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\begin{center}
{\LARGE Some Properties of Associated Stirling Numbers}
\end{center}

\begin{center}
Feng-Zhen Zhao\\
Department of Applied Mathematics\\
Dalian University of Technology\\
Dalian 116024 \\
China \\
\href{mailto:fengzhenzhao@yahoo.com.cn}{\tt fengzhenzhao@yahoo.com.cn}
\end{center}

\begin{abstract}
In this paper, we discuss the properties of associated Stirling
numbers. By means of the method of coefficients, we establish a series
of identities involving associated Stirling numbers, Bernoulli
numbers, harmonic numbers, and the Cauchy numbers of the first kind.
In addition, we give the asymptotic expansion of certain sums
involving $2$-associated Stirling numbers of the second kind by
Darboux's method.
\end{abstract}


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\section{Introduction}

Stirling numbers are generalized by many forms. See for instance
\cite{ref1, ref2, ref3, ref4, ref5} and \cite{ref9}. In this paper,
we are interested in associated Stirling numbers. The associated
Stirling numbers of the first kind $s_2(n, k)$ \cite{ref3} are given
by
\begin{eqnarray*}
\sum_{n=k}^{\infty}s_2(n, k)\frac{t^n}{n!}=\frac{[\ln(1+t)-t]^k}{k!}
\end{eqnarray*}
and the $r$-associated Stirling numbers of the second kind $S_r(n, k)$
\cite{ref3} are given by
\begin{eqnarray*}
\sum_{n=k}^{\infty}S_r(n,
k)\frac{t^n}{n!}=\frac{1}{k!}\bigg[e^t-\sum_{j=0}^{r-1}\frac{t^j}{j!}\bigg]^k,
\end{eqnarray*}
where $k$ and $r$ are positive integers. It is clear that
\begin{eqnarray*}
\sum_{n=k}^{\infty}S_2(n,
k)\frac{t^n}{n!}&amp;=&amp;\frac{(e^t-1-t)^k}{k!},\\
\sum_{n=k}^{\infty}S_3(n,
k)\frac{t^n}{n!}&amp;=&amp;\frac{(e^t-1-t-t^2/2)^k}{k!}.
\end{eqnarray*}
Like the ordinary Stirling numbers, the associated Stirling numbers also play
important roles in combinatorics. For example, $|s_2(n, k)|$
equals the number of
derangements of a set ${\it N}$ ($|{\it N}|=n$), with $k$ orbits, and
$S_r(n, k)$ is the number of partitions of the set ${\it N}$ ($|{\it
N}|=n$), into $k$ blocks, all of cardinality $\geq r$. It is clear
that $S_1(n, k)$ is the Stirling number of the second kind $S(n,
k)$. Therefore, associated Stirling numbers deserve to be
investigated. The aim of this paper is to investigate the properties
of associated Stirling numbers by making use of the method of
coefficients \cite{ref7}. We establish a series of identities
relating associated Stirling numbers with Bernoulli, harmonic, and
Cauchy numbers of the first kind. In addition, we give the
asymptotic expansion of certain sums involving $r$-associated
Stirling numbers by Darboux's method.

The paper is organized as follows. In Section 2, we establish a
series of identities involving associated Stirling, Bernoulli,
harmonic and Cauchy numbers of the first kind. In Section 3, we give
the asymptotic expansion of certain sums involving $r$-associated
Stirling numbers by Darboux's method.

For convenience, we recall some definitions of combinatorial numbers
involved in the paper. Throughout, we denote Stirling numbers of the
first kind by $s(n, k)$, and let $B_n$, $B_n^{(k)}$, and $E_n$ stand
for Bernoulli, generalized Bernoulli, and Euler numbers
respectively. That is,
\begin{eqnarray*}
&amp;&amp;\sum_{n=k}^{\infty}s(n,
k)\frac{t^n}{n!}=\frac{\ln^k(1+t)}{k!},\quad\quad
\sum_{n=0}^{\infty}B_n\frac{t^n}{n!}=\frac{t}{e^t-1}, \\
&amp;&amp;\sum_{n=0}^{\infty}B_n^{(k)}\frac{t^n}{n!}=\frac{t^k}{(e^t-1)^k}
\quad (k\geq 1),
\quad\sum_{n=0}^{\infty}E_n\frac{t^n}{n!}=\frac{2}{e^t+e^{-t}}.
\end{eqnarray*}
The Cauchy numbers of the first kind $a_n$ are given by
\begin{eqnarray*}
\sum_{n=0}^{\infty}a_n\frac{t^n}{n!}=\frac{t}{\ln(1+t)}.
\end{eqnarray*}
The harmonic numbers $H_n$ are given by
\begin{eqnarray*}
\sum_{n=1}^{\infty}H_nt^n=-\frac{\ln(1-t)}{1-t}.
\end{eqnarray*}
In this paper, $[t^n]f(t)$ denotes the coefficient of $t^n$ in
$f(t)$, where
$$
f(t)=\sum_{n=0}^{\infty}f_nt^n.
$$
The expression $[t^n]$  is called the {\it ``coefficient of'' functionals}
\cite{ref7}. If $f(t)$ and $g(t)$ are formal power series, the
following relations hold \cite{ref7}:
\begin{eqnarray}
&amp;&amp;[t^n](\alpha f(t)+\beta g(t))=\alpha[t^n]f(t)+\beta[t^n]g(t), \label{pf-1.1}\\
&amp;&amp;[t^n]tf(t)=[t^{n-1}]f(t), \label{pf-1.2}\\
&amp;&amp;[t^n]f^{\prime}(t)=(n+1)[t^{n+1}]f(t), \label{pf-1.3}\\
&amp;&amp;[t^n]f(t)g(t)=\sum_{k=0}^n([y^k]f(y))[t^{n-k}]g(t). \label{pf-1.4}
\end{eqnarray}
In Section 2, we obtain a series of identities related to associated
Stirling numbers by using (\ref{pf-1.1})-(\ref{pf-1.4}).
%\mbox{}\\[4pt]

%\begin{flushleft}
\section{Identities involving associated Stirling,
Bernoulli, and harmonic numbers}
%\end{flushleft}

Bernoulli numbers and harmonic numbers are important in
combinatorics, and Stirling numbers are related to them. From
\cite{ref3}, we know that Stirling numbers and Bernoulli numbers
satisfy
\begin{eqnarray*}
\sum_{j=0}^n\frac{(-1)^jj!S(n, j)}{j+1}=B_n, \ \ \sum_{j=0}^ns(n,
j)B_j=\frac{(-1)^nn!}{n+1}.
\end{eqnarray*}
By the generating functions of $S_2(n, k)$, $S(n, k)$, and $B_n$,
we observe that $S_2(n, k)$ is also related to $B_n$, and we have

\begin{thm} For $n\geq 1$ and $k\geq 1$, $S_2(n, k)$, $B_n$, and $S(n, k)$
satisfy the equations
\begin{eqnarray}
\sum_{j=0}^nS_2(n-j+k, k){n+k\choose
j}B_j&amp;=&amp;(n+k)\sum_{j=1}^k\frac{(-1)^{k-j}}{j}{n+k-1\choose
k-j}S(n+j-1,
j-1)\nonumber\\
&amp;&amp;+(-1)^k{n+k\choose k}B_n, \label{pf-2.1}
\end{eqnarray}
\begin{eqnarray}
\sum_{j=0}^n{n+k-1\choose j}S_2(n-j+k, k)B_j&amp;=&amp;(n+k-1)S_2(n+k-2,
k-1) \ \ k\geq 2. \label{pf-2.2}
\end{eqnarray}
\end{thm}

\begin{proof}
By the definitions of $S_2(n, k)$, $B_n$, and $S(n , k)$, we have
\begin{eqnarray*}
\sum_{j=0}^nS_2(n-j+k, k){n+k\choose
j}B_j&amp;=&amp;(n+k)!\sum_{j=0}^n\frac{S_2(n-j+k, k)}{(n-j+k)!}
\cdot\frac{B_j}{j!}
\\
&amp;=&amp;(n+k)!\sum_{j=0}^n[t^{n-j+k}]\frac{(e^t-1-t)^k}{k!}[t^j]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k)!\sum_{j=0}^n[t^{n-j}]\frac{(e^t-1-t)^k}{k!t^k}[t^j]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k)![t^n]\frac{(e^t-1-t)^kt}{k!t^k(e^t-1)}\\
&amp;=&amp;(n+k)![t^n]\sum_{j=0}^k(-1)^{k-j}{k\choose
j}\frac{(e^t-)^{j-1}t^{-j+1}}{k!}\\
&amp;=&amp;\frac{(-1)^k(n+k)!}{k!}[t^n]\frac{t}{e^t-1}+(n+k)!\sum_{j=1}^k[t^n]\frac{(-1)^{k-j}(e^t-1)^{j-1}
}{j(k-j)!(j-1)!t^{j-1}}\\
&amp;=&amp;(-1)^k{n+k\choose
k}B_n+(n+k)!\sum_{j=1}^k\frac{(-1)^{k-j}S(n+j-1,
j-1)}{j(k-j)!(n+j-1)!}.
\end{eqnarray*}
Then (\ref{pf-2.1}) holds.

Now we give the proof of (\ref{pf-2.2}).
\begin{eqnarray*}
\sum_{j=0}^n{n+k-1\choose j}S_2(n-j+k,
k)B_j&amp;=&amp;(n+k-1)!\sum_{j=0}^n\frac{S_2(n-j+k,
k)}{(n-j+k-1)!}\cdot\frac{B_j}{j!}\\
&amp;=&amp;(n+k-1)!\sum_{j=0}^n\frac{(n-j+k)S_2(n-j+k,
k)}{(n-j+k)!}\cdot\frac{B_j}{j!}\\
&amp;=&amp;(n+k-1)!\sum_{j=0}^n(n-j+k)[t^{n-j+k}]\frac{(e^t-1-t)^k}{k!}[t^j]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k-1)!\sum_{j=0}^n[t^{n-j+k-1}]\frac{(e^t-1-t)^{k-1}(e^t-1)}{(k-1)!}[t^j]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k-1)!\sum_{j=0}^n[t^{n-j}]\frac{k(e^t-1-t)^{k-1}(e^t-1)}{t^{k-1}k!}[t^j]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k-1)![t^n]\frac{(e^t-1-t)^{k-1}}{t^{k-2}(k-1)!}\\
&amp;=&amp;(n+k-1)![t^{n+k-2}]\frac{(e^t-1-t)^{k-1}}{(k-1)!}\\
&amp;=&amp;(n+k-1)S_2(n+k-2, k-1).
\end{eqnarray*}
This completes the proof.
\end{proof}

Formula (\ref{pf-2.1}) relates associated Stirling, Bernoulli, and
Stirling numbers of the second kind.

The generating functions of generalized Bernoulli numbers
$B_n^{(k)}$ implies that they are related to associated Stirling
numbers. For $S_2(n, k)$ and $B_n^{(k)}$, we get

\begin{cor} For $n\geq 1$ and $k\geq 1$, $2$-associated Stirling
numbers $S_2(n, k)$ and generalized Bernoulli numbers $B_n^{(k)}$
satisfy
\begin{eqnarray}
\sum_{j=0}^nS_2(n-j+k, k){n+k\choose j}B_j^{(k)}={n+k\choose
k}\sum_{j=0}^k(-1)^{k-j}{k\choose j}B_n^{(k-j)}. \label{pf-2.3}
\end{eqnarray}
\end{cor}
\begin{proof}
\begin{eqnarray*}
\sum_{j=0}^nS_2(n-j+k, k){n+k\choose
j}B_j^{(k)}&amp;=&amp;(n+k)!\sum_{j=0}^n\frac{S_2(n-j+k,
k)}{(n-j+k)!}\cdot\frac{B_j^{(k)}}{j!}\\
&amp;=&amp;(n+k)!\sum_{j=0}^n[t^{n-j+k}]\frac{(e^t-1-t)^k}{k!}[t^j]\frac{t^k}{(e^t-1)^k}\\
&amp;=&amp;\frac{(n+k)!}{k!}\sum_{j=0}^n[t^{n-j}]\frac{(e^t-1-t)^k}{t^k}[t^j]\frac{t^k}{(e^t-1)^k}\\
&amp;=&amp;\frac{(n+k)!}{k!}[t^n]\frac{(e^t-1-t)^k}{(e^t-1)^k}\\
&amp;=&amp;\frac{(n+k)!}{k!}\sum_{j=0}^k(-1)^{k-j}{k\choose
j}\frac{B_n^{(k-j)}}{n!}.
\end{eqnarray*}
Hence (\ref{pf-2.3}) holds.
\end{proof}

For $S_3(n, k)$ and Bernoulli numbers $B_n$, we have

\begin{thm} For $n\geq k$ and $k\geq 1$, $S_3(n, k)$ and $B_n$
satisfy
\begin{eqnarray}
\sum_{j=0}^n{n+k\choose j+k}S_3(j+k,
k)B_{n-j}&amp;=&amp;(n+k)!\sum_{j=1}^k\frac{(-1)^{k-j}}{j(k-j)!}\sum_{j_1=0}^{k-j}{k-j\choose
j_1}\frac{S(n-j_1+j-1, j-1)}{2^{j_1}(n-j_1+j-1)!}\nonumber\\
&amp;&amp;+\frac{(-1)^k(n+k)!}{k!}\sum_{j=0}^k\frac{B_{n-j}}{2^j(n-j)!}{k\choose
j}. \label{pf-2.4}
\end{eqnarray}
\end{thm}

\begin{proof}
From the generating functions of $S_3(n, k)$ and $B_n$, we have
\begin{eqnarray*}
\sum_{j=0}^n{n+k\choose j+k}S_3(j+k,
k)B_{n-j}&amp;=&amp;(n+k)!\sum_{j=0}^n\frac{S_3(j+k, k)}{(j+k)!}
\frac{B_{n-j}}{(n-j)!}\\
&amp;=&amp;(n+k)!\sum_{j=0}^n[t^{j+k}]\frac{(e^t-1-t-t^2/2)^k}{k!}[t^{n-j}]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k)!\sum_{j=0}^n[t^j]\frac{(e^t-1-t-t^2/2)^k}{t^kk!}[t^{n-j}]\frac{t}{e^t-1}\\
&amp;=&amp;(n+k)![t^n]\frac{(e^t-1-t-t^2/2)^kt}{k!t^k(e^t-1)}\\
&amp;=&amp;(n+k)![t^n]\bigg(\sum_{j=0}^k(-1)^{k-j}{k\choose
j}\frac{(e^t-1)^j(t+t^2/2)^{k-j}t}{k!(e^t-1)t^k}\bigg)\\
&amp;=&amp;(n+k)![t^n]\frac{(-1)^k(1+t/2)^kt}{k!(e^t-1)}\\
&amp;&amp;+[t^n]\sum_{j=1}^k(-1)^{k-j}{k\choose
j}\frac{(e^t-1)^{j-1}(t+t^2/2)^{k-j}t}{k!t^k}\\
&amp;=&amp;(-1)^k(n+k)!\sum_{j=0}^k{k\choose j}\frac{B_{n-j}}{2^j(n-j)!k!}\\
&amp;&amp;+\frac{(n+k)!}{k!}[t^n]\sum_{j=1}^k(-1)^{k-j}{k\choose
j}(e^t-1)^{j-1}\sum_{j_1=0}^{k-j}{k-j\choose
j_1}\frac{t^{j_1-j+1}}{2^{j_1}}\\
&amp;=&amp;\frac{(-1)^k(n+k)!}{k!}\sum_{j=0}^k{k\choose
j}\frac{B_{n-j}}{2^j(n-j)!}\\
&amp;&amp;+(n+k)!\sum_{j=1}^k\frac{(-1)^{k-j}}{j(k-j)!}\sum_{j_1=0}^{k-j}{k-j\choose
j_1}\frac{S(n-j_1+j-1, j-1)}{2^{j_1}(n-j_1+j-1)!}
\end{eqnarray*}
Then (\ref{pf-2.4}) holds.
\end{proof}

There are many identities relating Stirling numbers of the first
kind and harmonic numbers in \cite{ref3}. For example,
\begin{eqnarray*}
(-1)^{n+1}s(n+1, 2)&amp;=&amp;n!H_n, \\
(-1)^ns(n+1, 3)&amp;=&amp;\frac{n!}{2}(H^2_n-H_n^{(2)}),\\
(-1)^{n+1}s(n+1, 4)&amp;=&amp;\frac{n!}{6}(H^2_n-3H_nH_n^{(2)}+2H_n^{(3)}),
\end{eqnarray*}
where $H_n^{(s)}=1+2^{-s}+3^{-s}+\cdots+n^{-s}$.

For associated Stirling numbers of the first kind and harmonic
numbers, we can prove
\begin{thm} For $n\geq 1$ and $k\geq 1$, we have
\begin{eqnarray}
\sum_{j=0}^n\frac{(-1)^jH_{j+1}s_2(n-j+k,
k)}{(j+2)(n-j+k)!}&amp;=&amp;\frac{(-1)^k}{2}\sum_{j=0}^k\frac{(-1)^j(j+1)(j+2)s(n+j+2,
j+2)}{(k-j)!(n+j+2)!}.\nonumber \\
\label{pf-2.5}
\end{eqnarray}
\end{thm}
\begin{proof}
By integrating the generating function for $H_n$ we have
\begin{eqnarray*}
\sum_{n=0}^{\infty}\frac{H_{n+1}t^n}{n+2}=\frac{\ln^2(1-t)}{2t^2}.
\end{eqnarray*}
One can verify that
\begin{eqnarray*}
\frac{[\ln(1-t)+t]^k\ln^2(1-t)}{2(-1)^kk!t^{k+2}}=\frac{(-1)^k}{2k!}\sum_{j=0}^k{k\choose
j}\frac{\ln^{j+2}(1-t)}{t^{j+2}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
[t^n]\frac{[\ln(1-t)+t]^k\ln^2(1-t)}{2(-1)^kk!t^{k+2}}&amp;=&amp;\sum_{j=0}^n[t^{n-j+k}]
\frac{[\ln(1-t)+t]^k}{(-1)^kk!}[t^j]\frac{\ln^2(1-t)}{2t^2}\\
&amp;=&amp;\sum_{j=0}^n\frac{(-1)^{n-j}s_2(n-j+k,
k)H_{j+1}}{(n-j+k)!(j+2)}\\
&amp;=&amp;\frac{(-1)^k}{2k!}\sum_{j=0}^k{k\choose
j}[t^n]\frac{\ln^{j+2}(1-t)}{t^{j+2}},
\end{eqnarray*}
\begin{eqnarray*}
\sum_{j=0}^n\frac{(-1)^{n-j}H_{j+1}s_2(n-j+k,
k)}{(j+2)(n-j+k)!}=\frac{(-1)^{n+k}}{2}\sum_{j=0}^k\frac{(-1)^j(j+1)(j+2)s(n+j+2,
j+2)}{(k-j)!(n+j+2)!}.
\end{eqnarray*}
Hence (\ref{pf-2.5}) holds.
\end{proof}

There are some identities involving Stirling numbers and Cauchy
numbers of the first kind. For example
\begin{eqnarray*}
\sum_{j=0}^na_jS(n, j)=\frac{1}{n+1}, \ \ a_n=\sum_{j=0}^n\frac{s(n,
j)}{j+1}.
\end{eqnarray*}
See \cite{ref3, ref6} for more details. For associated Stirling
numbers of the first kind and the Cauchy numbers of the first kind,
we have
\begin{thm} For $n\geq 1$ and $k\geq 1$, $s_2(n, k)$ and $a_n$
satisfy
\begin{eqnarray}
\sum_{j=0}^ns_2(n-j+k, k){n+k\choose
j}a_j&amp;=&amp;(n+k)\sum_{j=1}^k\frac{(-1)^{k-j}}{j}{n+k-1\choose
k-j}s(n+j-1,
j-1)\nonumber\\
&amp;&amp;+(-1)^k{n+k\choose k}a_n. \label{pf-2.6}
\end{eqnarray}
\end{thm}

The proof of (\ref{pf-2.6}) is similar to that of (\ref{pf-2.1}) and
is omitted here.

 Formula (\ref{pf-2.6}) relates
associated Stirling numbers and Cauchy numbers.
%\mbox{}\\[4pt]

%\begin{flushleft}
\section{Asymptotic Expansion of Certain Sums Involving $2$-associated Stirling numbers of
the second kind, Bernoulli numbers, and Euler Numbers }
%\end{flushleft}

We know that it is difficult to compute the accurate values of
certain sums involving $r$-associated Stirling numbers. However,
sometimes we can give their asymptotic expansion. In this section,
we give asymptotic expansion of certain sums for $2$-associated
Stirling numbers of the second kind, Bernoulli numbers, and Euler
numbers by Darboux's
method. We first recall a lemma (see\cite{ref8}):\\
{\bf Lemma:} Assume that $f(t)=\sum_{n\geq 0}a_nt^n$ is an analytic
function for $|t|&lt;r$ and with a finite number of algebraic
singularities on the circle $|t|=r$. $\alpha_1$, $\alpha_2$,
$\cdots$, $\alpha_l$ are singularities of order $\omega$, where
$\omega$ is the highest order of all singularities. Then
\begin{eqnarray}
a_n=(n^{\omega-1}/\Gamma(\omega))\times\bigg(\sum_{k=1}^lg_k(\alpha_k)\alpha^{-n}_k+\mbox{o}(r^{-n})\bigg),
\label{pf-3.1}
\end{eqnarray}
where $\Gamma(\omega)$ is the gamma function, and
$$
g_k(\alpha_k)=\lim_{t\rightarrow
\alpha_k}(1-(t/\alpha_k))^{\omega}f(t).
$$
By using (\ref{pf-3.1}), we obtain
\begin{thm} Suppose that $n\geq 1$ and $k\geq 1$, where $k$ is fixed. When $n\to\infty$, we have
\begin{eqnarray}
\sum_{p+q=2n}\frac{S_2(p+k,
k)B_q}{(p+k)!q!}&amp;\sim&amp;\frac{2(-1)^{n+k+1}}{(2\pi)^{2n}k!},
\label{pf-3.2}\\
\sum_{p+q=n}\frac{S_2(p+k,
k)E_q}{(p+k)!q!}&amp;\sim&amp;\frac{2^{n+1}[(2+2i-\pi)^ki^{-n}
+(2-2i-\pi)^k(-i)^{-n}]}{\pi^{n+k+1}k!}.\label{pf-3.3}
\end{eqnarray}
\end{thm}
\begin{proof}
Because the proof of (\ref{pf-3.3}) is similar to that of
(\ref{pf-3.2}), we only prove that (\ref{pf-3.2}) holds. It is clear
that
\begin{eqnarray*}
\sum_{p=0}^{\infty}S_2(p+k,
k)\frac{t^p}{(p+k)!}\sum_{q=0}^{\infty}B_q\frac{t^q}{q!}&amp;=&amp;\frac{(e^t-1-t)^k}{k!t^{k-1}(e^t-1)}.
\end{eqnarray*}
Let
\begin{eqnarray*}
f(t)=\frac{(e^t-1-t)^k}{k!t^{k-1}(e^t-1)}.
\end{eqnarray*}
Then $f(t)$ is analytic for $|t|&lt;2\pi$ and with two algebraic
singularities on the circle $|t|=2\pi$. $\alpha_1=2\pi i$ and
$\alpha_2=-2\pi i$ are singularities of order $1$. One can compute
that
\begin{eqnarray*}
\lim_{t\to2\pi i}\bigg(1-\frac{t}{2\pi
i}\bigg)f(t)&amp;=&amp;\lim_{t\to-2\pi i}\bigg(1+\frac{t}{2\pi
i}\bigg)f(t)\\
&amp;=&amp;\frac{(-1)^{k+1}}{k!}.
\end{eqnarray*}
It follows from (\ref{pf-3.1}) that
\begin{eqnarray*}
\sum_{p+q=n}\frac{S_2(p+k,
k)B_q}{(p+k)!q!}=\frac{1}{\Gamma(1)}\bigg\{\frac{(-1)^{k+1}[(2\pi
i)^{-n}+(-2\pi i)^{-n}]}{k!}+\mbox{o}((2\pi)^{-n})\bigg\}.
\end{eqnarray*}
Then we have
$$
\sum_{p+q=2n}\frac{S_2(p+k,
k)B_q}{(p+k)!q!}\sim\frac{(-1)^{k+1}[i^{2n}+(-i)^{2n}]}{(2\pi)^{2n}k!}
$$
Hence (\ref{pf-3.2}) holds.
\end{proof}

\section{Acknowledgments}

The author would like to thank the
anonymous referee for his criticism and useful suggestions.

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\bibliographystyle{plain}

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\bibitem{ref6} D. Merlini, R. Sprugnoli, and M. C.
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 05A16, 05A19.

\noindent \emph{Keywords: }
Stirling numbers, Bernoulli numbers, harmonic
numbers, asymptotic expansion, Darboux's method.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  December 8 2007;
revised version received May 2 2008.
Published in {\it Journal of Integer Sequences},
May 6 2008. Minor revisions, June 11 2008, June 20 2008, July 5 2008.


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