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 $F_{2k}$, 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 $n\ge3$).

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 $n\ge3$).

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 $F_{2k}$, 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 sum in the middle in the obvious way

$$\alpha^{-2n+1}-\alpha^{-4n-1}=\sum_{k=n}^{2n} \frac1{\alpha^{2k}} <\sum_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}}\le$$ $$\le \frac1{1-\alpha^{-4n}} \sum_{k=n}^{2n}\frac1{\alpha^{2k}}= \frac{\alpha^{-2n+1}-\alpha^{-4n-1}}{1-\alpha^{-4n}} $$

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 $n\ge3$).

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 $F_{2k}$, 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 $n\ge3$).