Open Access

Sums of products of two reciprocal Fibonacci numbers

Advances in Difference Equations20162016:136

https://doi.org/10.1186/s13662-016-0860-0

Received: 4 February 2016

Accepted: 10 May 2016

Published: 18 May 2016

Abstract

In this paper, we employ elementary methods to investigate the reciprocal sums of the products of two Fibonacci numbers in several ways. First, we consider the sums of the reciprocals of the products of two Fibonacci numbers and establish five interesting families of identities. Then we extend such analysis to the alternating sums and obtain five analogous results.

Keywords

Fibonacci number reciprocal sum floor function

MSC

11B39

1 Introduction

For an integer \(n\geq0\), the Fibonacci number \(F_{n}\) is defined by
$$F_{n}=F_{n-1}+F_{n-2}\quad \mbox{for }n\geq2, $$
with \(F_{0}=0\) and \(F_{1}=1\). There exists a simple and nonobvious formula for the Fibonacci numbers:
$$ F_{n}=\frac{1}{\sqrt{5}} \biggl(\frac{1+\sqrt{5}}{2} \biggr)^{n}- \frac{1}{\sqrt{5}} \biggl(\frac{1-\sqrt{5}}{2} \biggr)^{n}. $$

The Fibonacci numbers play an important role in the theory and applications of mathematics, and its various properties have been investigated by many authors; see [14].

In recent years, there has been an increasing interest in studying the reciprocal sums of the Fibonacci numbers. For example, Elsner, Shimomura, and Shiokawa [58] investigated algebraic relations for reciprocal sums of the Fibonacci numbers. Ohtsuka and Nakamura [9] studied the partial infinite sums of the reciprocal Fibonacci numbers. They established the following results, where \(\lfloor\cdot\rfloor\) denotes the floor function.

Theorem 1.1

For all \(n\geq2\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{n-2} &\textit{if }n \textit{ is even}; \\ F_{n-2}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Theorem 1.2

For each \(n\geq1\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac {1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{n}F_{n-1}-1 &\textit{if }n \textit{ is even}; \\ F_{n}F_{n-1} &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Recently, Wang and Wen [10] considered the partial finite sums of the reciprocal Fibonacci numbers and strengthened Theorem 1.1 and Theorem 1.2 to the finite-sum case.

Theorem 1.3

  1. (i)
    For all \(n\geq4\),
    $$ \Biggl\lfloor \Biggl(\sum_{k=n}^{2n} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor =F_{n-2}. $$
     
  2. (ii)
    If \(m\geq3\) and \(n\geq2\), then
    $$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{n-2} &\textit{if }n \textit{ is even}; \\ F_{n-2}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$
     

Theorem 1.4

For all \(m\geq2\) and \(n\geq1\), we have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{n}F_{n-1}-1 &\textit{if }n \textit{ is even}; \\ F_{n}F_{n-1} &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Furthermore, Wang and Zhang [11] studied the reciprocal sums of the Fibonacci numbers with even or odd indexes and obtained the following main results.

Theorem 1.5

We have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{2k}} \Biggr)^{-1} \Biggr\rfloor =\left \{ \textstyle\begin{array}{l@{\quad}l} F_{2n-1} &\textit{if }m=2\textit{ and }n\geq3; \\ F_{2n-1}-1 &\textit{if }m\geq3\textit{ and }n\geq1. \end{array}\displaystyle \right . $$

Theorem 1.6

For all \(n\geq1\) and \(m\geq2\), we have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{2k-1}} \Biggr)^{-1} \Biggr\rfloor =F_{2n-2}. $$

Theorem 1.7

If \(n\geq1\) and \(m\geq2\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{2k}^{2}} \Biggr)^{-1} \Biggr\rfloor =F_{4n-2}-1. $$

Theorem 1.8

For all \(n\geq1\) and \(m\geq2\), we have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {1}{F_{2k-1}^{2}} \Biggr)^{-1} \Biggr\rfloor =F_{4n-4}. $$

More recently, Wang and Zhang [12] proceeded with investigating the reciprocal sums of the Fibonacci numbers according to the subscripts modulo 3 and found many identities. Here are a few examples.

Theorem 1.9

  1. (i)
    For all \(n\geq2\),
    $$ \Biggl\lfloor \Biggl(\sum_{k=n}^{2n} \frac{1}{F_{3k}} \Biggr)^{-1} \Biggr\rfloor =2F_{3n-2}. $$
     
  2. (ii)
    If \(m\geq3\) and \(n\geq1\), then
    $$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{3k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} 2F_{3n-2} &\textit{if }n \textit{ is even}; \\ 2F_{3n-2}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$
     

Theorem 1.10

If \(n\geq2\) and \(m\geq2\), we have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{3k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{3n}^{2}-F_{3n-3}^{2} &\textit{if }n \textit{ is even}; \\ F_{3n}^{2}-F_{3n-3}^{2}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

In this article, we focus ourselves on the sums and alternating sums of the products of two reciprocal Fibonacci numbers. By evaluating the integer parts of these sums, we obtain several interesting families of identities concerning the Fibonacci numbers.

2 Main results I: reciprocal sums

We first introduce several well-known results on the Fibonacci numbers, which will be used throughout the article. The detailed proofs can be found in, for example, [4] and [13].

Lemma 2.1

For any positive integers m and n, we have
$$ F_{m}F_{n}+F_{m+1}F_{n+1}=F_{m+n+1}. $$
(2.1)

Lemma 2.2

If \(n\geq1\), then
$$\begin{aligned}& F_{2n} = F_{n+1}^{2}-F_{n-1}^{2}, \end{aligned}$$
(2.2)
$$\begin{aligned}& F_{2n+1} = F_{n+1}F_{n+2}-F_{n-1}F_{n}. \end{aligned}$$
(2.3)

Lemma 2.3

Let a, b, c, d be positive integers with \(a+b=c+d\) and \(b\geq\max\{c,d\}\). Then
$$ F_{a}F_{b}-F_{c}F_{d}=(-1)^{a+1}F_{b-c}F_{b-d}. $$
(2.4)

2.1 Reciprocal sum of \(F_{k}F_{k+1}\)

Lemma 2.4

For all \(n\geq1\), we have
$$ F_{2n+1}^{2}+1>\bigl(F_{n+1}^{2}+1 \bigr)^{2}>F_{n}F_{n+1}\bigl(F_{n+1}^{2}+1 \bigr). $$
(2.5)

Proof

It follows from (2.1) that \(F_{2n+1}=F_{n}^{2}+F_{n+1}^{2}\). Hence,
$$F_{2n+1}^{2}+1=F_{n}^{4}+2F_{n}^{2}F_{n+1}^{2}+F_{n+1}^{4}+1>F_{n+1}^{4}+2F_{n+1}^{2}+1= \bigl(F_{n+1}^{2}+1\bigr)^{2}. $$
It is clear that \(F_{n+1}^{2}\geq F_{n}F_{n+1}\); therefore, \(F_{n+1}^{2}+1>F_{n}F_{n+1}\), which yields the second inequality. □

Theorem 2.5

If \(m\geq2\) and \(n\geq1\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {1}{F_{k}F_{k+1}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{n}^{2} &\textit{if }n \textit{ is even}; \\ F_{n}^{2}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Proof

We first consider the case where n is even. By elementary manipulations and setting \(a=k-1\), \(b=k+1\), and \(c=d=k\) in (2.4), we obtain, for \(k\geq1\),
$$\begin{aligned} \frac{1}{F_{k}^{2}}-\frac{1}{F_{k}F_{k+1}}-\frac{1}{F_{k+1}^{2}} =& \frac{F_{k+1}^{2}-F_{k}F_{k+1}-F_{k}^{2}}{F_{k}^{2}F_{k+1}^{2}} \\ =&\frac{F_{k-1}F_{k+1}-F_{k}^{2}}{F_{k}^{2}F_{k+1}^{2}} \\ =&\frac{(-1)^{k}}{F_{k}^{2}F_{k+1}^{2}}. \end{aligned}$$
(2.6)
Now we have
$$ \sum_{k=n}^{mn}\frac{1}{F_{k}F_{k+1}}= \frac{1}{F_{n}^{2}}-\frac {1}{F_{mn+1}^{2}}+ \sum_{k=n}^{mn} \frac{(-1)^{k-1}}{F_{k}^{2}F_{k+1}^{2}}. $$
Since n is even, it is easy to see that
$$ \sum_{k=n}^{mn}\frac{(-1)^{k-1}}{F_{k}^{2}F_{k+1}^{2}}< 0, $$
which implies that
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}}< \frac{1}{F_{n}^{2}}. $$
(2.7)
A direct calculation shows that, for \(k\geq1\),
$$\begin{aligned}& \frac{1}{F_{k}^{2}+1}-\frac{1}{F_{k}F_{k+1}}-\frac{1}{F_{k+1}^{2}+1} \\& \quad =\frac {F_{k}F_{k+1}(F_{k+1}^{2}-F_{k}^{2})-(F_{k}^{2}+1)(F_{k+1}^{2}+1)}{(F_{k}^{2}+1)F_{k}F_{k+1}(F_{k+1}^{2}+1)} \\& \quad =\frac {F_{k}F_{k+1}(F_{k+1}^{2}-F_{k}^{2}-F_{k}F_{k+1})-F_{k}^{2}-F_{k+1}^{2}-1}{(F_{k}^{2}+1)F_{k}F_{k+1}(F_{k+1}^{2}+1)} \\& \quad =\frac {F_{k}F_{k+1}(F_{k-1}F_{k+1}-F_{k}^{2})-F_{k}^{2}-F_{k+1}^{2}-1}{(F_{k}^{2}+1)F_{k}F_{k+1}(F_{k+1}^{2}+1)} \\& \quad =\frac {(-1)^{k}F_{k}F_{k+1}-F_{k}^{2}-F_{k+1}^{2}-1}{(F_{k}^{2}+1)F_{k}F_{k+1}(F_{k+1}^{2}+1)} \\& \quad < 0. \end{aligned}$$
Therefore,
$$\begin{aligned} \sum_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}} =&\frac {1}{F_{n}^{2}+1}-\frac{1}{F_{mn+1}^{2}+1}+ \sum _{k=n}^{mn}\frac{(-1)^{k-1}F_{k}F_{k+1}+F_{k}^{2}+ F_{k+1}^{2}+1}{(F_{k}^{2}+1)F_{k}F_{k+1}(F_{k+1}^{2}+1)} \\ >&\frac{1}{F_{n}^{2}+1}-\frac{1}{F_{2n+1}^{2}+1}+ \frac {F_{n}^{2}+F_{n+1}^{2}+1-F_{n}F_{n+1}}{(F_{n}^{2}+1)F_{n}F_{n+1}(F_{n+1}^{2}+1)} \\ >&\frac{1}{F_{n}^{2}+1}+\frac{1}{F_{n}F_{n+1}(F_{n+1}^{2}+1)}-\frac {1}{F_{2n+1}^{2}+1} \\ >&\frac{1}{F_{n}^{2}+1}, \end{aligned}$$
(2.8)
where the last inequality follows from (2.5).
Combining (2.7) and (2.8), we have
$$ \frac{1}{F_{n}^{2}+1}< \sum_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}}< \frac {1}{F_{n}^{2}}, $$
which means that the statement is true when n is even.
We now concentrate ourselves on the case where n is odd. It is obviously true for \(n=1\). Now we assume that \(n \geq3\). A similar calculation shows that, for \(k\geq3\),
$$ \frac{1}{F_{k}^{2}-1}-\frac{1}{F_{k}F_{k+1}}-\frac{1}{F_{k+1}^{2}-1} =\frac {(-1)^{k}F_{k}F_{k+1}+F_{k}^{2}+F_{k+1}^{2}+1}{(F_{k}^{2}-1)F_{k}F_{k+1}(F_{k+1}^{2}-1)}>0, $$
from which we get
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}}< \frac{1}{F_{n}^{2}-1}-\frac {1}{F_{mn+1}^{2}-1}< \frac{1}{F_{n}^{2}-1}. $$
(2.9)
It follows from (2.1) that
$$ F_{2n+1}=F_{n-1}F_{n+1}+F_{n}F_{n+2}=F_{n}^{2}+F_{n+1}^{2}, $$
which implies that \(F_{2n+1}\geq F_{n}F_{n+2}\), and \(F_{2n+2}>F_{2n+1}>F_{n+1}^{2}\). Therefore,
$$ F_{2n+1}^{2}F_{2n+2}>F_{n}^{2}F_{n+1}^{2}F_{n+2}^{2}. $$
(2.10)
Invoking (2.6), (2.2), and the fact n is odd, we have
$$\begin{aligned} \sum_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}} =&\frac{1}{F_{n}^{2}}-\frac {1}{F_{mn+1}^{2}}+ \sum _{k=n}^{mn}\frac{(-1)^{k-1}}{F_{k}^{2}F_{k+1}^{2}} \\ >&\frac{1}{F_{n}^{2}}-\frac{1}{F_{2n+1}^{2}}+\frac {1}{F_{n}^{2}F_{n+1}^{2}}-\frac{1}{F_{n+1}^{2}F_{n+2}^{2}} \\ =&\frac{1}{F_{n}^{2}}+\frac {F_{n+2}^{2}-F_{n}^{2}}{F_{n}^{2}F_{n+1}^{2}F_{n+2}^{2}}-\frac {1}{F_{2n+1}^{2}} \\ =&\frac{1}{F_{n}^{2}}+\frac{F_{2n+2}}{F_{n}^{2}F_{n+1}^{2}F_{n+2}^{2}}-\frac {F_{2n+2}}{F_{2n+1}^{2}F_{2n+2}} \\ >&\frac{1}{F_{n}^{2}}, \end{aligned}$$
(2.11)
where the last inequality follows from (2.10).
Combining (2.9) and (2.11) yields that
$$ \frac{1}{F_{n}^{2}}< \sum_{k=n}^{mn} \frac{1}{F_{k}F_{k+1}}< \frac {1}{F_{n}^{2}-1}, $$
from which the desired result follows immediately. □

2.2 Reciprocal sum of \(F_{2k-1}F_{2k}\)

Lemma 2.6

For all \(n\geq1\), we have
$$ F_{8n+1}>F_{2n-1}F_{2n}(F_{4n+1}+1). $$

Proof

Applying (2.1) repeatedly, we have
$$ F_{8n+1}=F_{4n+1}^{2}+F_{4n}^{2}>F_{4n+1}^{2}+F_{4n+1}=F_{4n+1} (F_{4n+1}+1 )> F_{2n-1}F_{2n} (F_{4n+1}+1 ), $$
which completes the proof. □

Theorem 2.7

For all \(n\geq2\) and \(m\geq2\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {1}{F_{2k-1}F_{2k}} \Biggr)^{-1} \Biggr\rfloor =F_{4n-3}. $$

Proof

It follows from (2.3) that
$$\begin{aligned}& F_{4n+1} = F_{2n+1}F_{2n+2}-F_{2n-1}F_{2n}, \end{aligned}$$
(2.12)
$$\begin{aligned}& F_{4n-3} = F_{2n-1}F_{2n}-F_{2n-3}F_{2n-2}. \end{aligned}$$
(2.13)
Employing (2.4), we can easily get that
$$\begin{aligned}& F_{2n-2}F_{2n+1} = F_{2n-1}F_{2n}-1, \end{aligned}$$
(2.14)
$$\begin{aligned}& F_{2n-3}F_{2n+2} = F_{2n-1}F_{2n}+2. \end{aligned}$$
(2.15)
Applying (2.12), (2.13), (2.14), and (2.15), it is easy to see that, for all \(k\geq2\),
$$\begin{aligned} \frac{1}{F_{4k-3}}-\frac{1}{F_{2k-1}F_{2k}}-\frac{1}{F_{4k+1}} =&\frac{F_{2k-1}F_{2k} (F_{4k+1}-F_{4k-3} )-F_{4k-3}F_{4k+1}}{F_{4k-3}F_{2k-1}F_{2k}F_{4k+1}} \\ =&\frac {F_{2k-3}F_{2k-2}F_{2k+1}F_{2k+2}-F_{2k-1}^{2}F_{2k}^{2}}{F_{4k-3}F_{2k-1}F_{2k}F_{4k+1}} \\ =&\frac{F_{2k-1}F_{2k}-2}{F_{4k-3}F_{2k-1}F_{2k}F_{4k+1}} \\ >&0, \end{aligned}$$
which implies that
$$ \sum_{k=n}^{mn} \frac{1}{F_{2k-1}F_{2k}}< \sum_{k=n}^{mn} \biggl( \frac{1}{F_{4k-3}}-\frac{1}{F_{4k+1}} \biggr) =\frac{1}{F_{4n-3}}- \frac{1}{F_{4mn+1}}< \frac{1}{F_{4n-3}}. $$
(2.16)
It follows from (2.1) that
$$ F_{4n+1}>F_{4n}=F_{2n-1}F_{2n}+F_{2n}F_{2n+1}>F_{2n-1}F_{2n}+1. $$
Therefore, by (2.13) we obtain
$$ F_{4n+1}-F_{2n-3}F_{2n-2}>F_{2n-1}F_{2n}-F_{2n-3}F_{2n-2}+1=F_{4n-3}+1. $$
(2.17)
Elementary manipulations and (2.17) yield, for \(k\geq2\),
$$\begin{aligned} \frac{1}{F_{4k-3}+1}-\frac{1}{F_{2k-1}F_{2k}}-\frac{1}{F_{4k+1}+1} =&\frac {F_{2k-3}F_{2k-2}-F_{4k+1}-3}{(F_{4k-3}+1)F_{2k-1}F_{2k}(F_{4k+1}+1)} \\ < &\frac{-1}{F_{2k-1}F_{2k}(F_{4k+1}+1)}. \end{aligned}$$
Now we can deduce that
$$\begin{aligned} \sum_{k=n}^{mn}\frac{1}{F_{2k-1}F_{2k}} >& \frac {1}{F_{4n-3}+1}-\frac{1}{F_{4mn+1}+1}+\sum_{k=n}^{mn} \frac{1}{F_{2k-1}F_{2k}(F_{4k+1}+1)} \\ >&\frac{1}{F_{4n-3}+1}+\frac{1}{F_{2n-1}F_{2n}(F_{4n+1}+1)}-\frac {1}{F_{4mn+1}+1} \\ >&\frac{1}{F_{4n-3}+1}+\frac{1}{F_{2n-1}F_{2n}(F_{4n+1}+1)}-\frac {1}{F_{8n+1}+1} \\ >&\frac{1}{F_{4n-3}+1}, \end{aligned}$$
(2.18)
where the last inequality follows from Lemma 2.6.
Combining (2.16) and (2.18), we deduce
$$ \frac{1}{F_{4n-3}+1}< \sum_{k=n}^{mn} \frac {1}{F_{2k-1}F_{2k}}< \frac{1}{F_{4n-3}}, $$
which yields the desired identity. □

Similarly, we can prove the following result, whose proof is left as an exercise to the readers.

Theorem 2.8

For all \(n\geq2\) and \(m\geq2\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {1}{F_{2k}F_{2k+1}} \Biggr)^{-1} \Biggr\rfloor =F_{4n-1}-1. $$

2.3 Reciprocal sum of \(F_{2k-1}F_{2k+1}\)

Lemma 2.9

For all \(n\geq2\), we have
$$ F_{n+2}^{2}-F_{n-2}^{2}>4F_{n-1}F_{n+1}. $$

Proof

It is easy to check that
$$\begin{aligned} F_{n+2}^{2}-F_{n-2}^{2}&= (F_{n+2}-F_{n-2} ) (F_{n+2}+F_{n-2} ) \\ &> (F_{n+1}+F_{n-1} ) (F_{n+1}+F_{n-1} ) \\ &=F_{n+1}^{2}+F_{n-1}^{2}+2F_{n-1}F_{n+1} \\ &= (F_{n+1}-F_{n-1} )^{2}+4F_{n-1}F_{n+1} \\ &>4F_{n-1}F_{n+1}. \end{aligned}$$
 □

Lemma 2.10

If \(n\geq1\), then
$$ F_{8n+2}> (F_{4n-2}+1 ) (F_{4n+2}+1 ). $$

Proof

It follows from (2.2) that
$$ F_{8n+2}=F_{4n+2}^{2}-F_{4n}^{2}= (F_{4n+2}-F_{4n} ) (F_{4n+2}+F_{4n} ) =F_{4n+1} (F_{4n+2}+F_{4n} ). $$
It is obvious that \(F_{4n+1}>F_{4n-2}+1\) and \(F_{4n}>1\), which completes the proof. □

Theorem 2.11

For all \(n\geq1\) and \(m\geq2\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {1}{F_{2k-1}F_{2k+1}} \Biggr)^{-1} \Biggr\rfloor =F_{4n-2}. $$

Proof

Employing (2.2), we can readily see that
$$\begin{aligned}& F_{4n+2} = F_{2n+2}^{2}-F_{2n}^{2}=F_{2n+1} (F_{2n+2}+F_{2n} )=F_{2n+1}^{2}+2F_{2n}F_{2n+1}, \\& F_{4n-2} = F_{2n}^{2}-F_{2n-2}^{2}=F_{2n-1} (F_{2n}+F_{2n-2} )=2F_{2n-1}F_{2n}-F_{2n-1}^{2}. \end{aligned}$$
Applying (2.4), we can establish the following identities:
$$\begin{aligned}& F_{2n+1}^{2} = F_{2n}F_{2n+2}+1, \\& F_{2n-1}^{2} = F_{2n-2}F_{2n}+1, \\& F_{2n}^{2} = F_{2n-2}F_{2n+2}+1. \end{aligned}$$
With the help of these identities, we now arrive at
$$\begin{aligned} F_{4n+2}-F_{4n-2} =&F_{2n+1}^{2}+F_{2n-1}^{2}+2F_{2n}F_{2n+1}-2F_{2n-1}F_{2n} \\ =&F_{2n+1}^{2}+F_{2n-1}^{2}+2F_{2n}^{2} \\ =& (F_{2n}F_{2n+2}+1 )+ (F_{2n-2}F_{2n}+1 )+ (F_{2n-2}F_{2n+2}+1 )+F_{2n}^{2} \\ =& (F_{2n+2}+F_{2n} ) (F_{2n}+F_{2n-2} )+3 \\ =&\frac{F_{4n+2}F_{4n-2}}{F_{2n-1}F_{2n+1}}+3. \end{aligned}$$
(2.19)
Elementary manipulations and (2.19) yield that, for \(k\geq1\),
$$\begin{aligned} \frac{1}{F_{4k-2}}-\frac{1}{F_{2k-1}F_{2k+1}}-\frac{1}{F_{4k+2}} =&\frac{F_{2k-1}F_{2k+1} (F_{4k+2}-F_{4k-2} )- F_{4k-2}F_{4k+2}}{F_{4k-2}F_{2k-1}F_{2k+1}F_{4k+2}} \\ =&\frac{3F_{2k-1}F_{2k+1}}{F_{4k-2}F_{2k-1}F_{2k+1}F_{4k+2}} \\ >&0, \end{aligned}$$
(2.20)
which implies that
$$ \sum_{k=n}^{mn} \frac{1}{F_{2k-1}F_{2k+1}}< \sum_{k=n}^{mn} \biggl( \frac{1}{F_{4k-2}}-\frac{1}{F_{4k+2}} \biggr)=\frac {1}{F_{4n-2}}- \frac{1}{F_{4mn+2}}< \frac{1}{F_{4n-2}}. $$
(2.21)
Invoking (2.20) and Lemma 2.9, we have
$$\begin{aligned}& \frac{1}{F_{4k-2}+1}-\frac{1}{F_{2k-1}F_{2k+1}}-\frac{1}{F_{4k+2}+1} \\& \quad = \frac{F_{2k-1}F_{2k+1} (F_{4k+2}-F_{4k-2} )- F_{4k-2}F_{4k+2}}{ (F_{4k-2}+1 )F_{2k-1}F_{2k+1} (F_{4k+2}+1 )} \\& \qquad {}-\frac{F_{4k-2}+F_{4k+2}+1}{ (F_{4k-2}+1 )F_{2k-1}F_{2k+1} (F_{4k+2}+1 )} \\& \quad = \frac{3F_{2k-1}F_{2k+1}- (F_{2k+2}^{2}-F_{2k-2}^{2} )-1}{ (F_{4k-2}+1 )F_{2k-1}F_{2k+1} (F_{4k+2}+1 )} \\& \quad < \frac{3F_{2k-1}F_{2k+1}-4F_{2k-1}F_{2k+1}-1}{ (F_{4k-2}+1 )F_{2k-1}F_{2k+1} (F_{4k+2}+1 )} \\& \quad < -\frac{1}{ (F_{4k-2}+1 ) (F_{4k+2}+1 )}, \end{aligned}$$
from which we deduce that
$$\begin{aligned} \sum_{k=n}^{mn} \frac{1}{F_{2k-1}F_{2k+1}} >&\sum_{k=n}^{mn} \biggl( \frac{1}{F_{4k-2}+1}-\frac {1}{F_{4k+2}+1} \biggr)+ \sum_{k=n}^{mn} \frac{1}{ (F_{4k-2}+1 ) (F_{4k+2}+1 )} \\ =&\frac{1}{F_{4n-2}+1}-\frac{1}{F_{4mn+2}+1}+ \sum_{k=n}^{mn} \frac{1}{ (F_{4k-2}+1 ) (F_{4k+2}+1 )} \\ >&\frac{1}{F_{4n-2}+1}+\frac{1}{ (F_{4n-2}+1 ) (F_{4n+2}+1 )}- \frac{1}{F_{8n+2}+1} \\ >&\frac{1}{F_{4n-2}+1}, \end{aligned}$$
(2.22)
where the last inequality follows from Lemma 2.10.
Combining (2.21) and (2.22), we obtain
$$ \frac{1}{F_{4n-2}+1}< \sum_{k=n}^{mn} \frac {1}{F_{2k-1}F_{2k+1}}< \frac{1}{F_{4n-2}}, $$
from which the desired result follows. □

Similarly, we can obtain the following result, whose proof is omitted here.

Theorem 2.12

For all \(n\geq1\) and \(m\geq2\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {1}{F_{2k}F_{2k+2}} \Biggr)^{-1} \Biggr\rfloor =F_{4n}-1. $$

3 Main results II: alternating reciprocal sums

In this section, we extend the analysis of the sums of the products of two reciprocal Fibonacci numbers to alternating sums.

3.1 Alternating reciprocal sum of \(F_{k}F_{k+1}\)

Lemma 3.1

For \(n\geq1\), we have
$$\begin{aligned}& \frac{F_{n+1}}{F_{n}}-\frac{1}{F_{2n}} = \frac {F_{2n+1}+(-1)^{n}-1}{F_{2n}}, \end{aligned}$$
(3.1)
$$\begin{aligned}& \frac{F_{n+1}}{F_{n}}+\frac{1}{F_{2n}} = \frac {F_{2n+1}+(-1)^{n}+1}{F_{2n}}. \end{aligned}$$
(3.2)

Proof

Applying (2.1) repeatedly and (2.4), we derive that
$$\begin{aligned} \frac{F_{n+1}}{F_{n}}-\frac{1}{F_{2n}} =&\frac {F_{n+1}F_{2n}-F_{n}}{F_{n}F_{2n}} \\ =&\frac{F_{n+1} (F_{n-1}F_{n}+F_{n}F_{n+1} )-F_{n}}{F_{n}F_{2n}} \\ =&\frac{F_{n+1}F_{n-1}+F_{n+1}^{2}-1}{F_{2n}} \\ =&\frac{ (F_{n+1}F_{n-1}-F_{n}^{2} )+ (F_{n}^{2}+F_{n+1}^{2} )-1}{F_{2n}} \\ =&\frac{F_{2n+1}+(-1)^{n}-1}{F_{2n}}. \end{aligned}$$
Then (3.2) immediately follows from (3.1). □

Lemma 3.2

If \(n\geq2\), then
$$\begin{aligned}& \frac{1}{F_{2n}-1}-\frac{F_{n+1}}{F_{n}F_{2n+1}}>0, \end{aligned}$$
(3.3)
$$\begin{aligned}& \frac{1}{F_{2n}+1}-\frac{F_{n+1}}{F_{n}F_{2n+1}}< 0. \end{aligned}$$
(3.4)

Proof

It follows from (2.4) that
$$ \frac{1}{F_{2n}-1}-\frac{F_{n+1}}{F_{n}F_{2n+1}}=\frac {F_{n}F_{2n+1}-F_{n+1}F_{2n}+F_{n+1}}{ (F_{2n}-1 )F_{n}F_{2n+1}}= \frac {(-1)^{n+1}F_{n}+F_{n+1}}{ (F_{2n}-1 )F_{n}F_{2n+1}}>0. $$
Similarly, we can prove (3.4). □

Lemma 3.3

For \(n\geq1\) and \(m\geq2\), we have
$$\begin{aligned}& \frac{F_{2n+1}}{F_{2n}}-\frac{F_{mn+2}}{F_{mn+1}}>0, \end{aligned}$$
(3.5)
$$\begin{aligned}& \frac{F_{2n+2}}{F_{2n+1}}-\frac{F_{mn+2}}{F_{mn+1}}\leq0. \end{aligned}$$
(3.6)

Proof

With the help of (2.4), we see that
$$ \frac{F_{2n+1}}{F_{2n}}-\frac{F_{mn+2}}{F_{mn+1}}=\frac {F_{2n+1}F_{mn+1}-F_{2n}F_{mn+2}}{F_{2n}F_{mn+1}} =(-1)^{2n+2} \frac{F_{(m-2)n+1}}{F_{2n}F_{mn+1}}>0. $$
A similar analysis yields (3.6). □

Theorem 3.4

If \(m\geq2\) and \(n\geq1\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{k}F_{k+1}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} F_{2n}-1 &\textit{if }n \textit{ is even}; \\ -F_{2n}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Proof

Employing (2.4), we derive that
$$ \frac{(-1)^{k}}{F_{k}F_{k+1}}=\frac {F_{k+1}^{2}-F_{k}F_{k+2}}{F_{k}F_{k+1}}=\frac{F_{k+1}}{F_{k}}-\frac {F_{k+2}}{F_{k+1}}, $$
which implies that
$$ \sum_{k=n}^{mn} \frac{(-1)^{k}}{F_{k}F_{k+1}}=\frac {F_{n+1}}{F_{n}}-\frac{F_{mn+2}}{F_{mn+1}}. $$
(3.7)
Furthermore, it follows from (3.6) and (2.4) that
$$ \frac{F_{n+1}}{F_{n}}-\frac{F_{mn+2}}{F_{mn+1}}\leq\frac {F_{n+1}}{F_{n}}- \frac{F_{2n+2}}{F_{2n+1}} =\frac{F_{n+1}F_{2n+1}-F_{n}F_{2n+2}}{F_{n}F_{2n+1}}=(-1)^{n}\frac {F_{n+1}}{F_{n}F_{2n+1}}. $$
(3.8)
We first assume that n is even. Then combining (3.8) and (3.3), we obtain
$$ \frac{F_{n+1}}{F_{n}}-\frac{F_{mn+2}}{F_{mn+1}}\leq\frac {F_{n+1}}{F_{n}F_{2n+1}}< \frac{1}{F_{2n}-1}. $$
(3.9)
On the other hand, it follows from (3.1) and (3.5) that
$$ \frac{F_{n+1}}{F_{n}}-\frac{1}{F_{2n}}-\frac{F_{mn+2}}{F_{mn+1}}>0, $$
which means that
$$ \frac{F_{n+1}}{F_{n}}-\frac{F_{mn+2}}{F_{mn+1}}>\frac{1}{F_{2n}}. $$
(3.10)
Now combining (3.7), (3.9), and (3.10), we deduce that
$$ \frac{1}{F_{2n}}< \sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{k}F_{k+1}}< \frac{1}{F_{2n}-1}, $$
which shows that the statement is true when n is even.
We now consider the case where n is odd. It is clearly true for \(n=1\), so we assume that \(n\geq3\). Applying (3.8) and (3.4), we can see that
$$ \frac{F_{n+1}}{F_{n}}-\frac{F_{mn+2}}{F_{mn+1}}\leq-\frac {F_{n+1}}{F_{n}F_{2n+1}}< - \frac{1}{F_{2n}+1}. $$
(3.11)
It follows from (3.2) and (3.5) that
$$ \frac{F_{n+1}}{F_{n}}+\frac{1}{F_{2n}}-\frac{F_{mn+2}}{F_{mn+1}}=\frac {F_{2n+1}}{F_{2n}}- \frac{F_{mn+2}}{F_{mn+1}}>0, $$
which means that
$$ \frac{F_{n+1}}{F_{n}}-\frac{F_{mn+2}}{F_{mn+1}}>-\frac{1}{F_{2n}}. $$
(3.12)
Combining (3.7), (3.11), and (3.12), we get that
$$ -\frac{1}{F_{2n}}< \sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{k}F_{k+1}}< -\frac{1}{F_{2n}+1}, $$
which yields the desired identity. □

3.2 Alternating reciprocal sums of \(F_{2k-1}F_{2k}\)

For \(n\geq2\), we define
$$\begin{aligned}& f(n) = \frac{1}{3F_{2n-2}F_{2n-1}}-\frac {(-1)^{n}}{F_{2n-1}F_{2n}}-\frac{1}{3F_{2n}F_{2n+1}}, \\& g(n) = \frac{1}{3F_{2n-2}F_{2n-1}+1}-\frac {(-1)^{n}}{F_{2n-1}F_{2n}}-\frac{1}{3F_{2n}F_{2n+1}+1}, \\& s(n) = \frac{-1}{3F_{2n-2}F_{2n-1}+1}-\frac {(-1)^{n}}{F_{2n-1}F_{2n}}+\frac{1}{3F_{2n}F_{2n+1}+1}, \\& t(n) = \frac{-1}{3F_{2n-2}F_{2n-1}}-\frac {(-1)^{n}}{F_{2n-1}F_{2n}}+\frac{1}{3F_{2n}F_{2n+1}}. \end{aligned}$$
It is not hard to check that \(f(n)\), \(g(n)\), \(s(n)\), and \(t(n)\) are all negative if n is even and positive otherwise.

Lemma 3.5

For \(n\geq2\), we have
$$ f(n)+f(n+1)>0. $$
(3.13)

Proof

The statement is clearly true when n is odd, so we assume that n is even in the rest of the proof. Applying (2.3), we derive that
$$\begin{aligned} f(n)+f(n+1) =&\frac{1}{3F_{2n-2}F_{2n-1}}-\frac {1}{F_{2n-1}F_{2n}}+\frac{1}{F_{2n+1}F_{2n+2}} - \frac{1}{3F_{2n+2}F_{2n+3}} \\ =&\frac{1}{3} \biggl(\frac{1}{F_{2n-2}F_{2n-1}}-\frac {1}{F_{2n+2}F_{2n+3}} \biggr) - \biggl(\frac{1}{F_{2n-1}F_{2n}}-\frac{1}{F_{2n+1}F_{2n+2}} \biggr) \\ =&\frac{1}{3}\frac {F_{2n+2}F_{2n+3}-F_{2n-2}F_{2n-1}}{F_{2n-2}F_{2n-1}F_{2n+2}F_{2n+3}} -\frac {F_{2n+1}F_{2n+2}-F_{2n-1}F_{2n}}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}} \\ =&\frac{1}{3}\frac{F_{2n+2}F_{2n+3}-F_{2n}F_{2n+1}+ F_{2n}F_{2n+1}-F_{2n-2}F_{2n-1}}{F_{2n-2}F_{2n-1}F_{2n+2}F_{2n+3}} \\ &{}-\frac {F_{2n+1}F_{2n+2}-F_{2n-1}F_{2n}}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}} \\ =&\frac{1}{3}\frac{F_{4n+3}+F_{4n-1}}{F_{2n-2}F_{2n-1}F_{2n+2}F_{2n+3}}- \frac{F_{4n+1}}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}} \\ =&\frac{1}{3}\frac{3F_{4n+1}}{F_{2n-2}F_{2n-1}F_{2n+2}F_{2n+3}}- \frac{F_{4n+1}}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}} \\ =&\frac{F_{4n+1}}{F_{2n-1}F_{2n+2}} \biggl(\frac {1}{F_{2n-2}F_{2n+3}}-\frac{1}{F_{2n}F_{2n+1}} \biggr) \\ =&\frac{F_{4n+1}}{F_{2n-1}F_{2n+2}}\cdot \frac {F_{2n}F_{2n+1}-F_{2n-2}F_{2n+3}}{F_{2n-2}F_{2n}F_{2n+1}F_{2n+3}} \\ =&\frac{F_{4n+1}}{F_{2n-1}F_{2n+2}}\cdot\frac {2}{F_{2n-2}F_{2n}F_{2n+1}F_{2n+3}} \\ >&0, \end{aligned}$$
where the last equality follows from (2.4). □

Remark

From the proof of Lemma 3.5 we can easily derive that if n is odd, then
$$\begin{aligned} f(n)+f(n+1) =&\frac{F_{4n+1}}{F_{2n-1}F_{2n+2}} \biggl(\frac {1}{F_{2n-2}F_{2n+3}}+ \frac{1}{F_{2n}F_{2n+1}} \biggr) \\ >&\frac{F_{4n+1}}{F_{2n-1}F_{2n+2}}\cdot\frac {2}{F_{2n-2}F_{2n}F_{2n+1}F_{2n+3}}. \end{aligned}$$
Therefore, whether n is even or odd, we always have
$$ f(n)+f(n+1)\geq\frac{F_{4n+1}}{F_{2n-1}F_{2n+2}}\cdot\frac {2}{F_{2n-2}F_{2n}F_{2n+1}F_{2n+3}} > \frac{2}{F_{2n-2}F_{2n}F_{2n+1}F_{2n+3}}. $$
(3.14)

Lemma 3.6

If \(n\geq2\) and \(m\geq2\), we have
$$ f(n)+f(n+1)+f(mn)>0. $$

Proof

If mn is odd, then the result follows from (3.13) and the fact \(f(mn)>0\). Now we consider the case where mn is even. It is straightforward to check that
$$\begin{aligned} f(mn) =&\frac{1}{3F_{2mn-2}F_{2mn-1}}-\frac {1}{F_{2mn-1}F_{2mn}}-\frac{1}{3F_{2mn}F_{2mn+1}} \\ =&\frac {F_{2mn}F_{2mn+1}-F_{2mn-2}F_{2mn-1}-3F_{2mn-2}F_{2mn+1}}{3F_{2mn-2}F_{2mn-1}F_{2mn}F_{2mn+1}} \\ =&\frac{F_{2mn} (F_{2mn}+F_{2mn-1} )- (F_{2mn}-F_{2mn-1} )F_{2mn-1}-3F_{2mn-2}F_{2mn+1}}{ 3F_{2mn-2}F_{2mn-1}F_{2mn}F_{2mn+1}} \\ =&\frac{F_{2mn}^{2}+F_{2mn-1}^{2}-3 (F_{2mn}-F_{2mn-1} ) (F_{2mn}+F_{2mn-1} )}{ 3F_{2mn-2}F_{2mn-1}F_{2mn}F_{2mn+1}} \\ =&\frac{4F_{2mn-1}^{2}-2F_{2mn}^{2}}{3F_{2mn-2}F_{2mn-1}F_{2mn}F_{2mn+1}}. \end{aligned}$$
Invoking (2.4), we get
$$\begin{aligned} F_{2mn}^{2}-2F_{2mn-1}^{2} =&(F_{2mn-1}+F_{2mn-2})^{2}-2F_{2mn-1}^{2} \\ =&2F_{2mn-2}F_{2mn-1}+F_{2mn-2}^{2}-F_{2mn-1}^{2} \\ =&2F_{2mn-2}F_{2mn-1}-F_{2mn-3}F_{2mn} \\ =&F_{2mn-2}F_{2mn-1}-(F_{2mn-3}F_{2mn}-F_{2mn-2}F_{2mn-1}) \\ =&F_{2mn-2}F_{2mn-1}-(-1)^{2mn-2} \\ =&F_{2mn-2}F_{2mn-1}-1. \end{aligned}$$
Therefore, we have
$$ f(mn)=\frac{-2 (F_{2mn-2}F_{2mn-1}-1 )}{3F_{2mn-2}F_{2mn-1}F_{2mn}F_{2mn+1}}> \frac{-2}{3F_{2mn}F_{2mn+1}}. $$
(3.15)
Combining (3.14) and (3.15), we obtain
$$\begin{aligned} f(n)+f(n+1)+f(mn) >&\frac{2}{F_{2n-2}F_{2n}F_{2n+1}F_{2n+3}}-\frac {2}{3F_{2mn}F_{2mn+1}} \\ \geq&\frac{2}{F_{2n}F_{2n+1}F_{2n-2}F_{2n+3}}-\frac{2}{3F_{4n}F_{4n+1}}. \end{aligned}$$
Since \(F_{4n+1}>F_{4n}\) and
$$ F_{4n}=F_{2n-1}F_{2n}+F_{2n}F_{2n+1}=F_{2n-3}F_{2n+2}+F_{2n-2}F_{2n+3}, $$
we determine that
$$ f(n)+f(n+1)+f(mn)>0. $$
The proof is completed. □

Lemma 3.7

If \(n\geq2\) is even, we have
$$ g(n)+g(n+1)< 0. $$

Proof

From the proof of Lemma 3.5 we know that
$$\begin{aligned} g(n)+g(n+1) =&\frac{1}{3F_{2n-2}F_{2n-1}+1}-\frac {1}{F_{2n-1}F_{2n}}+\frac{1}{F_{2n+1}F_{2n+2}} - \frac{1}{3F_{2n+2}F_{2n+3}+1} \\ =&\frac{3 (F_{2n+2}F_{2n+3}-F_{2n-2}F_{2n-1} )}{ (3F_{2n-2}F_{2n-1}+1 ) (3F_{2n+2}F_{2n+3}+1 )} - \biggl(\frac{1}{F_{2n-1}F_{2n}}-\frac{1}{F_{2n+1}F_{2n+2}} \biggr) \\ =&\frac{9F_{4n+1}}{ (3F_{2n-2}F_{2n-1}+1 ) (3F_{2n+2}F_{2n+3}+1 )} -\frac{9F_{4n+1}}{9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}. \end{aligned}$$
For \(n\geq2\), we have
$$\begin{aligned}& (3F_{2n-2}F_{2n-1}+1 ) (3F_{2n+2}F_{2n+3}+1 ) \\& \quad > 9F_{2n-2}F_{2n-1}F_{2n+2}F_{2n+3} +3F_{2n+2}F_{2n+3} \\& \quad = 9F_{2n-1}F_{2n+2} (F_{2n-2}F_{2n+3}-F_{2n}F_{2n+1} ) \\& \qquad {} +3F_{2n+2}F_{2n+3}+9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2} \\& \quad = 9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}+3F_{2n+2}F_{2n+3} -18F_{2n-1}F_{2n+2} \\& \quad = 9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}+3F_{2n+2} (F_{2n+3}-6F_{2n-1} ) \\& \quad = 9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}+3F_{2n+2} (3F_{2n-2}-F_{2n-1} ) \\& \quad > 9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}, \end{aligned}$$
which implies
$$ g(n)+g(n+1)< 0. $$
This completes the proof. □

Lemma 3.8

If \(n>0\) is even, then
$$ g(n)+\frac{1}{3F_{2n}F_{2n+1}+1}< 0. $$

Proof

The result follows from the definition of \(g(n)\) and the fact \(3F_{n}>F_{n+2}\). □

To introduce the property of \(s(n)\), we need two preliminary results.

Lemma 3.9

If \(n\geq5\), then
$$ F_{2n}F_{2n+1}>3F_{n-1}F_{n}F_{n+1}F_{n+2}. $$

Proof

It is easy to see that \(2F_{n}>F_{n+1}\) for \(n\geq3\). We claim that \(5F_{n}\geq3F_{n+1}\) if \(n\geq3\). First, the claim holds for \(n=3\) and \(n=4\). Now we assume that \(n\geq5\). It is straightforward to verify that
$$ 5F_{n}=3F_{n}+2F_{n}=3F_{n}+2F_{n-1}+2F_{n-2}>3F_{n}+2F_{n-1}+F_{n-1}=3F_{n+1}. $$
Since \(5F_{n}\geq3F_{n+1}\) and \(2F_{n+1}>F_{n+2}\), we have \(5F_{n}>F_{n+3}\) for \(n\geq3\).
It follows from (2.1) that \(F_{2n}=F_{n-1}F_{n}+F_{n}F_{n+1}\) and
$$ F_{2n+1}=F_{n-2}F_{n+2}+F_{n-1}F_{n+3}=F_{n-2}F_{n+2}+F_{n-1}F_{n+2}+F_{n-1}F_{n+1}, $$
from which we derive that
$$\begin{aligned} \frac{F_{2n}F_{2n+1}}{F_{n-1}F_{n}F_{n+1}F_{n+2}} =&\frac {F_{2n}}{F_{n}F_{n+1}}\cdot \frac{F_{2n+1}}{F_{n-1}F_{n+2}} \\ =& \biggl(1+\frac{F_{n-1}}{F_{n+1}} \biggr) \biggl(1+\frac {F_{n-2}}{F_{n-1}}+ \frac{F_{n+1}}{F_{n+2}} \biggr) \\ =&1+\frac{F_{n-2}}{F_{n-1}}+\frac{F_{n+1}}{F_{n+2}}+\frac{F_{n-1}}{F_{n+1}} + \frac{F_{n-2}}{F_{n+1}}+\frac{F_{n-1}}{F_{n+2}} \\ =&1+\frac{F_{n-2}}{F_{n-1}}+\frac{F_{n+1}}{F_{n+2}}+\frac {F_{n}}{F_{n+1}}+ \frac{F_{n-1}}{F_{n+2}} \\ >&1+\frac{3}{5}+\frac{3}{5}+\frac{3}{5}+ \frac{1}{5} \\ =&3, \end{aligned}$$
which completes the proof. □

Lemma 3.10

For \(n\geq5\), we have
$$ F_{2n+1} (F_{n+3}-6F_{n-1} )> (F_{n-2}F_{n-1}+1 )F_{n+3}. $$

Proof

It is easy to see that \(F_{n-2}^{2}F_{n+2}>F_{n+3}\) for \(n\geq5\), and thus we have
$$\begin{aligned} F_{2n+1}F_{n-2} =& (F_{n-2}F_{n+2}+F_{n-1}F_{n+3} )F_{n-2} \\ =&F_{n-2}^{2}F_{n+2}+F_{n-2}F_{n-1}F_{n+3} \\ >&F_{n+3}+F_{n-2}F_{n-1}F_{n+3} \\ =& (F_{n-2}F_{n-1}+1 )F_{n+3}. \end{aligned}$$
It is straightforward to check that \(F_{n+3}=5F_{n-1}+3F_{n-2}\), from which we get
$$ F_{n+3}-6F_{n-1}=3F_{n-2}-F_{n-1}>F_{n-2}. $$

Combining the last two inequalities yields the desired result. □

Lemma 3.11

If \(n\geq3\) is odd, then
$$ s(n)+s(n+1)-\frac{1}{3F_{4n}F_{4n+1}+1}>0. $$

Proof

Since n is odd, we have
$$\begin{aligned} s(n)+s(n+1) =&\frac{-1}{3F_{2n-2}F_{2n-1}+1}+\frac {1}{F_{2n-1}F_{2n}}-\frac{1}{F_{2n+1}F_{2n+2}} + \frac{1}{3F_{2n+2}F_{2n+3}+1} \\ =& \biggl(\frac{1}{F_{2n-1}F_{2n}}-\frac{1}{F_{2n+1}F_{2n+2}} \biggr)- \biggl( \frac{1}{3F_{2n-2}F_{2n-1}+1}- \frac{1}{3F_{2n+2}F_{2n+3}+1} \biggr) \\ =&\frac{F_{4n+1}}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}- \frac{9F_{4n+1}}{ (3F_{2n-2}F_{2n-1}+1 ) (3F_{2n+2}F_{2n+3}+1 )} \\ >&\frac{F_{4n+1}}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}- \frac{3F_{4n+1}}{ (3F_{2n-2}F_{2n-1}+1 )F_{2n+2}F_{2n+3}} \\ =&\frac{1}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}\cdot \frac{F_{4n+1} (F_{2n+3}-6F_{2n-1} )}{ (3F_{2n-2}F_{2n-1}+1 )F_{2n+3}} \\ >&\frac{1}{F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}\cdot \frac{F_{4n+1} (F_{2n+3}-6F_{2n-1} )}{ (3F_{2n-2}F_{2n-1}+3 )F_{2n+3}} \\ >&\frac{1}{3F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}, \end{aligned}$$
where the last inequality follows from Lemma 3.10.
Applying Lemma 3.9, we have
$$ 3F_{4n}F_{4n+1}+1>3F_{4n}F_{4n+1}>9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}, $$
which implies that
$$\begin{aligned} s(n)+s(n+1)-\frac{1}{3F_{4n}F_{4n+1}+1} >&s(n)+s(n+1)-\frac {1}{9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}} \\ >&\frac{1}{3F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}}-\frac {1}{9F_{2n-1}F_{2n}F_{2n+1}F_{2n+2}} \\ >&0, \end{aligned}$$
which completes the proof. □

Applying a similar analysis of \(f(n)\), we can obtain the following properties of \(t(n)\), whose proofs are omitted here.

Lemma 3.12

For \(n\geq2\),
$$ t(n)+t(n+1)< 0. $$

Lemma 3.13

If \(n\geq2\) and \(m\geq2\), we have
$$ t(n)+t(n+1)+t(mn)< 0. $$

Theorem 3.14

If \(m\geq2\) and \(n\geq3\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k-1}F_{2k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} 3F_{2n-2}F_{2n-1} &\textit{if }n \textit{ is even}; \\ -3F_{2n-2}F_{2n-1}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Proof

We first consider the case where n is even. With the help of \(f(n)\), we have
$$ \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k}}= \frac {1}{3F_{2n-2}F_{2n-1}}-\frac{1}{3F_{2mn}F_{2mn+1}}- \sum_{k=n}^{mn}f(k). $$
Lemma 3.5 implies that
$$ \sum_{k=n+2}^{mn-1}f(k)>0. $$
Furthermore, applying Lemma 3.6, we get
$$ \sum_{k=n}^{mn}f(k)=f(n)+f(n+1)+f(mn)+\sum _{k=n+2}^{mn-1}f(k)>0. $$
Hence, we obtain
$$ \sum_{k=n}^{mn} \frac{(-1)^{k}}{F_{2k-1}F_{2k}}< \frac {1}{3F_{2n-2}F_{2n-1}}. $$
(3.16)
It follows from Lemma 3.7 and Lemma 3.8 that
$$\begin{aligned} \sum_{k=n}^{mn} \frac{(-1)^{k}}{F_{2k-1}F_{2k}} =&\frac {1}{3F_{2n-2}F_{2n-1}+1}-\frac{1}{3F_{2mn}F_{2mn+1}+1}- \sum _{k=n}^{mn}g(k) \\ =&\frac{1}{3F_{2n-2}F_{2n-1}+1}-\sum_{k=n}^{mn-1}g(k)- \biggl(g(mn)+\frac{1}{3F_{2mn}F_{2mn+1}+1} \biggr) \\ >&\frac{1}{3F_{2n-2}F_{2n-1}+1}. \end{aligned}$$
(3.17)
Combining (3.16) and (3.17) yields
$$ \frac{1}{3F_{2n-2}F_{2n-1}+1}< \sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k-1}F_{2k}}< \frac{1}{3F_{2n-2}F_{2n-1}}, $$
which shows that the statement is true when n is even.
Next, we turn to the case where n is odd. It follows from Lemma 3.11 that
$$\begin{aligned} \sum_{k=n}^{mn} \frac{(-1)^{k}}{F_{2k-1}F_{2k}}&=\frac {-1}{3F_{2n-2}F_{2n-1}+1}+\frac{1}{3F_{2mn}F_{2mn+1}+1}- \sum _{k=n}^{mn}s(k) \\ &=\frac{-1}{3F_{2n-2}F_{2n-1}+1}- \biggl(s(n)+s(n+1)-\frac {1}{3F_{2mn}F_{2mn+1}+1} \biggr) -\sum _{k=n+2}^{mn}s(k) \\ &< \frac{-1}{3F_{2n-2}F_{2n-1}+1}- \biggl(s(n)+s(n+1)-\frac {1}{3F_{4n}F_{4n+1}+1} \biggr) -\sum _{k=n+2}^{mn}s(k) \\ &< \frac{-1}{3F_{2n-2}F_{2n-1}+1}. \end{aligned}$$
(3.18)
If mn is even, then it follows from Lemma 3.12 that \(\sum_{k=n}^{mn}t(k)<0\), and hence
$$\begin{aligned} \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k}} =& \frac {-1}{3F_{2n-2}F_{2n-1}}+\frac{1}{3F_{2mn}F_{2mn+1}}- \sum_{k=n}^{mn}t(k) \\ >&\frac{-1}{3F_{2n-2}F_{2n-1}}. \end{aligned}$$
If mn is odd, then it follows from Lemma 3.13 that
$$\begin{aligned} \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k}} =& \frac {-1}{3F_{2n-2}F_{2n-1}}+\frac{1}{3F_{2mn}F_{2mn+1}}- \sum_{k=n}^{mn}t(k) \\ =&\frac{-1}{3F_{2n-2}F_{2n-1}}+\frac{1}{3F_{2mn}F_{2mn+1}}-\sum_{k=n+2}^{mn-1}t(k) \\ &{}- \bigl(t(n)+t(n+1)+t(mn) \bigr) \\ >&\frac{-1}{3F_{2n-2}F_{2n-1}}. \end{aligned}$$
Therefore, if n is odd, then we always have
$$ \sum_{k=n}^{mn} \frac{(-1)^{k}}{F_{2k-1}F_{2k}}>\frac {-1}{3F_{2n-2}F_{2n-1}}. $$
(3.19)
It follows from (3.18) and (3.19) that
$$ \frac{-1}{3F_{2n-2}F_{2n-1}}< \sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k-1}F_{2k}}< \frac{-1}{3F_{2n-2}F_{2n-1}+1}, $$
which shows that the assertion for odd n also holds. □

Similarly, we can consider the alternating reciprocal sums of \(F_{2k}F_{2k+1}\) and obtain the following result, whose proof is similar to that of Theorem 3.14 and is omitted here.

Theorem 3.15

If \(m\geq2\) and \(n\geq2\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k}F_{2k+1}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} 3F_{2n-1}F_{2n}-1 &\textit{if }n \textit{ is even}; \\ -3F_{2n-1}F_{2n}, &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

3.3 Alternating sums of \(F_{2k-1}F_{2k+1}\)

We first introduce the following notation:
$$\begin{aligned}& \alpha(n) = \frac{1}{3F_{2n-1}^{2}}-\frac {(-1)^{n}}{F_{2n-1}F_{2n+1}}-\frac{1}{3F_{2n+1}^{2}}, \\& \beta(n) = \frac{1}{3F_{2n-1}^{2}-1}-\frac {(-1)^{n}}{F_{2n-1}F_{2n+1}}-\frac{1}{3F_{2n+1}^{2}-1}, \\& \gamma(n) = \frac{-1}{3F_{2n-1}^{2}-1}-\frac {(-1)^{n}}{F_{2n-1}F_{2n+1}}+\frac{1}{3F_{2n+1}^{2}-1}, \\& \delta(n) = \frac{-1}{3F_{2n-1}^{2}}-\frac {(-1)^{n}}{F_{2n-1}F_{2n+1}}+\frac{1}{3F_{2n+1}^{2}}. \end{aligned}$$
It is not hard to check that \(\alpha(n)\), \(\beta(n)\), \(\gamma(n)\), and \(\delta(n)\) are all negative if n is even and positive otherwise.

Lemma 3.16

If \(n>0\) is even, then
$$ \alpha(n)+\alpha(n+1)< 0. $$

Proof

Since n is even, we have
$$\begin{aligned} \alpha(n)+\alpha(n+1) =&\frac{1}{3F_{2n-1}^{2}}-\frac {1}{F_{2n-1}F_{2n+1}}+ \frac{1}{F_{2n+1}F_{2n+3}} -\frac{1}{3F_{2n+3}^{2}} \\ =&\frac{1}{3} \biggl(\frac{1}{F_{2n-1}^{2}}-\frac{1}{F_{2n+3}^{2}} \biggr)- \frac{1}{F_{2n+1}} \biggl(\frac{1}{F_{2n-1}}-\frac{1}{F_{2n+3}} \biggr) \\ =&\frac{1}{3} \biggl(\frac{1}{F_{2n-1}}-\frac{1}{F_{2n+3}} \biggr) \biggl(\frac{1}{F_{2n-1}}+\frac{1}{F_{2n+3}}-\frac {3}{F_{2n+1}} \biggr) \\ =&\frac{1}{3} \biggl(\frac{1}{F_{2n-1}}-\frac{1}{F_{2n+3}} \biggr) \biggl(\frac{1}{F_{2n+3}}-\frac{F_{2n-3}}{F_{2n-1}F_{2n+1}} \biggr) \\ =&\frac{1}{3} \biggl(\frac{1}{F_{2n-1}}-\frac{1}{F_{2n+3}} \biggr) \biggl(\frac {F_{2n-1}F_{2n+1}-F_{2n-3}{F_{2n+3}}}{F_{2n-1}F_{2n+1}F_{2n+3}} \biggr) \\ =&- \biggl(\frac{1}{F_{2n-1}}-\frac{1}{F_{2n+3}} \biggr) \frac{1}{F_{2n-1}F_{2n+1}F_{2n+3}} \\ < &0, \end{aligned}$$
where the last equality follows from (2.4). □

Lemma 3.17

For \(n>0\),
$$ 6F_{2n-1}F_{2n+1}> \bigl(3F_{n-1}^{2}-1 \bigr)F_{n+3}^{2}. $$

Proof

It is easy to see that the result holds when \(n<5\). Next we show that, for \(n\geq5\),
$$ 2F_{2n-1}F_{2n+1}>F_{2n}F_{2n+1}>F_{n-1}^{2}F_{n+3}^{2}, $$
from which the desired result follows.
The first inequality is obvious. It follows from (2.1) that
$$\begin{aligned}& F_{2n} = F_{n-2}F_{n+3}+F_{n-3}F_{n+2}, \\& F_{2n+1} = F_{n-1}F_{n+3}+F_{n-2}F_{n+2}, \end{aligned}$$
which implies that
$$\begin{aligned} F_{2n}F_{2n+1} =&F_{n-2}F_{n-1}F_{n+3}^{2}+F_{n-2}^{2}F_{n+2}F_{n+3}+ F_{n-3}F_{n-1}F_{n+2}F_{n+3}+F_{n-3}F_{n-2}F_{n+2}^{2} \\ =&F_{n-2}F_{n-1}F_{n+3}^{2}+ \bigl(F_{n-3}F_{n-1}-(-1)^{n} \bigr)F_{n+2}F_{n+3}+F_{n-3}F_{n-1}F_{n+2}F_{n+3} \\ &{}+F_{n-3}F_{n-2}F_{n+2}^{2} \\ =&F_{n-2}F_{n-1}F_{n+3}^{2}+2F_{n-3}F_{n-1}F_{n+2}F_{n+3}+F_{n-3}F_{n-2}F_{n+2}^{2}-(-1)^{n}F_{n+2}F_{n+3} \\ >&F_{n-2}F_{n-1}F_{n+3}^{2}+F_{n-3}F_{n-1}F_{n+3}^{2}+F_{n-3}F_{n-2}F_{n+2}^{2}-(-1)^{n}F_{n+2}F_{n+3} \\ =&F_{n-1}^{2}F_{n+3}^{2}+F_{n-3}F_{n-2}F_{n+2}^{2}-(-1)^{n}F_{n+2}F_{n+3} \\ >&F_{n-1}^{2}F_{n+3}^{2}, \end{aligned}$$
where the last inequality follows from the fact that, for \(n\geq5\),
$$F_{n-3}F_{n-2}F_{n+2}\geq2F_{n+2}>F_{n+3}. $$
This completes the proof. □

Lemma 3.18

For \(n\geq1\),
$$ \beta(n)+\beta(n+1)>0. $$

Proof

It is obviously true when n is odd, so we assume that n is even. Now we have
$$\begin{aligned} \beta(n)+\beta(n+1) =& \biggl(\frac{1}{3F_{2n-1}^{2}-1}-\frac {1}{3F_{2n+3}^{2}-1} \biggr)- \biggl(\frac{1}{F_{2n-1}F_{2n+1}}-\frac{1}{F_{2n+1}F_{2n+3}} \biggr) \\ =&\frac{3 (F_{2n+3}^{2}-F_{2n-1}^{2} )}{ (3F_{2n-1}^{2}-1 ) (3F_{2n+3}^{2}-1 )} -\frac{F_{2n+3}-F_{2n-1}}{F_{2n-1}F_{2n+1}F_{2n+3}} \\ =&\frac{9F_{2n+1} (F_{2n+3}-F_{2n-1} )}{ (3F_{2n-1}^{2}-1 ) (3F_{2n+3}^{2}-1 )} -\frac{F_{2n+3}-F_{2n-1}}{F_{2n-1}F_{2n+1}F_{2n+3}}. \end{aligned}$$
Since
$$\begin{aligned}& \bigl(3F_{2n-1}^{2}-1 \bigr) \bigl(3F_{2n+3}^{2}-1 \bigr) \\& \quad = 9F_{2n-1}^{2}F_{2n+3}^{2}-3F_{2n-1}^{2}-3F_{2n+3}^{2}+1 \\& \quad = 9F_{2n-1}F_{2n+3} \bigl(F_{2n+1}^{2}+1 \bigr)-3F_{2n-1}^{2}-3F_{2n+3}^{2}+1 \\& \quad = 9F_{2n-1}F_{2n+1}^{2}F_{2n+3}+9F_{2n-1}F_{2n+3}-3F_{2n-1}^{2}-3F_{2n+3}^{2}+1 \\& \quad < 9F_{2n-1}F_{2n+1}^{2}F_{2n+3}+9F_{2n-1}F_{2n+3}-3F_{2n+3}^{2} \\& \quad = 9F_{2n-1}F_{2n+1}^{2}F_{2n+3}+3F_{2n+3} (3F_{2n-1}-F_{2n+3} ) \\& \quad = 9F_{2n-1}F_{2n+1}^{2}F_{2n+3}-3F_{2n+3} (2F_{2n-1}+3F_{2n-2} ) \\& \quad < 9F_{2n-1}F_{2n+1}^{2}F_{2n+3}, \end{aligned}$$
we have
$$ \beta(n)+\beta(n+1)>0, $$
which completes the proof. □

Remark

From the proof of Lemma 3.18 we can derive that if n is odd, then
$$\begin{aligned} \beta(n)+\beta(n+1) =& \biggl(\frac{1}{3F_{2n-1}^{2}-1}-\frac {1}{3F_{2n+3}^{2}-1} \biggr)+ \biggl(\frac{1}{F_{2n-1}F_{2n+1}}-\frac{1}{F_{2n+1}F_{2n+3}} \biggr) \\ >& \biggl(\frac{1}{3F_{2n-1}^{2}-1}-\frac{1}{3F_{2n+3}^{2}-1} \biggr)- \biggl( \frac{1}{F_{2n-1}F_{2n+1}}-\frac{1}{F_{2n+1}F_{2n+3}} \biggr) \\ =&\frac{9F_{2n+1} (F_{2n+3}-F_{2n-1} )}{ (3F_{2n-1}^{2}-1 ) (3F_{2n+3}^{2}-1 )} -\frac{F_{2n+3}-F_{2n-1}}{F_{2n-1}F_{2n+1}F_{2n+3}} \\ >&\frac{9F_{2n+1}^{2}}{ (3F_{2n-1}^{2}-1 ) (3F_{2n+3}^{2}-1 )} -\frac{F_{2n+1}}{F_{2n-1}F_{2n+1}F_{2n+3}} \\ =&\frac{9F_{2n+1}^{2}}{ (3F_{2n-1}^{2}-1 ) (3F_{2n+3}^{2}-1 )} -\frac{1}{F_{2n-1}F_{2n+3}} \\ >&\frac{3F_{2n+1}^{2}}{ (3F_{2n-1}^{2}-1 )F_{2n+3}^{2}} -\frac{1}{F_{2n-1}F_{2n+3}} \\ =&\frac{3F_{2n-1} (F_{2n+1}^{2}-F_{2n-1}F_{2n+3} )+F_{2n+3}}{ (3F_{2n-1}^{2}-1 )F_{2n-1}F_{2n+3}^{2}} \\ =&\frac{F_{2n+3}-3F_{2n-1}}{ (3F_{2n-1}^{2}-1 )F_{2n-1}F_{2n+3}^{2}} \\ =&\frac{2F_{2n}+F_{2n-2}}{ (3F_{2n-1}^{2}-1 )F_{2n-1}F_{2n+3}^{2}}. \end{aligned}$$
Thus, we have that, for all \(n>0\),
$$ \beta(n)+\beta(n+1)>\frac{2F_{2n}+F_{2n-2}}{ (3F_{2n-1}^{2}-1 )F_{2n-1}F_{2n+3}^{2}} >\frac{2}{ (3F_{2n-1}^{2}-1 )F_{2n+3}^{2}}. $$
(3.20)

Lemma 3.19

If \(n\geq1\) and \(m\geq2\), we have
$$ \beta(n)+\beta(n+1)+\beta(mn)>0. $$

Proof

If mn is odd, then the result follows from the facts \(\beta(mn)>0\) and \(\beta(n)+\beta(n+1)>0\). Next, we focus ourselves on the case where mn is even. It is easy to see that
$$\begin{aligned} \beta(mn) =&\frac{1}{3F_{2mn-1}^{2}-1}-\frac {1}{F_{2mn-1}F_{2mn+1}}-\frac{1}{3F_{2mn+1}^{2}-1} \\ =&\frac{3F_{2mn+1}^{2}-3F_{2mn-1}^{2}}{ (3F_{2mn-1}^{2}-1 ) (3F_{2mn+1}^{2}-1 )} -\frac{1}{F_{2mn-1}F_{2mn+1}} \\ >&\frac{3F_{2mn+1}^{2}-3F_{2mn-1}^{2}}{3F_{2mn-1}^{2}\cdot3F_{2mn+1}^{2}} -\frac{1}{F_{2mn-1}F_{2mn+1}} \\ =&\frac{F_{2mn+1}^{2}-F_{2mn-1}^{2}}{3F_{2mn-1}^{2}F_{2mn+1}^{2}}-\frac {1}{F_{2mn-1}F_{2mn+1}}. \end{aligned}$$
Furthermore, since
$$\begin{aligned} F_{2mn+1}^{2}-F_{2mn-1}^{2} =&F_{2mn+1} (F_{2mn-2}+2F_{2mn-1} )-F_{2mn-1}^{2} \\ =&2F_{2mn-1}F_{2mn+1}+F_{2mn-2} (F_{2mn-1}+F_{2mn} )-F_{2mn-1}^{2} \\ =&2F_{2mn-1}F_{2mn+1}+F_{2mn-2}F_{2mn-1}+ \bigl(F_{2mn-2}F_{2mn}-F_{2mn-1}^{2} \bigr) \\ =&2F_{2mn-1}F_{2mn+1}+F_{2mn-2}F_{2mn-1}+(-1)^{2mn-1} \\ >&2F_{2mn-1}F_{2mn+1}, \end{aligned}$$
we have
$$ \beta(mn)>\frac{2F_{2mn-1}F_{2mn+1}}{3F_{2mn-1}^{2}F_{2mn+1}^{2}}-\frac {1}{F_{2mn-1}F_{2mn+1}} =-\frac{1}{3F_{2mn-1}F_{2mn+1}}. $$
From (3.20) we see that
$$\begin{aligned} \beta(n)+\beta(n+1)+\beta(mn) >&\frac{2}{ (3F_{2n-1}^{2}-1 )F_{2n+3}^{2}} -\frac{1}{3F_{2mn-1}F_{2mn+1}} \\ \geq&\frac{2}{ (3F_{2n-1}^{2}-1 )F_{2n+3}^{2}} -\frac{1}{3F_{4n-1}F_{4n+1}} \\ >&0, \end{aligned}$$
where the last inequality follows from Lemma 3.17. □

Applying a similar analysis of \(\beta(n)\), we can obtain the following properties of \(\gamma(n)\), and the details are left as an exercise.

Lemma 3.20

For \(n\geq1\), we have
$$ \gamma(n)+\gamma(n+1)< 0. $$

Lemma 3.21

If \(n\geq1\) and \(m\geq2\), then
$$ \gamma(n)+\gamma(n+1)+\gamma(mn)< 0. $$

Lemma 3.22

If \(n\geq1\) is odd, then we have
$$ \delta(n)+\delta(n+1)-\frac{1}{3F_{4n+1}^{2}}>0. $$

Proof

Since n is odd, we have
$$ \delta(n)+\delta(n+1)=-\frac{1}{3F_{2n-1}^{2}}+\frac {1}{F_{2n-1}F_{2n+1}}- \frac{1}{F_{2n+1}F_{2n+3}} +\frac{1}{3F_{2n+3}^{2}}. $$
Applying the argument in the proof of Lemma 3.16, we obtain
$$ \delta(n)+\delta(n+1)=\frac {F_{2n+3}-F_{2n-1}}{F_{2n-1}^{2}F_{2n+1}F_{2n+3}^{2}}>\frac {1}{F_{2n-1}^{2}F_{2n+3}^{2}}. $$
Since \(F_{4n+1}=F_{2n-2}F_{2n+2}+F_{2n-1}F_{2n+3}\), we have \(F_{4n+1}\geq F_{2n-1}F_{2n+3}\). Thus,
$$ 3F_{4n+1}^{2}>F_{2n-1}^{2}F_{2n+3}^{2}. $$
Combining the last two inequalities yields the desired result. □

Theorem 3.23

If \(m\geq2\) and \(n\geq2\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k-1}F_{2k+1}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} 3F_{2n-1}^{2}-1 &\textit{if }n \textit{ is even}; \\ -3F_{2n-1}^{2} &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

Proof

We first consider the case where n is even. Now we have
$$ \alpha(mn)+\frac{1}{3F_{2mn+1}^{2}}=\frac{1}{3F_{2mn-1}^{2}}-\frac {1}{F_{2mn-1}F_{2mn+1}}< 0, $$
where the inequality follows from the fact \(3F_{n-1}>F_{n+1}\).
Combining Lemma 3.16 and the last inequality, we derive that
$$\begin{aligned} \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}} =& \frac {1}{3F_{2n-1}^{2}}-\frac{1}{3F_{2mn+1}^{2}} -\sum_{k=n}^{mn} \alpha(k) \\ =&\frac{1}{3F_{2n-1}^{2}} -\sum_{k=n}^{mn-1} \alpha(k)- \biggl(\alpha(mn)+\frac {1}{3F_{2mn+1}^{2}} \biggr) \\ >&\frac{1}{3F_{2n-1}^{2}}. \end{aligned}$$
With the help of \(\beta(n)\), Lemma 3.18, and Lemma 3.19, we get
$$\begin{aligned} \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}} =& \frac {1}{3F_{2n-1}^{2}-1}-\frac{1}{3F_{2mn+1}^{2}-1} -\sum_{k=n}^{mn} \beta(k) \\ < &\frac{1}{3F_{2n-1}^{2}-1}- \bigl(\beta(n)+\beta(n+1)+\beta (mn) \bigr)-\sum _{k=n+2}^{mn-1}\beta(k) \\ < &\frac{1}{3F_{2n-1}^{2}-1}. \end{aligned}$$
Therefore, we arrive at
$$ \frac{1}{3F_{2n-1}^{2}}< \sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k-1}F_{2k+1}}< \frac{1}{3F_{2n-1}^{2}-1}, $$
which shows that the statement is true when n is even.
We now turn to the case where n is odd. We have
$$ \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}}= \frac {-1}{3F_{2n-1}^{2}-1}+\frac{1}{3F_{2mn+1}^{2}-1} -\sum_{k=n}^{mn} \gamma(k). $$
If mn is even, we easily see that
$$ \sum_{k=n}^{mn}\gamma(k)< 0 $$
by Lemma 3.20. Therefore,
$$ \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}}> \frac {-1}{3F_{2n-1}^{2}-1}. $$
If mn is odd, then employing Lemma 3.20 and Lemma 3.21, we deduce
$$\begin{aligned} \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}} =& \frac {-1}{3F_{2n-1}^{2}-1}+\frac{1}{3F_{2mn+1}^{2}-1} -\sum_{k=n+2}^{mn-1} \gamma(k) \\ &{}- \bigl(\gamma(n)+\gamma(n+1)+\gamma(mn) \bigr) \\ >&\frac{-1}{3F_{2n-1}^{2}-1}. \end{aligned}$$
Thus, we always have
$$ \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}}> \frac {-1}{3F_{2n-1}^{2}-1}, $$
provided that n is odd.
Since n is odd, it follows from Lemma 3.22 that
$$ \sum_{k=n+2}^{mn}\delta(k)>0. $$
Furthermore, applying the last inequality and Lemma 3.22, we derive that
$$\begin{aligned} \sum_{k=n}^{mn}\frac{(-1)^{k}}{F_{2k-1}F_{2k+1}} =& \frac {-1}{3F_{2n-1}^{2}}+ \frac{1}{3F_{2mn+1}^{2}}-\sum_{k=n}^{mn} \delta(k) \\ =&\frac{-1}{3F_{2n-1}^{2}}-\sum_{k=n+2}^{mn} \delta(k)- \biggl(\delta(n)+\delta(n+1)- \frac{1}{3F_{2mn+1}^{2}} \biggr) \\ < &\frac{-1}{3F_{2n-1}^{2}}- \biggl(\delta(n)+\delta(n+1)-\frac {1}{3F_{4n+1}^{2}} \biggr) \\ < &\frac{-1}{3F_{2n-1}^{2}}. \end{aligned}$$
Therefore, when n is odd, we have
$$ \frac{-1}{3F_{2n-1}^{2}-1}< \sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k-1}F_{2k+1}}< \frac{-1}{3F_{2n-1}^{2}}, $$
which yields the desired identity. □

Similarly, we can prove the following result, whose proof is omitted here.

Theorem 3.24

If \(m\geq2\) and \(n\geq2\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac {(-1)^{k}}{F_{2k}F_{2k+2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \textstyle\begin{array}{l@{\quad}l} 3F_{2n}^{2} &\textit{if }n \textit{ is even}; \\ -3F_{2n}^{2}-1 &\textit{if }n\textit{ is odd}. \end{array}\displaystyle \right . $$

4 Conclusions

In this paper, we investigate the sums and alternating sums of the products of two reciprocal Fibonacci numbers in various ways. The results are new and interesting. In particular, the techniques for dealing with alternating sums can be applied to study other types of alternating sums, which will be presented in a future paper.

Declarations

Acknowledgements

The authors would like to thank the anonymous reviewers for their helpful comments. This work was supported by the National Natural Science Foundation of China (No. 11401080).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Communication and Information Engineering, University of Electronic Science and Technology of China
(2)
School of Mathematical Sciences, University of Electronic Science and Technology of China

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© Liu and Wang 2016