The question comes from On the Finite Sum of Reciprocal Fibonacci Sequences and On the finite sum of reciprocal Fibonacci sequences.
Althouth I cannot prove the identity $$\left\lfloor \left( \sum_{k=n}^{2n}{\frac{1}{F_{2k}}} \right) ^{-1} \right\rfloor =F_{2n-1}~~~~(n\ge 3)(*),$$ I seem to prove this identity
$$\lfloor \left( \sum_{k=n}^{+\infty}{\dfrac{1}{F_{2k+1}}} \right) ^{-1} \rfloor =F_{2n}.$$
where $\lfloor x \rfloor$ is th floor function.
Here's my attempts.
For any integer $n$, $F_{n+1}F_{n-1}-F^2_{n}=(-1)^n$
This is equivalent to proving that $$\dfrac{1}{F_{2n}+1}<\sum_{k=n}^{+\infty}{\dfrac{1}{F_{2k+1}}}<\dfrac{1}{F_{2n}}.$$
Firstly, we prove $$\dfrac{1}{F_{2k}+1}<\dfrac{1}{F_{2k+1}}+\dfrac{1}{F_{2k+2}+1}.$$
\begin{align*} \dfrac{1}{F_{2k}+1}-\dfrac{1}{F_{2k+1}}-\dfrac{1}{F_{2k+2}+1} &=\dfrac{F_{2k+1}^{2}-\left( F_{2k}+1 \right) \left( F_{2k+2}+1 \right)}{\left( F_{2k}+1 \right) F_{2k+1}\left( F_{2k+2}+1 \right)}\\ &=\dfrac{-F_{2k}-F_{2k+2}}{\left( F_{2k}+1 \right) F_{2k+1}\left( F_{2k+2}+1 \right)}\\&<0. \end{align*}
Secondly, we proof $$\dfrac{1}{F_{2k}}>\dfrac{1}{F_{2k+1}}+\dfrac{1}{F_{2k+2}}.$$
\begin{align*} \dfrac{1}{F_{2k}}-\dfrac{1}{F_{2k+1}}-\dfrac{1}{F_{2k+2}} &=\dfrac{F_{2k+1}F_{2k+1}-F_{2k}F_{2k+2}}{F_{2k}F_{2k+1}F_{2k+2}}\\ &=\dfrac{1}{F_{2k}F_{2k+1}F_{2k+2}}\\&>0. \end{align*}
$(1)$: $$\dfrac{1}{F_{2n}}>\dfrac{1}{F_{2n+1}}+\dfrac{1}{F_{2n+2}}>\dfrac{1}{F_{2n+1}}+\dfrac{1}{F_{2n+2}}+\dfrac{1}{F_{2n+4}}>\cdots>\sum_{k=n}^{+\infty}{\dfrac{1}{F_{2k+1}}}.$$
$(2):$ $$\dfrac{1}{F_{2n}+1}<\dfrac{1}{F_{2n+1}}+\dfrac{1}{F_{2n+2}+1}<\dfrac{1}{F_{2n+1}}+\dfrac{1}{F_{2n+3}}+\dfrac{1}{F_{2n+5}+1}<\cdots<\sum_{k=n}^{+\infty}{\dfrac{1}{F_{2k+1}}}.$$
So I can get $$\lfloor \left( \sum_{k=n}^{+\infty}{\dfrac{1}{F_{2k+1}}} \right) ^{-1} \rfloor =F_{2n}.$$
Do I try it right? Any help and references are greatly appreciated.
Thanks!