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\begin{center}
\vskip 1cm{\LARGE\bf Alternating Sums of the Reciprocal\\[5pt] Fibonacci Numbers}
\vskip 1cm
{\large  Andrew Yezhou Wang\\
School of Mathematical Sciences \\
University of Electronic Science and Technology of China\\
Chengdu 611731\\
P. R. China\\
\href{mailto:yzwang@uestc.edu.cn}{\tt yzwang@uestc.edu.cn}\\
\ \\
Tingrui Yuan\\
School of Communication and Information Engineering \\
University of Electronic Science and Technology of China\\
Chengdu 611731\\
P. R. China\\
\href{mailto:ytr2009@icloud.com}{\tt ytr2009@icloud.com}
}
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\vskip .2 in

\begin{abstract}
In this paper, we investigate the alternating sums of the reciprocal Fibonacci numbers $\sum\nolimits_{k=n}^{mn}{(-1)^k}/{F_{ak+b}}$, where $a\in\{1,2,3\}$ and $b<a$. The integer parts of the reciprocals of these sums are expressed explicitly in terms of the Fibonacci numbers.
\end{abstract}


\section{Introduction}

For an integer $n\geq 0$, the \emph{Fibonacci number} $F_n$ is defined recurrently by $F_n=F_{n-1}+F_{n-2}$ with $F_0=0$ and $F_1=1$.

Recently, Ohtsuka and Nakamura \cite{ON89} studied the infinite sums of the reciprocal Fibonacci numbers, and established the following result, where $\lfloor\cdot\rfloor$ denotes the floor function.
\begin{theorem}\label{thmONsum}
For all $n\geq 2$,
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{\infty}\frac{1}{F_k}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
F_{n-2}, &\mbox{if $n$ is even};\\[5pt]
F_{n-2}-1, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}

More recently, Wang and Wen \cite{Wang15} strengthened Theorem \ref{thmONsum} to the finite sum case.
\begin{theorem}
If $m\geq 3$ and $n\geq 2$, then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{1}{F_k}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
F_{n-2}, &\mbox{if $n$ is even};\\[5pt]
F_{n-2}-1, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}

In this article, we focus on the alternating sums of the reciprocal Fibonacci numbers \[\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{ak+b}},\]
where $a\in\{1,2,3\}$ and $b<a$. By evaluating the integer parts of these sums, we obtain several interesting families of identities concerning the Fibonacci numbers.


\section{Results for $a=1$}

We first introduce several well-known results, which will be used throughout the article. The detailed proofs can
be found in, for example, \cite[Thm.\ 7, p.\ 9]{NNV02} and \cite{RS75}.
\begin{lemma}\label{sq}
For any positive intergers $m$ and $n$, we have
\begin{align*}
F_mF_n+F_{m+1}F_{n+1}=F_{m+n+1}.
\end{align*}
\end{lemma}

\begin{lemma}\label{minus}
For all $n\geq 1$, we have
\begin{eqnarray*}
F_{2n+1}=F_{n+1}F_{n+2}-F_{n-1}F_{n}.
\end{eqnarray*}
\end{lemma}

\begin{lemma}\label{stone}
Let $a,b,c,d$ be positive integers with $a+b=c+d$ and $b\geq\mathrm{max}\{c,d\}$. Then
\begin{align*}
F_aF_b-F_cF_d=(-1)^{a+1}F_{b-c}F_{b-d}.
\end{align*}
\end{lemma}

For the sake of argument, we present four auxiliary functions
\begin{eqnarray*}
f_1(n)&=&\frac{1}{F_{n+1}}-\frac{(-1)^n}{F_n}-\frac{1}{F_{n+2}},\\[3pt]
f_2(n)&=&\frac{1}{F_{n+1}-1}-\frac{(-1)^n}{F_n}-\frac{1}{F_{n+2}-1},\\[3pt]
f_3(n)&=&\frac{-1}{F_{n+1}+1}-\frac{(-1)^n}{F_n}+\frac{1}{F_{n+2}+1},\\[3pt]
f_4(n)&=&\frac{-1}{F_{n+1}}-\frac{(-1)^n}{F_n}+\frac{1}{F_{n+2}}.
\end{eqnarray*}
It is clear that $f_i(n)$ ($1\leq i\leq4$) is positive if $n$ is odd, and negative otherwise.

\begin{lemma}\label{lemf1}
If $n\geq2$ is even, then
\begin{align*}
f_1(n)+f_1(n+1)<0.
\end{align*}
\end{lemma}
\begin{proof}
Since $n$ is even, it is straightforward to see
\begin{eqnarray*}
f_1(n)+f_1(n+1)&=&\frac{2}{F_{n+1}}-\frac{1}{F_n}-\frac{1}{F_{n+3}}\\[3pt]
&=&\frac{\left(2F_n-F_{n+1}\right)F_{n+3}-F_nF_{n+1}}{F_nF_{n+1}F_{n+3}}\\[3pt]
&=&\frac{F_{n-2}F_{n+3}-F_nF_{n+1}}{F_nF_{n+1}F_{n+3}}\\[3pt]
&=&\frac{-2}{F_nF_{n+1}F_{n+3}}\\[3pt]
&<&0,
\end{eqnarray*}
where the last equality follows from Lemma \ref{stone} and the fact that $n$ is even.
\end{proof}

\begin{lemma}\label{lemf2}
For all $n\geq 2$, we have
\begin{align*}
f_2(n)+f_2(n+1)>0.
\end{align*}
\end{lemma}
\begin{proof}
The statement is clearly true if $n$ is odd. Thus, we focus on the case where $n$ is even. It follows from the definition of $f_2(n)$ and Lemma \ref{stone} that
\begin{eqnarray*}
f_2(n)+f_2(n+1)&=&\left(\frac{1}{F_{n+1}-1}-\frac{1}{F_{n+3}-1}\right)-
\left(\frac{1}{F_{n}}-\frac{1}{F_{n+1}}\right)\\[3pt]
&=&\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}-\frac{F_{n-1}}{F_nF_{n+1}}\\[3pt]
&=&\frac{F_{n+1}\left(F_nF_{n+2}-F_{n-1}F_{n+3}\right)+F_{n-1}\left(F_{n+1}+F_{n+3}-1\right)}
{F_nF_{n+1}\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}\\[3pt]
&=&\frac{-2F_{n+1}+F_{n-1}\left(2F_{n+1}+F_{n+2}-1\right)}
{F_nF_{n+1}\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}\\[3pt]
&=&\frac{2\left(F_{n-1}-1\right)F_{n+1}+F_{n-1}\left(F_{n+2}-1\right)}
{F_nF_{n+1}\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}\\[3pt]
&>&0,
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lem4f2}
For all $n\geq 2$, we have
\begin{align*}
\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}-\frac{F_{n-1}}{F_nF_{n+1}}-\frac{1}{F_{2n+1}-1}\geq0.
\end{align*}
\end{lemma}
\begin{proof}
Applying Lemma \ref{sq}, it is easy to see that, for $n\geq 2$,
\begin{eqnarray*}
F_{2n+1}-1-2F_nF_{n+1}=F_n^2+F_{n+1}^2-2F_nF_{n+1}-1=\left(F_{n+1}-F_n\right)^2-1\geq0,
\end{eqnarray*}
from which we derive the conclusion that
\begin{align*}
\frac{1}{F_{2n+1}-1}\leq\frac{1}{2F_nF_{n+1}}.
\end{align*}
Therefore, we have
\begin{align*}
\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}&-\frac{F_{n-1}}{F_nF_{n+1}}-\frac{1}{F_{2n+1}-1}\\[3pt]
&\geq\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}
-\frac{F_{n-1}}{F_nF_{n+1}}-\frac{1}{2F_nF_{n+1}}\\[3pt]
&=\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}-\frac{2F_{n-1}+1}{2F_nF_{n+1}},
\end{align*}
whose numerator is
\begin{align*}
\psi(n):=2F_nF_{n+1}F_{n+2}-\left(2F_{n-1}+1\right)\left(F_{n+1}-1\right)\left(F_{n+3}-1\right).
\end{align*}
Applying Lemma \ref{stone} repeatedly and the fact $F_{n+3}=3F_{n+1}-F_{n-1}$, we can obtain
\begin{eqnarray*}
\psi(n)&=&2F_{n+1}\left(F_nF_{n+2}-F_{n-1}F_{n+3}\right)+2F_{n-1}F_{n+1}+2F_{n-1}F_{n+3}-F_{n+1}F_{n+3}\\[3pt]
&&{}+(F_{n+1}+F_{n+3})-2F_{n-1}-1\\[3pt]
&=&\left((-1)^{n+1}+1\right)4F_{n+1}+2F_{n-1}F_{n+1}+(2F_{n-1}-F_{n+1})F_{n+3}-3F_{n-1}-1\\[3pt]
&=&\left((-1)^{n+1}+1\right)4F_{n+1}+F_{n-1}(2F_{n+1}-F_{n+2})+(F_{n-1}F_{n+2}-F_{n-2}F_{n+3})\\[3pt]
&&{}-3F_{n-1}-1\\[3pt]
&=&\left((-1)^{n+1}+1\right)4F_{n+1}+F_{n-1}^2-3F_{n-1}-1+(-1)^n3.
\end{eqnarray*}
If $n$ is even, we have $\psi(n)=(F_{n-1}-1)(F_{n-1}-2)\geq 0$. If $n$ is odd, we have
\begin{align*}
\psi(n)=(F_{n-1}+1)(F_{n-1}+4)+8(F_n-1)>0.
\end{align*}
Therefore, $\psi(n)\geq 0$ always holds. This completes the proof.
\end{proof}

\begin{lemma}\label{lemf2mn}
If $n\geq 2$ and $m\geq 2$, then
\begin{align*}
f_2(n)+f_2(n+1)+f_2(mn)+\frac{1}{F_{mn+2}-1}>0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is odd, then the result follows from Lemma \ref{lemf2} and the fact $f_2(mn)>0$. So we assume that $mn$ is even. Now we have
\begin{eqnarray*}
f_2(mn)+\frac{1}{F_{mn+2}-1}=\frac{1}{F_{mn+1}-1}-\frac{1}{F_{mn}}=\frac{-(F_{mn-1}-1)}{F_{mn}(F_{mn+1}-1)}
>\frac{-1}{F_{mn+1}-1}.
\end{eqnarray*}
From the proof of Lemma \ref{lemf2} we know that whether $n$ is even or odd,
\begin{align*}
f_2(n)+f_2(n+1)\geq\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}-\frac{F_{n-1}}{F_nF_{n+1}}.
\end{align*}
Therefore,
\begin{align*}
f_2(n)+f_2(n+1)+f_2(mn)+\frac{1}{F_{mn+2}-1}&>
\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}-\frac{F_{n-1}}{F_nF_{n+1}}-\frac{1}{F_{mn+1}-1}\\[3pt]
&\geq\frac{F_{n+2}}{\left(F_{n+1}-1\right)\left(F_{n+3}-1\right)}-\frac{F_{n-1}}{F_nF_{n+1}}-\frac{1}{F_{2n+1}-1}\\[3pt]
&\geq0,
\end{align*}
where the last inequality follows from Lemma \ref{lem4f2}.
\end{proof}

Employing the fact $2(F_{2n+2}+1)\geq(F_{n+1}+1)(F_{n+3}+1)$ and similar arguments in the proof of Lemma \ref{lem4f2}, we have the following result, whose proof is omitted here.
\begin{lemma}\label{lemf3}
If $n\geq 5$ is odd, then
\begin{align*}
f_3(n)+f_3(n+1)>\frac{1}{F_{2n+2}+1}.
\end{align*}
\end{lemma}

Now we establish two properties about $f_4(n)$.
\begin{lemma}\label{lemf4}
For $n\geq1$, we have
\begin{align*}
f_4(n)+f_4(n+1)<0.
\end{align*}
\end{lemma}
\begin{proof}
If $n$ is even, the result follows from the definition of $f_4(n)$. Next we consider the case where $n$ is odd. Applying the argument in the proof of Lemma \ref{lemf1}, we can easily deduce that
\begin{align*}
f_4(n)+f_4(n+1)=\frac{-2}{F_{n+1}}+\frac{1}{F_n}+\frac{1}{F_{n+3}}=\frac{-2}{F_{n}F_{n+1}F_{n+3}}<0.
\end{align*}
This completes the proof.
\end{proof}

\begin{lemma}\label{lemf4mn}
If $n\geq 1$ and $m\geq 2$, then
\begin{align*}
f_4(n)+f_4(n+1)+f_4(mn)<0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is even, the result follows from Lemma \ref{lemf4} and the fact $f_4(mn)<0$. So we assume that $mn$ is odd, which implies that $m\geq 3$ and $n$ is odd. Since $mn$ is odd, we have
\begin{align*}
f_4(mn)=\frac{-1}{F_{mn+1}}+\frac{1}{F_{mn}}+\frac{1}{F_{mn+2}}<\frac{1}{F_{mn}}\leq\frac{1}{F_{3n}}.
\end{align*}
Now we have
\begin{align*}
f_4(n)+f_4(n+1)+f_4(mn)<\frac{-2}{F_nF_{n+1}F_{n+3}}+\frac{1}{F_{3n}}.
\end{align*}
To complete the proof, we only need to show that $2F_{3n}>F_nF_{n+1}F_{n+3}$.

It follows from Lemma \ref{sq} that $F_{2n+2}=F_{n-1}F_{n+2}+F_nF_{n+3}$, which implies $F_{n}F_{n+1}F_{n+3}<F_{n+1}F_{2n+2}$. Furthermore, employing Lemma \ref{sq} again, we can conclude that
\begin{eqnarray*}
F_{n+1}F_{2n+2}&=&(F_{n-1}+F_n)(F_{2n}+F_{2n+1})\\[3pt]
&=&(F_{n-1}F_{2n}+F_nF_{2n+1})+F_{n-1}F_{2n+1}+F_nF_{2n}\\[3pt]
&=&F_{3n}+F_{n-1}F_{2n+1}+F_{n+1}F_{2n}-F_{n-1}F_{2n}\\[3pt]
&=&F_{3n}+(F_{n-1}F_{2n-1}+F_{n+1}F_{2n})\\[3pt]
&<&2F_{3n},
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{theorem}
If $n\geq 4$ and $m\geq 2$, then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
F_{n+1}-1, &\mbox{if $n$ is even};\\[5pt]
-F_{n+1}-1, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}
\begin{proof}
We first consider the case where $n$ is even. It follows from Lemma \ref{lemf1} that
\begin{align*}
\sum\limits_{k=n}^{mn-1}f_1(k)<0.
\end{align*}
It is clear that $mn$ is even, which ensures that
\begin{align*}
f_1(mn)+\frac{1}{F_{mn+2}}<0.
\end{align*}
With the help of $f_1(n)$ and the above two inequalities, we can obtain
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}=\frac{1}{F_{n+1}}-\left(\frac{1}{F_{mn+2}}+f_1(mn)\right)
-\sum\limits_{k=n}^{mn-1}f_1(k)>\frac{1}{F_{n+1}}.
\end{eqnarray*}
Applying Lemma \ref{lemf2} and Lemma \ref{lemf2mn}, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}
&=&\frac{1}{F_{n+1}-1}-\left(f_2(n)+f_2(n+1)+f_2(mn)+\frac{1}{F_{mn+2}-1}\right)-\sum\limits_{k=n+2}^{mn-1}f_2(k)\\
&<&\frac{1}{F_{n+1}-1}.
\end{eqnarray*}
Therefore, we obtain
\begin{eqnarray*}
\frac{1}{F_{n+1}}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}<\frac{1}{F_{n+1}-1},
\end{eqnarray*}
which shows that the statement is true when $n$ is even.

We now turn to consider the case where $n\geq 5$ is odd. If $mn$ is odd, it is easy to see that
\begin{align*}
f_3(mn)-\frac{1}{F_{mn+2}+1}>0.
\end{align*}
Lemma \ref{lemf3} tells us that $f_3(n)+f_3(n+1)>0$. Therefore,
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}
=\frac{-1}{F_{n+1}+1}-\sum\limits_{k=n}^{mn-1}f_3(k)-\left(f_3(mn)-\frac{1}{F_{mn+2}+1}\right)<\frac{-1}{F_{n+1}+1}.
\end{eqnarray*}
If $mn$ is even, employing Lemma \ref{lemf3} again, we can deduce
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}
&=&\frac{-1}{F_{n+1}+1}-\sum\limits_{k=n+2}^{mn}f_3(k)-\left(f_3(n)+f_3(n+1)-\frac{1}{F_{mn+2}+1}\right)\\[3pt]
&\leq&\frac{-1}{F_{n+1}+1}-\sum\limits_{k=n+2}^{mn}f_3(k)-\left(f_3(n)+f_3(n+1)-\frac{1}{F_{2n+2}+1}\right)\\[3pt]
&<&\frac{-1}{F_{n+1}+1}.
\end{eqnarray*}
Now we can conclude that if $n\geq 5$ is odd, then
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}<\frac{-1}{F_{n+1}+1}.
\end{eqnarray*}
If $mn$ is even, then Lemma \ref{lemf4} implies that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}f_4(k)<0.
\end{eqnarray*}
If $mn$ is odd, invoking Lemma \ref{lemf4} and Lemma \ref{lemf4mn}, we can get
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}f_4(k)=\sum\limits_{k=n+2}^{mn-1}f_4(k)+\left(f_4(n)+f_4(n+1)+f_4(mn)\right)<0.
\end{eqnarray*}
Thus, we always have
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}f_4(k)<0,
\end{eqnarray*}
from which we obtain
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}&=&\frac{-1}{F_{n+1}}+\frac{1}{F_{mn+2}}-\sum\limits_{k=n}^{mn}f_4(k)
>\frac{-1}{F_{n+1}}.
\end{eqnarray*}
Therefore, we arrive at
\begin{eqnarray*}
\frac{-1}{F_{n+1}}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{k}}<\frac{-1}{F_{n+1}+1},
\end{eqnarray*}
which shows that the result holds for odd $n$.
\end{proof}


\section{Results for $a=2$}

We first introduce the following notations
\begin{eqnarray*}
g_1(n)&=&\frac{1}{F_{2n-2}+F_{2n}}-\frac{(-1)^n}{F_{2n}}-\frac{1}{F_{2n}+F_{2n+2}},\\[3pt]
g_2(n)&=&\frac{1}{F_{2n-2}+F_{2n}-1}-\frac{(-1)^n}{F_{2n}}-\frac{1}{F_{2n}+F_{2n+2}-1},\\[3pt]
g_3(n)&=&\frac{1}{F_{2n-2}+F_{2n}+1}-\frac{(-1)^n}{F_{2n}}-\frac{1}{F_{2n}+F_{2n+2}+1},\\[3pt]
g_4(n)&=&\frac{-1}{F_{2n-2}+F_{2n}}-\frac{(-1)^n}{F_{2n}}+\frac{1}{F_{2n}+F_{2n+2}},\\[3pt]
g_5(n)&=&\frac{-1}{F_{2n-2}+F_{2n}+1}-\frac{(-1)^n}{F_{2n}}+\frac{1}{F_{2n}+F_{2n+2}+1}.
\end{eqnarray*}
It is routine to check that for $1\leq i\leq 5$, $g_i(n)$ is positive if $n$ is odd, and negative otherwise.

\begin{lemma}\label{lemg1}
If $n\geq 1$, then $g_1(n)+g_1(n+1)>0$ and
\begin{align*}
g_1(n)+g_1(n+1)>g_1(n+2)+g_1(n+3).
\end{align*}
\end{lemma}
\begin{proof}
If $n$ is odd, we have
\begin{eqnarray*}
g_1(n)+g_1(n+1)&=&\left(\frac{1}{F_{2n-2}+F_{2n}}-\frac{1}{F_{2n+2}+F_{2n+4}}\right)+
\left(\frac{1}{F_{2n}}-\frac{1}{F_{2n+2}}\right)\\[3pt]
&=&\frac{5F_{2n+1}}{(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}+\frac{F_{2n+1}}{F_{2n}F_{2n+2}}\\[3pt]
&>&0.
\end{eqnarray*}
Applying the easily checked fact
\begin{eqnarray*}
\frac{F_{2n+1}}{F_{2n-2}+F_{2n}}&>&\frac{F_{2n+5}}{F_{2n+6}+F_{2n+8}},\\[3pt]
\frac{F_{2n+1}}{F_{2n}F_{2n+2}}&>&\frac{F_{2n+5}}{F_{2n+4}F_{2n+6}},
\end{eqnarray*}
we can conclude that $g_1(n)+g_1(n+1)>g_1(n+2)+g_1(n+3)$.

Now we consider the case where $n$ is even. Doing some elementary manipulations and using Lemma \ref{stone}, we have
\begin{eqnarray*}
g_1(n)+g_1(n+1)&=&\left(\frac{1}{F_{2n-2}+F_{2n}}-\frac{1}{F_{2n}}\right)+
\left(\frac{1}{F_{2n+2}}-\frac{1}{F_{2n+2}+F_{2n+4}}\right)\\[3pt]
&=&\frac{F_{2n-2}(F_{2n}F_{2n+4}-F_{2n+2}^2)+(F_{2n}^2-F_{2n-2}F_{2n+2})F_{2n+4}}
{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}\\[3pt]
&=&\frac{F_{2n+4}-F_{2n-2}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}\\[3pt]
&=&\frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}\\[3pt]
&>&0.
\end{eqnarray*}
Applying the above identity, we see that
\begin{eqnarray*}
\frac{g_1(n)+g_1(n+1)}{g_1(n+2)+g_1(n+3)}=\frac{F_{2n+1}F_{2n+4}F_{2n+6}}{F_{2n}F_{2n+2}F_{2n+5}}\cdot
\frac{F_{2n+6}+F_{2n+8}}{F_{2n-2}+F_{2n}}>1.
\end{eqnarray*}
Thus, $g_1(n)+g_1(n+1)>g_1(n+2)+g_1(n+3)$ also holds.
\end{proof}

\begin{lemma}\label{lemg1pre}
For $n\geq 1$, we have
\begin{align*}
F_{6n+2}>F_{2n}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4}).
\end{align*}
\end{lemma}
\begin{proof}
It follows from Lemma \ref{stone} that
\begin{eqnarray*}
F_{2n-1}F_{2n+3}-F_{2n-2}F_{2n+4}&=&5,\\[3pt]
F_{2n-1}F_{2n+1}-F_{2n}^2&=&1,\\[3pt]
F_{2n+1}F_{2n+3}-F_{2n}F_{2n+4}&=&2.
\end{eqnarray*}
Thus, $F_{2n-1}F_{2n+3}>F_{2n-2}F_{2n+4}$, $F_{2n-1}F_{2n+1}>F_{2n}^2$, and $F_{2n+1}F_{2n+3}>F_{2n}F_{2n+4}$.

Employing Lemma \ref{sq} repeatedly and the above three inequalities, we have
\begin{eqnarray*}
F_{6n+2}&=&F_{2n}F_{4n+1}+F_{2n+1}F_{4n+2}\\[3pt]
&=&F_{2n}(F_{2n-2}F_{2n+2}+F_{2n-1}F_{2n+3})+F_{2n+1}(F_{2n-1}F_{2n+2}+F_{2n}F_{2n+3})\\[3pt]
&>&F_{2n-2}F_{2n}F_{2n+2}+F_{2n-2}F_{2n+4}F_{2n}+F_{2n}^2F_{2n+2}+F_{2n}F_{2n+4}F_{2n}\\[3pt]
&=&F_{2n}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4}),
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lemg1mn}
If $n\geq 1$ and $m\geq 3$, then
\begin{align*}
g_1(n)+g_1(n+1)+g_1(mn)>0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is odd, then the result follows from Lemma \ref{lemg1} and the fact $g_1(mn)>0$. Thus we focus on the case where $mn$ is even. For $k\geq 1$,
\begin{eqnarray*}
\frac{1}{F_{2k-2}+F_{2k}}-\frac{1}{F_{2k}}&=&-\frac{F_{2k-2}}{(F_{2k-2}+F_{2k})F_{2k}}\\[3pt]
&=&-\frac{F_{2k-2}}{F_{2k-2}F_{2k}+F_{2k}^2}\\[3pt]
&>&-\frac{F_{2k-2}}{F_{2k-2}F_{2k+2}}\\[3pt]
&=&-\frac{1}{F_{2k+2}},
\end{eqnarray*}
where the inequality follows from $F_{2k}^2-F_{2k-2}F_{2k+2}=1$. Since $mn$ is even, employing the above inequality, we have
\begin{eqnarray*}
g_1(mn)>-\frac{1}{F_{2mn+2}}-\frac{1}{F_{2mn}+F_{2mn+2}}>-\frac{2}{F_{2mn+2}}\geq-\frac{2}{F_{6n+2}}.
\end{eqnarray*}

From the proof of Lemma \ref{lemg1} we know that whether $n$ is even or odd, we always have
\begin{align*}
g_1(n)+g_1(n+1)\geq\frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}.
\end{align*}
Therefore,
\begin{eqnarray*}
g_1(n)+g_1(n+1)+g_1(mn)&>&\frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}-\frac{2}{F_{6n+2}}\\[3pt]
&>&\frac{2}{F_{2n}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}-\frac{2}{F_{6n+2}}\\[3pt]
&>&0,
\end{eqnarray*}
where the last inequality follows from Lemma \ref{lemg1pre}.
\end{proof}

\begin{lemma}\label{lemg2pre}
If $n>0$, then
\begin{align*}
2F_{4n}(F_{4n}+F_{4n+2})>F_{2n+2}F_{4n+3}(F_{2n-2}+F_{2n}).
\end{align*}
\end{lemma}
\begin{proof}
It suffices to show that $2F_{4n}^2>F_{2n-2}F_{2n+2}F_{4n+3}$ and $2F_{4n}F_{4n+2}>F_{2n}F_{2n+2}F_{4n+3}$. These two inequalities can be proved using similar arguments, so we only prove the first one.

Applying Lemma \ref{stone} repeatedly and Lemma \ref{sq}, we can obtain
\begin{eqnarray*}
2F_{4n}^2&=&2F_{4n-3}F_{4n+3}-8\\[3pt]
&=&2(F_{2n-2}^2+F_{2n-1}^2)F_{4n+3}-8\\[3pt]
&>&(F_{2n-2}F_{2n-1}+2F_{2n-1}^2)F_{4n+3}-8\\[3pt]
&=&F_{2n-1}F_{2n+1}F_{4n+3}-8\\[3pt]
&=&(F_{2n-2}F_{2n+2}+2)F_{4n+3}-8\\[3pt]
&>&F_{2n-2}F_{2n+2}F_{4n+3}.
\end{eqnarray*}
The proof is completed.
\end{proof}

\begin{lemma}\label{lemg2}
For all $n\geq 2$, we have
\begin{align*}
g_2(n)+g_2(n+1)+g_2(2n)>0.
\end{align*}
\end{lemma}
\begin{proof}
It is straightforward to verify that $F_{2n-2}+F_{2n}+F_{2n+2}+F_{2n+4}=3(F_{2n}+F_{2n+2})$. Applying Lemma \ref{stone} repeatedly, we get
\begin{eqnarray*}
(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})&=&F_{2n-2}F_{2n+2}+F_{2n-2}F_{2n+4}+F_{2n}F_{2n+2}+F_{2n}F_{2n+4}\\[3pt]
&=&F_{2n-2}F_{2n+2}+(F_{2n}F_{2n+2}-3)+F_{2n}F_{2n+2}\\[3pt]
&&{}+F_{2n}(2F_{2n+2}+F_{2n+1})\\[3pt]
&=&(F_{2n-2}F_{2n+2}-F_{2n}^2)+(F_{2n}^2+F_{2n}F_{2n+1})\\[3pt]
&&{}+4F_{2n}F_{2n+2}-3\\[3pt]
&=&5F_{2n}F_{2n+2}-4.
\end{eqnarray*}

It follows from the definition of $g_2(n)$ and the above two equations that
\begin{eqnarray*}
g_2(n)+g_2(n+1)&\geq&\left(\frac{1}{F_{2n-2}+F_{2n}-1}-\frac{1}{F_{2n+2}+F_{2n+4}-1}\right)
-\left(\frac{1}{F_{2n}}-\frac{1}{F_{2n+2}}\right)\\[3pt]
&=&\frac{5F_{2n+1}}{(F_{2n-2}+F_{2n}-1)(F_{2n+2}+F_{2n+4}-1)}-\frac{F_{2n+1}}{F_{2n}F_{2n+2}}\\[3pt]
&=&\frac{3(F_{2n}+F_{2n+2}+1)F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n}-1)(F_{2n+2}+F_{2n+4}-1)}\\[3pt]
&>&\frac{1}{F_{2n+2}(F_{2n-2}+F_{2n}-1)},
\end{eqnarray*}
where the last inequality follows from $3F_{n}>F_{n+2}$.

It is routine to show
\begin{eqnarray*}
2(F_{4n+2}-F_{4n-2})&=&2(2F_{4n}+F_{4n-1}-F_{4n-2})\\[3pt]
&=&3F_{4n}+F_{4n}+2F_{4n-3}\\[3pt]
&>&3F_{4n}+(2F_{4n-2}+F_{4n-3})+F_{4n-2}\\[3pt]
&>&3(F_{4n-2}+F_{4n}),
\end{eqnarray*}
which means
\begin{align*}
F_{4n+2}-F_{4n-2}>\frac{3}{2}(F_{4n-2}+F_{4n}).
\end{align*}

Employing the above inequality, we can deduce that
\begin{eqnarray*}
g_2(2n)&=&\frac{F_{4n+2}-F_{4n-2}}{(F_{4n-2}+F_{4n}-1)(F_{4n}+F_{4n+2}-1)}-\frac{1}{F_{4n}}\\[3pt]
&>&\frac{3}{2(F_{4n}+F_{4n+2}-1)}-\frac{1}{F_{4n}}\\[3pt]
&=&\frac{-F_{4n+3}+2}{2F_{4n}(F_{4n}+F_{4n+2}-1)}\\[3pt]
&>&-\frac{F_{4n+3}}{2F_{4n}(F_{4n}+F_{4n+2}-1)}.
\end{eqnarray*}

Now we conclude that
\begin{eqnarray*}
g_2(n)+g_2(n+1)+g_2(2n)>\frac{1}{F_{2n+2}(F_{2n-2}+F_{2n}-1)}-\frac{F_{4n+3}}{2F_{4n}(F_{4n}+F_{4n+2}-1)}>0,
\end{eqnarray*}
where the last inequality follows from Lemma \ref{lemg2pre}.
\end{proof}

Applying the argument in the proof of Lemma \ref{lemg2}, it can be readily seen the following property of $g_3(n)$, whose proof is omitted here.
\begin{lemma}\label{lemg3}
If $n\geq 2$ is even, we have
\begin{align*}
g_3(n)+g_3(n+1)<0.
\end{align*}
\end{lemma}

Imitating the proof of Lemma \ref{lemg1} and Lemma \ref{lemg1mn} respectively, we can easily get the following results on $g_4(n)$.
\begin{lemma}\label{lemg4}
For $n\geq 1$, we have
\begin{align*}
g_4(n)+g_4(n+1)<0.
\end{align*}
\end{lemma}
\begin{lemma}\label{lemg4mn}
If $n\geq1$ and $m\geq2$, then
\begin{align*}
g_4(n)+g_4(n+1)+g_4(mn)<0.
\end{align*}
\end{lemma}

\begin{lemma}\label{lemg5}
If $n\geq 1$ is odd, we have
\begin{align*}
g_5(n)+g_5(n+1)>\frac{1}{F_{4n}+F_{4n+2}+1}.
\end{align*}
\end{lemma}
\begin{proof}
It is easy to see that the result is true for $n=1$, thus we assume that $n\geq3$. From the proof of Lemma \ref{lemg2}, we can easily obtain that if $n\geq 3$ is odd, then
\begin{eqnarray*}
g_5(n)+g_5(n+1)&=&\frac{3(F_{2n}+F_{2n+2}-1)F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n}+1)(F_{2n+2}+F_{2n+4}+1)}\\[3pt]
&>&\frac{1}{F_{2n+2}(F_{2n-2}+F_{2n}+1)}.
\end{eqnarray*}

Employing Lemma \ref{sq} repeatedly, it is easy to see that
\begin{eqnarray*}
F_{2n+2}(F_{2n-2}+F_{2n}+1)&<&F_{2n-2}F_{2n+3}+F_{2n}F_{2n+3}+F_{2n+2}\\[3pt]
&=&F_{4n}-F_{2n-3}F_{2n+2}+F_{4n+2}-F_{2n-1}F_{2n+2}+F_{2n+2}\\[3pt]
&<&F_{4n}+F_{4n+2}.
\end{eqnarray*}
Combining the above two inequalities yields the desired result.
\end{proof}

\begin{lemma}\label{lem2k2}
For $n\geq 2$, we have
\begin{align*}
F_{4n-2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})>F_{4n}(F_{4n-2}+F_{4n}).
\end{align*}
\end{lemma}
\begin{proof}
We first consider the right-hand side. Applying $F_{4n}^2-F_{4n-1}F_{4n+1}=-1$, we have
\begin{eqnarray*}
F_{4n}(F_{4n-2}+F_{4n})=F_{4n-2}F_{4n}+F_{4n}^2=F_{4n-2}F_{4n}+F_{4n-1}F_{4n+1}-1=F_{8n-1}-1.
\end{eqnarray*}

For the left-hand side, we have that if $n\geq2$, then
\begin{eqnarray*}
(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})&=&F_{2n-2}F_{2n+2}+F_{2n}F_{2n+2}+F_{2n-2}F_{2n+4}+F_{2n}F_{2n+4}\\[3pt]
&>&(F_{2n-2}F_{2n+1}+F_{2n-1}F_{2n+2})+(F_{2n-2}F_{2n+3}\\[3pt]
&&{}+F_{2n-1}F_{2n+4})+F_{2n-2}F_{2n+4}\\[3pt]
&>&F_{4n}+F_{4n+2}+2.
\end{eqnarray*}
Therefore, using the fact $F_{4n-2}F_{4n+2}-F_{4n-1}F_{4n+1}=-2$, we have
\begin{eqnarray*}
F_{4n-2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})&>&F_{4n-2}F_{4n}+F_{4n-2}F_{4n+2}+2\\[3pt]
&=&F_{4n-2}F_{4n}+F_{4n-1}F_{4n+1}\\[3pt]
&=&F_{8n-1}.
\end{eqnarray*}
Thus the left-hand side is greater than the right-hand side.
\end{proof}

\begin{theorem}
If $n\geq 2$ is even and $m\geq 2$, then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
F_{2n-2}+F_{2n}-1, &\mbox{if $m=2$};\\[5pt]
F_{2n-2}+F_{2n}, &\mbox{if $m>2$}.
\end{array}
\right.
\end{equation*}
\end{theorem}
\begin{proof}
We first consider the case where $m=2$. From Lemma \ref{lemg1} we know that
\begin{eqnarray*}
\sum\limits_{k=n}^{2n-1}g_1(k)<\frac{4F_{2n+1}}{F_{2n}F_{2n+2}(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}\cdot\frac{n}{2}
<\frac{1}{(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}.
\end{eqnarray*}
In addition,
\begin{align*}
g_1(2n)+\frac{1}{F_{4n}+F_{4n+2}}=\frac{1}{F_{4n-2}+F_{4n}}-\frac{1}{F_{4n}}=\frac{-F_{4n-2}}{F_{4n}(F_{4n-2}+F_{4n})}.
\end{align*}
Therefore, invoking Lemma \ref{lem2k2}, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}g_1(k)+\frac{1}{F_{4n}+F_{4n+2}}<\frac{1}{(F_{2n-2}+F_{2n})(F_{2n+2}+F_{2n+4})}
-\frac{F_{4n-2}}{F_{4n}(F_{4n-2}+F_{4n})}<0.
\end{eqnarray*}

Now with the help of $g_1(n)$, we can obtain
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{2k}}=\frac{1}{F_{2n-2}+F_{2n}}-\frac{1}{F_{4n}+F_{4n+2}}
-\sum\limits_{k=n}^{2n}g_1(k)>\frac{1}{F_{2n-2}+F_{2n}}.
\end{eqnarray*}

From the proof of Lemma \ref{lemg2}, we know that $g_2(n)+g_2(n+1)>0$. Moreover, applying Lemma \ref{lemg2}, we can deduce
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}g_2(k)=g_2(n)+g_2(n+1)+g_2(2n)+\sum\limits_{k=n+2}^{2n-1}g_2(k)>0.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{2k}}=\frac{1}{F_{2n-2}+F_{2n}-1}-\frac{1}{F_{4n}+F_{4n+2}-1}
-\sum\limits_{k=n}^{2n}g_2(k)<\frac{1}{F_{2n-2}+F_{2n}-1}.
\end{eqnarray*}

We now conclude that
\begin{eqnarray*}
\frac{1}{F_{2n-2}+F_{2n}}<\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{2k}}<\frac{1}{F_{2n-2}+F_{2n}-1},
\end{eqnarray*}
which shows that the statement for $m=2$ is true.

Next we turn to consider the case where $m>2$. First, employing Lemma \ref{lemg1} and Lemma \ref{lemg1mn}, we see that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}
<\frac{1}{F_{2n-2}+F_{2n}}-\left(g_1(n)+g_1(n+1)+g_1(mn)\right)
-\sum\limits_{k=n+2}^{mn-1}g_1(k)<\frac{1}{F_{2n-2}+F_{2n}}.
\end{eqnarray*}

We write the sum in terms of $g_3(n)$ as
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}
&=&\frac{1}{F_{2n-2}+F_{2n}+1}-\sum\limits_{k=n}^{mn-1}g_3(k)-\left(g_3(mn)+\frac{1}{F_{2mn}+F_{2mn+2}+1}\right)\\[3pt]
&=&\frac{1}{F_{2n-2}+F_{2n}+1}-\sum\limits_{k=n}^{mn-1}g_3(k)
-\left(\frac{1}{F_{2mn-2}+F_{2mn}+1}-\frac{1}{F_{2mn}}\right)\\[3pt]
&>&\frac{1}{F_{2n-2}+F_{2n}+1},
\end{eqnarray*}
where the last inequality follows from Lemma \ref{lemg3}. Now we get
\begin{eqnarray*}
\frac{1}{F_{2n-2}+F_{2n}+1}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}<\frac{1}{F_{2n-2}+F_{2n}},
\end{eqnarray*}
which yields the desired identity.
\end{proof}

\begin{theorem}
If $n\geq 1$ is odd and $m\geq 2$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}\right)^{-1}\right\rfloor=
-F_{2n-2}-F_{2n}-1.
\end{equation*}
\end{theorem}
\begin{proof}
If $mn$ is even, it follows from Lemma \ref{lemg4} that
\begin{align*}
\sum\limits_{k=n}^{mn}g_4(k)<0.
\end{align*}
If $mn$ is odd, then Lemma \ref{lemg4} and Lemma \ref{lemg4mn} ensure that
\begin{align*}
\sum\limits_{k=n}^{mn}g_4(k)=\sum\limits_{k=n+2}^{mn-1}g_4(k)+\left(g_4(n)+g_4(n+1)+g_4(mn)\right)<0.
\end{align*}
Therefore, we always have
\begin{align*}
\sum\limits_{k=n}^{mn}g_4(k)<0.
\end{align*}

With the help of $g_4(n)$, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}=\frac{-1}{F_{2n-2}+F_{2n}}+\frac{1}{F_{2mn-2}+F_{2mn}}
-\sum\limits_{k=n}^{mn}g_4(k)>\frac{-1}{F_{2n-2}+F_{2n}}.
\end{eqnarray*}

From Lemma \ref{lemg5} we know that if $n$ is odd, then $g_5(n)+g_5(n+1)>0$. Now we claim that
\begin{align*}
\sum\limits_{k=n}^{mn}g_5(k)>\frac{1}{F_{2mn}+F_{2mn+2}+1}.
\end{align*}
If $mn$ is even, employing Lemma \ref{lemg5}, we obtain
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}g_5(k)-\frac{1}{F_{2mn}+F_{2mn+2}+1}&\geq&\sum\limits_{k=n}^{mn}g_5(k)-
\frac{1}{F_{4n}+F_{4n+2}+1}\\[3pt]
&\geq&g_5(n)+g_5(n+1)-\frac{1}{F_{4n}+F_{4n+2}+1}\\[3pt]
&>&0.
\end{eqnarray*}
If $mn$ is odd, then
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}g_5(k)-\frac{1}{F_{2mn}+F_{2mn+2}+1}&=&\sum\limits_{k=n}^{mn-1}g_5(k)+
\left(g_5(mn)-\frac{1}{F_{2mn}+F_{2mn+2}+1}\right)\\[3pt]
&>&-\frac{1}{F_{2mn-2}+F_{2mn}+1}+\frac{1}{F_{2mn}}\\[3pt]
&>&0.
\end{eqnarray*}

Therefore, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}=\frac{-1}{F_{2n-2}+F_{2n}+1}+\frac{1}{F_{2mn-2}+F_{2mn}+1}
-\sum\limits_{k=n}^{mn}g_5(k)<\frac{-1}{F_{2n-2}+F_{2n}+1}.
\end{eqnarray*}

Now we can conclude that
\begin{eqnarray*}
\frac{-1}{F_{2n-2}+F_{2n}}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k}}<\frac{-1}{F_{2n-2}+F_{2n}+1},
\end{eqnarray*}
from which the desired result follows.
\end{proof}

Similarly, we can prove the following results.
\begin{theorem}
If $n\geq 4$ is even and $m\geq 2$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k-1}}\right)^{-1}\right\rfloor=
F_{2n-3}+F_{2n-1}-1.
\end{equation*}
\end{theorem}

\begin{theorem}
If $n\geq 3$ is odd and $m\geq 2$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{2k-1}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
-F_{2n-3}-F_{2n-1}-1, &\mbox{if $m=2$};\\[5pt]
-F_{2n-3}-F_{2n-1}, &\mbox{if $m>2$}.
\end{array}
\right.
\end{equation*}
\end{theorem}


\section{Results for $a=3$}

We first introduce the following notations:
\begin{eqnarray*}
s_1(n)&=&\frac{1}{2F_{3n-1}}-\frac{(-1)^n}{F_{3n}}-\frac{1}{2F_{3n+2}},\\[3pt]
s_2(n)&=&\frac{1}{2F_{3n-1}-1}-\frac{(-1)^n}{F_{3n}}-\frac{1}{2F_{3n+2}-1},\\[3pt]
s_3(n)&=&\frac{-1}{2F_{3n-1}}-\frac{(-1)^n}{F_{3n}}+\frac{1}{2F_{3n+2}},\\[3pt]
s_4(n)&=&\frac{-1}{2F_{3n-1}+1}-\frac{(-1)^n}{F_{3n}}+\frac{1}{2F_{3n+2}+1}.
\end{eqnarray*}
It is easy to see that for each $i$, $s_i(n)$ is positive if $n$ is odd, and negative otherwise.

\begin{lemma}\label{lems1}
If $n\geq 2$ is even, then
\begin{align*}
s_1(n)+s_1(n+1)<0.
\end{align*}
\end{lemma}
\begin{proof}
Since $n$ is even, applying Lemma \ref{stone} twice, we have
\begin{eqnarray*}
s_1(n)+s_1(n+1)&=&\left(\frac{1}{2F_{3n-1}}-\frac{1}{2F_{3n+5}}\right)-
\left(\frac{1}{F_{3n}}-\frac{1}{F_{3n+3}}\right)\\[3pt]
&=&\frac{2F_{3n+2}}{F_{3n-1}F_{3n+5}}-\frac{2F_{3n+1}}{F_{3n}F_{3n+3}}\\[3pt]
&=&2\cdot\frac{F_{3n}F_{3n+2}F_{3n+3}-F_{3n-1}F_{3n+1}F_{3n+5}}{F_{3n-1}F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&=&2\cdot\frac{F_{3n}F_{3n+2}F_{3n+3}-(F_{3n}^2+1)F_{3n+5}}{F_{3n-1}F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&=&2\cdot\frac{F_{3n}(F_{3n+2}F_{3n+3}-F_{3n}F_{3n+5})-F_{3n+5}}{F_{3n-1}F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&=&2\cdot\frac{2F_{3n}-F_{3n+5}}{F_{3n-1}F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&<&0,
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lems2}
For all $n\geq 1$, we have
\begin{align*}
s_2(n)+s_2(n+1)>0.
\end{align*}
\end{lemma}
\begin{proof}
It is clear that the result holds if $n$ is odd. In the rest, we assume that $n$ is even. Applying the analysis in the proof of Lemma \ref{lems1}, we can easily obtain
\begin{align*}
s_2(n)+s_2(n+1)&=\left(\frac{1}{2F_{3n-1}-1}-\frac{1}{2F_{3n+5}-1}\right)-
\left(\frac{1}{F_{3n}}-\frac{1}{F_{3n+3}}\right)\\[3pt]
&=\frac{8F_{3n+2}}{(2F_{3n-1}-1)(2F_{3n+5}-1)}-\frac{2F_{3n+1}}{F_{3n}F_{3n+3}}\\[3pt]
&=\frac{8(F_{3n}F_{3n+2}F_{3n+3}-F_{3n-1}F_{3n+1}F_{3n+5})+2F_{3n+1}(2F_{3n-1}+2F_{3n+5}-1)}
{(2F_{3n-1}-1)(2F_{3n+5}-1)F_{3n}F_{3n+3}}\\[3pt]
&=\frac{16F_{3n}-8F_{3n+5}+2F_{3n+1}(2F_{3n-1}+2F_{3n+5}-1)}{(2F_{3n-1}-1)(2F_{3n+5}-1)F_{3n}F_{3n+3}}\\[3pt]
&>\frac{4F_{3n+1}F_{3n+5}-8F_{3n+5}}{(2F_{3n-1}-1)(2F_{3n+5}-1)F_{3n}F_{3n+3}}\\[3pt]
&>0.
\end{align*}
The proof is completed.
\end{proof}

\begin{lemma}\label{lems2mn}
If $n\geq 1$ and $m\geq2$, then
\begin{align*}
s_2(n)+s_2(n+1)+s_2(mn)>0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is odd, then the result follows from Lemma \ref{lems2} and the fact $s_2(mn)>0$. So we assume that $mn$ is even. Now it is clear that
\begin{align*}
s_2(mn)=\frac{1}{2F_{3mn-1}-1}-\frac{1}{F_{3mn}}-\frac{1}{2F_{3mn+2}-1}>-\frac{1}{F_{3mn}}\geq-\frac{1}{F_{6n}}.
\end{align*}
If $n$ is odd, we have
\begin{eqnarray*}
s_2(n)+s_2(n+1)&>&\frac{1}{F_{3n}}-\frac{1}{F_{3n+3}}=\frac{2F_{3n+1}}{F_{3n}F_{3n+3}}>\frac{2}{F_{3n}F_{3n+3}}.
\end{eqnarray*}
If $n$ is even, then from Lemma \ref{lems2} we know that
\begin{eqnarray*}
s_2(n)+s_2(n+1)&>&\frac{4F_{3n+1}F_{3n+5}-8F_{3n+5}}{(2F_{3n-1}-1)(2F_{3n+5}-1)F_{3n}F_{3n+3}}\\[3pt]
&=&\frac{4F_{3n+5}(2F_{3n-1}+F_{3n-2}-2)}{(2F_{3n-1}-1)(2F_{3n+5}-1)F_{3n}F_{3n+3}}\\[3pt]
&>&\frac{2}{F_{3n}F_{3n+3}}.
\end{eqnarray*}

Now we can derive the conclusion that
\begin{eqnarray*}
s_2(n)+s_2(n+1)+s_2(mn)>\frac{2}{F_{3n}F_{3n+3}}-\frac{1}{F_{6n}}\geq0,
\end{eqnarray*}
where the last inequality follows from
\begin{eqnarray*}
2F_{6n}=F_{3n}(2F_{3n-1}+2F_{3n+1})>F_{3n}(F_{3n}+2F_{3n+1})=F_{3n}F_{3n+3}.
\end{eqnarray*}
This completes the proof.
\end{proof}


\begin{lemma}\label{lems3}
For all $n\geq 1$,
\begin{align*}
s_3(n)+s_3(n+1)<0.
\end{align*}
\end{lemma}
\begin{proof}
The result clearly holds when $n$ is even. If $n$ is odd, applying similar analysis in the proof of Lemma \ref{lems1}, we can easily derive
\begin{eqnarray*}
s_3(n)+s_3(n+1)=2\cdot\frac{2F_{3n}-F_{3n+5}}{F_{3n-1}F_{3n}F_{3n+3}F_{3n+5}}<0,
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lems3mn}
If $n\geq 1$ and $m\geq 2$, then
\begin{align*}
s_3(n)+s_3(n+1)+s_3(mn)<0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is even, then the result follows from Lemma \ref{lems3} and the fact $s_3(mn)<0$. Now we assume that $mn$ is odd, which implies that $n$ is odd and $m\geq 3$. First we have
\begin{eqnarray*}
s_3(mn)=\frac{-1}{2F_{3mn-1}}+\frac{1}{F_{3mn}}+\frac{1}{2F_{3mn+2}}<\frac{1}{F_{3mn}}\leq\frac{1}{F_{9n}}.
\end{eqnarray*}
Moreover, from the proof of Lemma \ref{lems3} we know
\begin{eqnarray*}
s_3(n)+s_3(n+1)=-\frac{2(F_{3n+5}-2F_{3n})}{F_{3n-1}F_{3n}F_{3n+3}F_{3n+5}}
<-\frac{1}{F_{3n-1}F_{3n}F_{3n+3}}.
\end{eqnarray*}
Now we arrive at
\begin{eqnarray*}
s_3(n)+s_3(n+1)+s_3(mn)<-\frac{1}{F_{3n-1}F_{3n}F_{3n+3}}+\frac{1}{F_{9n}}<0,
\end{eqnarray*}
where the last inequality follows from
\begin{eqnarray*}
F_{9n}=F_{3n-2}F_{6n+1}+F_{3n-1}F_{6n+2}>F_{3n-1}(F_{3n-1}F_{3n+2}+F_{3n}F_{3n+3})>F_{3n-1}F_{3n}F_{3n+3}.
\end{eqnarray*}
The proof is completed.
\end{proof}

\begin{lemma}\label{lems4}
If $n\geq 1$ is odd, then
\begin{align*}
s_4(n)+s_4(n+1)>\frac{1}{2F_{6n+2}+1}.
\end{align*}
\end{lemma}
\begin{proof}
It is easy to check that the result holds for $n=1$, so we assume that $n\geq 3$. Applying the similar analysis in the proof of Lemma \ref{lems1}, we have that, for $n\geq 3$,
\begin{eqnarray*}
s_4(n)+s_4(n+1)
&=&-\left(\frac{1}{2F_{3n-1}+1}-\frac{1}{2F_{3n+5}+1}\right)+\left(\frac{1}{F_{3n}}-\frac{1}{F_{3n+3}}\right)\\[3pt]
&=&-\frac{2F_{3n+5}-2F_{3n-1}}{(2F_{3n-1}+1)(2F_{3n+5}+1)}+\frac{2F_{3n+1}}{F_{3n}F_{3n+3}}\\[3pt]
&>&-\frac{F_{3n+5}-F_{3n-1}}{(2F_{3n-1}+1)F_{3n+5}}+\frac{2F_{3n+1}}{F_{3n}F_{3n+3}}\\[3pt]
&=&-\frac{4F_{3n+2}}{(2F_{3n-1}+1)F_{3n+5}}+\frac{2F_{3n+1}}{F_{3n}F_{3n+3}}\\[3pt]
&=&\frac{4(F_{3n-1}F_{3n+1}F_{3n+5}-F_{3n}F_{3n+2}F_{3n+3})
+2F_{3n+1}F_{3n+5}}{(2F_{3n-1}+1)F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&=&\frac{4(2F_{3n}-F_{3n+5})+2F_{3n+1}F_{3n+5}}{(2F_{3n-1}+1)F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&>&\frac{(2F_{3n+1}-4)F_{3n+5}}{(2F_{3n-1}+1)F_{3n}F_{3n+3}F_{3n+5}}\\[3pt]
&>&\frac{1}{F_{3n}F_{3n+3}}.
\end{eqnarray*}
In addition, we have
\begin{align*}
F_{2n+2}=F_{n-1}F_{n+2}+F_{n}F_{n+3}>F_{n}F_{n+3}.
\end{align*}
Combining the above two inequalities together yields the desired result.
\end{proof}

\begin{theorem}\label{thm3k}
If $n\geq 1$ and $m\geq 2$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
2F_{3n-1}-1, &\mbox{if $n$ is even};\\[5pt]
-2F_{3n-1}-1, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}
\begin{proof}
We first consider the case where $n$ is even. With the help of $s_1(n)$, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}
&=&\frac{1}{2F_{3n-1}}-\sum\limits_{k=n}^{mn-1}s_1(k)-\left(s_1(mn)+\frac{1}{2F_{3mn+2}}\right)\\[3pt]
&=&\frac{1}{2F_{3n-1}}-\sum\limits_{k=n}^{mn-1}s_1(k)-\left(\frac{1}{2F_{3mn+2}}-\frac{1}{F_{3mn}}\right)\\[3pt]
&>&\frac{1}{2F_{3n-1}}-\sum\limits_{k=n}^{mn-1}s_1(k)\\[3pt]
&>&\frac{1}{2F_{3n-1}},
\end{eqnarray*}
where the last inequality follows from Lemma \ref{lems1}.

Employing Lemma \ref{lems2} and Lemma \ref{lems2mn}, we can deduce that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}
&=&\frac{1}{2F_{3n-1}-1}-\frac{1}{2F_{3mn+2}-1}-\sum\limits_{k=n+2}^{mn-1}s_2(k)
-\left(s_2(n)+s_2(n+1)+s_2(mn)\right)\\[3pt]
&<&\frac{1}{2F_{3n-1}-1}.
\end{eqnarray*}

Therefore, we obtain
\begin{eqnarray*}
\frac{1}{2F_{3n-1}}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}<\frac{1}{2F_{3n-1}-1},
\end{eqnarray*}
which shows that the statement is true when $n$ is even.

We now turn to consider the case where $n$ is odd. If $m$ is even, applying Lemma \ref{lems3} and Lemma \ref{lems4}, we can deduce that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}=\frac{-1}{2F_{3n-1}}+\frac{1}{2F_{3mn+2}}
-\sum\limits_{k=n}^{mn}s_3(k)>\frac{-1}{2F_{3n-1}},
\end{eqnarray*}
and
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}
&=&\frac{-1}{2F_{3n-1}+1}-\sum\limits_{k=n+2}^{mn}s_4(k)-\left(s_4(n)+s_4(n+1)-\frac{1}{2F_{3mn+2}+1}\right)\\[3pt]
&\leq&\frac{-1}{2F_{3n-1}+1}-\sum\limits_{k=n+2}^{mn}s_4(k)-\left(s_4(n)+s_4(n+1)-\frac{1}{2F_{6n+2}+1}\right)\\[3pt]
&<&\frac{-1}{2F_{3n-1}+1}.
\end{eqnarray*}
Thus, if $n$ is odd and $m$ is even, we have
\begin{eqnarray*}
\frac{-1}{2F_{3n-1}}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}<\frac{-1}{2F_{3n-1}+1}.
\end{eqnarray*}
If $m$ is odd, then Lemma \ref{lems3} and Lemma \ref{lems3mn} implies that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}
=\frac{-1}{2F_{3n-1}}+\frac{1}{2F_{3mn+2}}-\sum\limits_{k=n+2}^{mn-1}s_3(k)-\left(s_3(n)+s_3(n+1)+s_3(mn)\right)
>\frac{-1}{2F_{3n-1}}.
\end{eqnarray*}
And it follows from Lemma \ref{lems4} that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}
&=&\frac{-1}{2F_{3n-1}+1}-\sum\limits_{k=n}^{mn-1}s_4(k)-\left(s_4(mn)-\frac{1}{2F_{3mn+2}+1}\right)\\[3pt]
&=&\frac{-1}{2F_{3n-1}+1}-\sum\limits_{k=n}^{mn-1}s_4(k)-\left(\frac{1}{F_{3mn}}-\frac{1}{2F_{3mn-1}+1}\right)\\[3pt]
&<&\frac{-1}{2F_{3n-1}+1}.
\end{eqnarray*}
Thus, if $n$ and $m$ are both odd, then
\begin{eqnarray*}
\frac{-1}{2F_{3n-1}}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k}}<\frac{-1}{2F_{3n-1}+1}
\end{eqnarray*}
also holds. Hence, the statement is true when $n$ is odd.
\end{proof}

\begin{theorem}\label{thm3k+12}
If $n\geq 2$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
2F_{3n}-1, &\mbox{if $n$ is even};\\[5pt]
-2F_{3n}-1, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}

\begin{theorem}\label{thm3k+13}
If $n\geq 1$ and $m\geq 3$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
2F_{3n}, &\mbox{if $n$ is even};\\[5pt]
-2F_{3n}, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}

\begin{theorem}\label{thm3k+2}
If $n\geq 1$ and $m\geq 2$,  then
\begin{equation*}
\left\lfloor\left(\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+2}}\right)^{-1}\right\rfloor=
\left\{
\begin{array}{ll}
2F_{3n+1}-1, &\mbox{if $n$ is even};\\[5pt]
-2F_{3n+1}-1, &\mbox{if $n$ is odd}.
\end{array}
\right.
\end{equation*}
\end{theorem}

\begin{remark}
We will prove Theorem \ref{thm3k+12} and Theorem \ref{thm3k+13} in detail in the next section. The proof of Theorem \ref{thm3k+2} is very similar to that of Theorem \ref{thm3k}, thus omitted here.
\end{remark}


\section{Proof of Theorem \ref{thm3k+12} and Theorem \ref{thm3k+13}}

We begin with introducing the following auxiliary functions:
\begin{eqnarray*}
t_1(n)&=&\frac{1}{2F_{3n}}-\frac{(-1)^n}{F_{3n+1}}-\frac{1}{2F_{3n+3}},\\[3pt]
t_2(n)&=&\frac{1}{2F_{3n}-1}-\frac{(-1)^n}{F_{3n+1}}-\frac{1}{2F_{3n+3}-1},\\[3pt]
t_3(n)&=&\frac{1}{2F_{3n}+1}-\frac{(-1)^n}{F_{3n+1}}-\frac{1}{2F_{3n+3}+1},\\[3pt]
t_4(n)&=&\frac{-1}{2F_{3n}}-\frac{(-1)^n}{F_{3n+1}}+\frac{1}{2F_{3n+3}},\\[3pt]
t_5(n)&=&\frac{-1}{2F_{3n}+1}-\frac{(-1)^n}{F_{3n+1}}+\frac{1}{2F_{3n+3}+1},\\[3pt]
t_6(n)&=&\frac{-1}{2F_{3n}-1}-\frac{(-1)^n}{F_{3n+1}}+\frac{1}{2F_{3n+3}-1}.
\end{eqnarray*}
It is straightforward to check that each $t_i(n)$ is positive if $n$ is odd, and negative otherwise.

\begin{lemma}\label{lemt1geq}
For all $n\geq 1$, we have $t_1(n)+t_1(n+1)>0$ and
\begin{align*}
t_1(n)+t_1(n+1)>t_1(n+2)+t_1(n+3).
\end{align*}
\end{lemma}
\begin{proof}
If $n$ is odd, we have
\begin{eqnarray*}
t_1(n)+t_1(n+1)=\left(\frac{1}{2F_{3n}}-\frac{1}{2F_{3n+6}}\right)
+\left(\frac{1}{F_{3n+1}}-\frac{1}{F_{3n+4}}\right)
=\frac{2F_{3n+3}}{F_{3n}F_{3n+6}}+\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}>0.
\end{eqnarray*}
Since
\begin{eqnarray*}
\frac{F_{3n+3}}{F_{3n}F_{3n+6}}&>&\frac{F_{3n+9}}{F_{3n+6}F_{3n+12}},\\[3pt]
\frac{F_{3n+2}}{F_{3n+1}F_{3n+4}}&>&\frac{F_{3n+8}}{F_{3n+7}F_{3n+10}},
\end{eqnarray*}
we can conclude that $t_1(n)+t_1(n+1)>t_1(n+2)+t_1(n+3)$.

Now we consider the case where $n$ is even. Applying Lemma \ref{stone} repeatedly, we have
\begin{eqnarray*}
t_1(n)+t_1(n+1)
&=&\left(\frac{1}{2F_{3n}}-\frac{1}{2F_{3n+6}}\right)-\left(\frac{1}{F_{3n+1}}-\frac{1}{F_{3n+4}}\right)\\[3pt]
&=&\frac{2F_{3n+3}}{F_{3n}F_{3n+6}}-\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}\\[3pt]
&=&2\cdot\frac{F_{3n+1}F_{3n+3}F_{3n+4}-F_{3n}F_{3n+2}F_{3n+6}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&2\cdot\frac{F_{3n+1}F_{3n+2}F_{3n+3}+F_{3n+1}F_{3n+3}^2}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&&{}-2\cdot\frac{F_{3n}F_{3n+2}F_{3n+4}+F_{3n}F_{3n+2}F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&2\cdot\frac{F_{3n+2}(F_{3n+1}F_{3n+3}-F_{3n}F_{3n+4})
+F_{3n+1}(F_{3n+1}F_{3n+5}-1)}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&&{}-2\cdot\frac{F_{3n}F_{3n+2}F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&2\cdot\frac{2F_{3n+2}+F_{3n+5}(F_{3n+1}^2-F_{3n}F_{3n+2})-F_{3n+1}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&2\cdot\frac{2F_{3n+2}+F_{3n+5}-F_{3n+1}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&2\cdot\frac{F_{3n}+F_{3n+2}+F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&>&0.
\end{eqnarray*}

In addition, it is easy to see that $F_{3n}+F_{3n+2}+F_{3n+5}=3F_{3n}+3F_{3n+1}+F_{3n+4}$, thus
\begin{align*}
t_1(n)+t_1(n+1)=\frac{6}{F_{3n+1}F_{3n+4}F_{3n+6}}+\frac{6}{F_{3n}F_{3n+4}F_{3n+6}}+
\frac{2}{F_{3n}F_{3n+1}F_{3n+6}},
\end{align*}
which decreases as $n$ grows.
\end{proof}

\begin{lemma}\label{lemt1pro}
For all $n\geq1$, we have
\begin{align*}
2F_{3n+3}>F_{n}F_{n+1}F_{n+6}.
\end{align*}
\end{lemma}
\begin{proof}
Applying Lemma \ref{sq} repeatedly, we obtain
\begin{eqnarray*}
F_{3n+3}&=&F_{n}F_{2n+2}+F_{n+1}F_{2n+3}\\[3pt]
&=&F_{n}(F_{n}F_{n+1}+F_{n+1}F_{n+2})+F_{n+1}(F_{n}F_{n+2}+F_{n+1}F_{n+3})\\[3pt]
&=&F_{n}F_{n+1}(F_{n}+2F_{n+2})+F_{n+1}^2(F_{n}+2F_{n+1})\\[3pt]
&=&F_{n}F_{n+1}(F_{n}+F_{n+1}+2F_{n+2})+2F_{n+1}^3\\[3pt]
&>&F_{n}F_{n+1}(3F_{n+2}+2F_{n+1})\\[3pt]
&=&F_{n}F_{n+1}F_{n+5}.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
2F_{3n+3}-F_{n}F_{n+1}F_{n+6}>2F_{n}F_{n+1}F_{n+5}-F_{n}F_{n+1}F_{n+6}=F_{n}F_{n+1}(2F_{n+5}-F_{n+6})>0,
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lemt1mn}
If $n\geq 1$ and $m\geq 3$, then
\begin{align*}
t_1(n)+t_1(n+1)+t_1(mn)>0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is odd, then the result follows from Lemma \ref{lemt1geq} and the fact $t_1(mn)>0$. Now we assume that $mn$ is even. It follows from Lemma \ref{stone} that $F_{3mn}F_{3mn+1}=F_{3mn-2}F_{3mn+3}+2$, from which we get
\begin{eqnarray*}
t_1(mn)&=&\frac{1}{2F_{3mn}}-\frac{1}{F_{3mn+1}}-\frac{1}{2F_{3mn+3}}\\[3pt]
&=&-\frac{F_{3mn-2}}{2(F_{3mn-2}F_{3mn+3}+2)}-\frac{1}{2F_{3mn+3}}\\[3pt]
&>&-\frac{F_{3mn-2}}{2F_{3mn-2}F_{3mn+3}}-\frac{1}{2F_{3mn+3}}\\[3pt]
&=&-\frac{1}{F_{3mn+3}}\\[3pt]
&\geq&-\frac{1}{F_{9n+3}}.
\end{eqnarray*}

On the other hand, it follows from the proof of Lemma \ref{lemt1geq} that
\begin{align*}
t_1(n)+t_1(n+1)>\frac{2}{F_{3n}F_{3n+1}F_{3n+6}}.
\end{align*}
Now we arrive at
\begin{eqnarray*}
t_1(n)+t_1(n+1)+t_1(mn)>\frac{2}{F_{3n}F_{3n+1}F_{3n+6}}-\frac{1}{F_{9n+3}}>0,
\end{eqnarray*}
where the last inequality follows from Lemma \ref{lemt1pro}.
\end{proof}

\begin{lemma}\label{lemt1sumpre}
For all $n\geq 2$, we have
\begin{align*}
F_{2n}F_{2n+1}-F_{n+1}F_{n+4}F_{2n-2}<0.
\end{align*}
\end{lemma}
\begin{proof}
It follows from Lemma \ref{minus} and Lemma \ref{stone} respectively that
\begin{eqnarray*}
F_{n+2}F_{n+3}-F_{n}F_{n+1}&=&F_{2n+3},\\[3pt]
F_{n+1}F_{n+4}-F_{n+2}F_{n+3}&=&(-1)^{n},
\end{eqnarray*}
from which we can deduce that
\begin{eqnarray*}
F_{n+1}F_{n+4}=F_{n}F_{n+1}+F_{2n+3}+(-1)^n>F_{2n+3}+2.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
F_{2n}F_{2n+1}-F_{n+1}F_{n+4}F_{2n-2}&<&F_{2n}F_{2n+1}-(F_{2n+3}+2)F_{2n-2}\\[3pt]
&=&(F_{2n}F_{2n+1}-F_{2n-2}F_{2n+3})-2F_{2n-2}\\[3pt]
&=&2-2F_{2n-2}\\[3pt]
&\leq&0,
\end{eqnarray*}
where the last equality follows from Lemma \ref{stone}.
\end{proof}

\begin{lemma}\label{lemt1sum}
If $n\geq 2$ is even, then
\begin{align*}
\sum\limits_{k=n}^{2n}t_1(k)+\frac{1}{2F_{6n+3}}<0.
\end{align*}
\end{lemma}
\begin{proof}
From the proof of Lemma \ref{lemt1geq} we know that if $n$ is even, then
\begin{eqnarray*}
t_1(n)+t_1(n+1)=2\cdot\frac{F_{3n}+F_{3n+2}+F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
<\frac{2F_{3n+6}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}=\frac{2}{F_{3n}F_{3n+1}F_{3n+4}}.
\end{eqnarray*}
Applying Lemma \ref{lemt1geq} again and the above inequality, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}t_1(k)+\frac{1}{2F_{6n+3}}
&=&\sum\limits_{k=n}^{2n-1}t_1(k)+\left(t_1(2n)+\frac{1}{2F_{6n+3}}\right)\\[3pt]
&<&\frac{2}{F_{3n}F_{3n+1}F_{3n+4}}\cdot\frac{n}{2}+\left(\frac{1}{2F_{6n}}-\frac{1}{F_{6n+1}}\right)\\[3pt]
&=&\frac{n}{F_{3n}F_{3n+1}F_{3n+4}}-\frac{F_{6n-2}}{2F_{6n}F_{6n+1}}\\[3pt]
&<&\frac{1}{2F_{3n+1}F_{3n+4}}-\frac{F_{6n-2}}{2F_{6n}F_{6n+1}}\\[3pt]
&<&0,
\end{eqnarray*}
where the last inequality follows from Lemma \ref{lemt1sumpre}.
\end{proof}

\begin{lemma}\label{lemt2}
For all $n\geq 1$, we have
\begin{align*}
t_2(n)+t_2(n+1)>0.
\end{align*}
\end{lemma}
\begin{proof}
It is easy to see that the result is true when $n$ is odd. So we assume that $n$ is even. It follows from the definition of $t_2(n)$ that
\begin{eqnarray*}
t_2(n)+t_2(n+1)&=&\left(\frac{1}{2F_{3n}-1}-\frac{1}{2F_{3n+6}-1}\right)-\frac{1}{F_{3n+1}}+\frac{1}{F_{3n+4}}\\[3pt]
&=&\frac{2F_{3n+6}-2F_{3n}}{(2F_{3n}-1)(2F_{3n+6}-1)}-\frac{1}{F_{3n+1}}+\frac{1}{F_{3n+4}}\\[3pt]
&>&\frac{F_{3n+6}-F_{3n}}{2F_{3n}F_{3n+6}}-\frac{1}{F_{3n+1}}+\frac{1}{F_{3n+4}}\\[3pt]
&=&\frac{1}{2F_{3n}}-\frac{1}{F_{3n+1}}+\frac{1}{F_{3n+4}}-\frac{1}{2F_{3n+6}}\\[3pt]
&=&t_1(n)+t_1(n+1)\\[3pt]
&>&0,
\end{eqnarray*}
where the last inequality follows from the proof of Lemma \ref{lemt1geq}.
\end{proof}

\begin{lemma}\label{lemt2mn}
If $n\geq 1$ and $m\geq 2$, then
\begin{align*}
t_2(n)+t_2(n+1)+t_2(mn)>0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is odd, then the result follows from Lemma \ref{lemt2} and the fact $t_2(mn)>0$. Thus we assume that $mn$ is even in the rest. Applying the argument in the proof of Lemma \ref{lemt2} and Lemma \ref{lemt1mn}, we can easily obtain
\begin{align*}
t_2(mn)>t_1(mn)>-\frac{1}{F_{3mn+3}}\geq-\frac{1}{F_{6n+3}}.
\end{align*}
If $n$ is odd, we have
\begin{eqnarray*}
t_2(n)+t_2(n+1)>\frac{1}{F_{3n+1}}-\frac{1}{F_{3n+4}}=\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}
>\frac{2}{(2F_{3n}-1)F_{3n+4}}.
\end{eqnarray*}
If $n$ is even, then from the proof of Lemma \ref{lemt2} and Lemma \ref{lemt1geq} we know that
\begin{eqnarray*}
t_2(n)+t_2(n+1)&>&\frac{F_{3n+6}-F_{3n}}{(2F_{3n}-1)F_{3n+6}}-\frac{1}{F_{3n+1}}+\frac{1}{F_{3n+4}}\\[3pt]
&=&\frac{4F_{3n+3}}{(2F_{3n}-1)F_{3n+6}}-\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}\\[3pt]
&=&\frac{4(F_{3n+1}F_{3n+3}F_{3n+4}-F_{3n}F_{3n+2}F_{3n+6})
+2F_{3n+2}F_{3n+6}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&>&\frac{2F_{3n+2}F_{3n+6}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&>&\frac{2}{(2F_{3n}-1)F_{3n+4}}.
\end{eqnarray*}
Therefore, we always have
\begin{eqnarray*}
t_2(n)+t_2(n+1)>\frac{2}{(2F_{3n}-1)F_{3n+4}},
\end{eqnarray*}
from which we get
\begin{eqnarray*}
t_2(n)+t_2(n+1)+t_2(mn)>\frac{2}{(2F_{3n}-1)F_{3n+4}}-\frac{1}{F_{6n+3}}>0,
\end{eqnarray*}
where the last inequality follows from the fact $F_{6n+3}=F_{3n-1}F_{3n+3}+F_{3n}F_{3n+4}$.
\end{proof}

\begin{lemma}\label{lemt3}
If $n\geq 2$ is even, then
\begin{align*}
t_3(n)+t_3(n+1)<0.
\end{align*}
\end{lemma}
\begin{proof}
Applying the analysis in the proof of Lemma \ref{lemt1geq}, we can deduce that
\begin{eqnarray*}
t_3(n)+t_3(n+1)
&=&\left(\frac{1}{2F_{3n}+1}-\frac{1}{2F_{3n+6}+1}\right)-\left(\frac{1}{F_{3n+1}}-\frac{1}{F_{3n+4}}\right)\\[3pt]
&=&\frac{2F_{3n+6}-2F_{3n}}{(2F_{3n}+1)(2F_{3n+6}+1)}-\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}\\[3pt]
&<&\frac{F_{3n+6}-F_{3n}}{(2F_{3n}+1)F_{3n+6}}-\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}\\[3pt]
&=&\frac{4F_{3n+3}}{(2F_{3n}+1)F_{3n+6}}-\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}\\[3pt]
&=&\frac{4(F_{3n+1}F_{3n+3}F_{3n+4}-F_{3n}F_{3n+2}F_{3n+6})
-2F_{3n+2}F_{3n+6}}{(2F_{3n}+1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&\frac{4(F_{3n}+F_{3n+2}+F_{3n+5})-2F_{3n+2}F_{3n+6}}{(2F_{3n}+1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&<&\frac{4F_{3n+6}-2F_{3n+2}F_{3n+6}}{(2F_{3n}+1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&<&0.
\end{eqnarray*}
The proof is completed.
\end{proof}

\begin{lemma}\label{lemt4}
If $n\geq 1$ is odd, then
\begin{align*}
t_4(n)+t_4(n+1)>\frac{1}{2F_{9n+3}}.
\end{align*}
\end{lemma}
\begin{proof}
Applying similar arguments in the proof of Lemma \ref{lemt1geq}, we obtain that if $n$ is odd,
\begin{eqnarray*}
t_4(n)+t_4(n+1)=2\cdot\frac{F_{3n}+F_{3n+2}+F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
>\frac{2}{F_{3n}F_{3n+1}F_{3n+6}}.
\end{eqnarray*}
It follows from Lemma \ref{lemt1pro} that
\begin{eqnarray*}
\frac{1}{F_{3n}F_{3n+1}F_{3n+6}}>\frac{1}{2F_{9n+3}}.
\end{eqnarray*}
Combining the above two inequalities yields the desired result.
\end{proof}

\begin{lemma}\label{lemt4sum}
For all $n\geq 2$, we have
\begin{align*}
\sum\limits_{k=n}^{2n}t_4(k)<\frac{1}{2F_{6n+3}}.
\end{align*}
\end{lemma}
\begin{proof}
If $n$ is even, it is easy to see that $t_4(n)+t_4(n+1)<0$. Thus,
\begin{align*}
\sum\limits_{k=n}^{2n}t_4(k)=\sum\limits_{k=n}^{2n-1}t_4(k)+t_4(2n)<0<\frac{1}{2F_{6n+3}}.
\end{align*}
If $n$ is odd,
\begin{eqnarray*}
t_4(n)+t_4(n+1)=2\cdot\frac{F_{3n}+F_{3n+2}+F_{3n+5}}{F_{3n}F_{3n+1}F_{3n+4}F_{3n+6}}
<\frac{2}{F_{3n}F_{3n+1}F_{3n+4}},
\end{eqnarray*}
which implies that
\begin{align*}
\sum\limits_{k=n}^{2n}t_4(k)<\frac{2}{F_{3n}F_{3n+1}F_{3n+4}}\cdot\frac{n}{2}<\frac{1}{2F_{3n+1}F_{3n+4}}
<\frac{1}{2F_{6n+3}},
\end{align*}
where the last inequality follows from that for $n\geq 1$,
\begin{align*}
F_{3n+1}F_{3n+4}=F_{3n-1}F_{3n+4}+F_{3n}F_{3n+4}>F_{3n-1}F_{3n+3}+F_{3n}F_{3n+4}=F_{6n+3}.
\end{align*}
This completes the proof.
\end{proof}

\begin{lemma}\label{lemt5}
If $n\geq 1$ is odd, then
\begin{align*}
t_5(n)+t_5(n+1)>\frac{1}{2F_{6n+3}+1}.
\end{align*}
\end{lemma}
\begin{proof}
Imitating the proof of Lemma \ref{lemt3}, we can easily obtain
\begin{eqnarray*}
t_5(n)+t_5(n+1)&>&\frac{4(F_{3n}+F_{3n+2}+F_{3n+5})+2F_{3n+2}F_{3n+6}}{(2F_{3n}+1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&>&\frac{2F_{3n+2}F_{3n+6}}{(2F_{3n}+1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&>&\frac{1}{F_{3n+1}F_{3n+4}}.
\end{eqnarray*}
In addition, we have
\begin{align*}
2F_{2n+3}=2F_{n-1}F_{n+3}+2F_{n}F_{n+4}>F_{n+1}F_{n+4}.
\end{align*}
Combining the above two inequalities together yields the desired result.
\end{proof}

\begin{lemma}\label{lemt6}
For all $n\geq 1$, we have
\begin{align*}
t_6(n)+t_6(n+1)<0.
\end{align*}
\end{lemma}
\begin{proof}
It is clear that the result holds if $n$ is even. Now we assume that $n$ is odd. Applying the telescoping technique in the proof of Lemma \ref{lemt2mn} and the similar analysis in the proof of Lemma \ref{lemt1geq}, we obtain
\begin{eqnarray*}
t_6(n)+t_6(n+1)&<&-\left\{\frac{4F_{3n+3}}{(2F_{3n}-1)F_{3n+6}}-\frac{2F_{3n+2}}{F_{3n+1}F_{3n+4}}\right\}\\[3pt]
&=&-\frac{4(F_{3n+1}F_{3n+3}F_{3n+4}-F_{3n}F_{3n+2}F_{3n+6})
+2F_{3n+2}F_{3n+6}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&-\frac{-4(F_{3n}+F_{3n+2}+F_{3n+5})+2F_{3n+2}F_{3n+6}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&=&\frac{4(F_{3n}+F_{3n+2}+F_{3n+5})-2F_{3n+2}F_{3n+6}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&<&\frac{4F_{3n+6}-2F_{3n+2}F_{3n+6}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}F_{3n+6}}\\[3pt]
&<&0,
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{lemt6mn}
If $n\geq 1$ and $m\geq 2$, we have
\begin{align*}
t_6(n)+t_6(n+1)+t_6(mn)<0.
\end{align*}
\end{lemma}
\begin{proof}
If $mn$ is even, then the result follows from Lemma \ref{lemt6} and the fact $t_6(mn)<0$, so we assume that $mn$ is odd in the rest. Now we have
\begin{eqnarray*}
t_6(mn)&=&\frac{-1}{2F_{3mn}-1}+\frac{1}{F_{3mn+1}}+\frac{1}{2F_{3mn+3}-1}\\[3pt]
&=&\frac{-4F_{3mn+1}}{(2F_{3mn}-1)(2F_{3mn+3}-1)}+\frac{1}{F_{3mn+1}}\\[3pt]
&<&\frac{-2F_{3mn+1}}{(2F_{3mn}-1)F_{3mn+3}}+\frac{1}{F_{3mn+1}}\\[3pt]
&=&\frac{-2F_{3mn+1}^2+2F_{3mn}F_{3mn+3}-F_{3mn+3}}{(2F_{3mn}-1)F_{3mn+1}F_{3mn+3}}\\[3pt]
&=&\frac{-2(F_{3mn+1}^2-F_{3mn}F_{3mn+2})+2F_{3mn}F_{3mn+1}-F_{3mn+1}-F_{3mn+2}}{(2F_{3mn}-1)F_{3mn+1}F_{3mn+3}}\\[3pt]
&=&\frac{2+(2F_{3mn}-1)F_{3mn+1}-F_{3mn+2}}{(2F_{3mn}-1)F_{3mn+1}F_{3mn+3}}\\[3pt]
&<&\frac{(2F_{3mn}-1)F_{3mn+1}}{(2F_{3mn}-1)F_{3mn+1}F_{3mn+3}}\\[3pt]
&=&\frac{1}{F_{3mn+3}}.
\end{eqnarray*}
Since $mn$ is odd, we must have that $n$ is odd and $m\geq 3$. Therefore,
\begin{align*}
t_6(mn)<\frac{1}{F_{9n+3}}.
\end{align*}
It follows from the proof of Lemma \ref{lemt6} that if $n$ is odd,
\begin{align*}
t_6(n)+t_6(n+1)<\frac{4-2F_{3n+2}}{(2F_{3n}-1)F_{3n+1}F_{3n+4}}<\frac{2-F_{3n+2}}{F_{3n}F_{3n+1}F_{3n+4}}
<-\frac{1}{F_{3n}F_{3n+1}F_{3n+4}}.
\end{align*}
Now we arrive at
\begin{align*}
t_6(n)+t_6(n+1)+t_6(mn)<\frac{1}{F_{9n+3}}-\frac{1}{F_{3n}F_{3n+1}F_{3n+4}}.
\end{align*}
Employing Lemma \ref{sq}, we easily see that $F_{3n+3}>F_{2n}F_{n+4}$ and $F_{2n}>F_{n}F_{n+1}$, which implies
\begin{align*}
F_{9n+3}>F_{3n}F_{3n+1}F_{3n+4}.
\end{align*}
Therefore,
\begin{align*}
t_6(n)+t_6(n+1)+t_6(mn)<\frac{1}{F_{9n+3}}-\frac{1}{F_{3n}F_{3n+1}F_{3n+4}}<0.
\end{align*}
The proof is completed.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3k+12}.]
We first consider the case where $n$ is even. Applying Lemma \ref{lemt1sum}, we have
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}=\frac{1}{2F_{3n}}-\frac{1}{2F_{6n+3}}
-\sum\limits_{k=n}^{2n}t_1(k)>\frac{1}{2F_{3n}}.
\end{eqnarray*}
It follows from Lemma \ref{lemt2} and Lemma \ref{lemt2mn} that
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}
=\frac{1}{2F_{3n}-1}-\frac{1}{2F_{6n+3}-1}-\left(t_2(n)+t_2(n+1)+t_2(2n)\right)-\sum\limits_{k=n+2}^{2n-1}t_2(k)
<\frac{1}{2F_{3n}-1}.
\end{eqnarray*}
Therefore, we have
\begin{align*}
\frac{1}{2F_{3n}}<\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}<\frac{1}{2F_{3n}-1},
\end{align*}
which means that the result holds when $n$ is even.

We now turn to consider the case where $n$ is odd. From Lemma \ref{lemt4sum}, we know that
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}=\frac{-1}{2F_{3n}}+\frac{1}{2F_{6n+3}}
-\sum\limits_{k=n}^{2n}t_4(k)>\frac{-1}{2F_{3n}}.
\end{eqnarray*}
With the help of Lemma \ref{lemt5}, it is easy to see that
\begin{eqnarray*}
\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}
=\frac{-1}{2F_{3n}+1}-\left(t_5(n)+t_5(n+1)-\frac{1}{2F_{6n+3}+1}\right)-\sum\limits_{k=n+2}^{2n}t_5(k)
<\frac{-1}{2F_{3n}+1}.
\end{eqnarray*}
Thus, we obtain
\begin{align*}
\frac{-1}{2F_{3n}}<\sum\limits_{k=n}^{2n}\frac{(-1)^k}{F_{3k+1}}<\frac{-1}{2F_{3n}+1},
\end{align*}
which yields the desired identity.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3k+13}.]
We first consider the case where $n$ is even. Applying Lemma \ref{lemt1geq} and Lemma \ref{lemt1mn}, we see
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}
=\frac{1}{2F_{3n}}-\frac{1}{2F_{3mn+3}}-\sum\limits_{k=n+2}^{mn-1}t_1(k)-\left(t_1(n)+t_1(n+1)+t_1(mn)\right)
<\frac{1}{2F_{3n}}.
\end{eqnarray*}
It follows from Lemma \ref{lemt3} that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}
&=&\frac{1}{2F_{3n}+1}-\sum\limits_{k=n}^{mn-1}t_3(k)-\left(t_3(mn)+\frac{1}{2F_{3mn+3}+1}\right)\\[3pt]
&=&\frac{1}{2F_{3n}+1}-\sum\limits_{k=n}^{mn-1}t_3(k)-\left(\frac{1}{2F_{3mn}+1}-\frac{1}{F_{3mn+1}}\right)\\[3pt]
&>&\frac{1}{2F_{3n}+1}.
\end{eqnarray*}
Therefore, we obtain
\begin{eqnarray*}
\frac{1}{2F_{3n}+1}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}<\frac{1}{2F_{3n}},
\end{eqnarray*}
which shows that the statement is true when $n$ is even.

Now we turn to consider the case where $n$ is odd. Lemma \ref{lemt4} tells us that
\[t_4(n)+t_4(n+1)>0.\]
Hence if $mn$ is odd,
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}
&=&\frac{-1}{2F_{3n}}-\sum\limits_{k=n}^{mn-1}t_4(k)-\left(t_4(mn)-\frac{1}{2F_{3mn+3}}\right)\\[3pt]
&=&\frac{-1}{2F_{3n}}-\sum\limits_{k=n}^{mn-1}t_4(k)-\left(\frac{1}{F_{3mn+1}}-\frac{1}{2F_{3mn}}\right)\\[3pt]
&<&\frac{-1}{2F_{3n}}.
\end{eqnarray*}
And it follows from Lemma \ref{lemt6} and Lemma \ref{lemt6mn} that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}
&=&\frac{-1}{2F_{3n}-1}+\frac{1}{2F_{3mn+3}-1}-\sum\limits_{k=n+2}^{mn-1}t_6(k)
-\left(t_6(n)+t_6(n+1)+t_6(mn)\right)\\[3pt]
&>&\frac{-1}{2F_{3n}-1}.
\end{eqnarray*}
Therefore, we have
\begin{eqnarray*}
\frac{-1}{2F_{3n}-1}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}<\frac{-1}{2F_{3n}}.
\end{eqnarray*}


If $mn$ is even, then Lemma \ref{lemt4} implies that
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}&=&\frac{-1}{2F_{3n}}-\sum\limits_{k=n+2}^{mn}t_4(k)-
\left(t_4(n)+t_4(n+1)-\frac{1}{2F_{3mn+3}}\right)\\[3pt]
&<&\frac{-1}{2F_{3n}}-\sum\limits_{k=n+2}^{mn}t_4(k)-\left(t_4(n)+t_4(n+1)-\frac{1}{2F_{9n+3}}\right)\\[3pt]
&<&\frac{-1}{2F_{3n}},
\end{eqnarray*}
and from Lemma \ref{lemt6} we obtain
\begin{eqnarray*}
\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}=\frac{-1}{2F_{3n}-1}+\frac{1}{2F_{3mn+3}-1}-\sum\limits_{k=n}^{mn}t_6(k)
>\frac{-1}{2F_{3n}-1}.
\end{eqnarray*}
Hence, we also have
\begin{eqnarray*}
\frac{-1}{2F_{3n}-1}<\sum\limits_{k=n}^{mn}\frac{(-1)^k}{F_{3k+1}}<\frac{-1}{2F_{3n}},
\end{eqnarray*}
which yields the desired identity.
\end{proof}


\section{Acknowledgments}

The authors would like to thank the anonymous reviewers for their helpful comments. This work was supported by the Teaching Reform Research Project of University of Electronic Science and Technology of China (No.~2016XJYYB039).


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\bibitem{ON89}
H. Ohtsuka and S. Nakamura, On the sum of reciprocal Fibonacci numbers,
{\it Fibonacci Quart.} \textbf{46/47} (2008/2009), 153--159.

\bibitem{RS75}
R. R. Stone, General identities for Fibonacci and Lucas numbers with
polynomial subscripts in several variables, {\it Fibonacci Quart.}
\textbf{13} (1975), 289--294.

\bibitem{NNV02}
N. N. Vorobiev, {\it Fibonacci Numbers}, Springer, 2002.

\bibitem{Wang15}
A. Y. Z. Wang and P. B. Wen, On the partial finite sums of the
reciprocals of the Fibonacci numbers, {\it J. Inequal. Appl.}
\textbf{2015} (2015), Article 73.

\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: }
Fibonacci number, alternating sum, reciprocal.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000108}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 9 2016;
revised version received  December 24 2016.
Published in {\it Journal of Integer Sequences}, December 26 2016.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                



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