A Fibonacci sum

Let $F_n$ denote the $n$-th Fibonacci sum. Evaluate (in a closed form) the sum:

$$\sum_{n=0}^{N} \frac{1}{F_{2^n}}$$

Solution

First of all we will evaluate the following sum:$\displaystyle  \sum_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}$ . Indeed , we have that:

\begin{align*}
 \sum_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} &= \sum_{n=1}^\infty z^{2^n}\sum_{k=0}^\infty z^{k2^{n+1}} \\
&= \sum_{n=1}^\infty \sum_{k=0}^\infty z^{2^n+k2^{n+1}} \\
&= \sum_{n=1}^\infty \sum_{k=0}^\infty z^{2^n(1+2k)} \\
&= \sum_{m=1}^\infty z^{2m} = \frac{z^2}{1-z^2}
\end{align*}

This means that this number is actually algebraic. Hence for the initial sum , one can easily establish that:

$$\sum\limits_{n=0}^{N}\frac{1}{F_{2^n}}=3-\frac{F_{2^N-1}}{F_{2^N}}$$