I want to check if $$\left\lfloor \left( \sum\_{k=n}^{2n}{\frac{1}{F\_{2k}}} \right)^{-1} \right\rfloor =F\_{2n-1}~~(n\ge 3) \tag{$\*$}$$ where $\lfloor x \rfloor$ is th floor function.


The 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 use a Python program to check $(\*).$
> He says it is true for $n\le 35$. When $n=36$, the Python says it is not true, 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 $$\left( \sum\_{k=n}^{2n}{\frac{1}{F\_{2k}}} \right)^{-1} =F\_{2n-1}~~(n\ge 3) \tag{$\*\*$}.$$
> 
> 
> 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 do not 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 MSE as [On the finite sum of reciprocal Fibonacci sequences](https://math.stackexchange.com/questions/4747937/on-the-finite-sum-of-reciprocal-fibonacci-sequences)



On the finite sum of reciprocal Fibonacci sequences
We need to prove, equivalently  

$$\frac1{F\_{2n-1}+1}<\sum\_{k=n}^{2n}\frac1{F\_{2k}}\le \frac1{F\_{2n-1}}, $$
that is, by the above expression for $F\_{k}$, since $\beta=-\alpha^{-1}$, we need to check the double inequality
$$\frac1{\alpha^{2n-1}+\alpha^{-2n+1}+\sqrt5}<\sum\_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}}\le \frac1{\alpha^{2n-1}+\alpha^{-2n+1}}. $$


To do so we bound below and above the middle sum the obvious way


$$\sum\_{k=n}^{2n} \frac1{\alpha^{2k}} <\sum\_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}} \le \frac1{1-\alpha^{-4n}} \sum\_{k=n}^{2n}\frac1{\alpha^{2k}}.$$


By computation $\displaystyle\sum\_{k=n}^{2n} \frac1{\alpha^{2k}}= \alpha^{-2n+1}-\alpha^{-4n-1}$, so everything follows from


$$\frac1{\alpha^{2n-1}+\alpha^{-2n+1}+\sqrt5}\le \alpha^{-2n+1}-\alpha^{-4n-1} $$


and
$$\frac{\alpha^{-2n+1}-\alpha^{-4n-1}}{1-\alpha^{-4n}} \le \frac1{\alpha^{2n-1}+\alpha^{-2n+1}} $$
both easily checked (the former is true for any $n\ge0$, the latter for $n\ge3$).


