How find this Fibonacci sequence sum $\sum_{k=0}^{\infty}\frac{1}{F_{2^k}}$ let sequence $\{F_{n}\}$ such $$F_{1}=1,F_{2}=1,F_{m+1}=F_{m}+F_{m-1},m\ge 2$$
Find this value 

$$I=\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}}$$

My try: I know this 

$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$$

so

$$\dfrac{1}{F_{2^k}}=\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$
  so
  we only find this sum
  $$I=\sum_{k=0}^{\infty}\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$

But I can't.Thank you 
 A: In fact, this is summable. This surprises me, because of results like those in this question at MO, which in short discusses how we don't know if the sum of the reciprocals of all the Fibonacci numbers is transcendental, and other related unknown results.
This question boils down to the following:

Claim: $\displaystyle \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_4} + \ldots + \frac{1}{F_{2^n}} = 3 - \frac{F_{2^n-1}}{F_{2^n}}$ for $n \geq 2$.

Proof: Induct, and use $F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$ as you mentioned. $\diamondsuit$
Then take the limit as $n \to \infty$ to get that 
$$\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}} = \frac{7-\sqrt 5}{2}.$$
A: let
$$a=\dfrac{1+\sqrt{5}}{2}\Longrightarrow \dfrac{\sqrt{5}-1}{2}=a^{-1}$$
so
$$I=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}$$
so let $$a^{2^n}=x$$
then
$$\dfrac{1}{a^{2^n}-a^{-2^n}}=\dfrac{x}{x^2-1}=\dfrac{1}{x-1}-\dfrac{1}{x^2-1}=\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}$$
so
\begin{align*}I&=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}=\sqrt{5}\sum_{n=0}^{\infty}\left(\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}\right)\\
&=\lim_{n\to\infty}\left(\dfrac{\sqrt{5}}{a-1}-\dfrac{1}{a^{2^{n+1}}-1}\right)\\
&=\dfrac{\sqrt{5}}{a-1}\\
&=\dfrac{7-\sqrt{5}}{2}
\end{align*}
