Closed Form for Millin Series/Proof 1

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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

First we will prove that:

$\ds \sum_{r \mathop = 0}^n \frac 1 {F_{2^r} } = 3 - \frac {F_{2^n - 1} } {F_{2^n} }$

for $n \ge 1$.


We see that:

$\dfrac 1 {F_1} + \dfrac 1 {F_2} = 2 = 3 - \dfrac {F_1} {F_2}$

so the proposition holds for $n = 1$.

Suppose the proposition is true for $n = k$.

Then:

\(\ds \sum_{r \mathop = 0}^{k + 1} \frac 1 {F_{2^r} }\) \(=\) \(\ds 3 - \frac {F_{2^k - 1} } {F_{2^k} } + \frac 1 {F_{2^{k + 1} } }\)
\(\ds \) \(=\) \(\ds 3 - \frac {F_{2^k - 1} F_{2^{k + 1} } - F_{2^k} } {F_{2^k} F_{2^{k + 1} } }\)
\(\ds \) \(=\) \(\ds 3 - \frac {F_{2^k - 1} F_{2^k} \paren {F_{2^k + 1} + F_{2^k - 1} } - F_{2^k} } {F_{2^k} F_{2^{k + 1} } }\) repeated Fibonacci recurrence formula on $F_{2^{k + 1} }$
\(\ds \) \(=\) \(\ds 3 - \frac {F_{2^k - 1} F_{2^k + 1} + F_{2^k - 1}^2 - 1} {F_{2^{k + 1} } }\)
\(\ds \) \(=\) \(\ds 3 - \frac {F_{2^k}^2 + F_{2^k - 1}^2} {F_{2^{k + 1} } }\) Cassini's Identity
\(\ds \) \(=\) \(\ds 3 - \frac {F_{2^{k + 1} - 1} } {F_{2^{k + 1} } }\) repeated Fibonacci recurrence formula on $F_{2^{k + 1} - 1}$


Thus by Principle of Mathematical Induction, the proof is complete.

$\Box$


Now taking the limit, we have:

\(\ds \sum_{r \mathop = 0}^\infty \frac 1 {F_{2^r} }\) \(=\) \(\ds \lim_{n \mathop \to \infty} \sum_{r \mathop = 0}^n \frac 1 {F_{2^r} }\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {3 - \frac {F_{2^n - 1} } {F_{2^n } } }\)
\(\ds \) \(=\) \(\ds 3 - \frac 2 {\sqrt 5 + 1 }\) Ratio of Consecutive Fibonacci Numbers
\(\ds \) \(=\) \(\ds 3 - \frac 2 {\sqrt 5 + 1 } \paren {\frac {\sqrt 5 - 1 } {\sqrt 5 - 1 } }\) multiplying by $1$
\(\ds \) \(=\) \(\ds 3 - \frac {2 \paren {\sqrt 5 - 1 } } 4\)
\(\ds \) \(=\) \(\ds 3 - \frac {\sqrt 5 - 1} 2\)
\(\ds \) \(=\) \(\ds \frac {7 - \sqrt 5} 2\)

as required.

$\blacksquare$


Sources

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