1 Introduction
The Fibonacci sequence [1], Sequence A000045 is defined by the linear recurrence relation
where \(F_{n}\) is the nth Fibonacci number with \(F_{0}=0\) and \(F_{1}=1\). There exists a simple and non-obvious formula for the Fibonacci numbers,
$$F_{n}=F_{n-1}+F_{n-2}\quad \mbox{for }n\geq2, $$
$$F_{n}=\frac{1}{\sqrt{5}} \biggl(\frac{1+\sqrt{5}}{2} \biggr)^{n}- \frac{1}{\sqrt{5}} \biggl(\frac{1-\sqrt{5}}{2} \biggr)^{n}. $$
The Fibonacci sequence plays an important role in the theory and applications of mathematics, and its various properties have been investigated by many authors; see [2‐5].
Anzeige
In recent years, there has been an increasing interest in studying the reciprocal sums of the Fibonacci numbers. For example, Elsner et al. [6‐9] investigated the algebraic relations for reciprocal sums of the Fibonacci numbers. In [10], the partial infinite sums of the reciprocal Fibonacci numbers were studied by Ohtsuka and Nakamura. They established the following results, where \(\lfloor\cdot\rfloor\) denotes the floor function.
Theorem 1.1
For all
\(n\geq2\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \begin{array}{l@{\quad}l} F_{n-2}, &\textit{if }n\textit{ is even}; \\ F_{n-2}-1, &\textit{if }n\textit{ is odd}. \end{array} \right . $$
(1.1)
Theorem 1.2
For each
\(n\geq1\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \begin{array}{l@{\quad}l} F_{n}F_{n-1}-1, &\textit{if }n\textit{ is even}; \\ F_{n}F_{n-1}, &\textit{if }n\textit{ is odd}. \end{array} \right . $$
(1.2)
Further, Wu and Zhang [11, 12] generalized these identities to the Fibonacci polynomials and Lucas polynomials and various properties of such polynomials were obtained.
Anzeige
Recently, Holliday and Komatsu [13] considered the generalized Fibonacci numbers which are defined by
with \(G_{0}=0\) and \(G_{1}=1\), and a is a positive integer. They showed that
and
$$G_{n+2}=aG_{n+1}+G_{n},\quad n\geq0 $$
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{G_{k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \begin{array}{l@{\quad}l} G_{n}-G_{n-1}, &\mbox{if }n\mbox{ is even and }n\geq2; \\ G_{n}-G_{n-1}-1, &\mbox{if }n\mbox{ is odd and }n\geq1 \end{array} \right . $$
(1.3)
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{\infty} \frac{1}{G_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \begin{array}{l@{\quad}l} aG_{n}G_{n-1}-1, &\mbox{if }n\mbox{ is even and }n\geq2; \\ aG_{n}G_{n-1}, &\mbox{if }n\mbox{ is odd and }n\geq1. \end{array} \right . $$
(1.4)
More recently, Wu and Wang [14] studied the partial finite sum of the reciprocal Fibonacci numbers and deduced that, for all \(n\geq4\),
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{2n} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor =F_{n-2}. $$
(1.5)
2 Reciprocal sum of the Fibonacci numbers
We first present several well known results on Fibonacci numbers, which will be used throughout the article. The detailed proofs can be found in [5].
Lemma 2.1
Let
\(n\geq1\), we have
and
if
a
and
b
are positive integers.
$$ F_{n}^{2}-F_{n-1}F_{n+1}=(-1)^{n-1} $$
(2.1)
$$ F_{a}F_{b}+F_{a+1}F_{b+1}=F_{a+b+1} $$
(2.2)
As a consequence of (2.2), we have the following result.
Corollary 2.2
For all
\(n\geq1\), we have
$$\begin{aligned}& F_{2n} = F_{n-1}F_{n}+F_{n}F_{n+1}, \end{aligned}$$
(2.3)
$$\begin{aligned}& F_{2n+1} = F_{n}^{2}+F_{n+1}^{2}, \end{aligned}$$
(2.4)
$$\begin{aligned}& F_{2n+1} = F_{n-1}F_{n+1}+F_{n}F_{n+2}. \end{aligned}$$
(2.5)
It is easy to derive the following lemma and we leave the proof as a simple exercise.
Lemma 2.3
For each
\(n\geq1\), we have
$$ F_{n+1}F_{n+2}-F_{n-1}F_{n}=F_{2n+1}. $$
(2.6)
We now establish two inequalities on Fibonacci numbers which will be used later.
Lemma 2.4
If
\(n\geq6\), then
$$ F_{n-2}F_{n-1}>F_{n+1}. $$
(2.7)
Proof
It is easy to see that
Since \(n\geq6\), \(F_{n-2}-2>1\). So
which completes the proof. □
$$\begin{aligned} F_{n-2}F_{n-1}-F_{n+1} =&F_{n-2}F_{n-1}-(F_{n-1}+F_{n}) \\ =&F_{n-2}F_{n-1}-F_{n-1}-(F_{n-2}+F_{n-1}) \\ =&(F_{n-2}-2)F_{n-1}-F_{n-2}. \end{aligned}$$
$$F_{n-2}F_{n-1}-F_{n+1}>F_{n-1}-F_{n-2}>0, $$
Lemma 2.5
For each
\(n\geq3\), we have
$$ F_{3n-1}(F_{n}+F_{n-3})>F_{n-2}F_{n-1}F_{n}F_{n+1}. $$
(2.8)
Proof
The following are some inequalities on the sum of reciprocal Fibonacci numbers.
Proposition 2.6
For all
\(n\geq2\), we have
$$ \sum_{k=n}^{2n} \frac{1}{F_{k}}> \frac{1}{F_{n-2}+1}. $$
(2.9)
Proof
For all \(k\geq2\),
Invoking (2.1), we obtain \(F_{k-1}^{2}-F_{k}F_{k-2}=(-1)^{k}\). Therefore,
Now we have
Because of (2.5), we have
Thus, we arrive at
This completes the proof. □
$$\begin{aligned} \frac{1}{F_{k-2}+1}-\frac{1}{F_{k}}-\frac{1}{F_{k-1}+1} =&\frac {F_{k}-F_{k-2}-1}{F_{k}(F_{k-2}+1)} - \frac{1}{F_{k-1}+1} \\ =&\frac {(F_{k-1}-1)(F_{k-1}+1)-F_{k}F_{k-2}-F_{k}}{(F_{k-2}+1)(F_{k-1}+1)F_{k}} \\ =&\frac{F_{k-1}^{2}-F_{k}F_{k-2}-1-F_{k}}{(F_{k-2}+1)(F_{k-1}+1)F_{k}}. \end{aligned}$$
$$ \frac{1}{F_{k-2}+1}-\frac{1}{F_{k}}-\frac{1}{F_{k-1}+1}= \frac{(-1)^{k}-1-F_{k}}{(F_{k-2}+1)(F_{k-1}+1)F_{k}}. $$
$$\begin{aligned} \sum_{k=n}^{2n}\frac{1}{F_{k}} =& \frac{1}{F_{n-2}+1}-\frac{1}{F_{2n-1}+1}+ \sum_{k=n}^{2n} \frac{(-1)^{k-1}+1+F_{k}}{(F_{k-2}+1)(F_{k-1}+1)F_{k}} \\ >&\frac{1}{F_{n-2}+1}-\frac{1}{F_{2n-1}+1}+ \sum_{k=n}^{2n} \frac{1}{(F_{k-2}+1)(F_{k-1}+1)} \\ >&\frac{1}{F_{n-2}+1}+\frac{1}{(F_{n-2}+1)(F_{n-1}+1)}-\frac{1}{F_{2n-1}+1}. \end{aligned}$$
$$\begin{aligned} F_{2n-1}+1-(F_{n-2}+1) (F_{n-1}+1) =&F_{2n-1}-F_{n-2}F_{n-1}-F_{n-2}-F_{n-1} \\ =&F_{n-2}^{2}+F_{n-1}F_{n+1}-F_{n} \\ >&0. \end{aligned}$$
$$ \sum_{k=n}^{2n}\frac{1}{F_{k}}> \frac{1}{F_{n-2}+1}. $$
Proposition 2.7
Assume that
\(m\geq2\). Then, for all even integers
\(n\geq4\), we have
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}}< \frac{1}{F_{n-2}}. $$
(2.10)
Proof
By elementary manipulations and (2.1), we deduce that
Hence, for \(n\geq3\), we have
Since n is even,
from which we conclude that
The proof is complete. □
$$ \frac{1}{F_{k-2}}-\frac{1}{F_{k}}-\frac{1}{F_{k-1}}=\frac {(-1)^{k}}{F_{k-2}F_{k-1}F_{k}},\quad k\geq3. $$
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}}=\frac{1}{F_{n-2}}-\frac{1}{F_{mn-1}}+ \sum _{k=n}^{mn}\frac{(-1)^{k-1}}{F_{k-2}F_{k-1}F_{k}}. $$
(2.11)
$$\sum_{k=n}^{mn}\frac{(-1)^{k-1}}{F_{k-2}F_{k-1}F_{k}}< 0, $$
$$ \sum_{k=n}^{mn}\frac{1}{F_{k}}< \frac{1}{F_{n-2}}. $$
Proposition 2.8
If
\(n\geq5\)
is odd, then
$$ \sum_{k=n}^{2n} \frac{1}{F_{k}}< \frac{1}{F_{n-2}}. $$
(2.12)
Proof
It is straightforward to check that the statement is true when \(n=5\).
Now we assume that \(n\geq7\). Since n is odd, we have
Applying (2.7) and (2.6) yields
Employing (2.11) and the above two inequalities, (2.12) follows immediately. □
$$\sum_{k=n+1}^{2n}\frac{(-1)^{k-1}}{F_{k-2}F_{k-1}F_{k}}< 0. $$
$$\begin{aligned} \frac{1}{F_{n-2}F_{n-1}F_{n}}-\frac{1}{F_{2n-1}} < &\frac{1}{F_{n}F_{n+1}}-\frac{1}{F_{2n-1}} \\ =&\frac{F_{2n-1}-F_{n}F_{n+1}}{F_{n}F_{n+1}F_{2n-1}} \\ =&-\frac{F_{n-2}F_{n-1}}{F_{n}F_{n+1}F_{2n-1}} \\ <&0. \end{aligned}$$
Proposition 2.9
Let
\(m\geq3\)
be given. If
\(n\geq3\)
is odd, we have
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}}>\frac{1}{F_{n-2}}. $$
(2.13)
Proof
It is easy to see that
thus (2.13) holds for \(n=3\). Now we assume that \(n\geq 5\).
$$\sum_{k=3}^{3m}\frac{1}{F_{k}}> \frac{1}{ F_{1}}, $$
Based on (2.11) and using the fact n is odd, we have
It is clear that
Since \(m\geq3\) and invoking (2.8), we obtain
which implies
Therefore, (2.13) also holds for \(n\geq5\). □
$$\sum_{k=n}^{mn}\frac{1}{F_{k}}= \frac{1}{F_{n-2}}+\frac{1}{F_{n-2}F_{n-1}F_{n}}- \frac{1}{F_{n-1}F_{n}F_{n+1}}-\frac{1}{F_{mn-1}}+ \sum _{k=n+2}^{mn}\frac{(-1)^{k-1}}{F_{k-2}F_{k-1}F_{k}}. $$
$$\sum_{k=n+2}^{mn}\frac{(-1)^{k-1}}{F_{k-2}F_{k-1}F_{k}}>0. $$
$$F_{mn-1}(F_{n-3}+F_{n})\geq F_{3n-1}(F_{n-3}+F_{n})>F_{n-2}F_{n-1}F_{n}F_{n+1}, $$
$$\begin{aligned} \frac{1}{F_{n-2}F_{n-1}F_{n}}- \frac{1}{F_{n-1}F_{n}F_{n+1}}-\frac{1}{F_{mn-1}} =& \frac{F_{n+1}-F_{n-2}}{F_{n-2}F_{n-1}F_{n}F_{n+1}}- \frac{1}{F_{mn-1}} \\ =&\frac{F_{n-3}+F_{n}}{F_{n-2}F_{n-1}F_{n}F_{n+1}}- \frac{F_{n-3}+F_{n}}{F_{mn-1}(F_{n-3}+F_{n})} \\ >&0. \end{aligned}$$
Now we state our main results on the sum of reciprocal Fibonacci numbers.
Theorem 2.10
For all
\(n\geq4\), we have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{2n} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor =F_{n-2}. $$
(2.14)
Proof
Remark
Theorem 2.11
If
\(m\geq3\)
and
\(n\geq2\), then
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \begin{array}{l@{\quad}l} F_{n-2}, &\textit{if }n\textit{ is even}; \\ F_{n-2}-1, &\textit{if }n\textit{ is odd}. \end{array} \right . $$
(2.15)
Proof
It is clear that
Combining (2.9) and (2.10), we find that, for all even integers \(n\geq4\),
Thus (2.16) and (2.17) show that, for all \(m\geq3\),
provided that \(n\geq2\) is even.
$$ \Biggl\lfloor \Biggl(\sum_{k=2}^{2m} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor =F_{0}. $$
(2.16)
$$ \frac{1}{F_{n-2}+1}< \sum_{k=n}^{mn} \frac{1}{F_{k}}<\frac{1}{F_{n-2}}. $$
(2.17)
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor =F_{n-2}, $$
Next we aim to prove that, for \(m\geq3\) and all odd integers \(n\geq 3\),
If \(n=3\), we can readily see that
thus (2.18) holds for \(n=3\). So in the rest of the proof we assume that \(n\geq5\).
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}} \Biggr)^{-1} \Biggr\rfloor =F_{n-2}-1. $$
(2.18)
$$\sum_{k=3}^{3m}\frac{1}{F_{k}}>1, $$
It is not hard to derive that, for all \(k\geq5\),
Hence, we get
Finally, combining (2.19) with (2.13) yields (2.18). □
$$\frac{1}{F_{k-2}-1}-\frac{1}{F_{k}}-\frac{1}{F_{k-1}-1}= \frac{(-1)^{k}-1+F_{k}}{F_{k}(F_{k-2}-1)(F_{k-1}-1)}>0. $$
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}}< \frac{1}{F_{n-2}-1}-\frac{1}{F_{mn-1}-1} <\frac{1}{F_{n-2}-1}. $$
(2.19)
3 Reciprocal square sum of the Fibonacci numbers
We first give several preliminary results which will be used in our later proofs.
Lemma 3.1
For all
\(n\geq1\),
$$ F_{n}F_{n+1}-F_{n-1}F_{n+2}=(-1)^{n-1}. $$
(3.1)
Proof
It is easy to show that
Employing (2.1), the desired result follows. □
$$\begin{aligned} F_{n}F_{n+1}-F_{n-1}F_{n+2} =&F_{n}F_{n+1}-F_{n-1}(F_{n}+F_{n+1}) \\ =&F_{n}F_{n+1}-F_{n}F_{n-1}-F_{n-1}F_{n+1} \\ =&F_{n}(F_{n+1}-F_{n-1})-F_{n-1}F_{n+1} \\ =&F_{n}^{2}-F_{n-1}F_{n+1}. \end{aligned}$$
Proposition 3.2
Given an integer
\(m\geq2\)
and let
\(n\geq3\)
be odd, we have
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}^{2}}< \frac{1}{F_{n-1}F_{n}}. $$
(3.2)
Proof
It is straightforward to check that, for each \(k\geq2\),
where the last equality follows from (3.1).
$$\begin{aligned}& \frac{1}{F_{k-1}F_{k}}-\frac{1}{F_{k}^{2}} - \frac{1}{F_{k+1}^{2}}-\frac {1}{F_{k+1}F_{k+2}} \\& \quad = \frac {F_{k}F_{k+1}^{2}F_{k+2}-F_{k-1}F_{k+1}^{2}F_{k+2}-F_{k-1}F_{k}^{2}F_{k+2}-F_{k-1}F_{k}^{2} F_{k+1}}{F_{k-1}F_{k}^{2}F_{k+1}^{2}F_{k+2}} \\& \quad = \frac{F_{k}F_{k+1}(F_{k+1}F_{k+2}-F_{k-1}F_{k})-F_{k-1}F_{k+2}(F_{k+1}^{2}+F_{k}^{2})}{ F_{k-1}F_{k}^{2}F_{k+1}^{2}F_{k+2}} \\& \quad = \frac{(F_{k}F_{k+1}-F_{k-1}F_{k+2})F_{2k+1}}{ F_{k-1}F_{k}^{2}F_{k+1}^{2}F_{k+2}} \\& \quad = \frac{(-1)^{k-1}F_{2k+1}}{F_{k-1}F_{k}^{2}F_{k+1}^{2}F_{k+2}}, \end{aligned}$$
Since n is odd, we have
If m is even, then
If m is odd, then
Thus, (3.2) always holds. □
$$\frac{1}{F_{n-1}F_{n}}-\frac{1}{F_{n}^{2}}-\frac{1}{F_{n+1}^{2}}-\frac {1}{F_{n+1}F_{n+2}}>0. $$
$$\sum_{k=n}^{mn}\frac{1}{F_{k}^{2}}< \frac{1}{F_{n-1}F_{n}}-\frac{1}{F_{mn}F_{mn+1}} <\frac{1}{F_{n-1}F_{n}}. $$
$$\sum_{k=n}^{mn}\frac{1}{F_{k}^{2}}< \frac{1}{F_{n-1}F_{n}}-\frac{1}{F_{mn-1}F_{mn}}+ \frac{1}{F_{mn}^{2}}<\frac{1}{F_{n-1}F_{n}}. $$
Proposition 3.3
Let
n
be odd, then we have
$$ \sum_{k=n}^{2n} \frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}+1}. $$
(3.3)
Proof
Invoking (2.1), we can readily derive that
Now we have
It is obvious that \(2F_{k-1}F_{k}\geq F_{k-1}F_{k+1}\). From (2.1) and the fact that n is odd, we obtain
which implies that
By (2.3) and (2.4), we have
from which we conclude that
The proof is complete. □
$$ \frac{1}{F_{k-1}F_{k}+1}-\frac{1}{F_{k}^{2}}-\frac{1}{F_{k}F_{k+1}+1}=\left \{ \begin{array}{l@{\quad}l} -\frac{2F_{k-1}F_{k}+1}{F_{k}^{2}(F_{k-1}F_{k}+1)(F_{k}F_{k+1}+1)}, &\mbox{if }k \mbox{ is odd}; \\ -\frac{2F_{k}F_{k+1}+1}{F_{k}^{2}(F_{k-1}F_{k}+1)(F_{k}F_{k+1}+1)}, &\mbox{if } k\mbox{ is even}. \end{array} \right . $$
$$\begin{aligned} \sum_{k=n}^{2n}\frac{1}{F_{k}^{2}} =& \frac{1}{F_{n-1}F_{n}+1}+\biggl( \frac{2F_{n-1}F_{n}+1}{F_{n}^{2}(F_{n-1}F_{n}+1)(F_{n}F_{n+1}+1)} \\ &{}+\frac {2F_{n+1}F_{n+2}+1}{F_{n+1}^{2}(F_{n}F_{n+1}+1)(F_{n+1}F_{n+2}+1)}+\cdots \\ &{}+\frac {2F_{2n}F_{2n+1}+1}{F_{2n}^{2}(F_{2n-1}F_{2n}+1)(F_{2n}F_{2n+1}+1)}\biggr)- \frac{1}{F_{2n}F_{2n+1}+1} \\ >&\frac{1}{F_{n-1}F_{n}+1}+\frac {2F_{n-1}F_{n}+1}{F_{n}^{2}(F_{n-1}F_{n}+1)(F_{n}F_{n+1}+1)}- \frac{1}{F_{2n}F_{2n+1}+1}. \end{aligned}$$
$$\frac{2F_{n-1}F_{n}+1}{F_{n}^{2}}\geq\frac{F_{n-1}F_{n+1}+1}{F_{n}^{2}}=\frac {F^{2}_{n}}{F_{n}^{2}}=1, $$
$$\sum_{k=n}^{2n}\frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}+1}+ \frac{1}{(F_{n-1}F_{n}+1)(F_{n}F_{n+1}+1)}-\frac{1}{F_{2n}F_{2n+1}+1}. $$
$$\begin{aligned} F_{2n}F_{2n+1}+1 >&(F_{n-1}F_{n}+F_{n}F_{n+1}) \bigl(F_{n}^{2}+F_{n+1}^{2}\bigr) \\ >&(F_{n-1}F_{n}+1) (F_{n}F_{n+1}+1), \end{aligned}$$
$$\sum_{k=n}^{2n}\frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}+1}. $$
Proposition 3.4
Suppose that
\(m\geq2\)
and
\(n>0\)
is even. Then
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}^{2}}< \frac{1}{F_{n-1}F_{n}-1}. $$
(3.4)
Proof
Applying (2.1), we can rewrite \(F_{k}^{4}\) as
In addition,
Combining (3.5) and (3.6) yields
Therefore,
which completes the proof. □
$$ F_{k}^{4}=F_{k-1}F_{k}^{2}F_{k+1}+(-1)^{k-1}F_{k}^{2}. $$
(3.5)
$$\begin{aligned} (F_{k-1}F_{k}-1) (F_{k}F_{k+1}-1) =&F_{k-1}F_{k}^{2}F_{k+1}-F_{k-1}F_{k}-F_{k}F_{k+1}+1 \\ =&F_{k-1}F_{k}^{2}F_{k+1}-2F_{k-1}F_{k}-F_{k}^{2}+1. \end{aligned}$$
(3.6)
$$\begin{aligned} \frac{1}{F_{k-1}F_{k}-1}-\frac{1}{F_{k}^{2}}-\frac{1}{F_{k}F_{k+1}-1} =& \frac{F_{k}^{2}}{(F_{k-1}F_{k}-1)(F_{k}F_{k+1}-1)}- \frac{1}{F_{k}^{2}} \\ =&\frac {2F_{k-1}F_{k}-1+F_{k}^{2}+(-1)^{k-1}F_{k}^{2}}{(F_{k-1}F_{k}-1)(F_{k}F_{k+1}-1)F_{k}^{2}} \\ \geq&\frac{2F_{k-1}F_{k}-1}{(F_{k-1}F_{k}-1)(F_{k}F_{k+1}-1)F_{k}^{2}} \\ >&0. \end{aligned}$$
$$\sum_{k=n}^{mn}\frac{1}{F_{k}^{2}}< \frac{1}{F_{n-1}F_{n}-1}- \frac{1}{F_{mn}F_{mn+1}-1}<\frac{1}{F_{n-1}F_{n}-1}, $$
Proposition 3.5
If
\(n>0\)
is even, then
$$ \sum_{k=n}^{2n} \frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}}. $$
(3.7)
Proof
Employing (2.1), we can deduce that
Hence, since n is even, we have
It is easy to see that
thus
We claim that
First, by (2.6), we have
It follows from (2.3), (2.4), and (2.5) that
which implies that
Thus we obtain
which yields the desired (3.7). □
$$ \frac{1}{F_{k-1}F_{k}}-\frac{1}{F_{k}^{2}}-\frac{1}{F_{k}F_{k+1}}= \frac{(-1)^{k-1}}{F_{k-1}F_{k}^{2}F_{k+1}}. $$
$$\begin{aligned} \begin{aligned} \sum_{k=n}^{2n}\frac{1}{F_{k}^{2}}={}& \frac{1}{F_{n-1}F_{n}}+ \sum_{k=n}^{2n} \frac{(-1)^{k}}{F_{k-1}F_{k}^{2}F_{k+1}}-\frac {1}{F_{2n}F_{2n+1}} \\ ={}&\frac{1}{F_{n-1}F_{n}}+ \biggl(\frac{1}{{F_{n-1}F_{n}^{2}F_{n+1}}}-\frac {1}{F_{n}F_{n+1}^{2}F_{n+2}} - \frac{1}{F_{2n}F_{2n+1}} \biggr) \\ &{}+\sum_{k=n+2}^{2n-1}\frac{(-1)^{k}}{F_{k-1}F_{k}^{2}F_{k+1}}+ \frac{1}{F_{2n-1}F_{2n}^{2}F_{2n+1}}. \end{aligned} \end{aligned}$$
$$\sum_{k=n+2}^{2n-1}\frac{(-1)^{k-1}}{F_{k-1}F_{k}^{2}F_{k+1}}>0, $$
$$\sum_{k=n}^{2n}\frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}}+ \biggl( \frac{1}{F_{n-1}F_{n}^{2}F_{n+1}}-\frac{1}{F_{n}F_{n+1}^{2}F_{n+2}} - \frac{1}{F_{2n}F_{2n+1}} \biggr). $$
$$ \frac{1}{F_{n-1}F_{n}^{2}F_{n+1}}-\frac{1}{F_{n}F_{n+1}^{2}F_{n+2}} -\frac{1}{F_{2n}F_{2n+1}}>0. $$
$$ \frac{1}{F_{n-1}F_{n}^{2}F_{n+1}}-\frac{1}{F_{n}F_{n+1}^{2}F_{n+2}}-\frac {1}{F_{2n}F_{2n+1}}= \frac{F_{2n+1}}{F_{n-1}F_{n}^{2}F_{n+1}^{2}F_{n+2}}- \frac{F_{2n+1}}{F_{2n}F_{2n+1}^{2}}. $$
$$\begin{aligned}& F_{2n} > F_{n-1}F_{n}, \\& F_{2n+1} > F_{n+1}^{2}, \\& F_{2n+1} > F_{n}F_{n+2}, \end{aligned}$$
$$F_{2n}F_{2n+1}^{2}>F_{n-1}F_{n}^{2}F_{n+1}^{2}F_{n+2}. $$
$$\frac{1}{F_{n-1}F_{n}^{2}F_{n+1}}-\frac{1}{F_{n}F_{n+1}^{2}F_{n+2}} -\frac{1}{F_{2n}F_{2n+1}}>0, $$
Now we introduce our main result on the square sum of reciprocal Fibonacci numbers.
Theorem 3.6
For all
\(n\geq1\)
and
\(m\geq2\), we have
$$ \Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor = \left \{ \begin{array}{l@{\quad}l} F_{n}F_{n-1}, &\textit{if }n\textit{ is odd}; \\ F_{n}F_{n-1}-1, &\textit{if }n\textit{ is even}. \end{array} \right . $$
(3.8)
Proof
We first consider the case when n is odd. If \(n=1\), the result is clearly true. So we assume that \(n\geq3\).
It follows from (3.3) that
Employing (3.2) and (3.9) yields
which implies that, if \(n>0\) is odd, we have
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}^{2}}\geq\sum_{k=n}^{2n} \frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}+1}. $$
(3.9)
$$\frac{1}{F_{n-1}F_{n}+1}< \sum_{k=n}^{mn} \frac{1}{F_{k}^{2}}<\frac{1}{F_{n-1}F_{n}}, $$
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor =F_{n}F_{n-1}. $$
We now consider the case where \(n>0\) is even. It follows from (3.7) that
Combining (3.4) and (3.10), we arrive at
from which we find that, if \(n>0\) is even,
This completes the proof. □
$$ \sum_{k=n}^{mn} \frac{1}{F_{k}^{2}}\geq\sum_{k=n}^{2n} \frac{1}{F_{k}^{2}}> \frac{1}{F_{n-1}F_{n}}. $$
(3.10)
$$\frac{1}{F_{n-1}F_{n}}< \sum_{k=n}^{mn} \frac{1}{F_{k}^{2}}<\frac{1}{F_{n-1}F_{n}-1}, $$
$$\Biggl\lfloor \Biggl(\sum_{k=n}^{mn} \frac{1}{F_{k}^{2}} \Biggr)^{-1} \Biggr\rfloor =F_{n}F_{n-1}-1. $$
Acknowledgements
This work was supported by the National Natural Science Foundation of China (No. 11401080) and the Research Project of Education Teaching Reform of University of Electronic Science and Technology of China (No. 2013XJYEL032). The authors would like to thank the referees for helpful comments leading to an improvement of an earlier version.
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to deriving all the results of this article, and read and approved the final manuscript.