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\begin{center}
\vskip 1cm{\LARGE\bf Identities Involving Generalized Harmonic \\
\vskip 0.01in
Numbers and Other Special Combinatorial \\
\vskip 0.01in
Sequences \\
\vskip .10in } \vskip 1cm \large Huyile Liang\footnote{This work was
supported by the Science Research Foundation of Inner Mongolia
(2012MS0118) and the National Natural Science Foundation of China
(NSFC Grant \#11061020).} and Wuyungaowa\\
Department of Mathematics\\
College of Sciences and Technology\\
Inner Mongolia University\\
Hohhot 010021 \\
P. R. China\\
\href{mailto:lianghuyile@gmail.com}{\tt lianghuyile@gmail.com } \\
\href{mailto:wuyungw@163.com}{\tt wuyungw@163.com} \\
\end{center}


\vskip .2 in

\begin{abstract}
In this paper, we study the properties of the generalized harmonic
numbers $H_{n,k,r}(\alpha,\beta)$. In particular, by means of the
method of coefficients, generating functions and Riordan arrays, we
establish some identities involving the numbers
$H_{n,k,r}(\alpha,\beta)$, Cauchy numbers, generalized Stirling
numbers, Genocchi numbers and  higher order Bernoulli numbers.
Furthermore, we obtain the asymptotic values of some summations
associated with the numbers $H_{n,k,r}(\alpha,\beta)$ by Darboux's
method and Laplace's method.
\end{abstract}

\section{Introduction}
Harmonic numbers are important in various branches of combinatorics
and number theory, and they also frequently appear in the analysis
of algorithms and expressions for special functions. Recently, many
papers have been devoted to the study of harmonic number identities
by various methods; see, for instance, \cite{ref1, ref2, ref3, ref4,
ref5, ref6, ref7}. We recall the definition of harmonic
numbers $H_n=\sum_{k=1}^{n}\frac{1}{k}$ for $n \geq 0$.
The generating function of $H_n$ is
$\sum_{n=1}^{\infty}H_nt^n=-\frac{\ln(1-t)}{1-t}$. In this paper, we
discuss a class of generalized harmonic numbers
$H_{n,k,r}(\alpha,\beta)$.  We refer to Zhao and Wuyungaowa
\cite{ref10} for this topic. The definition of
 $H_{n,k,r}(\alpha,\beta)$ is
\begin{equation}
 \sum_{n=0}^{\infty}H_{n,k,r}(\alpha,\beta)t^n=\frac{(-\ln(1-\alpha t))^r}{(1-\beta
 t)^k} ,   \label{af-1.1}
\end{equation}
where ${k\geq}1$ and ${r\geq}1$ are integers. Let~$(\alpha,\beta)$
be a pair of real numbers and (${\alpha\beta\neq0}$). From the
generating function of $H_{n,k,r}(\alpha,\beta)$, we know that
$H_{n,1,1}(1,1)=H_n ({n\geq0})$.

From (\ref{af-1.1}) we obtain
\begin{align}
   &{(-\ln(1-\alpha t))^r=(1-\beta t)^k\sum_{n=0}^{\infty}H_{n,k,r}(\alpha,\beta)t^n}\nonumber\\
  &=\sum_{i=0}^{k}\binom{k}{i}(-\beta t)^i\sum_{n=0}^{\infty}H_{n,k,r}(\alpha,\beta)t^n\nonumber\\
  &=\sum_{n=0}^{\infty}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^hH_{n-h,k,r}(\alpha,\beta)t^n. \label{af-1.2}
\end{align}


For convenience, let us recall some definitions and notations.
Denote the generalized Stirling
numbers of the first kind by $s(n, k; r)$, and the generalized
Stirling numbers of the second kind
 by $S(n, k; r)$. Denote further $C_n^{(k)}$,
$\hat{C}_n^{(k)}$, $B_n^{(r)}$, $G_n^{(x)}$, $G_n^{(k)}$ be the
higher order Cauchy numbers of both kinds, higher order Bernoulli
numbers, the generalized Genocchi numbers, the higher order Genocchi
numbers and the generalized Lah numbers. These numbers satisfy the
following generating functions respectively:
\begin{align}
& \sum_{n=k}^{\infty}s(n, k; r)\frac{(-1)^{n-k}t^n}{n!}=\frac{\ln^k(1+t)}{(1+t)^rk!},\quad k=0,1,2,\ldots, \label{af-1.3}\\
& \sum_{n=k}^{\infty}|s(n, k; r)|\frac{t^n}{n!}=\frac{(-\ln(1-t))^k}{(1-t)^rk!},\quad k=0,1,2,\ldots, \label{af-1.4}\\
& \sum_{n=k}^{\infty}S(n, k; r)\frac{t^n}{n!}=\frac{e^{rt}(e^t-1)^k}{k!},\quad k=0,1,2,\ldots, \label{af-1.5}\\
&  \sum_{n=0}^{\infty}C_n^{(k)}\frac{t^n}{n!}=\bigg(\frac{t}{\ln(1+t)}\bigg)^k,\label{af-1.6}\\
 &\sum_{n=k}^{\infty}\hat{C}_n^{(k)}\frac{t^n}{n!}=\bigg(\frac{t}{(1+t)\ln(1+t)}\bigg)^k,\label{af-1.7}\\
& \sum_{n=0}^{\infty}G_n^{(k)}\frac{t^n}{n!}=\bigg(\frac{2t}{e^t+1}\bigg)^k,\label{af-1.8}\\
& \sum_{n=0}^{\infty}\frac{G_n^{(x)}}{2^n}\frac{t^n}{n!}=\bigg(\frac{2}{e^t+1}\bigg)^x,\label{af-1.9}
\end{align}
\begin{align}
& \sum_{n=0}^{\infty}B_n^{(r)}\frac{t^n}{n!}=\bigg(\frac{t}{e^t-1}\bigg)^r,\label{af-1.10}\\
& \sum_{n=k}^{\infty}L(n,k;r)\frac{t^n}{n!}=(1+t)^r\frac{1}{k!}\bigg(\frac{-t}{1+t}\bigg)^k,\label{af-1.11}\\
 &\sum_{n=0}^{\infty}H_{n}^{(r)}(z)t^n=\frac{(-\ln(1-t))^{r+1}}{t(1-t)^{1-z}},\label{af-1.12}\\
&\sum_{n=1}^{\infty}H_{n}^{(r)}t^n=\frac{-\ln(1-t)}{(1-t)^r}.\label{af-1.13}
\end{align}

Let $[t^n]f(t)$ be the coefficient of $t^n$ in the formal power
series of $f(t)$, where $ f(t)=\sum_{n=0}^{\infty}f_nt^n$. (See
Merlini, Sprugnoli, and Verri \cite{ref8} for related topics.) If
$f(t)$ and $g(t)$ are formal power series, we get the following
relations:

\begin{align}
&[t^n](\alpha f(t)+\beta g(t)) =\alpha [t^n]f(t)+\beta [t^n]g(t), \label{af-1.14}\\
&[t^n]f(t)=[t^{n-1}]f(t), \label{af-1.15}\\
&[t^n]f(t)g(t)=\sum_{j=0}^{n}[y^j]f(y)[t^{n-j}]g(t). \label{af-1.16}
\end{align}

A {\it Riordan array} is a pair $(d(t),h(t))$ of formal power series with
$h_0=h(0)=0$. It defines an infinite lower triangular array
$(d_{n,k})_{n,k\in N}$ according to the rule
\[
d_{n,k}=[t^n]d(t)(h(t))^k\,.
\]
Hence we write $\{d_{n,k}\}=(d(t),h(t))$. Moreover, if $(d(t),h(t))$
is a Riordan array and $f(t)$ is the generating function of the
sequence $\{f_k\}_{k\in N}$, i.e., $f(t)=\sum_{k=0}^{\infty}f_kt^k$,
then we have
\begin{align}
 &\sum_{k=0}^{\infty}d_{n,k}f_k=[t^n]d(t)f(h(t))=[t^n]d(t)[f(y)\mid y=h(t)]. \label{af-1.17}
\end{align}

Furthermore, based on the generating function (\ref{af-1.1}) we
obtain the next three Riordan arrays:
\begin{align}
 &\{H_{n,k,r}(\alpha, \beta)\}=\bigg(\frac{1}{(1-\beta t)^k}, \frac{-\ln(1-\alpha t)}{t}\bigg), \label{af-1.18}\\
 &\{H_{n,k,{r+1}}(\alpha, \beta)\}=\bigg(\frac{-\ln(1-\alpha t)}{(1-\beta t)^k}, \frac{-\ln(1-\alpha t)}{t}\bigg), \label{af-1.19}\\
 &\{H_{n,k,r}(\alpha, \alpha)\}=\bigg(\frac{1}{(1-\alpha t)^k}, \frac{-\ln(1-\alpha t)}{t}\bigg). \label{af-1.20}
\end{align}
In this paper, we pay particular attention to the three
Riordan arrays above.

\section{Identities involving {$H_{n,k,r}(\alpha,\beta)$}, $s(n, k; r)$, $S(n, k; r)$, $B_n^{(r)}$ and $L(n,k;r)$}

\begin{theorem}
Let $ k,~r,~m\geq1$, $l\geq0$ be integers. Then
\begin{align}
   &\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)|s(n-j, m; l)|\frac{\alpha^{n-j}}{(n-j)!}
   =\frac{1}{m!}H_{n,l,{m+r}}(\alpha,\alpha). \label{af-2.1}
\end{align}
\end{theorem}
\begin{proof}
By applying (\ref{af-1.2}), (\ref{af-1.4}) and (\ref{af-1.16}), we
get
\begin{align*}
&\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)|s(n-j, m; l)|\frac{\alpha^{n-j}}{(n-j)!}\\
   &=[t^n]\frac{(-\ln(1-\alpha t))^{m+r}}{(1-\alpha t)^l
   m!}=\frac{1}{m!}H_{n,l,m+r}(\alpha,\alpha)\,.
\end{align*}
\end{proof}


\begin{theorem}
Let $n,~k,~j\geq 1$, $l\geq 0$ be integers. Then
\begin{equation}
  \sum_{j=m}^{n}H_{n,k,j}(\alpha,\beta)S(j, m; l)\frac{m!}{j!}=\sum_{i=0}^{n-m}\binom{i+k-1}{i}\binom{n-i+l-1}{n-m-i}\beta^i \alpha^{n-i}. \label{af-2.2}
\end{equation}
\end{theorem}

\begin{proof} By using (\ref{af-1.5}), (\ref{af-1.17}) and (\ref{af-1.18}), we
obtain
\begin{align*}
&\sum_{j=m}^{n}H_{n,k,j}(\alpha,\beta)S(j, m; l)\frac{m!}{j!}
   =[t^n]\frac{1}{(1-\beta t)^k} [(e^y-1)^me^{yl}\mid y=-\ln(1-\alpha t)]\\
 &=\alpha^m[t^{n-m}]\frac{1}{(1-\beta t)^k}\frac{1}{(1-\alpha t)^{l+m}}
 =\sum_{i=0}^{n-m}\binom{i+k-1}{i}\binom{n-i+l-1}{n-m-i}\beta^i \alpha^{n-i}\,.
\end{align*}
\end{proof}

\begin{corollary}
The following relations hold:
\begin{align}
&\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)|s(n-j,
m)|\frac{\alpha^{n-j}}{(n-j)!}
=\frac{\alpha^n}{m!}|s(n, m+r)|\frac{(m+r)!}{n!}, \label{af-2.3}\\
  &\sum_{j=m}^{n}H_{n,k,j}(\alpha,\alpha)S(j, m; l)\frac{m!}{j!}=\binom{n+l+k-1}{n-m}\alpha^n, \label{af-2.4}\\
   &\sum_{j=m}^{n}H(n,r-1)S(j, m; l)\frac{m!}{j!}=\binom{n+l}{n-m}. \label{af-2.5}
\end{align}
\end{corollary}
\begin{proof}
Setting $l=0$ in (\ref{af-2.1}), we get (\ref{af-2.3}). Setting
$\beta=\alpha$ in (\ref{af-2.2}), we have (\ref{af-2.4}). 
Setting $\beta=\alpha=k=1$ in (\ref{af-2.2}), we obtain
(\ref{af-2.5}).
\end{proof}

\begin{theorem}
Let $n,~k\geq 1$ and $l,~j\geq 0$ be integers. Then
\begin{align}
  &\sum_{j=m}^{n}H_{n,k,j+1}(\alpha,\beta)S(j, m;
  l)\frac{m!}{j!}=\sum_{i=0}^{n-m}H_{i,k,1}(\alpha,\beta)\binom{n-i+l-1}{n-m-i}\alpha^{n-i}. \label{af-2.6}
\end{align}
\end{theorem}
\begin{proof} By applying (\ref{af-1.5}), (\ref{af-1.17}) and (\ref{af-1.19}), we have
\begin{align*}
 &\sum_{j=m}^{n}H_{n,k,j+1}(\alpha,\beta)S(j, m; l)\frac{m!}{j!}
=[t^n]\frac{-\ln(1-\alpha t)}{(1-\beta t)^k}
    [(e^y-1)^me^{yl}\mid y=-\ln(1-\alpha t)]\\
 &=\alpha^m[t^{n-m}]\frac{-\ln(1-\alpha t)}{(1-\beta t)^k}\frac{1}{(1-\alpha t)^{l+m}}
 =\sum_{i=0}^{n-m}H_{i,k,1}(\alpha,\beta)\binom{n-i+l-1}{n-m-i}\alpha^{n-i}\,.
\end{align*}
\end{proof}


\begin{corollary}
The following relations hold:
\begin{align}
 &\sum_{j=m}^{n}H_{n,k,j+1}(\alpha,\alpha)S(j, m; l)\frac{m!}{j!}=\alpha^n H_{n-m-1}(1-m-l-k), \label{af-2.7}\\
 &\sum_{j=m}^{n}H_{n,k,j+1}(\alpha,\alpha)S(j, m; l)\frac{m!}{j!}=\alpha^n H_{n-m}^{(m+l+k)}. \label{af-2.8}
\end{align}
\end{corollary}

\begin{proof}
Setting $\beta=\alpha$ in (\ref{af-2.6}), we obtain (\ref{af-2.7})
and (\ref{af-2.8}).
\end{proof}

\begin{theorem}
Let $k,~r\geq 1$ and $m,~l\geq 0$ be integers. Then
\begin{align}
   \sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\frac{L(n-j,m;l)}{(n-j)!}(-\alpha)^{n-j}\nonumber \\
    =\begin{cases}
            \sum_{i=r}^{l-m}\binom{l-m}{i}|s(n-m-i, r)|\frac{(-1)^i \alpha^n  r!}{m!(n-m-i)!}, & \text{if $l>m$; }\\
            \frac{\alpha^n }{m!}|s(n-m, r)|\frac{r!}{(n-m)!}, & \text{if $l=m$; }\\
                \frac{\alpha^m}{m!}{H}_{n-m,m-l,r}(\alpha,\alpha), & \text{if $l<m$. }
   \end{cases}\label{af-2.9}
\end{align}
\end{theorem}


\begin{proof} By applying (\ref{af-1.2}), (\ref{af-1.11}) and (\ref{af-1.16}), we get
\begin{align*}
&\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\frac{L(n-j,m;l)}{(n-j)!}(-\alpha)^{n-j}\\
&=[t^n](-\ln(1-\alpha t))^{r}\frac{(1-\alpha t)^l}{m!}\bigg(\frac{\alpha t}{1-\alpha t}\bigg)^m\\
  & =\begin{cases}
        \sum_{i=r}^{l-m}\binom{l-m}{i}|s(n-m-i, r)|\frac{(-1)^i \alpha^n r!}{m!(n-m-i)!}, & \text{if $l>m$; }\\
             \frac{\alpha^n }{m!}|s(n-m, r)|\frac{r!}{(n-m)!}, & \text{if $l=m$; }\\
                \frac{\alpha^m}{m!}{H}_{n-m,m-l,r}(\alpha,\alpha), & \text{if $l<m$. }
   \end{cases}
\end{align*}
\end{proof}


\begin{theorem}
Let $n,~j,~k,~m\geq 1$ be integers. Then
\begin{align}
 \sum_{j=1}^{n}{H}_{n,k,j}(\alpha,\alpha)\frac{B_j^{(m)}}{j!}
=\begin{cases}
       \frac{1}{\alpha^m}{H}_{n+m,k-m,m}(\alpha,\alpha), & \text{if $k>m$; }\\
            \alpha^n |s(n+m, m)|\frac{m!}{(n+m)!}, & \text{if $k=m$; }\\
                \sum_{i=0}^{m-k}\binom{m-k}{i}|s(n+m-i, m)|\frac{(-1)^i\alpha^n m! }{(n+m-i)!}, & \text{if $k<m$. }\label{af-2.10}
    \end{cases}
\end{align}
\end{theorem}
\begin{proof} By applying (\ref{af-1.10}), (\ref{af-1.17}) and (\ref{af-1.20}), we get
 \begin{align*}
\sum_{j=1}^{n}{H}_{n,k,j}(\alpha,\alpha)\frac{B_j^{(m)}}{j!}
  &=[t^n]\frac{1}{(1-\alpha t)^k}\left[\bigg(\frac{y}{e^y-1}\bigg)^m \mid y=-\ln(1-\alpha  t)\right]\\
&=\begin{cases}
       \frac{1}{\alpha^m}{H}_{n+m,k-m,m}(\alpha,\alpha), & \text{if $k>m$; }\\
            \alpha^n |s(n+m, m)|\frac{m!}{(n+m)!}, & \text{if $k=m$; }\\
                \sum_{i=0}^{m-k}\binom{m-k}{i}|s(n+m-i, m)|\frac{(-1)^i\alpha^n m! }{(n+m-i)!}, & \text{if $k<m$. }
    \end{cases}
\end{align*}
\end{proof}


\section{Identities involving {$ H_{n,k,r}(\alpha,\beta)$}, Genocchi numbers and Cauchy numbers }
\begin{theorem}
 Let $x$ be a real number, $ k,~i\geq 1$, $n\geq0$ be integers. Then
\begin{equation}
\sum_{i=1}^{n}H_{n,k,i}(\alpha,\beta)\frac{G_{i}^{(x)}}{2^ii!}=\sum_{h=0}^{x}\sum_{j=0}^{n-h}
 \binom{n-h-j+k-1}{k-1}\binom{j+x-1}{x-1}\binom{x}{h}\frac{(-1)^h}{2^j}\beta^{n-j-h}\alpha^{j+h}. \label{af-3.1}
\end{equation}
\end{theorem}
\begin{proof} By using (\ref{af-1.9}), (\ref{af-1.17}) and (\ref{af-1.18}), we get
\begin{align*}
\sum_{i=1}^{n}H_{n,k,i}(\alpha,\beta)\frac{G_{i}^{(x)}}{2^ii!}
&=[t^n]\frac{1}{(1-\beta t)^k}
    \left[\bigg(\frac{2}{e^y+1}\bigg)^x\ \Big|\ y=-\ln(1-\alpha    t)\right]\\
    &=[t^n]\frac{1}{(1-\beta t)^k}\frac{1}{(1-\frac{\alpha t}{2})^x}(1-\alpha t)^x\\
    &=\sum_{h=0}^{x}\sum_{j=0}^{n-h}\binom{n-h-j+k-1}{k-1}\binom{j+x-1}{x-1}\binom{x}{h}\frac{(-1)^h}{2^j}\beta^{n-j-h}\alpha^{j+h}\,.\\
\end{align*}
\end{proof}

\begin{theorem}
Let  $x$ be a real number, $k,~j\geq 1$ be integers. Then
\begin{align}
\sum_{j=1}^{\infty}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(x)}}{2^jj!}
    =\begin{cases}
            \sum_{i=0}^{n}\binom{i+k-x-1}{i}\binom{x+n-i-1}{n-i}\frac{\alpha^n}{2^{n-i}}, & \text{if $k>x$; }\\
            \binom{n+k-1}{n} \frac{\alpha^n}{2^n}, & \text{if $k=x$; }\\
                \sum_{i=0}^{x-k}\binom{x-k}{i}\binom{x+n-i-1}{n-i}\frac{(-1)^i\alpha^n}{2^{n-i}}, & \text{if $k<x$. }\label{af-3.2}
        \end{cases}
\end{align}
\end{theorem}

\begin{proof} By using (\ref{af-1.9}), (\ref{af-1.17}) and (\ref{af-1.20}), we have
\begin{align*}
\sum_{j=1}^{\infty}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(x)}}{2^jj!}
&=[t^n]\frac{1}{(1-\alpha t)^k}
    \left[\bigg(\frac{2}{e^y+1}\bigg)^x\ \Big|\ y=-\ln(1-\alpha t)\right]\\
 &=\begin{cases}
            \sum_{i=0}^{n}\binom{i+k-x-1}{i}\binom{x+n-i-1}{n-i}\frac{\alpha^n}{2^{n-i}}, & \text{if $k>x$; }\\
            \binom{n+k-1}{n} \frac{\alpha^n}{2^n}, & \text{if $k=x$; }\\
                \sum_{i=0}^{x-k}\binom{x-k}{i}\binom{x+n-i-1}{n-i}\frac{(-1)^i\alpha^n}{2^{n-i}}, & \text{if $k<x$. }
        \end{cases}
\end{align*}
\end{proof}


\begin{corollary}
The following relations hold:
\begin{align}
 &\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}}{2^jj!}= \sum_{i=0}^{n}\binom{i+k-2}{i}\frac{\alpha^n}{2^{n-i}}, \label{af-3.3}\\
 &\sum_{j=1}^{n}H(n,j-1)\frac{G_{j}}{2^jj!}=\frac{1}{2^n}. \label{af-3.4}
\end{align}
\end{corollary}

\begin{proof}
Setting $x=1$ in (\ref{af-3.2}), we have (\ref{af-3.3}). Setting
$x=k=1$ in (\ref{af-3.2}), we obtain (\ref{af-3.4}).
\end{proof}

\begin{theorem}
Let $k,~l,~m\geq 1$ be integers. Then
\begin{equation}
\sum_{l=1}^{n}H_{n,k,l}(\alpha,\beta)\frac{G_{l}^{(m)}}{l!}=\sum_{j=0}^{m}\sum_{i=0}^{n-j}H_{i,k,m}(\alpha,\beta)\binom{m}{j}
\binom{n-i-j+m-1}{m-1}\frac{(-1)^j\alpha^{n-i}}{2^{n-i-j}}. \label{af-3.5}\\
\end{equation}
\end{theorem}
\begin{proof} By applying (\ref{af-1.8}), (\ref{af-1.17}) and (\ref{af-1.18}), we have
\begin{align*}
\sum_{l=1}^{n}H_{n,k,l}(\alpha,\beta)\frac{G_{l}^{(m)}}{l!}
&=[t^n]\frac{1}{(1-\beta t)^k}
    \left[\bigg(\frac{2y}{e^y+1}\bigg)^m\ \Big|\ y=-\ln(1-\alpha    t)\right]\\
    &=[t^n]\frac{(-\ln(1-\alpha t))^m}{(1-\beta t)^k}\frac{1}{(1-\frac{\alpha t}{2})^m}(1-\alpha t)^m\\
    &=\sum_{j=0}^{m}\sum_{i=0}^{n-j}H_{i,k,m}(\alpha,\beta)\binom{m}{j}
\binom{n-i-j+m-1}{m-1}\frac{(-1)^j\alpha^{n-i}}{2^{n-i-j}}\,.\\
\end{align*}
\end{proof}

\begin{theorem}
Let $n,~k,~j,~m\geq 1$ be integers. Then
\begin{align}
\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(m)}}{j!}
    =\begin{cases}
            \sum_{i=0}^{n}H_{n-i,k-m,m}(\alpha,\alpha)\binom{i+m-1}{m-1}\frac{\alpha^i}{2^i}, & \text{if $k>m$; }\\
            \sum_{i=m}^{n}|s(i, m)|\frac{m!}{i!}\binom{n-i+m-1}{m-1} \frac{\alpha^n}{2^{n-i}}, & \text{if $k=m$; }\\
                \sum_{i=0}^{m-k}\sum_{j=0}^{n-i}\binom{m-k}{i}\binom{j+m-1}{m-1}|s(n-i-j, m)|\frac{(-1)^i\alpha^nm!}{2^j (n-i-j)!}, & \text{if $k<m$. }
        \end{cases}\label{af-3.6}
\end{align}
\end{theorem}

\begin{proof} By using (\ref{af-1.8}), (\ref{af-1.17}) and (\ref{af-1.20}), we have
\begin{align*}
 \sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(m)}}{j!}&=[t^n]\frac{1}{(1-\alpha t)^k}
    \left[\bigg(\frac{2y}{e^y+1}\bigg)^m\ \Big|\ y=-\ln(1-\alpha t)\right]\\
 &=\begin{cases}
            \sum_{i=0}^{n}H_{n-i,k-m,m}(\alpha,\alpha)\binom{i+m-1}{m-1}\frac{\alpha^i}{2^i}, & \text{if $k>m$; }\\
            \sum_{i=m}^{n}|s(i, m)|\frac{m!}{i!}\binom{n-i+m-1}{m-1} \frac{\alpha^n}{2^{n-i}}, & \text{if $k=m$; }\\
                \sum_{i=0}^{m-k}\sum_{j=0}^{n-i}\binom{m-k}{i}\binom{j+m-1}{m-1}|s(n-i-j, m)|\frac{(-1)^i\alpha^nm!}{2^j (n-i-j)!}, & \text{if $k<m$. }
        \end{cases}
\end{align*}
\end{proof}


\begin{theorem}
Let $n,~k,~r,~m\geq 1$ be integers. Then
\begin{align}
 \sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\frac{(-\alpha)^{n-j}{C}_{n-j}^{(m)}}{(n-j)!}
=\begin{cases}
            \alpha^n|s(n-m, r-m)|\frac{(r-m)!}{(n-m)!}, & \text{if $r>m$; }\\
            \alpha^m\delta_{n,m}, & \text{if $r=m$; }\\
                \frac{(-1)^{n-r}\alpha^n}{(n-r)!}{C}_{n-r}^{(m-r)}, & \text{if $r<m$. }
 \end{cases}\label{af-3.7}
\end{align}
\end{theorem}

\begin{proof}By using (\ref{af-1.2}), (\ref{af-1.6}) and (\ref{af-1.16}), we have
\begin{align*}
\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\frac{(-\alpha)^{n-j}{C}_{n-j}^{(m)}}{(n-j)!}
&=[t^n](-\ln(1-\alpha t))^{r}\bigg(\frac{-\alpha t}{\ln(1-\alpha t)}\bigg)^m\\
 &=\begin{cases}
            \alpha^n|s(n-m, r-m)|\frac{(r-m)!}{(n-m)!}, & \text{if $r>m$; }\\
            \alpha^m\delta_{n,m}, & \text{if $r=m$; }\\
                \frac{(-1)^{n-r}\alpha^n}{(n-r)!}{C}_{n-r}^{(m-r)}, & \text{if $r<m$. }
 \end{cases}
\end{align*}
\end{proof}

\begin{theorem}
Let $n,~k,~r,~m\geq 1$ be integers. Then
\begin{align}
 \sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\frac{(-\alpha)^{n-j}\hat{C}_{n-j}^{(m)}}{(n-j)!}\nonumber\\
        =\begin{cases}
            \alpha^m H_{n-m,m,r-m}(\alpha,\alpha), & \text{if $r>m$; }\\
            \alpha^n\binom{n-1}{n-m}, & \text{if $r=m$; }\\
                 \sum_{i=m-r}^{n-r}\hat{C}_{n-r-i}^{(m-r)}\frac{(-1)^{n-r-i}\alpha^n}{(n-r-i)!}\binom{i+r-1}{r-1}, & \text{if $r<m$. }
 \end{cases}
\label{af-3.8}
\end{align}
\end{theorem}

\begin{proof}The proof of (\ref{af-3.8}) is similar to that of (\ref{af-3.7}),
and it is omitted here.
\end{proof}

\begin{theorem}
Let $ k,~r\geq 1$ be integers. Then
\begin{align}
 &\sum_{j=0}^{n}{H}_{j,k,r}(\alpha,\beta)\frac{(-\alpha)^{n-j}{C}_{n-j}}{(n-j)!}=\alpha {H}_{n-1,k,r-1}(\alpha,\beta),\label{af-3.9}\\
 &\sum_{j=0}^{n}{H}_{j,k,r}(\alpha,\beta)\frac{(-\alpha)^{n-j}\hat{C}_{n-j}}{(n-j)!}=\sum_{i=0}^{n-1}{H}_{i,k,r-1}(\alpha,\beta)\alpha^{n-i}.\label{af-3.10}
\end{align}
\end{theorem}

\begin{proof}
By applying (\ref{af-1.1}), (\ref{af-1.6}) and (\ref{af-1.16}), we get
\begin{align*}
 &\sum_{j=0}^{n}{H}_{j,k,r}(\alpha,\beta)\frac{(-\alpha)^{n-j}{C}_{n-j}}{(n-j)!}
 =[t^n]\frac{(-\ln(1-\alpha t))^{r}}{(1-\beta t)^{k}}\frac{\alpha t}{-\ln(1-\alpha t)}=\alpha {H}_{n-1,k,r-1}(\alpha,\beta)\,.
\end{align*}
Hence (\ref{af-3.9}) holds. The proof of (\ref{af-3.10}) is similar
to that of (\ref{af-3.9}), and it is omitted here.
\end{proof}

\begin{corollary} Let $n,~r\geq2$. Then
\begin{align}
 &\sum_{j=r}^{n}\frac{H(j,r-1)(-\alpha)^{n-j}{C}_{n-j}}{(n-j)!}=\alpha^nH(n-1,r-2),\label{af-3.11}\\
 &\sum_{j=r}^{n}\frac{H(j,r-1)(-\alpha)^{n-j}\hat{C}_{n-j}}{(n-j)!}=\alpha^{n}\sum_{i=r}^{n-1}H(i-1,r-2).\label{af-3.12}
\end{align}
\end{corollary}

\begin{proof}
Setting $\beta=\alpha$ and $k=1$ in (\ref{af-3.9}) and
(\ref{af-3.10}), we obtain (\ref{af-3.11}) and (\ref{af-3.12}).
\end{proof}



\section{Asymptotics}

In this section, we give the asymptotic expansion of certain
sums involving $H_{n,k,r}(\alpha, \beta)$. We first recall three
lemmas.

A singularity of $f(z)$ at $|z|=w$ is called {\it algebraic} if $f(z)$ can
be written as the sum of a function analytic in a neighborhood of
$w$ and a finite number of terms of the form
\begin{align}
(1-\frac{z}{w})^\alpha g(z), \label{af-4.1}
\end{align}
where $g(z)$ is analytic near $w$, $g(w)\neq 0$ and $\alpha\not\in\{
0,1,2, \ldots \}$.

\begin{lemma} \label{af-4.2} (See \cite{ref8}.) 
Suppose that $f(z)$ is analytic for $|z|<R$, $R>0$
and has only algebraic singularities on $|z|=R$. Let $a$ be the
minimum of $\Re(\alpha)$ for the terms of the form at the
singularity of $f(z)$ on $|z|=R$, and let $w_j$, $\alpha_j$ and
$g_j(z)$ be the $w,\alpha$ and $g(z)$ for those terms of the form
(\ref{af-4.1}) for which $Re(\alpha)=a$. Then, as
$n\rightarrow\infty$,
\begin{align*}
[z^n]f(z)=\sum_{j}\frac{g_j(w_j)n^{-\alpha_j-1}}
    {\Gamma(-\alpha_j)w_j^n}+o(R^{-n}n^{-a-1}).
\end{align*}
\end{lemma}

\begin{lemma}\label{af-4.3} (see \cite{ref11})
Let $\alpha$ be a real number and
\[
L(z)=\ln(\frac{1}{1-z})\,.
\]
When $n\rightarrow\infty$,
\begin{align*}
&[z^n](1-z)^{\alpha}L^k(z)\sim
    \frac{1}{\Gamma(-\alpha)}n^{-\alpha-1}\ln^{k}n\,,
    \quad(\alpha\not\in\{0,1,2,\ldots\})\\
&[z^n](1-z)^{m}L^{k}(z)\sim(-1)^{m}k{m!}n^{-m-1}\ln^{k-1}n\,,
    \quad(m\in\mathbb{Z}_{\geq0},~k\in\mathbb{Z}_{\geq1}).
\end{align*}
\end{lemma}


\begin{lemma} \label{af-4.4} (see \cite{ref11}) Suppose that $a(z)=\sum a_nz^n$ and $b(z)=\sum
b_nz^n$ are power series with radii of convergence
$\alpha>\beta\geq0$, respectively. Suppose that
$\frac{b_{n-1}}{b_n}\rightarrow \beta$ as $n\rightarrow\infty$. If
$a(\beta)\neq0$ and $\sum c_nz^n=a(z)b(z)$, then
\begin{align*}
 & c_n\sim a(\beta)b_n\quad\text{as }n\rightarrow\infty.
\end{align*}
\end{lemma}

\begin{theorem}
Let $k,~r\geq1$. As $n\rightarrow \infty$, we get
\begin{equation*}
   \sum_{h=0}^{n}\binom{k}{h}(-\beta)^h{H}_{n-h,k,r}(\alpha,\beta)\sim{\frac{\alpha^n r}{n}}\ln^{r-1}n\,.
\end{equation*}
\end{theorem}

\begin{proof}
By Eq.~(\ref{af-1.2}) and Lemma~\ref{af-4.3}, we get
\begin{align*}
  &\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{n-h,k,r}(\alpha,\beta)
  =[t^n](-\ln(1-\alpha t))^r\sim{\frac{\alpha^n r}{n}}\ln^{r-1}n\,.
 \end{align*}
\end{proof}

\begin{theorem}
Let $k,~r,~m\geq 1$. As $n\rightarrow\infty$, we have
\begin{align*}
&\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h
{H}_{j-h,k,r}(\alpha,\beta)|s(n-j,
m)|\frac{\alpha^{n-j}}{(n-j)!}\sim\frac{\alpha^n (m+r)}{m! n}(\ln
n)^{m+r-1}\,.
\end{align*}
\end{theorem}

\begin{proof}
By Eq.~(\ref{af-2.3}) and Lemma \ref{af-4.3}, we obtain
 \begin{align*}
&\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h
  {H}_{j-h,k,r}(\alpha,\beta)|s(n-j, m)|\frac{\alpha^{n-j}}{(n-j)!}=\frac{1}{m!}[t^n](-\ln(1-\alpha
  t))^{m+r}\\
 &\sim\frac{\alpha^n (m+r)}{m! n}(\ln
n)^{m+r-1}\,.
\end{align*}
\end{proof}

\begin{theorem}
Let $k,~j,~m\geq 1$ and $l\geq 0$. As $n\rightarrow\infty$, we have
\begin{align*}
&\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)S(j, m;
l)\frac{m!}{j!}\sim\frac{\alpha ^n n^{k+m+l-1}}{\Gamma(k+l+m)}\,.
\end{align*}
\end{theorem}
\begin{proof}By Eq. (\ref{af-2.4}) and Lemma \ref{af-4.2}, we obtain
 \begin{align*}
 \sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)S(j, m; l)\frac{m!}{j!}=\alpha^m[t^n]\frac{t^m}{(1-\alpha t)^{k+l+m}}
 \sim\frac{\alpha ^n n^{k+m+l-1}}{\Gamma(k+l+m)}\,.
\end{align*}
\end{proof}


\begin{theorem}
Let $k,~r\geq1$. As $n\rightarrow \infty$, we get
\begin{equation*}
   \sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\binom{n-j+l}{n-j}\beta^{n-j}\sim{\frac{\alpha^n n^l}{l!}\ln^r n}\,.
\end{equation*}
\end{theorem}

\begin{proof}
It is well known that
\begin{align}
 &\sum_{n=0}^{\infty}\binom{n+l}{n}t^n=\frac{1}{(1-t)^{l+1}}. \label{af-4.5}
 \end{align}
By Lemma \ref{af-4.3} and Eq.~(\ref{af-4.5}), we get
\begin{align*}
  &\sum_{j=0}^{n}\sum_{h=0}^{k}\binom{k}{h}(-\beta)^h{H}_{j-h,k,r}(\alpha,\beta)\binom{n-j+l}{n-j}\alpha^{n-j}=[t^n]\frac{(-\ln(1-\alpha t))^r}{(1-\alpha t)^{l+1}}\sim{\frac{\alpha^n n^l}{l!}\ln^r n}\,.
 \end{align*}
\end{proof}

\begin{theorem}
Let $j,~k,~m\geq 1$. As $n\rightarrow\infty$, we have
\begin{align*}
 &\sum_{j=1}^{n}{H}_{n,k,j}(\alpha,\alpha)\frac{B_j^{(m)}}{j!}
\sim\begin{cases}
            (-1)^{m-k} \alpha^n m (m-k)!(n+m)^{k-m-1}(\ln(n+m))^{m-1}, & \text{if $m-k\in\mathbb{Z}_{\geq0}$; }\\
               \frac{\alpha^n (n+m)^{k-m-1}}{\Gamma(m-k)}(\ln(n+m))^m, & \text{if $m-k\not\in\mathbb{Z}_{\geq0}$. }
    \end{cases}
\end{align*}
\end{theorem}
\begin{proof}
By the proof of Eq.~(\ref{af-2.8}) and Lemma \ref{af-4.3}, we obtain
 \begin{align*}
\sum_{j=1}^{n}{H}_{n,k,j}(\alpha,\alpha)\frac{B_j^{(m)}}{j!} =[t^n]\frac{1}{(1-\alpha t)^k}\left[\bigg(\frac{y}{e^y-1}\bigg)^m \mid y=-\ln(1-\alpha  t)\right]\\
\sim\begin{cases}
            (-1)^{m-k} \alpha^n m (m-k)!(n+m)^{k-m-1}(\ln(n+m))^{m-1}, & \text{if $m-k\in\mathbb{Z}_{\geq0}$; }\\
               \frac{\alpha^n (n+m)^{k-m-1}}{\Gamma(m-k)}(\ln(n+m))^m, & \text{if $m-k\not\in\mathbb{Z}_{\geq0}$. }
    \end{cases}
\end{align*}
\end{proof}

\begin{theorem}
Let $j$ be fixed and $x$ be a real number. As $n\rightarrow\infty$,
we have
\begin{equation*}
\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(x)}}{2^jj!}
    \sim\begin{cases}
            \frac{2^x \alpha^n n^{k-x-1}}{\Gamma(k-x)}, & \text{if $k>x$; }\\
            \frac{\alpha^n n^{k-1}}{2^n \Gamma(k)}, & \text{if $k=x$; }\\
                \frac{(-1)^{x-k}\alpha^n n^{x-1}}{2^n \Gamma(x)}, & \text{if $k<x$. }
        \end{cases}
\end{equation*}
\end{theorem}

\begin{proof} 
By the proof of (\ref{af-3.2}) and Lemma \ref{af-4.2}, we obtain
\begin{align*}
\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(x)}}{2^jj!}
=[t^n]\begin{cases}
        \frac{1}{(1-\alpha t)^{k-x}}\frac{1}{(1-\frac{\alpha t}{2})^x}, & \text{if $k>x$; }\\
        (1-\frac{\alpha t}{2})^x, & \text{if $k=x$; }\\
        \frac{(1-\alpha t)^{x-k}}{(1-\frac{\alpha t}{2})^x}, & \text{if $k<x$. }
    \end{cases}
\sim\begin{cases}
            \frac{2^x \alpha^n n^{k-x-1}}{\Gamma(k-x)}, & \text{if $k>x$; }\\
            \frac{\alpha^n n^{k-1}}{2^n \Gamma(k)}, & \text{if $k=x$; }\\
                \frac{(-1)^{x-k}\alpha^n n^{x-1}}{2^n \Gamma(x)}, & \text{if $k<x$. }
        \end{cases}
\end{align*}
\end{proof}

\begin{theorem}Let $j,~k,~m\geq 1$, $j$ be fixed.
As $n\rightarrow\infty$, we have
\begin{equation*}
\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(m)}}{j!}
    \sim\begin{cases}
        \frac{\alpha^n 2^m n^{k-m-1}\ln^{m+1}n}{\Gamma(k-m)}, & \text{if $k>m$; }\\
        \frac{2^m \alpha^n m}{n}\ln^{m-1}n, & \text{if $k=m$; }\\
        \frac{(-1)^{m-k}2^m \alpha^n(m-k)!}{n^{m+1-k}}\ln^m n, & \text{if $k<m$. }
    \end{cases}
\end{equation*}
\end{theorem}

\begin{proof}
By (\ref{af-3.6}), we see that
\begin{align*}
\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(m)}}{j!}
    =[t^n]\begin{cases}
          \frac{1}{(1-\alpha t)^{k-m}}\frac{1}{(1-\frac{\alpha t}{2})^m}(-\ln(1-\alpha t))^m, & \text{if $k>m$; }\\
        \frac{1}{(1-\frac{\alpha t}{2})^m}(-\ln(1-\alpha t))^m, & \text{if $k=m$; }\\
        (1-\alpha t)^{m-k}\frac{1}{(1-\frac{\alpha t}{2})^m}(-\ln(1-\alpha t))^m, & \text{if $k<m$. }
        \end{cases}
\end{align*}
Let $k=m $, and set
\[
a(t)=\frac{1}{(1-\frac{\alpha t}{2})^m }=\sum_{n=0}^{\infty}
\binom{n+m-1}{m-1}(\frac{\alpha t}{2})^n \text{ and }
b(t)=(-\ln(1-\alpha t))^m=\sum_{n=m}^{\infty} |s(n,
m)|\frac{m!}{n!}\alpha ^nt^n\,.
\]
in Lemma \ref{af-4.4}. Since $\frac{|s(n-1,
m)|\frac{m!}{(n-1)!}}{|s(n, m)|\frac{m!}{n!}}\rightarrow
\frac{1}{\alpha}$ as $n\rightarrow\infty$,
$a(\frac{1}{\alpha})=2^m\neq 0$.  Then we have
\[
\sum_{j=1}^{n}H_{n,k,j}(\alpha,\alpha)\frac{G_{j}^{(m)}}{j!}
    \sim\begin{cases}
        \frac{\alpha^n 2^m n^{k-m-1}\ln^{m+1}n}{\Gamma(k-m)}, & \text{if $k>m$; }\\
        \frac{2^m \alpha^n m}{n}\ln^{m-1}n, & \text{if $k=m$; }\\
        \frac{(-1)^{m-k}2^m \alpha^n(m-k)!}{n^{m+1-k}}\ln^m n, & \text{if $k<m$. }
    \end{cases}
\]
This gives the result for the case $k=m$. The proofs of the cases
$k>m$ and $k<m$ are similar to that of $k=m$.
\end{proof}

In the final result of this section, we give the asymptotic
expansion of certain sums for binomial coefficients and
$H_{n,k,r}(\alpha,\beta)$ by Laplace's method.

\begin{theorem}
Let $k,~r\geq1$, $0<\beta$, $(b+1)^{b+1}\neq\beta b^b$,
$(b+1)^{b+1}>\alpha b^b $ and b be a positive integer. As
$k\rightarrow \infty$, we get
\begin{align}
 &\sum_{n=0}^{\infty}\frac{H_{n,k,r}(\alpha,\beta)}{[(b+1)n+1]\binom{(b+1)n}{n}}\nonumber\\
 &\sim{\bigg(\frac{(b+1)^{b+1}}{(b+1)^{b+1}-\beta b^b}\bigg)^{k-\frac{1}{2}}
 \bigg(\ln\frac{(b+1)^{b+1}}{(b+1)^{b+1}-\alpha b^b}\bigg)^{r}\sqrt{\frac{2\pi(b+1)^{b-2}}{\beta
 kb^{b-1}}}}. \label{af-4.6}
\end{align}
\end{theorem}

\begin{proof}
From Tiberiu \cite{ref9}, we know that the inverse of a binomial
coefficient is related to an integral as follows:
\begin{equation}
 \binom{n}{m}^{-1}=(n+1)\int^{1}_{0}t^{m}(1-t)^{n-m}dt. \label{af-4.7}
\end{equation}

Owing to (\ref{af-4.7}), we have
\begin{align*}
&\sum_{n=0}^{\infty}\frac{H_{n,k,r}(\alpha,\beta)}{[(b+1)n+1]\binom{(b+1)n}{n}}
=\sum_{n=0}^{\infty}H_{n,k,r}(\alpha,\beta)\int^{1}_{0}t^n(1-t)^{nb}dt\\
&=\int^{1}_{0}\sum_{n=0}^{\infty}H_{n,k,r}(\alpha,\beta)(t(1-t)^b)^ndt
 =\int^{1}_{0}\frac{(-\ln(1-\alpha t(1-t)^b))^r}{(1-\beta
 t(1-t)^b)^k}dt\\
& =\int^{1}_{0}(-\ln(1-\alpha t(1-t)^b)^r e^{-k\ln(1-\beta
t(1-t)^b)} dt\,.
\end{align*}

Let $\varphi(t)=(-\ln(1-\alpha t(1-t)^b)^r $ and $h(t)=-\ln(1-\beta
t(1-t)^b)$. Then $h(t)$ reaches a maximum at $t=\frac{1}{b+1}$,
$h^{\prime}(\frac{1}{b+1})=0$ and $
h^{\prime\prime}(\frac{1}{b+1})<0$. By applying Laplace's method, we
have
\begin{align*}
  &\sum_{n=0}^{\infty}\frac{H_{n,k,r}(\alpha,\beta)}{[(b+1)n+1]\binom{(b+1)n}{n}}
  =\int^{1}_{0}(-\ln(1-\alpha t(1-t)^b)^r e^{-k\ln(1-\beta t(1-t)^b}dt\,\\
  & \sim{\varphi(\frac{1}{b+1})e^{kh(\frac{1}{b+1})}\sqrt{\frac{-2\pi}{k h^{\prime\prime}(\frac{1}{b+1})}}}\,.
\end{align*}
\end{proof}



\begin{corollary} 
Let $k,~r\geq1$, $\alpha<4$, $0<\beta$, $\beta\neq4 $ and $b$ be a
positive integer. As $k\rightarrow \infty$, we get
\begin{align}
    &\sum_{n=0}^{\infty}\frac{H_{n,k,r}(\alpha,\beta)}{(2n+1)\binom{2n}{n}}
         \sim{\bigg(\frac{4}{4-\beta}\bigg)^{k-\frac{1}{2}}\bigg(\ln\frac{4}{4-\alpha}\bigg)^r\sqrt{\frac{\pi}{k\beta}}}. \label{af-4.8}
\end{align}
\end{corollary}
\begin{proof}
Setting $b=1$ in (\ref{af-4.6}), we obtain (\ref{af-4.8}).
\end{proof}

\begin{theorem}
Let $k,~ r\geq1$, $0<\alpha$, $(b+1)^{b+1}\neq -\beta b^b$ and 
$b$ be a positive integer. As $r\rightarrow \infty$, we get
\begin{align}
 &\sum_{n=0}^{\infty}\frac{(-1)^nH_{n,k,r}(\alpha,\beta)}{[(b+1)n+1]\binom{(b+1)n}{n}}\nonumber\\
 &\sim(-1)^r{\bigg(\frac{(b+1)^{b+1}}{(b+1)^{b+1}+\beta b^b}\bigg)^k
 \bigg(\ln\frac{(b+1)^{b+1}+\alpha b^b}{(b+1)^{b+1}}\bigg)^{r+\frac{1}{2}}\sqrt{\frac{2\pi(b+1)^{b-2}((b+1)^{b+1}+\alpha b^b)}{ \alpha r
 b^{b-1}(b+1)^{b+1}}}}. \label{af-4.9}
\end{align}
\end{theorem}

\begin{proof} Owing to (\ref{af-4.7}), we get
\begin{align*}
&\sum_{n=0}^{\infty}\frac{(-1)^nH_{n,k,r}(\alpha,\beta)}{[(b+1)n+1]\binom{(b+1)n}{n}}
=\sum_{n=0}^{\infty}(-1)^nH_{n,k,r}(\alpha,\beta)\int^{1}_{0}t^n(1-t)^{nb}dt\\
&=\int^{1}_{0}\sum_{n=0}^{\infty}H_{n,k,r}(\alpha,\beta)(-t(1-t)^b)^ndt
 =\int^{1}_{0}\frac{(-\ln(1+\alpha t(1-t)^b))^r}{(1+\beta
 t(1-t)^b)^k}dt\\
& =(-1)^r\int^{1}_{0}\frac{e^{r\ln\ln(1+\alpha t(1-t)^b}} {(1+\beta
 t(1-t)^b)^k} dt\,.
\end{align*}

Let $\varphi(t)=\frac{1}{(1+\beta t(1-t)^b)^k} $ and
$h(t)=\ln\ln(1+\alpha t(1-t)^b)$. Then $h(t)$ reaches the maximum at
$t=\frac{1}{b+1},$ $h^{\prime}(\frac{1}{b+1})=0$ and $
h^{\prime\prime}(\frac{1}{b+1})<0.~ $ By applying Laplace's method,
we have
\begin{align*}
  &\sum_{n=0}^{\infty}\frac{(-1)^nH_{n,k,r}(\alpha,\beta)}{[(b+1)n+1]\binom{(b+1)n}{n}}
  =\int^{1}_{0}(-\ln(1-\alpha t(1-t)^b)^r e^{-k\ln(1-\beta t(1-t)^b}dt\,\\
  &\sim{\varphi(\frac{1}{b+1})e^{rh(\frac{1}{b+1})}\sqrt{\frac{-2\pi}{rh^{\prime\prime}(\frac{1}{b+1})}}}\,.
\end{align*}
\end{proof}

\begin{corollary} Let $k,~r\geq1$, $0<\alpha$, $\beta\neq-4 $. As $r\rightarrow \infty$, we get
\begin{align}
 &\sum_{n=0}^{\infty}\frac{(-1)^nH_{n,k,r}(\alpha,\beta)}{(2n+1)\binom{2n}{n}}
 \sim(-1)^r\bigg(\frac{4}{4+\beta}\bigg)^{k}\bigg(\ln\frac{4+\alpha}{4}\bigg)^{r+\frac{1}{2}}\sqrt{\frac{\pi(4+\alpha)}{4\alpha r
 }}. \label{af-4.10}
\end{align}
\end{corollary}

\begin{proof}
Setting $b=1$ in (\ref{af-4.9}), we derive (\ref{af-4.10}).
\end{proof}

\section{Acknowledgment}
The authors are grateful to the anonymous referee for his/her
helpful comments.

\begin{thebibliography}{99}

\bibitem{ref1}
V. S. Adamic,
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\noindent 2010 {\it Mathematics Subject Classification}: Primary
05A16, 05A19; Secondary 05A15.

\noindent \emph{Keywords: } generalized harmonic number, Genocchi
number, Stirling number, Cauchy number, Riordan array, method of
coefficients.

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\vspace*{+.1in}
\noindent
Received  October 25 2012;
revised version received  November 20 2012.
Published in {\it Journal of Integer Sequences}, December 27 2012.

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