Closed Form for Millin Series/Proof 2

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Theorem

The Millin series has the closed form expression:

$\ds \sum_{n \mathop = 0}^\infty \frac 1 {F_{2^n} } = \frac {7 - \sqrt 5} 2$


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:

\(\ds \map F {\alpha x} + \map F {\beta x}\) \(=\) \(\ds \sum_{k \mathop = 1}^\infty \frac {\alpha^{2^{k - 1} } + \beta^{2^{k - 1} } } {F_{2^k} } x^{2^{k - 1} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^\infty \frac {L_{2^{k - 1} } } {F_{2^k} } x^{2^{k - 1} }\) Closed Form for Lucas Numbers
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^\infty \frac {x^{2^{k - 1} } } {F_{2^{k - 1} } }\) Fibonacci Number 2n equals Fibonacci Number n by Lucas Number n
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {x^{2^k} } {F_{2^k} }\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \frac {x^{2^0} } {F_{2^0} } + \sum_{k \mathop = 1}^\infty \frac {\paren {x^2}^{2^{k - 1} } } {F_{2^k} }\)
\(\ds \) \(=\) \(\ds x + \map F {x^2}\)

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

\(\ds -\beta + \map F {\beta^2}\) \(=\) \(\ds \map F {-\beta^2} + \map F {-\alpha \beta}\)
\(\ds \leadsto \ \ \) \(\ds \map F 1\) \(=\) \(\ds -\beta + 2 \beta^2\) as $\alpha \beta = -1$ and $\map F {\beta^2} = \map F {-\beta^2} + 2 \beta^2$
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^\infty \frac 1 {F_{2^k} }\) \(=\) \(\ds 1 + \map F 1\)
\(\ds \) \(=\) \(\ds 1 - \beta + 2 \beta^2\)
\(\ds \) \(=\) \(\ds 2 - \paren {1 + \beta - \beta^2} + \beta^2\)
\(\ds \) \(=\) \(\ds 2 + \dfrac {3 - \sqrt 5} 2\)
\(\ds \) \(=\) \(\ds \dfrac {7 - \sqrt 5} 2\)

$\blacksquare$


Sources

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