TITLE: How find this Fibonacci sequence sum $\sum_{k=0}^{\infty}\frac{1}{F_{2^k}}$
QUESTION [12 upvotes]: 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

REPLY [12 votes]: 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*}