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\begin{center}
\vskip 1cm{\LARGE\bf Sums Involving Moments of 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}\\
\
\ \\
\
\large 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}\\
\end{center}

\vskip .2 in

\begin{abstract}
We investigate sums of the form $\sum_{0\leq k\leq
n}k^{m}\binom{n}{k}^{-1}.$ We establish a recurrence relation and
compute its ordinary generating function. As application we give the
asymptotic expansion. The results extend the earlier works by
various authors. In the last section, we establish that $\sum_{0\leq
k\leq n} \frac{k^{m}}{n^m} \binom{n}{k}^{-1}$ tends to $1$ as $n
\rightarrow \infty$ and that $\sum_{0\leq k\leq
n-m}k^{m}\binom{n}{k}^{-1}$ tends to $m!$ as $n \rightarrow \infty$.
\end{abstract}

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\section{Introduction}
For all nonnegative integers $n,m$, let%
\begin{equation}
S_{n}^{\left(  m\right)  }:=%
{\displaystyle\sum\limits_{k=0}^{n}}
k^{m}\binom{n}{k}^{-1}.\label{for1}%
\end{equation}


There are several papers in the literature dealing with sums
involving inverses of binomial coefficients 
\cite{juan,rockett,sofo,sprungoli,sury04,trif,yang06,zhao}.
The cases $m=0$ and $m=1$ were intensively studied.

In 1947, Staver \cite{staver}, using the identity
\[%
{\displaystyle\sum\limits_{k=0}^{n}}
k\binom{n}{k}^{-1}=%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  n-k\right)  \binom{n}{k}^{-1},
\]
obtained a relation between $S_{n}^{\left(  1\right)  }$ and
$S_{n}^{\left(
0\right)  }$:
\begin{equation}
S_{n}^{\left(  1\right)  }=\frac{n}{2}S_{n}^{\left(  0\right)  }%
,\label{relation1}%
\end{equation}
and established a recurrence relation for $S_{n}^{\left(  0\right)  }$%
\begin{equation}
S_{n+1}^{\left(  0\right)  }=\frac{n+2}{2(n+1)}S_{n}^{\left(  0\right)
}+1.\label{rec1}%
\end{equation}
Staver also proved for the first time the well-known formula%
\begin{equation}
S_{n}^{\left(  0\right)  }=\frac{n+1}{2^{n+1}}%
{\displaystyle\sum\limits_{k=1}^{n+1}}
\frac{2^{k}}{k}.\label{clo1}%
\end{equation}

For applications of (\ref{clo1}), see Nedemeyer and Smorodinsky
\cite{Nedemeyer}, Mansour and West \cite{mansour2}, and for a
probabilistic application, see Letac \cite[p.\ 14]{letac}. Finally,
for the asymptotic expansion, see Comtet \cite[p.\ 294]{comtet} and
Yang and Zhao \cite{yang10}.

In $1981$, using induction and the relation
\[
\binom{n}{k}^{-1}=\binom{n-1}{k-1}^{-1}-\frac{n-k}{n-k+1}\binom{n}{k-1}^{-1}%
\]
Rockett \cite{rockett} proved (\ref{rec1}) and (\ref{clo1}).

In 1993, Sury \cite{sury93}
connected the inverse binomial coefficients to the beta function as follows%
\[
\binom{n}{k}^{-1}=\left(  n+1\right)
{\displaystyle\int\limits_{0}^{1}}
x^{k}\left(  1-x\right)  ^{n-k}dx
\]
and proved the relation (\ref{clo1}) (see also \cite{trif}). Some years
later, Mansour \cite{mansour}, generalized the idea of Sury
\cite{sury93}, and gave an approach based on calculus to obtain the
generating function for related combinatorial identities.

\begin{theorem}[Mansour \cite{mansour}]\label{Ma}
Let $r,n\geq k$ be any nonnegative integer numbers, and let $f\left(
n,k\right)  $ be given by
\begin{equation*}
f\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{1}%
\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\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}
In particular,%
\[%
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  0\right)  }x^{n}=\frac{2}{\left(  x-1\right)  \left(
x-2\right)  }-\frac{2\ln\left(  1-x\right)  }{\left(  x-2\right)  ^{2}},
\]
and%
\[%
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  1\right)  }x^{n}=-\frac{x\left(  3x-4\right)  }{\left(
x-1\right)  ^{2}\left(  x-2\right)  ^{2}}+\frac{2x\ln\left(  1-x\right)
}{\left(  x-2\right)  ^{3}}.
\]
We shall use the following well-known basic tools \cite{knuth}:

\begin{itemize}
\item[$(i)$] The Stirling numbers of the second kind $%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
$ (\seqnum{A008277}), can be defined by the generating function
\[%
{\displaystyle\prod\limits_{j=1}^{k}}
\frac{x}{1-jx}=%
{\displaystyle\sum\limits_{n\geq k}}
\genfrac{\{}{\}}{0pt}{}{n}{k}%
x^{n}.
\]
The most basic recurrence relation is%
\[%
\genfrac{\{}{\}}{0pt}{}{n+1}{k}%
=%
\genfrac{\{}{\}}{0pt}{}{n}{k-1}%
+k%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
,
\]
with $%
\genfrac{\{}{\}}{0pt}{}{n}{1}%
=%
\genfrac{\{}{\}}{0pt}{}{n}{n}%
$ $=1.$ An important relation involving $%
\genfrac{\{}{\}}{0pt}{}{n}{k}%
$ is
\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[$\left(  ii\right)  $] The Eulerian numbers $%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
$ (\seqnum{A008292}) are defined by
\[%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
=%
{\displaystyle\sum\limits_{i=0}^{k}}
\left(  -1\right)  ^{i}\left(  k-i\right)  ^{n}\binom{n+1}{i},\text{ \ }n\geq
k\geq1.
\]
Which also satisfy the recursive relation%
\[%
\genfrac{\langle}{\rangle}{0pt}{}{n}{k}%
=k%
\genfrac{\langle}{\rangle}{0pt}{}{n-1}{k}%
+\left(  n-k+1\right)
\genfrac{\langle}{\rangle}{0pt}{}{n-1}{k-1}%
,
\]
with $%
\genfrac{\langle}{\rangle}{0pt}{}{1}{1}%
=1.$

\item[$\left(  iii\right)  $] The Worpitzky numbers $W_{n,k}$ (\seqnum{A028246}), 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}
\end{itemize}

\bigskip

Some simple properties related to these three remarkable sequences are
%
\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}

\section{Some results on $ S_{n}^{\left(  2\right)  }$}
In this section, we give some results concerned $S_{n}^{\left(  2\right)  }$.
Applying Theorem \ref{Ma}, for $a_{n}=n^{2}$ and $b_{n}=1$, we obtain, for $|x|&lt;1$,
$$A(x)=\frac{x(x+1)}{(1-x)^{3}},$$ and $$B(x)=\frac{1}{1-x},$$ from (\ref{2}) we get the generating function for  $S_{n}^{\left(  2\right)  }$
\begin{align}
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  2\right)  }x^{n}&amp; =\frac{d}{dx}\left[  x%
{\displaystyle\int\limits_{0}^{1}}
\frac{xt\left(  xt+1\right)  }{\left(  1-xt\right)  ^{3}\left(  1-x+xt\right)
}dt\right]\nonumber\\  &amp;  =\frac{2x\left(  2x^{3}-3x^{2}-3x+5\right)
}{(x-2)^{3}(x-1)^{3}}-2\left(  x^{2}+2x-2\right)  \frac{\ln(1-x)}{(x-2)^{4}%
}.\label{m3}%
\end{align}
In addition, we have a relation between $S_{n}^{\left(  2\right)  }$ and $S_{n}^{\left(  0\right)  }$ given by the following theorem.

\begin{theorem}
If $n$ is a nonnegative integer, then $S_{n}^{\left(  2\right)  }$ satisfies
the recursion relation%
\begin{equation}
S_{n+1}^{\left(  2\right)  }=\frac{\left(  n-1\right)  \left(  n+2\right)
^{2}}{2\left(  n-2\right)  \left(  n+1\right)  ^{2}}S_{n}^{\left(  2\right)
}+\frac{\allowbreak\left(  n+2\right)  \left(  n^{2}-2n-2\right)  }{2\left(
n-2\right)  },\label{m1}%
\end{equation}
and%
\begin{equation}
S_{n}^{\left(  2\right)  }=\frac{1}{4}\left(  n+1\right)  \left(  n-2\right)
S_{n}^{\left(  0\right)  }+\frac{1}{2}\left(  n+1\right)  ^{2}.\label{m2}%
\end{equation}

\end{theorem}

\begin{proof}
We use the WZ method \cite{PetWilZeil96}. Denote the summand in $S_{n}^{\left(  2\right)  }$ by
$L\left(  n,k\right)  :=k^{2}\binom{n}{k}^{-1},$ by the Zeilberger's Maple
package {\small\textsf{EKHAD}}\normalsize, we construct the function
\begin{align*}
G\left(  n,k\right)   &amp;  =\left(  n^{2}k^{2}+n+3nk-nk^{2}+2+2k-2k^{2}\right)
\left(  n+1-k\right)  \dbinom{n}{k}^{-1}
\end{align*}
such that
\[
\left(  n-1\right)  \left(  n+2\right)  ^{2}L\left(  n,k\right)  -2\left(
n-2\right)  \left(  n+1\right)  ^{2}L\left(  n+1,k\right)  =G\left(
n,k+1\right)  -G\left(  n,k\right).
\]
By summing the above telescoping equation over $k$ from $0$ to $n-1$, we obtain the following recurrence relation
\[
\left(  n-1\right)  \left(  n+2\right)  ^{2}%
{\displaystyle\sum\limits_{k=0}^{n-1}}
L\left(  n,k\right)  -2\left(  n-2\right)  \left(  n+1\right)  ^{2}%
{\displaystyle\sum\limits_{k=0}^{n-1}}
L\left(  n+1,k\right)  =G\left(  n,n\right)  -G\left(  n,0\right),
\]%
that we can rewrite
\begin{multline*}
\left(  n-1\right)  \left(  n+2\right)  ^{2}S_{n}^{\left(  2\right)  }-\left(
n-1\right)  \left(  n+2\right)  ^{2}n^{2}-2\left(  n-2\right)  \left(
n+1\right)  ^{2}S_{n+1}^{\left(  2\right)  }\\
+2\left(  n-2\right)  \left(  n+1\right)  ^{2}\left(  \frac{n^{2}}%
{n+1}+\left(  n+1\right)  ^{2}\right)  =n^{3}(n-1),
\end{multline*}
as desired.
We prove the relation (\ref{m2}) by induction on $n$, the result clearly holds
for $n=0$, we now show that the formula for $n+1$ follows from (\ref{m1}) and induction hypothesis

\begin{multline*}
S_{n+1}^{\left(  2\right)  }=\frac{\left(  n-1\right)  \left(  n+2\right)
^{2}}{2\left(  n-2\right)  \left(  n+1\right)  ^{2}}\left(  \frac{1}{4}\left(
n+1\right)  \left(  n-2\right)  S_{n}^{\left(  0\right)  }+\frac{1}{2}\left(
n+1\right)  ^{2}\right)  \\
+\frac{\allowbreak\left(  n+2\right)  \left(  n^{2}-2n-2\right)  }{2\left(
n-2\right)  }.
\end{multline*}
Applying (\ref{rec1}), we obtain
\[
S_{n+1}^{\left(  2\right)  }=\frac{1}{4}\left(  n+2\right)  \left(
n-1\right)  S_{n+1}^{\left(  0\right)  }+\frac{1}{2}\left(  n+2\right)  ^{2}.
\]
This completes the proof.
\end{proof}

Using the relation  $%
{\displaystyle\sum\limits_{k=0}^{n}}
k^{3}\binom{n}{k}^{-1}=%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  n-k\right)  ^{3}\binom{n}{k}^{-1}$ and relations (\ref{relation1}) and (\ref{m2}), we
easily obtain the following identity between $S_{n}^{\left(  3\right)  }$ and
$S_{n}^{\left(  0\right)  }.$

\begin{corollary}
For any nonnegative integers $n$, we have%
\[
S_{n}^{\left(  3\right)  }=\frac{1}{8}n\left(  n^{2}-3n-6\right)
S_{n}^{\left(  0\right)  }+\frac{3}{4}n\left(  n+1\right)  ^{2}.
\]

\end{corollary}
\section{Generalization}
Before stating the main result of this section, we need a lemma.
\begin{lemma}%
\begin{equation}
S_{n+j}^{\left(  0\right)  }=\left(  n+j+1\right)  \left(  \frac{1}%
{2^{j}\left(  n+1\right)  }S_{n}^{\left(  0\right)  }+%
{\displaystyle\sum\limits_{r=0}^{j-1}}
\frac{1}{2^{r}\left(  n+j-r+1\right)  }\right)  ,\label{l1}%
\end{equation}

\end{lemma}

\begin{proof}
We proceed by induction on $j.$ For $j=0$ the identity (\ref{l1}) holds. Now
suppose that (\ref{l1}) holds for some $j$ and replace $n$ by $n+1$ in
(\ref{l1}), then using relation (\ref{rec1}) the result follows.
\end{proof}
\begin{theorem}
For any nonnegative integers $m$ and $n$, we have
\begin{equation}
S_{n}^{\left(  m\right)  }={\displaystyle\sum\limits_{j=0}^{m}}\left(
-1\right)  ^{m+j}\dbinom{n+j}{n}W_{m,j}\left(  S_{n+j}^{\left(  0\right)
}-\frac{1}{\left(  n+j\right)  !}\sum\limits_{r=0}^{j-1}r!\left(
n+j-r\right)  !\right)  ,\label{sah}%
\end{equation}
with the usual convention that the empty sum is $0$.
\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}}
\left(  \left(  k+1\right)  -1\right)  ^{m}\dbinom{n}{k}^{-1},\\
&amp;  =%
{\displaystyle\sum\limits_{k=0}^{n}}
\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}}
\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}{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)!.
\end{align*}
After some rearrangement,%
\begin{align*}
S_{n}^{\left(  m\right)  } &amp;  =%
{\displaystyle\sum\limits_{k=0}^{n}}
{\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)  \dbinom{n+j}{k+j}^{-1},\\
&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}%
j!\frac{\left(  n+j\right)  !}{n!j!}%
{\displaystyle\sum\limits_{r=j}^{n+j}}
\dbinom{n+j}{r}^{-1}.
\end{align*}
Now, from (\ref{Ho1}) and (\ref{Ho2}), the result holds.
\end{proof}

Setting $m=4$ in (\ref{sah}), we have the following

\begin{corollary}
If $n$ is a nonnegative integer, then
\[
S_{n}^{(4)}=\frac{1}{16}\left(  n+1\right)  \left(  n^{3}-7n^{2}-2n+16\right)
S_{n}^{\left(  0\right)  }+\frac{1}{8}\left(  7n-8\right)  \left(  n+1\right)
^{3}.
\]

\end{corollary}

\begin{theorem}\label{recur}
For any nonnegative integers $m$ and $n$
\begin{equation}
S_{n+1}^{\left(  m\right)  }=\delta_{0m}+\frac{1}{n+1}%
%TCIMACRO{\dsum \limits_{j=0}^{m+1}}%
%BeginExpansion
{\displaystyle\sum\limits_{j=0}^{m+1}}
%EndExpansion
\binom{m+1}{j}S_{n}^{\left(  j\right)  },\label{mi}%
\end{equation}
where $\delta_{ij}$ is the Kronecker symbol.
\end{theorem}

\begin{proof}
Recall that  $\binom{n+1}{k}=\frac{n+1}{k}\binom{n}{k-1},$ we have
\begin{align*}%
{\displaystyle\sum\limits_{k=0}^{n+1}}
\frac{k^{m}}{\binom{n+1}{k}}  &amp; =\delta_{0m}+%
{\displaystyle\sum\limits_{k=1}^{n+1}}
k^{m}\binom{n+1}{k}^{-1}\\
&amp; =\delta_{0m}+\frac{1}{n+1}%
{\displaystyle\sum\limits_{k=1}^{n+1}}
k^{m+1}\binom{n}{k-1}^{-1}\\
&amp; =\delta_{0m}+\frac{1}{n+1}%
{\displaystyle\sum\limits_{k=0}^{n}}
\left(  k+1\right)  ^{m+1}\binom{n}{k}^{-1}\\
&amp; =\delta_{0m}+\frac{1}{n+1}%
{\displaystyle\sum\limits_{j=0}^{m+1}}
\binom{m+1}{j}%
{\displaystyle\sum\limits_{k=0}^{n}}
k^{j}\binom{n}{k}^{-1}.%
\end{align*}
This proves the theorem.
\end{proof}
Setting $m=1$ in (\ref{mi}) and using (\ref{m2}), we have the following


\begin{corollary}
If $n$ is a nonnegative integer, then%
\[
S_{n+1}^{\left(  1\right)  }=\frac{n+2}{2n}S_{n}^{\left(  1\right)  }+\frac
{1}{2}\left(  n+1\right)  .
\]

\end{corollary}
\section{Ordinary generating function}

We apply Theorem \ref{Ma}, for $a_{n}=n^{m}$ $(m\geq1)$, $b_{n}=1,$ and for
$\left\vert x\right\vert &lt;1$ 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}%
x^{k+1}=\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*}
From (\ref{2}), we get
\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
\[%
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  m\right)  }x^{n}=\frac{d}{dx}\left[
{\displaystyle\int\limits_{0}^{x}}
\frac{\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
y^{k+1}}{\left(  1-y\right)  ^{m+1}\left(  1-x+y\right)  }dy\right].
\]
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}%
y^{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}%
y^{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}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 $y=0$ for $\left\vert x\right\vert &lt;1$, we obtain the following 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_{k=0}%
^{m}\sum\limits_{i=0}^{s}\frac{\left(  -1\right)  ^{i+s}}{\left(  2-x\right)
^{i+1}}%
\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}}{\left(
2-x\right)  ^{s-m+1+j}}W_{m,j}.\nonumber
\end{align}
and
\begin{align}
\alpha^{\left(  m\right)  }\left(  x\right)   &amp;  =\frac{1}{\left(  2-x\right)
^{m+1}}\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  x-1\right)  ^{k+1}\label{alphas}\\
&amp;  =\sum\limits_{j=0}^{m}\frac{\left(  -1\right)  ^{m+j}}{\left(  2-x\right)
^{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 $R^{\left(  m\right)  }(y)$%
\[
R^{\left(  m\right)  }(y)=\left(  1-y\right)  ^{m+1}\alpha^{\left(  m\right)
}\left(  x\right)  +\left(  1-x+y\right)  \sum\limits_{s=0}^{m}\left(
1-y\right)  ^{s}\alpha_{s}^{\left(  m\right)  }\left(  x\right).
\]
After some rearrangement, we get%
\begin{multline*}
R^{\left(  m\right)  }(y)=\frac{\left(  1-y\right)  ^{m+1}}{\left(
2-x\right)  ^{m+1}}\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  x-1\right)  ^{k+1}+\\
\left(  1-x+y\right)  \sum\limits_{s=0}^{m}\left(  1-y\right)  ^{s}%
\sum\limits_{k=0}^{m}\sum\limits_{i=0}^{s}\frac{\left(  -1\right)  ^{i+s}%
}{\left(  2-x\right)  ^{i+1}}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\dbinom{k+1}{s-i}%
\end{multline*}%
\begin{multline*}
=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left[  \frac{\left(  1-y\right)  ^{m+1}}{\left(  2-x\right)  ^{m+1}}\left(
x-1\right)  ^{k+1}\right.  +\\
\left.  \frac{1-x+y}{2-x}\sum\limits_{s=0}^{m}\frac{\left(  1-y\right)  ^{s}%
}{\left(  2-x\right)  ^{s}}\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(
-1\right)  ^{j}\left(  2-x\right)  ^{j}\right]  ,
\end{multline*}
using binomial formula, we obtain%
\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}}{\left(  2-x\right)  ^{m+1}}%
\sum\limits_{j=0}^{k+1}\binom{k+1}{j}(2-x)^{j}(-1)^{j}\right.  +\\
\sum\limits_{s=0}^{m}\frac{\left(  1-y\right)  ^{s}}{\left(  2-x\right)  ^{s}%
}\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)  ^{j}\left(
2-x\right)  ^{j}-\\
\left.  \sum\limits_{s=0}^{m}\frac{\left(  1-y\right)  ^{s+1}}{\left(
2-x\right)  ^{s+1}}\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)
^{j}\left(  2-x\right)  ^{j}\right]  .
\end{multline*}
Now, for $k\leq m$%
\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}}{\left(
2-x\right)  ^{s}}\sum\limits_{j=0}^{m+1}\binom{k+1}{j}(2-x)^{j}(-1)^{j}%
\right.  +\\
\sum\limits_{s=0}^{m}\frac{\left(  1-y\right)  ^{s}}{\left(  2-x\right)  ^{s}%
}\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)  ^{j}\left(
2-x\right)  ^{j}-\\
\left.  \sum\limits_{s=0}^{m}\frac{\left(  1-y\right)  ^{s+1}}{\left(
2-x\right)  ^{s+1}}\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)
^{j}\left(  2-x\right)  ^{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}}{\left(
2-x\right)  ^{s}}\sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(  -1\right)
^{j}\left(  2-x\right)  ^{j}\right.  -\\
\left.  \sum\limits_{s=1}^{m+1}\frac{\left(  1-y\right)  ^{s}}{\left(
2-x\right)  ^{s}}\sum\limits_{j=0}^{s-1}\dbinom{k+1}{j}\left(  -1\right)
^{j}\left(  2-x\right)  ^{j}\right]
\end{multline*}%
\begin{multline*}
=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left[  1+\sum\limits_{s=1}^{m+1}\frac{\left(  1-y\right)  ^{s}}{\left(
2-x\right)  ^{s}}\left(  \sum\limits_{j=0}^{s}\dbinom{k+1}{j}\left(
-1\right)  ^{j}\left(  2-x\right)  ^{j}\right.  \right.  -\\
\left.  \left.  \sum\limits_{j=0}^{s-1}\dbinom{k+1}{j}\left(  -1\right)
^{j}\left(  2-x\right)  ^{j}\right)  \right]  .
\end{multline*}
Finally,%
\[
R^{\left(  m\right)  }(y)=\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
\left(  1+\sum\limits_{s=1}^{k+1}\dbinom{k+1}{s}\left(  y-1\right)
^{s}\right)  =\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
y^{k+1}.
\]


Now, according to (\ref{Ho2}) and (\ref{Ho5}), we have%
\begin{align*}
\alpha_{s}^{\left(  m\right)  }\left(  x\right)   &amp;  =\sum\limits_{i=0}%
^{s}\frac{\left(  -1\right)  ^{i+s}}{\left(  2-x\right)  ^{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}(m-s+i)!}{\left(
2-x\right)  ^{i+1}}%
\genfrac{\{}{\}}{0pt}{}{m+1}{m-s+i+1}%
\\
&amp;  =\sum\limits_{i=0}^{s}\frac{\left(  -1\right)  ^{i+s}}{\left(  2-x\right)
^{i+1}}W_{m,m-s+i}\\
&amp;  =\sum\limits_{j=m-s}^{m}\frac{\left(  -1\right)  ^{m+j}}{\left(
2-x\right)  ^{s-m+1+j}}W_{m,j},
\end{align*}
on the other side, it follows from (\ref{Ho3}) that
\begin{align*}
\alpha^{\left(  m\right)  }\left(  x\right)    &amp; =\sum\limits_{j=0}%
^{m-1}\left(  -1\right)  ^{m+j+1}\frac{(j+1)\left(  x-1\right)  }{\left(
2-x\right)  ^{j+2}}W_{m-1,j}\\
&amp; =\sum\limits_{j=0}^{m-1}\frac{\left(  -1\right)  ^{m+j}}{\left(  2-x\right)
^{j+1}}\left(  j+1\right)  W_{m-1,j}-\sum\limits_{j=0}^{m-1}\frac{\left(
-1\right)  ^{m+j}}{\left(  2-x\right)  ^{j+2}}\left(  j+1\right)  W_{m-1,j}\\
&amp; =\sum\limits_{j=0}^{m}\frac{\left(  -1\right)  ^{m+j}}{\left(  2-x\right)
^{j+1}}\left(  j+1\right)  W_{m-1,j}+\sum\limits_{j=0}^{m}\frac{\left(
-1\right)  ^{m+j}}{\left(  2-x\right)  ^{j+1}}jW_{m-1,j-1}.
\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{multline}%
{\displaystyle\int\limits_{0}^{x}}
\frac{\sum\limits_{k=0}^{m}%
\genfrac{\langle}{\rangle}{0pt}{}{m}{k}%
y^{k+1}}{\left(  1-y\right)  ^{m+1}\left(  1-x+y\right)  }dy=\label{g4}\\
-2\alpha^{\left(  m\right)  }\left(  x\right)  \ln\left(  1-x\right)
+\sum\limits_{s=0}^{m-1}\frac{\alpha_{s}^{\left(  m\right)  }\left(  x\right)
}{s-m}\left(  1-\left(  1-x\right)  ^{s-m}\right)  .
\end{multline}
By differentiating (\ref{g4}) we get the ordinary generating function of
$S_{n}^{\left(  m\right)  }$

\begin{multline}%
{\displaystyle\sum\limits_{n\geq0}}
S_{n}^{\left(  m\right)  }x^{n}=-2\frac{d}{dx}\alpha^{\left(  m\right)
}\left(  x\right)  \ln\left(  1-x\right)  +\frac{\alpha^{\left(  m\right)
}\left(  x\right)  }{1-x}+\label{generating}\\
\sum\limits_{s=0}^{m-1}\frac{\frac{d}{dx}\alpha_{s}^{\left(  m\right)
}\left(  x\right)  }{s-m}\left(  1-\left(  1-x\right)  ^{s-m}\right)
+\sum\limits_{s=0}^{m}\alpha_{s}^{\left(  m\right)  }\left(  x\right)  \left(
1-x\right)  ^{s-m-1}
\end{multline}
with
\[
\frac{d}{dx}\alpha_{s}^{\left(  m\right)  }\left(  x\right)  =\sum
\limits_{j=m-s}^{m}\frac{\left(  s-m+j+1\right)  \left(  -1\right)  ^{m+j}%
}{\left(  2-x\right)  ^{s-m+j+2}}W_{m,j}.%
\]
With Proposition \ref{gl1}, we can now rewrite (\ref{generating}) as follows
\begin{theorem}\label{rqh}
For any real number $x$ such that $\left\vert x\right\vert &lt;1$ and for all
positive integers $m\geq1$, we have
\begin{multline*}
{\displaystyle\sum\limits_{n\geq0}}S_{n}^{\left(  m\right)  }x^{n}=\left(
\sum\limits_{j=0}^{m}\frac{2\left(  j+1\right)  \left(  -1\right)  ^{m+j+1}%
}{\left(  2-x\right)  ^{j+2}}W_{m,j}\right)  \ln\left(  1-x\right)  \\
+\sum\limits_{0\leq j\leq s\leq m-1}\frac{\left(  -1\right)  ^{j}W_{m,m-j}%
}{\left(  2-x\right)  ^{s-j+1}}\left(  \dfrac{(s-j+1)\left(  1-\left(
1-x\right)  ^{s-m}\right)  }{\left(  s-m\right)  \left(  2-x\right)  }+\left(
1-x\right)  ^{s-m-1}\right)  \\
+\frac{1}{1-x}\sum\limits_{j=0}^{m}\frac{2\left(  -1\right)  ^{j+m}}{\left(
2-x\right)  ^{j+1}}W_{m,j}.
\end{multline*}

\end{theorem}
\section{Asymptotic expansion}

Yang and Zhao \cite{yang10} proved recently the asymptotic
expansions for $S_{n}^{(0)}$ and $S_{n}^{(1)}$ of the following type:%

\begin{equation}
S_{n}^{\left(  0\right)  }\sim2+\frac{2}{n-1}-\frac{1}{2^{n-1}},n\rightarrow
\infty,\label{wx}%
\end{equation}
and
\[
S_{n}^{\left(  1\right)  }\sim\frac{n}{2}\left(  2+\frac{2}{n-1}-\frac
{1}{2^{n-1}}\right)  ,n\rightarrow\infty.
\]

\bigskip

\noindent However, there are finer asymptotics which we proceed to
discuss now. We note that the following claim is consistent with
Theorem \ref{recur}.

\[
S_{n}^{\left(  m\right)  }\sim n^{m},\ \ n\rightarrow\infty.
\]

\noindent
It is not difficult to observe that $ T_{n}^{(m)}:=
{\displaystyle\sum\limits_{k=0}^{n-m}} k^{m}\binom{n}{k}^{-1}$ also
converges as $n \rightarrow \infty$ and, we prove alongside the
above asymptotic formula that this limit is $m!$.

\begin{theorem} For $m&gt;0$, we have
\[
\underset{n\rightarrow\infty}{\lim} T_{n}^{(m)}=m!,
\]
and
\[
\underset{n\rightarrow\infty}{\lim}
\frac{S_{n}^{(m)}}{n^m} =1.
\]

\end{theorem}

\begin{proof}
Let us denote the sum $S_{n}^{\left(  0\right)  }$ by $S_n$ for simplicity just in this proof. One has
the recurrence relation $S_n = \frac{n+1}{2n} S_{n-1} + 1$ (see
\cite{sury93}, for instance). Using this, it can be shown that
\[
S_n \rightarrow 2,\]
\[
n(S_n-2) \rightarrow 2,\]
\[
 n(n(S_n-2)-2) = n^2S_n-2n^2-2n \rightarrow 4,
\]
and so on. More generally, there
are constants $a_0,a_1,a_2,\ldots$ such that, recursively
\[
u^{(k)}_n = nu^{(k-1)}_n-a_{k-1},
\]
\[ u^{(k)}_n \rightarrow 0\ as\ n
\rightarrow \infty,
\]
\noindent
where $u^{(0)}_n = S_n-2$.\\
In other words, for all $k \geq 1$,
$$n^kS_n-(2n^k+a_0n^{k-1}+ \cdots
+ a_{k-2}n) \rightarrow a_k,$$ as $n \rightarrow \infty$, for some
$a_k$.\\
In fact, the sequence $a_0,a_1,\ldots$ is $2,4,16,88,616,5224,\ldots$ is
described by the generating function
$$\sum_{n\geq 0} a_n \frac{x^n}{n!} = \frac{2}{(2-e^x)^2}.$$
The constants can also be defined as
\[
a_{k}={\displaystyle\sum\limits_{r=0}^{k}}
\left(  r+2\right)  W_{k,r}-\left(  r+1\right)  W_{k-1,r},
\]
\noindent
or recursively as
$$a_k = 4 + \sum_{r=1}^k \left({k+1 \choose r} + {k \choose r}\right)(-1)^r
(2-a_0+a_1- \cdots \pm{a_{r-1}}) + (-1)^{k+1}(2-a_0+ \cdots
\pm{a_{k-1}}).$$
This is seen from
\begin{align*}
2n^{k+1}S_n &amp; = n^k(2nS_n) \\
&amp; =n^k((n+1)S_{n-1}+2n) \\
&amp; = (n^{k+1}+n^k)S_{n-1}+2n^{k+1} \\
&amp; = \sum_{r=1}^{k}
\left( {k+1 \choose r} + {k \choose r}\right) (n-1)^rS_{n-1}+ 2n^{k+1}.
\end{align*}

\noindent Write
\begin{equation}\label{decomp}
X^m = c_0+ c_1(X+1) + c_2(X+1)(X+2) + \cdots
+ c_m(X+1)(X+2) \cdots (X+m),
\end{equation}
where $c_i$'s depend on $m$ ($c_m=1$).
\smallskip
\noindent We have
\begin{align*}
T_{n}^{(m)} &amp; = \sum_{k=0}^{n-m}
\left( c_0+ c_1(k+1) + c_2(k+1)(k+2) + \cdots + c_m(k+1)(k+2) \cdots
(k+m)\right) \binom{n}{k}^{-1}\\
&amp; =
c_0 \sum_{k=0}^{n-m} \binom{n}{k}^{-1} + c_1(n+1) \sum_{k=0}^{n-m}
\binom{n+1}{k+1}^{-1} +\cdots + c_m(n+1) \cdots (n+m)
\sum_{k=0}^{n-m} \binom{n+m}{k+m}^{-1}\\
&amp; = c_0\left(S_n - \sum_{k=n-m+1}^n {n \choose k}^{-1}\right)
+ c_1(n+1)\left(S_{n+1}- \sum_{k=n-m+2}^{n+1} {n+1 \choose k}^{-1} - 1\right) +
\cdots \\ &amp; \quad \cdots + c_m(n+1) \cdots (n+m)\left(S_{n+m} - \sum_{k=n+1}^{n+m} {n+m
\choose k}^{-1} - \sum_{k=0}^{m-1} {n+m \choose k}^{-1}\right).
\end{align*}

\noindent A computation using the above result
$$n^kS_n - (2n^k+a_0n^{k-1} + \cdots + a_{k-1}n+a_k) \rightarrow 0,$$
shows
$$\lim_{n \rightarrow \infty} T_{n}^{(m)}= 0!c_0 + 1!c_1+ 2!c_2+ \cdots + (m-1)!c_{m-1}+ 2(m!)c_m.$$
This can be shown by induction on $m$. The proof is finished by
observing that $c_m=1$ and
$$0!c_0 + 1!c_1 + 2!c_2 + \cdots + m!c_m = 0,$$
which follows by looking at the constant term of (\ref{decomp}).
This finishes the proof of the first assertion. For
the second one, we proceed similarly to the above discussion. We
have
\begin{align*}
S_n^{(m)} &amp; = \sum_{k=0}^{n}
\left( c_0+ c_1(k+1) + c_2(k+1)(k+2) + \cdots + (k+1)(k+2) \cdots
(k+m)\right) \binom{n}{k}^{-1} \\
&amp; = c_0 S_n + c_1(n+1)\left(S_{n+1}-2\right)+ c_2(n+1)(n+2)\left(S_{n+2}-2 \sum_{k=0}^1
{n+2 \choose k}^{-1}\right) +
\cdots  \\&amp; \ \cdots  + (n+1) \cdots (n+m)\left(S_{n+m}-2 \sum_{k=0}^{m-1} {n+m \choose
k}^{-1}\right).
\end{align*}

Thus, looking at the largest degree factor, we obtain
$$S_n^{(m)} \sim n^m.$$

\end{proof}

\smallskip

\noindent A corollary of the above proof is that the sum $$S_n^{(0)}
= 2+ \frac{2}{n} + \frac{4}{n^2} + \frac{16}{n^3} + \frac{88}{n^4} +
\frac{616}{n^5} + \frac{5224}{n^6} + \cdots ,$$ as $n \rightarrow
\infty $.

\bigskip

\section{Acknowledgment}

The authors are indebted to the referee for making
constructive suggestions for improvement. The asymptotic result
$S_n^{(m)} \sim n^m$ was pointed out by the referee and we have
revised the paper to replace the earlier asymptotics by it. The
referee has a rather different approach to the results of this paper
and we look forward to seeing his proof in print soon.

The first and the second author were partially supported by CMEP
project 09 MDU 765.

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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, recurrence
relation, generating function, asymptotic expansion.

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

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\vspace*{+.1in}
\noindent
Received December 1 2010;
revised versions received December 8 2010; May 28 2011.
Published in {\it Journal of Integer Sequences}, June 10 2011.

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\noindent
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