How can I show that $\sum\limits_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}$ is algebraic? Show that $$\sum_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}$$ is algebraic. 
More specifically, solve this and get exact values. 
Then use the result to evaluate $$\sum_{n=0}^\infty \frac{1}{F_{2^n}}$$ where $$F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$$  and $\alpha=\frac{1+\sqrt{5}}{2}$  and $\beta=\frac{1-\sqrt{5}}{2}$.

There is in fact a nice closed form for the partial-sums that you can find just by calculating a bit:
$$ \begin{align*} f(z) &= \frac{z^2}{1-z^4} + \frac{z^4}{1-z^8} + \frac{z^8}{1-z^{16}} + \dots  \\ &= \frac{z^2(1+z^4)+z^4}{1-z^8} + \frac{z^8}{1-z^{16}} +\dots \\ &= \frac{z^2 + z^4 + z^6}{1-z^8} + \frac{z^8}{1-z^{16}} +\dots \\ &= \frac{(z^2 + z^4 + z^6)(1+z^8)+z^8}{1-z^{16}}+ \dots \\ &= \frac{z^2+z^4 + z^6+z^8+z^{10} + z^{12}+z^{14}}{1-z^{16}}+\dots \end{align*} $$
So first few partial sums are given by
$$\begin{align*} f_1(z)&= \frac{z^2}{1-z^4} = \frac{z^2-z^4}{1-z^2}\frac1{1-z^4} \\ f_2(z) &= \frac{z^2+z^4+z^6}{1-z^8} = \frac{z^2-z^8}{1-z^2}\cdot \frac1{1-z^8} \\ f_3(z) &= \dots  = \frac{z^2-z^{16}}{1-z^2}\cdot \frac1{1-z^{16}}\end{align*} $$
It should then be easy to show by induction that
$$ f_n(z) = \frac{z^2}{1-z^2}\cdot \frac{1-z^{2^{n-1}}}{1-z^{2^n}}. $$
The second factor is $1 + O(z^{2^{n-1}})\to  1$, as $n\to\infty$.
thus $$\boxed{f(z) = \displaystyle\lim_{n\to\infty} f_n(z) = \frac{z^2}{1-z^2}}$$
Unfortunately, I can't get WolframAlpha to find possible mistakes, so I hope someone else will.
