<s>[INST] 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 $\varphi- \dfrac{F(2n+2)
}{F(2n+1)}$?

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 [/INST] From a different expansion for the expression given by @achillehui $\frac{\alpha-\beta}{\alpha^{2m}+1}$, as a geometric series, particular cases are
$$\varphi = 1+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{2k}$$
$$\varphi = \frac{3}{2}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{3·2k}$$
$$\varphi = \frac{8}{5}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{5·2k}$$
$$\varphi = \frac{21}{13}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{7·2k}$$
so the general form seems to be

$$\varphi = \frac{F(2n+2)}{F(2n+1)} + \sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{(2n+1)2k}$$

This is alternating, an increasing version would be nice.
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