Egyptian fraction for $\varphi- {F(2n+2) \over F(2n+1)}$ The sum of the reciprocals of the ${2^n}$th Fibonacci numbers is known to be $\dfrac{3-\sqrt{5}}{2}$.
https://math.stackexchange.com/a/746678/134791
This may be written as the following closed form for an Egyptian fraction.
$$\varphi=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}$$
where $\varphi$ is the golden ratio
$$\varphi = \frac{1+\sqrt{5}}{2}$$
and $F(n)$ are the Fibonacci numbers as described by 
$$F(n)=F(n-1)+F(n-2)$$
with $F(0)=0, F(1)=1$.
The result generalizes to other samplings.

$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$

https://math.stackexchange.com/a/2307929/134791

Is there a similar formula for $\varphi- \dfrac{F(2n+2)
}{F(2n+1)}$?

Related questions
Numbers $p-\sqrt{q}$ having regular egyptian fraction expansions?
Egyptian fraction series for $\frac{99}{70}-\sqrt{2}$

Let $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2} = -\alpha^{-1}$. 
In terms of $\alpha, \beta$, we have the Binet's formula:
$$F(n) = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$
For any odd integer $m$, this leads to
$$\varphi - \frac{F(m+1)}{F(m)} = \alpha - \frac{\alpha^{m+1} - \beta^{m+1}}{\alpha^m - \beta^m}
= \frac{\beta^{m-1} + \beta^{m+1}}{\alpha^m - \beta^m}
= \frac{(\beta-\alpha)\beta^m}{\alpha^m - \beta^m}
= \frac{\alpha-\beta}{\alpha^{2m} + 1}
$$
Notice $\displaystyle\;\frac{1}{\alpha^{2m}+1}\;$ can be rewritten as
$$\begin{align}
 & \left(\frac{1}{\alpha^{2m}+1} + \frac{1}{\alpha^{4m}-1}\right)
 -\left(\frac{1}{\alpha^{4m}-1} - \frac{1}{\alpha^{8m}-1}\right)
 -\left(\frac{1}{\alpha^{8m}-1} - \frac{1}{\alpha^{16m}-1}\right)
 -\cdots\\
= & \frac{\alpha^{2m}}{\alpha^{4m}-1} 
- \frac{\alpha^{4m}}{\alpha^{8m}-1}
- \frac{\alpha^{8m}}{\alpha^{16m}-1}
- \cdots\\
= & \frac{1}{\alpha^{2m} - \beta^{2m}}
- \frac{1}{\alpha^{4m} - \beta^{4m}}
- \frac{1}{\alpha^{8m} - \beta^{8m}}
- \cdots
\end{align}
$$
Combine what is already known for even $m$, we obtain following formula for general $m$.
$$\varphi - \frac{F(m+1)}{F(m)} = 
\begin{cases}
\displaystyle\;\frac{1}{F(2m)} - \sum_{k=2}^\infty \frac{1}{F(2^km)}, & m \equiv 1 \pmod 2\\
\displaystyle\;-\sum_{k=1}^\infty \frac{1}{F(2^km)}, & m \equiv 0 \pmod 2
\end{cases}
$$
One may wonder what happens if we replace Fibonacci numbers by Lucas numbers.
Using a similar approach, one can show that
$$\varphi - \frac{L(m+1)}{L(m)} = 
\begin{cases}
\displaystyle\;-\sum_{k=1}^\infty \frac{1}{F(2^km)},& m \equiv 1 \pmod 2\\
\displaystyle\;\frac{1}{F(2m)} - \sum_{k=2}^\infty \frac{1}{F(2^km)}, & m \equiv 0\pmod 2
\end{cases}
$$
