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\begin{center}
\vskip 1cm{\LARGE\bf On the Sums of Reciprocal Hyperfibonacci  \\
\vskip .10in
 Numbers and Hyperlucas Numbers} \vskip 1cm
\large Rui Liu\footnote{This work was supported by the Science
Research Foundation of Dalian University of Technology (2008) and
the National Natural Science Foundation of China (NSFC Grant \#
11061020).} and Feng-Zhen Zhao\\
Department of Mathematics\\
 Dalian University of Technology \\
Dalian, Liaoning 116024  \\
P. R. China\\
\href{mailto:liurui1515@gmail.com}{\tt liurui1515@gmail.com}\\
\href{mailto:fengzhenzhao@yahoo.com.cn}{\tt fengzhenzhao@yahoo.com.cn}\\
\end{center}

\def\DD{\mathcal{D}}
\def\SS{\mathcal{S}}

\vskip .2 in

\begin{abstract}
In this paper, we discuss the properties of hyperfibonacci numbers
and hyperlucas numbers. We investigate the sums of reciprocal
hyperfibonacci numbers and hyperlucas numbers. In addition, we
establish some identities related to reciprocal hyperfibonacci
numbers and hyperlucas numbers.
\end{abstract}

\section{Introduction}
Fibonacci and Lucas sequences $\{F_n\}$ and $\{L_n\}$ have
fascinated both amateurs and professional mathematicians for
centuries. They are generalized to many forms. Dil and Mez\"{o}
\cite{ref4} introduced the definition of ``hyperfibonacci" numbers
$F_n^{(r)}$ and ``hyperlucas" numbers $L_n^{(r)}$ :
\begin{eqnarray*}
F_n^{(r)} &amp;=&amp; \sum_{j=0}^n F_j^{(r-1)}, \quad  with \quad
F_n^{(0)} = F_n ,\quad  F_0^{(r)} =0 , \quad F_1^{(r)} =1 ,
\\
  L_n^{(r)} &amp;=&amp;
\sum_{j=0}^n L_j^{(r-1)},  \quad with \quad L_n^{(0)} = L_n ,\quad
L_0^{(r)} = 2,\quad L_1^{(r)} = 2r+1,
\end{eqnarray*}
where r is a positive integer. It is well known that the Binet
forms of $\{F_n\}$ and $\{L_n\}$ are
\begin{eqnarray}
F_n=\frac{\alpha^n-(-1)^n\alpha^{-n}}{\sqrt 5}, \quad
L_n=\alpha^n+(-1)^n\alpha^{-n}, \label{af-1.1}
\end{eqnarray}
where $\alpha=(1+\sqrt 5)/2$. The sequences $\{F_n\}$ and $\{L_n\}$
satisfy the linear recurrence relation
\begin{eqnarray}
W_n=W_{n-1}+W_{n-2}, \quad n\geq 2. \label{af-1.2}
\end{eqnarray}
It is clear that
\begin{eqnarray}
F_n^{(1)}=F_{n+2}-1, \quad L_n^{(1)}=L_{n+2}-1.  \label{af-1.3}
\end{eqnarray}
Some values of $\{F_n^{(1)}\}$ and $\{L_n^{(1)}\}$ are given below.

\bigskip

\begin{center}
\begin{tabular}{c|ccccccccccccccc}
\hline {\em n} &amp; $0$ &amp; $1$ &amp; $2$ &amp;
 $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ &amp; $8$ &amp; $9$ &amp; $10$ &amp; $11$ &amp; $12$ &amp; $13$ &amp; $14$ \\
  \hline
\\
  $F_n^{(1)}$ &amp; $0$ &amp; $1$ &amp; $2$ &amp;
  $4$ &amp; $7$ &amp; $12$ &amp; $20$ &amp; $33$ &amp; $54$ &amp; $88$ &amp; $143$ &amp; $232$ &amp; $376$ &amp; $609$ &amp; $986$ \\
  \hline
\\
 $L_n^{(1)}$ &amp; $2$ &amp; $3$ &amp; $6$ &amp; $10$ &amp; $17$ &amp; $28$ &amp; $46$ &amp; $75$ &amp; $122$ &amp; $198$ &amp; $321$ &amp; $520$ &amp; $842$ &amp; $1363$ &amp; $2206$ \\
  \hline
\end{tabular}
\end{center}

\bigskip

These are sequences \seqnum{A000071} and \seqnum{A001610} in
Sloane's {\it Encylopedia} \cite{ref11}.
Some properties of $\{F_n^{(r)}\}$ and
$\{L_n^{(r)}\}$ are studied in the paper of Ning-Ning Cao and Feng-Zhen Zhao
\cite{ref3}. In this paper, we investigate the sums of
reciprocal hyperfibonacci numbers and hyperlucas numbers.

Now we recall some definitions involved in this paper. The
Fibonacci and Lucas zeta functions are defined by
$$
\zeta_F(s)=\sum_{n=1}^{\infty}\frac{1}{F_n^s} \quad {\rm and}
\quad \zeta_L(s)=\sum_{n=1}^{\infty}\frac{1}{L_n^s},
$$
where $\{F_n\}$ and $\{L_n\}$ are the Fibonacci and Lucas sequences,
respectively. Recently, properties of $\zeta_F(s)$ and $\zeta_L(s)$
are investigated in several different ways, see for instance
\cite{ref5, ref6, ref7,ref9}. In \cite{ref5}, the partial infinite
sums of reciprocal Fibonacci numbers were studied by Ohtsuka and
Nakamura, \cite{ref10}. They proved that
\begin{eqnarray}
\bigg\lfloor\bigg(\sum_{k=n}^{\infty}\frac{1}{F_k}
\bigg)^{-1}\bigg\rfloor=\begin{cases}F_{n-2}, &amp; \text{if $n$
is even and $n\geq 2$}; \\
F_{n-2}-1, &amp; \text{if $n$ is odd and $n\geq 1$},
\end{cases}
\label{af-1.4}
\end{eqnarray}
where $\lfloor\cdot\rfloor$ denotes the floor function. In
\cite{ref8}, Holliday and Komatsu generalize (\ref{af-1.4})
 to the generalized Fibonacci sequence. They showed that
\begin{eqnarray*}
\bigg\lfloor\bigg(\sum_{k= 1000 n}^{\infty}\frac{1}{G_k}
\bigg)^{-1}\bigg\rfloor=
\begin{cases}G_n-G_{n-1}, &amp; \text{if $n$ is even and $n\geq 2$}; \\
G_n-G_{n-1}-1, &amp; \text{if $n$ is odd and $n\geq 1$},
\end{cases}
\end{eqnarray*}
where $\{G_n\}$ is generalized Fibonacci sequence defined by
$G_{k+2}=aG_{k+1}+G_k (k\geq 0)$ with $G_0=0, G_1=1$, and $a$ is a
positive integer. In this paper, we discuss the partial infinite
sums of reciprocal hyperfibonacci numbers
 and hyperlucas numbers. In the next section, we investigate the sums of the following forms
\begin{eqnarray*}
\bigg\lfloor\bigg(\sum_{k=n}^{\infty}\frac{1}{F_k^{(1)}}\bigg)^{-1}\bigg\rfloor,
\quad
\bigg\lfloor\bigg(\sum_{k=n}^{\infty}\frac{1}{L_k^{(1)}}\bigg)^{-1}\bigg\rfloor.
\quad
\end{eqnarray*}
In addition, we establish some identities related to reciprocal
hyperfibonacci numbers and hyperlucas numbers.

\section{The partial infinite sums of reciprocal hyperfibonacci
numbers and hyperlucas numbers}

In this section, we discuss the partial infinite sums of
reciprocal hyperfibonacci numbers and hyperlucas numbers.
\begin{lemma} For $\{F_n\}$ and $\{L_n\}$, the following formulas hold:
\begin{eqnarray}
F_{n+1}F_{n+3}-F_nF_{n+4}&amp;=&amp;2(-1)^n, \label{af-2.1} \\
L_{n+1}L_{n+3}-L_nL_{n+4}&amp;=&amp;10(-1)^{n+1},\label{af-2.2} \\
F_{n+2}^2-F_{n+1}F_{n+3}&amp;=&amp;(-1)^{n+1},\label{af-2.3} \\
L_{n+2}^2-L_{n+1}L_{n+3}&amp;=&amp;5(-1)^n. \label{af-2.4}
\end{eqnarray}
\end{lemma}
\begin{proof}
From (\ref{af-1.1}), we can verify that
(\ref{af-2.1})-(\ref{af-2.4}) hold.
\end{proof}

\begin{theorem}
For $\{F_n^{(1)}\}$ and $\{L_n^{(1)}\}$ ($n\geq 3$), we have
\begin{eqnarray}
\bigg\lfloor\bigg(\sum_{k=n}^{\infty}\frac{1}{F_k^{(1)}}\bigg)^{-1}\bigg\rfloor&amp;=&amp;F_n-1, \label{af-2.5}\\
\bigg\lfloor\bigg(\sum_{k=n}^{\infty}\frac{1}{L_k^{(1)}}\bigg)^{-1}\bigg\rfloor&amp;=&amp;L_n-1,
\quad n\geq 4. \label{af-2.6}
\end{eqnarray}
\end{theorem}
\begin{proof} By using (\ref{af-1.2})--(\ref{af-1.3}) and
(\ref{af-2.1}), we get
\begin{eqnarray*}
\frac{1}{F_n^{(1)}-F_{n-1}^{(1)}}-\frac{1}{F_n^{(1)}}-\frac{1}{F_{n+1}^{(1)}}-\frac{1}{F_{n+2}^{(1)}-F_{n+1}^{(1)}}
&amp;=&amp;\frac{1}{F_n}-\frac{1}{F_n^{(1)}}-\frac{1}{F_{n+1}^{(1)}}-\frac{1}{F_{n+2}}\\
&amp;=&amp;\frac{F_n^{(1)}(F_{n+1}F_{n+3}-F_nF_{n+2}-F_{n+1})-F_nF_{n+2}F_{n+1}^{(1)}}{F_nF_{n+2}F_n^{(1)}F_{n+1}^{(1)}}\\
&amp;=&amp;\frac{F_n^{(1)}[2(-1)^n+F_nF_{n+3}-F_{n+1}]-F_nF_{n+2}F_{n+1}^{(1)}}{F_nF_{n+2}F_n^{(1)}F_{n+1}^{(1)}}\\
&amp;=&amp;\frac{F_n^{(1)}[2(-1)^n-F_{n+1}]+F_n[F_n^{(1)}F_{n+3}-F_{n+2}F_{n+1}^{(1)}]}{F_nF_{n+2}F_n^{(1)}F_{n+1}^{(1)}}\\
&amp;=&amp;\frac{F_n^{(1)}[2(-1)^n-F_{n+1}]-F_nF_{n+1}}{F_nF_{n+2}F_n^{(1)}F_{n+1}^{(1)}},
\quad n\geq 2,
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{1}{F_n^{(1)}-F_{n-1}^{(1)}-1}-\frac{1}{F_n^{(1)}}-\frac{1}{F_{n+1}^{(1)}}-\frac{1}{F_{n+2}^{(1)}-F_{n+1}^{(1)}-1}
&amp;=&amp;\frac{F_{n+1}F_{n+1}^{(1)}-(F_n-1)F_n^{(1)}-(F_n-1)F_{n+1}^{(1)}}{(F_n-1)F_n^{(1)}F_{n+1}^{(1)}}
\end{eqnarray*}
\begin{eqnarray*}
\quad\quad\quad &amp;=&amp;\frac{F_{n+1}F_{n+3}-F_{n+1}-F_nF_{n+3}-F_{n+1}+2F_n-F_nF_{n+2}+F_n^{(1)}+F_{n+1}^{(1)}}{(F_n-1)F_n^{(1)}F_{n+1}^{(1)}}\\
&amp;=&amp;\frac{F_nF_{n+4}+2(-1)^n-F_nF_{n+3}-F_nF_{n+2}-F_{n+1}+2F_n+F_{n+2}+F_{n+3}-2}{(F_n-1)F_n^{(1)}F_{n+1}^{(1)}}\\
&amp;= &amp;\frac{2((-1)^n+F_n+F_{n+2}-1)}{(F_n-1)F_n^{(1)}F_{n+1}^{(1)}},
\quad n\geq 3.
\end{eqnarray*}

By using (\ref{af-1.2})--(\ref{af-1.3}) and (\ref{af-2.2}), we get
\begin{eqnarray*}
\frac{1}{L_n^{(1)}-L_{n-1}^{(1)}}-\frac{1}{L_n^{(1)}}-\frac{1}{L_{n+1}^{(1)}}-\frac{1}{L_{n+2}^{(1)}-L_{n+1}^{(1)}}
&amp;=&amp;\frac{(10(-1)^{n+1}-L_{n+1})L_n^{(1)}-L_nL_{n+1}}{L_nL_{n+2}L_n^{(1)}L_{n+1}^{(1)}}
\end{eqnarray*}
for $n \geq 4$ and
\begin{eqnarray*}
\frac{1}{L_n^{(1)}-L_{n-1}^{(1)}-1}-\frac{1}{L_n^{(1)}}-\frac{1}{L_{n+1}^{(1)}}-\frac{1}{L_{n+2}^{(1)}-L_{n+1}^{(1)}-1}
&amp;=&amp;\frac{10(-1)^{n+1}+2(L_n+L_{n+2}-1)}{(L_n-1)L_n^{(1)}L_{n+1}^{(1)}},
\end{eqnarray*} for
$n \geq 2$.

From the inequalities
\begin{eqnarray*}
(2(-1)^n-F_{n+1})F_n^{(1)}-F_nF_{n+1}&amp;&lt;&amp;0, \quad n\geq 2, \\
(10(-1)^{n+1}-L_{n+1})L_n^{(1)}-L_nL_{n+1}&amp;&lt;&amp;0, \quad n\geq 4,\\
(-1)^n+F_n+F_{n+2}-1&amp;&gt;&amp;0, \quad n\geq 3,\\
10(-1)^{n+1}+2(L_n+L_{n+2}-1)&amp;&gt;&amp;0,\quad n\geq 2,
\end{eqnarray*}
we obtain
\begin{alignat*}{2}
\frac{1}{F_n^{(1)}-F_{n-1}^{(1)}}\ &amp;&lt; \quad\quad \sum_{k=n}^{\infty}\frac{1}{F_k^{(1)}}&amp;&lt;&amp; \ \frac{1}{F_n^{(1)}-F_{n-1}^{(1)}-1}, \quad n\geq 3, \\
F_n-1&amp;&lt;\bigg(\sum_{k=n}^{\infty}\frac{1}{F_k^{(1)}}\bigg)^{-1}&amp;&lt;&amp; \ F_n, \quad n\geq 3,\\
\frac{1}{L_n^{(1)}-L_{n-1}^{(1)}}&amp;&lt; \quad\quad \sum_{k=n}^{\infty}\frac{1}{L_k^{(1)}}&amp;&lt;&amp;\ \frac{1}{L_n^{(1)}-L_{n-1}^{(1)}-1}, \quad n\geq 4,\\
L_n-1&amp;&lt;\bigg(\sum_{k=n}^{\infty}\frac{1}{L_k^{(1)}}\bigg)^{-1}&amp;&lt;&amp;\ L_n,
\quad n\geq 4.
\end{alignat*}
Hence the relations (\ref{af-2.5})--(\ref{af-2.6}) hold.
\end{proof}

\begin{theorem}
For $\{F_n^{(1)}\}$ $(n\geq2) $ we have
\begin{eqnarray}
\bigg \lfloor
\bigg(\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}\bigg)\bigg
\rfloor&amp;=&amp;\begin{cases}F_{n-1}^{(1)}F_{n}^{(1)}+F_{n-1}^{(1)}-1, &amp;
\text{if $n$ is even and $n\geq 2$}; \\
F_{n-1}^{(1)}F_{n}^{(1)}+F_{n-1}^{(1)},  &amp;
\text{if $n$ is odd and $n\geq 1$}.
\end{cases}
\label{af-2.7} \\
\bigg \lfloor
\bigg(\sum_{k=n}^{\infty}\frac{1}{(L_k^{(1)})^2}\bigg)\bigg
\rfloor&amp;=&amp;L_{n-1}^{(1)}L_n^{(1)}+L_{n-1}^{(1)}-1, \quad\quad
\text{if $n$ is odd and $n&gt;1$}. \label{af-2.8}
\end{eqnarray}
\end{theorem}
\begin{proof}
By applying (\ref{af-1.2})--(\ref{af-1.3}) and (\ref{af-2.3}), we
get
\begin{eqnarray*}
\frac{1}{F_{n-1}^{(1)}F_n^{(1)}+F_{n-1}^{(1)}-1}-\frac{1}{(F_n^{(1)})^2}-\frac{1}{F_n^{(1)}F_{n+1}^{(1)}+F_n^{(1)}-1}
&amp;=&amp;\frac{1}{F_{n-1}^{(1)}F_{n+2}-1}-\frac{1}{(F_n^{(1)})^2}-\frac{1}{F_n^{(1)}F_{n+3}-1}
\end{eqnarray*}
\begin{eqnarray*}
\quad\quad\quad &amp;=&amp;\frac{(F_n^{(1)})^2(F_n^{(1)}F_{n+3}-F_{n-1}^{(1)}F_{n+2})-(F_{n-1}^{(1)}F_{n+2}-1)(F_n^{(1)}F_{n+3}-1)}{(F_{n-1}^{(1)}F_{n+2}-1)(F_n^{(1)})^2
(F_n^{(1)}F_{n+3}-1)}
\\
&amp;=&amp;\frac{{F_n^{(1)}}^2(F_{n+2}^2-F_{n+1})-F_{n-1}^{(1)}F_n^{(1)}F_{n+2}F_{n+3}+F_{n-1}^{(1)}F_{n+2}+F_n^{(1)}F_{n+3}-1}{(F_{n-1}^{(1)}F_{n+2}-1)(F_n^{(1)})^2
(F_n^{(1)}F_{n+3}-1)}
\\
&amp;=&amp;\frac{(F_{n+3}+(-1)^{n+1}F_{n+2})F_n^{(1)}+F_n^{(1)}F_{n+1}+F_{n-1}^{(1)}F_{n+2}-1}{(F_{n-1}^{(1)}F_{n+2}-1)(F_n^{(1)})^2
(F_n^{(1)}F_{n+3}-1)}&gt;0.
\end{eqnarray*}
Thus, we have
\begin{eqnarray*}
\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}&lt;\frac{1}{F_{n-1}^{(1)}F_n^{(1)}+F_{n-1}^{(1)}-1}.
\end{eqnarray*}
Similarly, we can prove that
\begin{eqnarray*}
\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}&gt;\frac{1}{F_{n-1}^{(1)}F_n^{(1)}+F_{n-1}^{(1)}+1}.
\end{eqnarray*}
On the other hand, we have
\begin{eqnarray*}
\frac{1}{F_{n-1}^{(1)}F_n^{(1)}+F_{n-1}^{(1)}}-\frac{1}{(F_n^{(1)})^2}-\frac{1}{F_n^{(1)}F_{n+1}^{(1)}+F_n^{(1)}}
&amp;=&amp;\frac{1}{F_{n-1}^{(1)}F_{n+2}}-\frac{1}{(F_n^{(1)})^2}-\frac{1}{F_n^{(1)}F_{n+3}}\\
&amp;=&amp;\frac{(F_n^{(1)})^2F_{n+3}-F_{n-1}^{(1)}F_n^{(1)}F_{n+2}-F_{n-1}^{(1)}F_{n+2}F_{n+3}}{F_{n-1}^{(1)}(F_n^{(1)})^2F_{n+2}F_{n+3}}\\
&amp;=&amp;\frac{F_n^{(1)}(F_n^{(1)}F_{n+3}-F_{n-1}^{(1)}F_{n+2})-F_{n-1}^{(1)}F_{n+2}F_{n+3}}{F_{n-1}^{(1)}(F_n^{(1)})^2F_{n+2}F_{n+3}}\\
&amp;=&amp;\frac{(-1)^{n+1}F_{n+2}+F_{n+1}}{F_{n-1}^{(1)}(F_n^{(1)})^2F_{n+2}F_{n+3}}.
\end{eqnarray*}
When $n\geq2$ is even, we can verify that
$$
\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}&gt;\frac{1}{F_{n-1}^{(1)}F_n^{(1)}+F_{n-1}^{(1)}},
$$
and when $n\geq1$ is odd
$$
\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}&lt;\frac{1}{F_{n-1}^{(1)}F_n^{(1)}+F_{n-1}^{(1)}}.
$$
Hence when $n$ is even, we obtain
$$
F_{n-1}^{(1)}F_{n}^{(1)}+F_{n-1}^{(1)}-1&lt;\bigg(\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}\bigg)^{-1}
&lt;F_{n-1}^{(1)}F_{n}^{(1)}+F_{n-1}^{(1)},
$$
and when $n$ is odd, we have
$$
F_{n-1}^{(1)}F_{n}^{(1)}+F_{n-1}^{(1)}&lt;\bigg(\sum_{k=n}^{\infty}\frac{1}{(F_k^{(1)})^2}\bigg)^{-1}&lt;F_{n-1}^{(1)}F_{n}^{(1)}+F_{n-1}^{(1)}+1.
$$
Then (\ref{af-2.7}) holds.

By applying (\ref{af-1.2})--(\ref{af-1.3}) and (\ref{af-2.4}), we
get
\begin{eqnarray*}
\sum_{k=n}^{\infty}\frac{1}{(L_k^{(1)})^2}&amp;&lt;&amp;\frac{1}{L_{n-1}^{(1)}L_n^{(1)}+L_{n-1}^{(1)}-1},\\
\sum_{k=n}^{\infty}\frac{1}{(L_k^{(1)})^2}&amp;&gt;&amp;\frac{1}{L_{n-1}^{(1)}L_n^{(1)}+L_{n-1}^{(1))}},\quad
\text{$n$ is odd}.
\end{eqnarray*}
Then (\ref{af-2.8}) holds.
\end{proof}


 In the final part of this section, we consider the generalized hyperfibonacci numbers $\{U_n^{(r)}\}$:
\begin{eqnarray*}
 U_n^{(r)}&amp;=&amp;\sum_{j=0}^n
U_j^{(r-1)},\quad with \quad U_n^{(0)}=U_n,\quad U_0^{(r)}=0,\quad
U_1^{(r)}=1.
\end{eqnarray*}
where
\begin{eqnarray*}
U_n=\frac{\tau^n-(-1)^n\tau^{-n}}{\sqrt{\Delta}},\quad
 \tau=(p+\sqrt{\Delta})/2,\quad
\Delta=p^2+4,
\end{eqnarray*}
and $p$ is a positive integer. It is evident that
\begin{eqnarray}
U_n^{(1)}=\frac{U_n+U_{n+1}-1}{p}. \label{af-2.9}
\end{eqnarray}
And $\{U_n\}$  satisfy that
\begin{eqnarray}
W_n=pW_{n-1}+W_{n-2}, \quad n\geq 2. \label{af-2.10}
\end{eqnarray}
When $p=1$, $U_n^{(1)}=F_n^{(1)}.$

Now we discuss the partial infinite sum of reciprocal generalized
hyperfibonacci numbers.
\begin{lemma} For $\{U_n\}$, the following formulas hold:
\begin{eqnarray}
U_{n+1}^2-U_nU_{n+2}&amp;=&amp;(-1)^n, \label{af-2.11}
\end{eqnarray}
\end{lemma}
From the definition of $\{U_n\}$ , we can prove that
(\ref{af-2.11}) holds.

\begin{theorem} When $n\geq 2$,
\begin{eqnarray}
\bigg\lfloor\bigg(\sum_{k=n}^{\infty}\frac{1}{U_k^{(1)}}\bigg)^{-1}\bigg\rfloor&amp;=&amp;U_n-1,
\label{af-2.12}
\end{eqnarray}
\end{theorem}
\begin{proof} It follows from (\ref{af-2.9}) and (\ref{af-2.10})--(\ref{af-2.11}) that
\begin{eqnarray*}
\frac{1}{U_n^{(1)}-U_{n-1}^{(1)}}-\frac{1}{U_n^{(1)}}-\frac{1}{U_{n+1}^{(1)}}-\frac{1}{U_{n+2}^{(1)}-U_{n+1}^{(1)}}
&amp;=&amp;\frac{pU_{n+1}}{U_nU_{n+2}}-\frac{U_n^{(1)}+U_{n+1}^{(1)}}{U_n^{(1)}U_{n+1}^{(1)}}
\end{eqnarray*}
\begin{eqnarray*}
\quad\quad\quad &amp;=&amp;\frac{U_{n+1}(U_n+U_{n+1}-1)(U_{n+1}+U_{n+2}-1)-U_nU_{n+2}(U_n+2U_{n+1}+U_{n+2}-2)}{pU_nU_{n+2}U_n^{(1)}U_{n+1}^{(1)}}\\
&amp;=&amp;\frac{(U_n+U_{n+1}-1)((-1)^n+U_{n+1}U_{n+2}-U_{n+1})-U_nU_{n+2}(U_{n+1}+U_{n+2}-1)}{pU_nU_{n+2}U_n^{(1)}U_{n+1}^{(1)}}\\
&amp;=&amp;\frac{(U_n+U_{n+1}-1)((-1)^n-U_{n+1})+U_{n+2}(U_{n+1}^2-U_{n+1}-U_nU_{n+2}+U_n)}{pU_nU_{n+2}U_n^{(1)}U_{n+1}^{(1)}}\\
&amp;=&amp;\frac{(U_n+U_{n+1}-1)((-1)^n-U_{n+1})+U_{n+2}((-1)^n-(p-1)U_n-U_{n-1})}{pU_nU_{n+2}U_n^{(1)}U_{n+1}^{(1)}}\\
&amp;&lt;&amp;0,
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{1}{U_n^{(1)}-U_{n-1}^{(1)}-1}-\frac{1}{U_n^{(1)}}-\frac{1}{U_{n+1}^{(1)}}-\frac{1}{U_{n+2}^{(1)}-U_{n+1}^{(1)}-1}
&amp;=&amp;\frac{pU_{n+1}}{(U_n-1)(U_{n+2}-1)}-\frac{1}{U_n^{(1)}}-\frac{1}{U_{n+1}^{(1)}}
\end{eqnarray*}
\begin{eqnarray*}
&amp;=&amp;\frac{U_{n+1}^{(1)}(U_{n+2}+U_n-U_{n+1}-1+(-1)^n)+U_n^{(1)}(U_n+U_{n+2}-1)}{(U_n-1)(U_{n+2}-1)U_n^{(1)}U_{n+1}^{(1)}}
+ \frac{U_n(U_{n+1}^{(1)}U_{n+1}-U_n^{(1)}U_{n+2})}{(U_n-1)(U_{n+2}-1)U_n^{(1)}U_{n+1}^{(1)}}\\
&amp;=&amp;\frac{U_{n+1}^{(1)}(U_{n+2}+U_n-U_{n+1}-1+(-1)^n)+U_n^{(1)}(U_n+U_{n+2}-1)}{(U_n-1)(U_{n+2}-1)U_n^{(1)}U_{n+1}^{(1)}}
+ \frac{U_n((-1)^n-U_{n+1}+U_{n+2})}{p(U_n-1)(U_{n+2}-1)U_n^{(1)}U_{n+1}^{(1)}}\\
&amp;&gt;&amp;0.
\end{eqnarray*}

Then we obtain
\begin{eqnarray*}
\frac{1}{U_n^{(1)}-U_{n-1}^{(1)}}&amp;&lt;&amp;\sum_{k=n}^{\infty}\frac{1}{U_k^{(1)}}&lt;\frac{1}{U_n^{(1)}-U_{n-1}^{(1)}-1}.
\end{eqnarray*}
Hence (\ref{af-2.12}) holds.
\end{proof}

\section{Some identities related to reciprocal hyperfibonacci numbers
and hyperlucas numbers}

In this section, we give some identities related to inverse of
hyperfibonacci and hyperlucas numbers. There are some identities
containing the reciprocals
of Fibonacci and Lucas numbers (see \cite{ref1,ref2}):
\begin{align*}
\sum_{n=0}^{\infty}\frac{1}{F_{2n+1}+F_s}&amp;=\frac{\sqrt 5
s}{2L_s},\quad \text{$s$ odd}, \\
\sum_{n=0}^{\infty}\frac{1}{F_{2n+1}+L_s/\sqrt5}&amp;=\frac{s}{2F_s},
\quad \text{$s &gt; 0$ is even}.
\end{align*}
For hyperfibonacci and hyperlucas numbers $F_n^{(1)}$ $L_n^{(1)}$,
we have
\begin{theorem}
Let m be a positive integer. For $F_n^{(1)}$ and $L_n^{(1)}$, we
have
\begin{eqnarray}
\sum\limits_{n=1}^{\infty}\frac{L_{n+2m+2}}
{F_n^{(1)}F_{n+4m}^{(1)}}&amp;=&amp;\frac{1}{F_{2m}}\sum \limits _{k=1}^{4m} \frac{1}{F_k^{(1)}}, \label{af-3.1}\\
 \sum\limits _{n=1}^{\infty}\frac{F_{n+2m+2}}{L_n^{(1)}L_{n+4m}^{(1)}}&amp;=&amp;
\frac{1}{5F_{2m}}\sum\limits _{n=1}^{4m}\frac{1}{L_k^{(1)}}, \label{af-3.2}\\
\sum_{n=1}^{\infty}\frac{F_{n+2m+3}}{F_n^{(1)}F_{n+4m+2}^{(1)}}&amp;=&amp;\frac{1}{L_{2m+1}}\sum_{k=1}^{4m+2}\frac{1}{F_k^{(1)}},
\label{af-3.3}\\
\sum_{n=1}^{\infty}\frac{L_{n+2m+3}}{L_n^{(1)}L_{n+4m+2}^{(1)}}&amp;=&amp;
\frac{1}{L_{2m+1}}\sum_{k=1}^{4m+2}\frac{1}{L_k^{(1)}}.\label{af-3.4}
\end{eqnarray}
\end{theorem}
\begin{proof}
It follows from (\ref{af-1.2}) that
\begin{eqnarray*}
\frac{1}{F_n^{(1)}}-\frac{1}{F_{n+4m}^{(1)}}&amp;=&amp;\frac{F_{n+4m}^{(1)}-F_n^{(1)}}{F_n^{(1)}F_{n+4m}^{(1)}},\\
\frac{1}{L_n^{(1)}}-\frac{1}{L_{n+4m}^{(1)}}&amp;=&amp;\frac{L_{n+4m}^{(1)}-L_n^{(1)}}{L_n^{(1)}-L_{n+4m}^{(1)}},\\
\frac{1}{F_n^{(1)}}-\frac{1}{F_{n+4m+2}^{(1)}}&amp;=&amp;\frac{F_{n+4m+4}-F_{n+2}}{F_n^{(1)}F_{n+4m+2}^{(1)}},\\
\frac{1}{L_n^{(1)}}-\frac{1}{L_{n+4m+2}^{(1)}}&amp;=&amp;\frac{L_{n+4m+4}-L_{n+2}}{L_n^{(1)}L_{n+4m+2}^{(1)}}.
\end{eqnarray*}
From
\begin{eqnarray*}
F_{n+4m}^{(1)}-F_n^{(1)}&amp;=&amp;F_{2m}L_{n+2m+2},\\
L_{n+4m}^{(1)}-L_n^{(1)}&amp;=&amp;5F_{2m}F_{n+2m+2},\\
F_{n+4m+4}-F_{n+2}&amp;=&amp;F_{n+2m+3}L_{2m+1},\\
L_{n+4m+4}-L_{n+2}&amp;=&amp;L_{n+2m+3}L_{2m+1}.
\end{eqnarray*}
we obtain the formula (\ref{af-3.1})--(\ref{af-3.4}).
\end{proof}

We can give other identities for $F_n^{(1)}$ and $L_n^{(1)}$. The
following lemma will be used (see \cite{ref12}).
\begin{lemma}
Let t be a real number with $|t|&gt;1$, $s$ and $a$ be positive
integers, and $b$ be a nonnegative integer. Then one has that
\begin{eqnarray}
\displaystyle
\sum_{n=0}^{\infty}\frac{1}{t^{2an+b}+t^{-2an-b}-(t^{as}+t^{-as})}&amp;=&amp;
\frac{1}{t^{as}-t^{-as}}\sum\limits_{n=0}^{s-1}\frac{1}{1-t^{2an+b-as}}
\label{af-3.5}
\end{eqnarray}
\end{lemma}
\begin{theorem}

Suppose that $a, b$ and $s$ are positive integers with $b&gt;as$. For $F_n^{(1)}$ and $L_n^{(1)}$, we have:\\
(i) when $a, b$ and $s$ are odd,
\begin{eqnarray}
\sum_{n=0}^{\infty}\frac{1}{F_{2an+b-2}^{(1)}-F_{as-2}^{(1)}}&amp;=&amp;
\frac{\sqrt{5}}{L_{as}}\sum_{n=0}^{s-1}\frac{1}{1-\alpha^{2an+b-as}},\label{af-3.6}
\end{eqnarray}
(ii) when $b$ and $s$ are both even,
\begin{eqnarray}
\sum_{n=0}^{\infty}\frac{1}{L_{2an+b-2}^{(1)}-L_{as-2}^{(1)}}=\frac{1}{\sqrt{5}F_{as}}\sum_{n=0}^{s-1}\frac{1}{1-\alpha^{2an+b-as}}.
\label{af-3.7}
\end{eqnarray}
\end{theorem}
\begin{proof}
By means of (\ref{af-3.6}) and $F_n^{(1)}=F_{n+2}-1$,
$L_n^{(1)}=L_{n+2}-1$, we can easily prove that (\ref{af-3.6}) and
(\ref{af-3.7}) hold.
\end{proof}

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



%==============================================================================================================
\begin{thebibliography}{99}
\bibliographystyle{plain}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B39, 11B37; Secondary 05A19.

\noindent \emph{Keywords: } Fibonacci numbers, Lucas numbers.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000071} and
\seqnum{A001610}.)

\bigskip
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\bigskip

\vspace*{+.1in}
\noindent
Received December 29 2011;
revised version received  March 26 2012.
Published in {\it Journal of Integer Sequences}, April 9 2012.

\bigskip
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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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\end{document}



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