Closed Form for Millin Series/Proof 2

Proof
Let:
 * $\ds \map F x := \sum_{k \mathop = 1}^\infty \frac {x^{2^{k - 1} } } {F_{2^k} }$
 * $\alpha := \dfrac {1 + \sqrt 5} 2$
 * $\beta := \dfrac {1 - \sqrt 5} 2$

Note that for $\size x \le 1$ we have:
 * $\size {\dfrac {x^{2^{k - 1} } } {F_{2^k} } } \le \dfrac 1 {F_{2^k} }$

Given that the Millin Series converges, by Comparison Test:
 * $\ds \map F x = \sum_{k \mathop = 1}^\infty \frac {x^{2^{k - 1} } } {F_{2^k} }$ converges for $\size x \le 1$

We have:


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

Hence:

Since $\beta^2 = \dfrac {3 - \sqrt 5} 2 \le 1$, we can substitute $x = -\beta$: