I want to check if $$\left \lfloor \left( \sum_{k=n}^{2n}{\dfrac{1}{F_{2k}}} \right) ^{-1} \right\rfloor =F_{2n-1}~~(n\ge 3)\qquad(*)$$ where $\lfloor x \rfloor$ is a floor function.
Fibonacci sequence is defined by $F_1=1$, $F_2=1$, $F_{n+1}=F_n+F_{n-1}~(n\ge 2)$. Then we can get $$F_n=\dfrac{\alpha ^n-\beta ^n}{\sqrt{5}}$$ where $\alpha=\dfrac{1+\sqrt{5}}{2}$ and $\beta=\dfrac{1-\sqrt{5}}{2}.$
The following are some of my attempts:
For some example:
$n=3$, the left hand is $5$, the right hand is $5.$
$n=4$, the left hand is $13$, the right hand is $13.$
$$\vdots$$
$n=15$, the left hand is $514229$, the right hand is $514229.$
It is all true. But as $n$ increases, the order of magnitude grows very rapidly.
I ask one of my good friends to ues a Python program to check $(*).$ He says it is true for $n\le 35$. When $n=36$, the Python says it is not ture, But when $n= 37$, it is true again.
Thus I change one way and I ask my fiend to use a Python program to check $$\left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right) ^{-1} =F_{2n-1}~~(n\ge 3)\qquad(**).$$
Then the program shows it is true at least for $31\le n\le 51.$
But as you see, the left hand of $(**)$ is a decimal and the right hand of $(**)$ is an integer.
So I donot know if it is because the order of magnitude on the left hand of $(*)$ is growing very fast, $(*)$ becomes not true due to some computer shortcomings.
Finally I wonder if $(*)$ is true or false? Any help and references are greatly appreciated.
Thanks!
I have also posted it on On the Finite Sum of Reciprocal Fibonacci Sequences.
Update: This identity comes from similiar identities.
$(1):$ In Hideyuki Ohtsuka and Shigeru Nakamura's paper "ON THE SUM OF RECIPROCAL FIBONACCI NUMBERS", they found $$\lfloor \left( \sum_{k=n}^{+\infty}{\dfrac{1}{F_k}} \right) ^{-1} \rfloor =\begin{cases} F_{n-2}, n\,\,\mathrm{is}\,\,\mathrm{even}\,\,\mathrm{and}\,\,n\geqslant 2.\\ F_{n-2}-1, n\,\,\mathrm{is}\,\,\mathrm{odd}\,\,\mathrm{and}\,\,n\geqslant 1.\\ \end{cases}.$$
$(2):$ In a master's thesis "On the sum of reciprocal fibonacci numbers" written by Guojie Zhang, he found
$$\lfloor \left( \sum_{k=n}^{+\infty}{\dfrac{1}{F_{3k}}} \right) ^{-1} \rfloor =\begin{cases} F_{3n-1}+F_{3n-4}, n\,\,\mathrm{is}\,\,\mathrm{even}\,\,\mathrm{and}\,\,n\geqslant 2.\\ F_{3n-1}+F_{3n-4}-1, n\,\,\mathrm{is}\,\,\mathrm{odd}\,\,\mathrm{and}\,\,n\geqslant 3.\\ \end{cases}$$
$$\lfloor \left( \sum_{k=n}^{+\infty}{\dfrac{1}{F_{4k}}} \right) ^{-1} \rfloor =F_{4n-1}+F_{4n-3}-1.$$
$(3):$ In a paper "The finite sum of reciprocal of the Fibonacci numbers" written by Wu, ZG Wang, they found $$\lfloor \left( \sum_{k=n}^{2n}{\frac{1}{F_k}} \right) ^{-1} \rfloor =F_{n-2}.$$
There are many other similar results and so on.
So I guess why there is no $2k$ identity and I use computer to get $(*)$, but I donnot know how to prove it for any integer $n\ge 3$.
Here is sketch of how such a result might accrue. We are asked if
$$\left \lfloor\left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right) ^{-1}\right\rfloor =F_{2n-1}~~(n\ge 3)$$
We shall assume that $n$ is a large number, so that we may take advantage of certain simplifications. Consider first, that
$$F_n=\frac{\varphi^n-\psi^n}{\varphi-\psi}\approx\frac{\varphi^n}{\varphi-\psi}\quad \text{ for large }n$$
where $\varphi,\psi=(1\pm\sqrt{5})/2$ . Then
$$\left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right)=(\varphi-\psi)\left( \sum_{k=n}^{2n}{\frac{1}{\varphi^{2k}}} \right)=\frac{\varphi-\psi}{\varphi^{2n}}\left(\sum_{k=0}^{2n-1}{\frac{1}{\varphi^{2k}}} \right)$$
Now, let $x=1/\varphi^2$, so that the summation becomes
$$\sum_{k=0}^{2n-1}{\frac{1}{\varphi^{2k}}}=1+x+x^2+x^3+...x^{n-1}\approx\frac{1}{1-x}=\frac{\varphi^2}{\varphi^2-1}=\varphi$$
again, for sufficiently large $n$. Then we can say
$$\left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right)=\frac{(\varphi-\psi)}{\varphi^{2n}}\varphi\approx\frac{1}{F_{2n-1}}$$
and finally
$$\left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right) ^{-1} =F_{2n-1}\quad(n>>3)$$