TITLE: How can I show that $\sum\limits_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}$ is algebraic?
QUESTION [8 upvotes]: 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 $\beta=\frac{1-\sqrt{5}}{2}$.

REPLY [5 votes]: Robert means this...
$$\begin{align}
\sum_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}} &=
\sum_{n=1}^\infty z^{2^n}\sum_{k=0}^\infty z^{k2^{n+1}} =
\sum_{n=1}^\infty \sum_{k=0}^\infty z^{2^n+k2^{n+1}} \\&=
\sum_{n=1}^\infty \sum_{k=0}^\infty z^{2^n(1+2k)} =
\sum_{m=1}^\infty z^{2m} =
\frac{z^2}{1-z^2}
\end{align}$$
Can you figure out the reason to go from the "double sum" in $n,k$ to the "single sum" in $m$?