
&amp;  =%
{\displaystyle\sum\limits_{k=0}^{m}}
W_{m,k}\left(  \frac{z}{1-z}\right)  ^{k+1}-z^{n+1}%
{\displaystyle\sum\limits_{i=0}^{\infty}}
{\displaystyle\sum\limits_{s=0}^{m}}
\binom{m}{s}\left(  n+1\right)  ^{m-s}i^{s}z^{i}\\
&amp;  =%
{\displaystyle\sum\limits_{k=0}^{m}}
W_{m,k}\left(  \frac{z}{1-z}\right)  ^{k+1}-z^{n+1}\text{ }%
{\displaystyle\sum\limits_{s=0}^{m}}
\binom{m}{s}\left(  n+1\right)  ^{m-s}%
{\displaystyle\sum\limits_{i=0}^{\infty}}
i^{s}z^{i},
\end{align*}
as desired.
\end{proof}

For an alternative proof see Boyadzhiev \cite{boyadzhiev}. The main result of
this section is to prove the following theorem which expresses explicitly the
alternating sums of the reciprocals of binomial coefficients, $T_{n}^{\left(
m,p\right)  }$, in terms of the Akiyama-Tanigawa matrix $A_{n,k}$.

\begin{theorem}
For nonnegative integers $n,m$ and $p$, we have%
\begin{multline}
T_{n}^{\left(  m,p\right)  }=\binom{n+p}{p}^{-1}\delta_{0m}+\left(
n+p+1\right)
{\displaystyle\sum\limits_{s=0}^{m}}
\left(  -1\right)  ^{n+s}\binom{m}{s}\left(  n+1\right)  ^{m-s}A_{s,n+p+1}%
\label{thfin}\\
-\frac{n+p+1}{n+1}%
{\displaystyle\sum\limits_{s=0}^{m}}
\left(  -1\right)  ^{s}\binom{n+s+p+2}{p+s+1}^{-1}W_{m,s},
\end{multline}
where $A_{i,j}$ is the Akiyama-Tanigawa matrix.
\end{theorem}

\begin{proof}
By the Beta function we can write%

\begin{align*}
T_{n}^{\left(  m,p\right)  }  &amp;  =%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  -1\right)  ^{k}k^{m}\left(  p+n+1\right)
{\displaystyle\int\limits_{0}^{1}}
x^{p+k}\left(  1-x\right)  ^{n-k}dx\\
&amp;  =\left(  p+n+1\right)
{\displaystyle\int\limits_{0}^{1}}
x^{p}\left(  1-x\right)  ^{n}%
{\displaystyle\sum\limits_{k=0}^{n}}
k^{m}\left(  \frac{-x}{1-x}\right)  ^{k}dx.
\end{align*}


Using the lemma, we get%
\begin{align*}
T_{n}^{\left(  m,p\right)  }  &amp;  =\left(  p+n+1\right)
{\displaystyle\int\limits_{0}^{1}}
x^{p}\left(  1-x\right)  ^{n}\left(
{\displaystyle\sum\limits_{k=0}^{m}}
W_{m,k}\left(  -x\right)  ^{k+1}\right. \\
&amp;  -\text{ }\left.
{\displaystyle\sum\limits_{s=0}^{m}}
\binom{m}{s}\left(  n+1\right)  ^{m-s}%
{\displaystyle\sum\limits_{k=0}^{s}}
\left(  -1\right)  ^{s+k}W_{s,k}\left(  -x\right)  ^{n+1}(1-x)^{k-n}\right)
dx\\
&amp;  =\frac{\left(  p+n+1\right)  }{n+1}%
{\displaystyle\sum\limits_{k=0}^{m}}
\left(  -1\right)  ^{k+1}\binom{n+k+p+2}{p+k+1}^{-1}W_{m,k}\text{ }\\
&amp;  -\left(  p+n+1\right)
{\displaystyle\sum\limits_{s=0}^{m}}
{\displaystyle\sum\limits_{k=0}^{s}}
\binom{m}{s}\left(  n+1\right)  ^{m-s}\frac{\left(  -1\right)  ^{n+s+k+1}%
}{k+1}\binom{n+k+2+p}{k+1+p}^{-1}W_{s,k}.
\end{align*}


Finally, from (\ref{Ber}) we obtain%

\begin{align*}
T_{n}^{\left(  m,p\right)  } &amp;  =\frac{\left(  p+n+1\right)  }{n+1}%
{\displaystyle\sum\limits_{k=0}^{m}}
\left(  -1\right)  ^{k+1}\binom{n+k+p+2}{p+k+1}^{-1}W_{m,k}\text{ }\\
&amp;  +\left(  n+p+1\right)
{\displaystyle\sum\limits_{s=0}^{m}}
\binom{m}{s}\left(  n+1\right)  ^{m-s}\frac{\left(  -1\right)  ^{n+s}}{\left(
n+p+1\right)  !}%
{\displaystyle\sum\limits_{k=0}^{n+p+1}}
\left(  -1\right)  ^{k}%
s(n+p+2,k+1)%
B_{s+k}.
\end{align*}
As desired, this completes the proof.
\end{proof}

Setting $p=0$ in (\ref{thfin}) we can rewrite (\ref{exppp}) as follows

\begin{corollary}%
\[
S_{n}^{\left(  m\right)  }=\delta_{0m}-A_{m+1,n}+%
{\displaystyle\sum\limits_{s=0}^{m}}
\left(  -1\right)  ^{n+s}\binom{m}{s}\left(  n+1\right)  ^{m-s+1}A_{s,n+1}.
\]

\end{corollary}


\section{A recurrence relation For $T_{n}^{\left(  m,p\right)  }$}

\begin{theorem}
For any nonnegative integers $m,n$ and $p$%
\begin{equation}
T_{n+1}^{\left(  m,p\right)  }=\frac{n+1}{n+p+1}T_{n}^{\left(  m,p\right)
}-\frac{1}{n+p+1}T_{n}^{\left(  m+1,p\right)  }-\left(  -1\right)  ^{n}\left(
n+1\right)  ^{m}.\label{recg}
\end{equation}


\end{theorem}

\begin{proof}
Using the identity
\[
\dbinom{n+p+1}{k+p}^{-1}=\frac{n+1}{n+p+1}\dbinom{n+p}{k+p}^{-1}-\frac
{k}{n+p+1}\dbinom{n+p}{k+p}^{-1},%
\]
we get the relation (\ref{recg}).
\end{proof}
Now, from (\ref{thfin}) and (\ref{recg}), we have the recurrence relation for $T_{n}^{\left(  m,p\right)  }$
\begin{theorem}
For nonnegative integers $n,m$ and $p$, we have
\begin{multline*}
T_{n+1}^{\left(  m,p\right)  }=\frac{n+1}{n+p+1}T_{n}^{\left(  m,p\right)
}-(-1)^{n}\left(  n+1\right)  ^{m+1}%
{\displaystyle\sum\limits_{s=0}^{m+1}}
\frac{\left(  -1\right)  ^{s}}{\left(  n+1\right)  ^{s}}\dbinom{m+1}%
{s}A_{s,n+p+1}\\
+\frac{1}{n+1}%
{\displaystyle\sum\limits_{s=0}^{m+1}}
\left(  -1\right)  ^{s}\dbinom{n+s+p+2}{p+s+1}^{-1}W_{m+1,s}-\left(  -1\right)
^{n}\left(  n+1\right)  ^{m},
\end{multline*}
with the initial condition $T_{0}^{\left(  m,p\right)  }=\binom{n+p}{p}^{-1}\delta_{0m}$.
\end{theorem}

\section{Acknowledgments}

The authors are grateful to the anonymous referee for providing a
number of constructive comments and illuminating suggestions, almost
all of which we have tried to carry out. The first and the second
author were partially supported by PNR project 8/u160/664.

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\end{thebibliography}


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\noindent 2000 {\it Mathematics Subject Classification}:  Primary
11B65; Secondary 05A10, 05A16.

\noindent {\it Keywords}: Binomial coefficient, Akiyama-Tanigawa
matrix, recurrence relation, generating function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A008275},
\seqnum{A008277},
\seqnum{A008292}, and
\seqnum{A028246}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received July 27 2011; 
revised version received January 13 2012.
Published in {\it Journal of Integer Sequences}, January 14 2012.

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