Closed Form for Millin Series/Proof 2

Proof
Let:
 * $\ds \map F x := \sum_{k \mathop = 0}^\infty \frac {x^{2^{k - 1} } } {F_{2^k} }$

Then:


 * $\ds \map F {\alpha x} := \sum_{k \mathop = 0}^\infty \frac {\alpha^{2^{k - 1} } x^{2^{k - 1} } } {F_{2^k} }$

Hence:

Putting $x = -\beta$: