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\begin{center}
\vskip 1cm{\LARGE\bf On the Sums of Reciprocal Generalized \\
\vskip .1in
Fibonacci Numbers
}
\vskip 1cm
\large
Kantaphon Kuhapatanakul\footnote{Supported by the Kasetsart University Research and Development Institute (KURDI), Thailand.}\\
Department of Mathematics\\
Faculty of Science\\
Kasetsart University\\
Bangkok 10900\\
Thailand\\
\href{mailto:fscikpkk@ku.ac.th}{\tt fscikpkk@ku.ac.th}
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\begin{abstract}
In this article we consider the infinite sums of reciprocal generalized
Fibonacci numbers and the infinite sums of reciprocal generalized
Fibonacci sums. Applying the floor function to the reciprocals of
these sums, our results generalize some identities of Holliday and
Komatsu and extend some results of Liu and Zhao.
\end{abstract}

\section{Introduction}
Let $a,b$ be two positive integers and $c$ non-negative integer. Define the generalized Fibonacci numbers $V_n(c;a,b)$, briefly $V_n$, by the following relation
$$V_0=c,\ \ \ V_1=1\ \ \ \ \text{and}\ \ \ \ \ V_{n+1}=aV_{n}+bV_{n-1}\ \ \ \ (n\ge1).$$ 

Here note that $V_n(0;1,1)=F_n$ are the Fibonacci numbers, $V_n(2;1,1)=L_n$ are the Lucas numbers and $V_n(0;2,1)=P_n$ are the Pell numbers.

Ohtsuka and Nakamura \cite{on} derived a formula for infinite sums of reciprocal Fibonacci numbers, as follows,
\begin{equation}  \label{ohna}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{F_k} \right)^{-1}\right\rfloor}=\begin{cases}
                            F_{n-2},  &amp;\text{if}~n~\text{is\ even~and}~n\ge2;\\
                            F_{n-2}-1,\ \ &amp;\text{if}~n~\text{is\ odd~and}~n\ge1,
                           \end{cases} 
\end{equation}    
where ${\lfloor \cdot\rfloor}$ is the floor function. 

Wenpeng and Tingting \cite{wt} gave analogue of the identity \eqref{ohna} for the Pell numbers:
\begin{equation}    \label{weti}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{P_k} \right)^{-1}\right\rfloor}=
\begin{cases}
     P_n-P_{n-1},  &amp;\text{if}~n~\text{is\ even~and}~n\ge2;\\
     P_n-P_{n-1}-1,\ \ &amp;\text{if}~n~\text{is\ odd~and}~n\ge1.
\end{cases} 
\end{equation}

Holliday and Komatsu \cite{hk} generalized \eqref{ohna} and \eqref{weti} to the generalized Fibonacci numbers $V_n(0;a,1)$, briefly $u_n$, and $V_n(c;1,1)$ for $c\ge1$, briefly $G_n$.
They showed that
\begin{equation}    \label{hoko}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{u_k} \right)^{-1}\right\rfloor}=\begin{cases}
                            u_{n}-u_{n-1},  &amp;\text{if}~n~\text{is\ even~and}~n\ge2;\\
                            u_{n}-u_{n-1}-1,\ \ &amp;\text{if}~n~\text{is\ odd~and}~n\ge1,
                           \end{cases}
\end{equation}
and
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{G_k} \right)^{-1}\right\rfloor}=\begin{cases}
                            G_{n-2},  &amp;\text{if}~n~\text{is\ odd~and}~n\ge n_1;\\
                            G_{n-2}-1,\ \ &amp;\text{if}~n~\text{is\ even~and}~n\ge n_2,
                           \end{cases}$$ 
where $n_1$ and $n_2$ are determined depending only on the value of $c$. For example, if $G_n=L_n$ or $c=2$, then $n_1=2$ and $n_2=3$.

By the same proof as the one for \eqref{hoko}, it is easy to verify that the identity \eqref{hoko} still holds for $V_n(0;a,b)$ in place of $V_n(0;a,1)$, provided that $1\le b\le a$. 

In this paper, we first give the analogue of the identity \eqref{hoko} for the alternating sums of reciprocals of $V_n(0;a,b)$ and prove the following result in the next section. 

\begin{theorem}    \label{main2}
Let $U_n:=V_n(0;a,b)$ with $1\le b\le a$. Then
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{(-1)^k}{U_k} \right)^{-1}\right\rfloor}=
  (-1)^n\left(U_n+U_{n-1}\right)-1\ \ \ \ \ (n\ge1).$$
\end{theorem}

We have a following corollary. 
\begin{corollary}    
For a positive integer $n$, we have
\begin{itemize}
  \item [$(1)$] ${\left\lfloor \left(\displaystyle\sum_{k=n}^\infty \frac{(-1)^k}{F_k} \right)^{-1}\right\rfloor}=
  (-1)^nF_{n+1}-1.$ 
  \item [$(2)$] ${\left\lfloor \left(\displaystyle\sum_{k=n}^\infty \frac{(-1)^k}{P_k} \right)^{-1}\right\rfloor}=(-1)^n\left(P_n+P_{n-1}\right)-1$.
\end{itemize}
\end{corollary}

In 2012, Liu and Zhao \cite{lz} showed the formulas for the infinite sums of reciprocal hyperfibonacci numbers and hyperlucas numbers as:   
\begin{align}
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=0}^k F_i} \right)^{-1}\right\rfloor}&amp;=                       
F_n-1\ \ \ \ (n\ge3),   \label{rf}\\
{\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=0}^k L_i} \right)^{-1}\right\rfloor}&amp;=                        L_n-1\ \ \ \ (n\ge4). \label{rl}
\end{align}
Especially, they also gave a following general result for $u_n=V_n(0;a,1)$ as:
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=0}^k u_i} \right)^{-1}\right\rfloor}=                       
u_n-1\ \ \ \ (n\ge3).$$

Next, we will extend the above results on the Fibonacci and Lucas numbers and $V_n(0;a,1)$ to the generalized Fibonacci numbers $V_n(c;1,1)$ and $V_n(0;a,b)$ and prove the following results in the next section.
\begin{theorem}    \label{main1}
Let $G_n=V_n(c;1,1)$ for $c\ge1$. We have
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=0}^k G_i} \right)^{-1}\right\rfloor}=G_n-1\ \ \ \ (n\ge n_0),$$
where $n_0$ is determined depending only on the value of $c$.
\end{theorem}

For example, we can determine $n_0$ for a fixed $c$ as follows:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline 
 $c$   &amp;1  &amp;2  &amp;3-5  &amp;6-13 &amp;14-34  &amp;35-89  &amp;90-233  &amp;234-610  &amp;611-1597  &amp;1598-4181\\ 
\hline 
 $n_0$ &amp;2  &amp;4  &amp;6    &amp;8     &amp;10    &amp;12     &amp;14     &amp;16   &amp;18       &amp;20\\
\hline       
\end{tabular}
\end{center}

Put $c=1,2$ in Theorem \ref{main1}, we can, respectively, deduce the identities \eqref{rf} and \eqref{rl}.
\begin{theorem}    \label{main3}
Let $a\ge b \ge1$ and $U_n=V_n(0;a,b)$. We have
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=0}^k U_i} \right)^{-1}\right\rfloor}=U_n-1\ \ \ \ (n\ge N_a),$$
where $N_a=3$ for $a=1$ and $N_a=2$ for $a\ge 2$.
\end{theorem}

We have the following corollary. 
\begin{corollary}
We have
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{1}{\sum_{i=0}^k P_i} \right)^{-1}\right\rfloor}=P_n-1\ \ \ \ (n\ge 2).$$
\end{corollary} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proofs of Theorems}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We begin with some identities of the generalized Fibonacci numbers $V_n(c;a,b)$ whose induction proofs are omitted.
\begin{lemma}   \label{lg}
Let $a,b$ be positive integers and $c$ non-negative integer. Then for $n\ge1$ we have 
\begin{itemize}
	\item [$(1)$] $V_{n+1}V_{n-1}-V^2_n=(-1)^nb^{n-1}(1-ac-bc^2)$.
	\item [$(2)$] $\displaystyle\sum_{i=0}^nV_i=\frac{1}{a+b-1}\left(V_{n+1}+bV_n+ac-c-1\right)$. 
\end{itemize}
\end{lemma}

{\sl Proof of Theorem \ref{main2}} :
By Lemma \ref{lg}, we have 
$$U_{n+1}U_{n-1}-U^2_n=(-1)^nb^{n-1}$$
and
$$\sum_{i=0}^nU_i=\frac{1}{a+b-1}\left(U_{n+1}+bU_n-1\right).$$
Since $a\ge b\ge1$, we have $U_{n+1}-U_{n-1}&gt;a^{n-1}\ge b^{n-1}$, so   
\begin{align*}
\frac{(-1)^n}{U_n+U_{n-1}-(-1)^n}&amp;-\frac{(-1)^{n+1}}{U_{n+1}+U_n-(-1)^{n+1}}-\frac{(-1)^n}{U_n}\\
&amp;=\frac{(-1)^{n+1}U_{n-1}+1}{U_n\left(U_n+U_{n-1}-(-1)^n\right)}-\frac{(-1)^{n+1}}{U_{n+1}+U_n-(-1)^{n+1}}\\
&amp;=\frac{(-1)^{n+1}U_{n-1}U_{n+1}+(-1)^nU^2_n+U_{n+1}-U_{n-1}+(-1)^n}{U_n\left(U_n+U_{n-1}-(-1)^n\right)(U_{n+1}+U_n-(-1)^{n+1})}\\
&amp;=\frac{(-1)^{n-1}b^{n-1}+U_{n+1}-U_{n-1}+(-1)^n}{U_n\left(U_n+U_{n-1}-(-1)^n\right)(U_{n+1}+U_n-(-1)^{n+1})}\\
&amp;&gt; 0.
\end{align*}
By applying the above inequality repeatedly, we obtain
\begin{align*}
\frac{1}{(-1)^n(U_n+U_{n-1})-1}&amp;&gt;\frac{(-1)^n}{U_n}+\frac{1}{(-1)^{n+1}(U_{n+1}+U_n)-1}\\
&amp;&gt;\frac{(-1)^n}{U_n}+\frac{(-1)^{n+1}}{U_{n+1}}+\frac{1}{(-1)^{n+2}(U_{n+2}+U_{n+1})-1}\\
&amp;\ \ \vdots\\
&amp;&gt;\frac{(-1)^n}{U_n}+\frac{(-1)^{n+1}}{U_{n+1}}+\frac{(-1)^{n+2}}{U_{n+2}}+\cdots.
\end{align*}
Therefore,
\begin{equation}    \label{ineq1}
\sum_{k=n}^\infty\frac{(-1)^k}{U_k}&lt;\frac{1}{(-1)^n(U_n+U_{n-1})-1}.
\end{equation}

Similarly, we have
\begin{align*}
\frac{(-1)^n}{U_n}-\frac{(-1)^n}{U_n+U_{n-1}}+\frac{(-1)^{n+1}}{U_{n+1}+U_n}
&amp;=(-1)^n\left(\frac{U_{n-1}}{U_n(U_n+U_{n-1})}+\frac{1}{U_{n+1}+U_n}\right)\\
&amp;=(-1)^n\left(\frac{U_{n-1}U_{n+1}-U^2_n}{U_n(U_n+U_{n-1})(U_{n+1}+U_n)}\right)\\
&amp;=\frac{b^{n-1}}{U_n(U_n+U_{n-1})(U_{n+1}+U_n)}\\
&amp;&gt;0,
\end{align*}
so 
$$\frac{(-1)^n}{U_n+U_{n-1}}&lt;\frac{(-1)^n}{U_n}+\frac{(-1)^{n+1}}{U_{n+1}+U_n}.$$
Repeating the above inequality, we obtain 
\begin{equation}    \label{ineq2}
\sum_{k=n}^\infty\frac{(-1)^k}{U_k}&gt;\frac{(-1)^n}{U_n+U_{n-1}}.
\end{equation}
Combining the \eqref{ineq1} and \eqref{ineq2}, we get
 $$\frac{1}{(-1)^n(U_n+U_{n-1})}&lt;\sum_{k=n}^\infty\frac{(-1)^k}{U_k}&lt;\frac{1}{(-1)^n(U_n+U_{n-1})-1},$$
it is equivalent to 
$${\left\lfloor \left(\sum_{k=n}^\infty \frac{(-1)^k}{U_k} \right)^{-1}\right\rfloor}=(-1)^n(U_n+U_{n-1})-1.$$\\


{\sl Proof of Theorem \ref{main1}} :
By Lemma \ref{lg}, we have
$$G_{n+1}G_{n-1}-G^2_n=(-1)^n(1-c-c^2)$$
and
$$\sum_{i=0}^nG_n=G_{n+2}-1.$$
Suppose $c\ge1$, we have
\begin{align*}
\frac{1}{G_n-1}-\frac{1}{G_{n+1}-1}-\frac{1}{\sum_{i=0}^n G_i}&amp;=\frac{1}{G_n-1}-\frac{1}{G_{n+1}-1}-\frac{1}{G_{n+2}-1}\\
&amp;=\frac{G_{n+1}}{(G_n-1)(G_{n+2}-1)}-\frac{1}{G_{n+1}-1}\\
&amp;=\frac{2G_n-1+(-1)^{n+1}(c^2+c-1)}{(G_n-1)(G_{n+1}-1)(G_{n+2}-1)}.
\end{align*}
Since $G_n$ is monotone increasing with $n$, we can take $n$ so large that $2G_n\ge (-1)^n(c^2+c)$ for a fixed $c$. Hence, the numerator of the right-hand side of the above identity is positive if $n\ge N_1$ for some positive integer $N_1$, so we get 
\begin{align*}
\frac{1}{G_n-1}&amp;\ge \frac{1}{\sum_{i=0}^nG_i}+\frac{1}{G_{n+1}-1}\\
&amp;\ge \frac{1}{\sum_{i=0}^nG_i}+\frac{1}{\sum_{i=0}^{n+1}G_i}+\frac{1}{G_{n+2}-1}\\
&amp;\ge \frac{1}{\sum_{i=0}^nG_i}+\frac{1}{\sum_{i=0}^{n+1}G_i}+\frac{1}{\sum_{i=0}^{n+2}G_i}+\cdots.
\end{align*}
Thus, 
\begin{equation}    \label{eq1}
\sum_{k=n}^\infty\frac{1}{\sum_{i=0}^kG_i}\le\frac{1}{G_n-1}\ \ \ \ \ (n\ge N_1).
\end{equation}

On the other hand, we have
\begin{align*}
\frac{1}{\sum_{i=0}^nG_i}-\frac{1}{G_n}+\frac{1}{G_{n+1}}&amp;=\frac{1}{G_{n+2}-1}-\frac{1}{G_n}+\frac{1}{G_{n+1}}\\
&amp;=\frac{1-G_{n+1}}{G_n(G_{n+2}-1)}+\frac{1}{G_{n+1}}\\
&amp;=\frac{G_{n-1}+(-1)^n(c^2+c-1)}{G_nG_{n+1}(G_{n+2}-1)}.
\end{align*}
Similarly, we can take $n$ so large that $G_{n-1}+(-1)^n(c^2+c-1)&gt;0$ for a fixed $c$. Hence, the numerator of the right-hand side of the above identity is positive if $n\ge N_2$ for some positive integer $N_2$, so we get
\begin{align*}
\frac{1}{G_n}&amp;&lt; \frac{1}{\sum_{i=0}^nG_i}+\frac{1}{G_{n+1}}\\
&amp;&lt; \frac{1}{\sum_{i=0}^nG_i}+\frac{1}{\sum_{i=0}^{n+1}G_i}+\frac{1}{G_{n+2}}\\
&amp;&lt; \frac{1}{\sum_{i=0}^nG_i}+\frac{1}{\sum_{i=0}^{n+1}G_i}+\frac{1}{\sum_{i=0}^{n+2}G_i}+\cdots.
\end{align*}
Thus,
\begin{equation}    \label{eq2}
\sum_{k=n}^\infty\frac{1}{\sum_{i=0}^kG_i}&gt;\frac{1}{G_n}\ \ \ \ \ (n\ge N_2).
\end{equation}
Combining the two inequalities \eqref{eq1} and \eqref{eq2} we obtain 
$$\frac{1}{G_n}&lt;\sum_{k=n}^\infty\frac{1}{\sum_{i=0}^kG_i}\le\frac{1}{G_n-1},$$
where $n\ge n_0=\max\{N_1,N_2\}$, which completes the proof.\\


{\sl Proof of Theorem \ref{main3}} :
The case $a=1$ has already been proved, it suffices to show the case $a\ge2$. Defining $S_n=\sum_{i=0}^nU_i$.
We have 
\begin{align*}
\frac{1}{S_n}-\frac{1}{U_n}+\frac{1}{U_{n+1}}&amp;=\frac{1}{U_{n+1}}-\frac{S_{n-1}}{U_nS_n}\\
 &amp;=\frac{U_nS_n-U_{n+1}S_{n-1}}{U_nU_{n+1}S_n}\\
 &amp;=\frac{U_{n+1}-U_n+(-1)^{n+1}}{aU_nU_{n+1}S_n}&gt;0,
\end{align*}
and for $n\ge2$ 
\begin{align*}
\frac{1}{U_n-1}-\frac{1}{U_{n+1}-1}-\frac{1}{S_n}&amp;=\frac{S_{n-1}+1}{(U_n-1)S_n}-\frac{1}{U_{n+1}-1}\\
&amp;=\frac{U_{n+1}S_{n-1}-U_nS_n+aS_n}{(U_n-1)(U_{n+1}-1)S_n}\\
&amp;=\frac{2U_n+(-1)^n-1}{a(U_n-1)(U_{n+1}-1)S_n}&gt;0.
\end{align*}
Then, we obtain
$$\frac{1}{U_n}&lt;\sum_{k=n}^\infty\frac{1}{S_k}\le \frac{1}{U_n-1}\ \ \ \ \ (n\ge2),$$
which completes the proof.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Acknowledgement} 

The author is grateful to the anonymous referee for his/her helpful comments. 


\begin{thebibliography}{9}
\bibitem{hk}
S. H. Holliday and T. Komatsu, On the sum of reciprocal generalized Fibonacci numbers, \emph{Integers} \textbf{11A} (2011).

\bibitem{on}
H. Ohtsuka and S. Nakamura, On the sum of reciprocal sums of Fibonacci numbers, \emph{Fibonacci Quart.} \textbf{46/47} (2008/2009), 153--159.

\bibitem{lz}
R. Liu and F. Z. Zhao, On the sums of reciprocal hyperfibonacci numbers
and hyperlucas numbers, \emph{J. Integer Seq.} \textbf{15} (2012),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Liu/liu10.html}{Article 12.4.5}. 

\bibitem{wt}
Z. Wenpeng and W. Tingting, The infinite sum of reciprocal Pell
numbers, \emph{Appl. Math. Comput.} \textbf{218} (2012), 6164--6167.

\end{thebibliography}


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

\noindent \emph{Keywords: } 
second-order recurrence, Fibonacci number, reciprocal sum.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000071},
\seqnum{A000129}, and
\seqnum{A001610}.)

\bigskip
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\vspace*{+.1in}
\noindent
Received  April 26 2013;
revised version received   July 3 2013.
Published in {\it Journal of Integer Sequences}, July 30 2013.

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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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