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Thread: another fibonacci infinite sum

  1. #1
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    another fibonacci infinite sum

    Soroban, here's another fun fibonacci sum for ya'.

    Prove that:

    $\displaystyle \frac{1}{F_{1}}+\frac{1}{F_{2}}+\frac{1}{F_{4}}+\f rac{1}{F_{8}}+\frac{1}{F_{16}}+............... = 4-{\tau}$

    In other words, prove: $\displaystyle \sum_{n=0}^{\infty}\frac{1}{F_{2^{n}}}=4-{\tau}$

    As you probably know, $\displaystyle \tau$ is the golden ratio.

    golden ratio = $\displaystyle \frac{1+\sqrt{5}}{2}$
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  2. #2
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    Rather anyone cares or not, but here is how I tackled this.

    I started with:

    $\displaystyle \frac{F_{2^{n}+1}+2F_{2^{n}-2}}{F_{2^{n}}}=\frac{F_{2^{n}+1}}{F_{2^{n}}}+\frac {2F_{2^{n}-2}}{F_{2^{n}}}$

    Now, $\displaystyle \lim_{n\to\infty}\frac{F_{2^{n}+1}}{F_{2^{n}}}={\t au}$

    So it's left to show that $\displaystyle \lim_{n\to\infty}\frac{2F_{2^{n}-2}}{F^{2^{n}}}=3-\sqrt{5}$.

    Because $\displaystyle {\tau}+3-\sqrt{5}=4-{\tau}$

    Based on the observation that $\displaystyle \frac{{\tau}^{n}-{\mu}^{n}}{{\tau}-{\mu}}$ = the Fibonacci sequence, and $\displaystyle {\mu}=1-{\tau}$

    Let $\displaystyle k=2^{n}$, then $\displaystyle \frac{2F_{k-2}}{F_{k}}$

    $\displaystyle F_{k}=\frac{{\tau}^{k}-{\mu}^{k}}{{\tau}-{\mu}}$

    and $\displaystyle F_{k-2}=\frac{{\tau}^{k-2}-{\mu}^{k-2}}{{\tau}-{\mu}}$

    Then we get: $\displaystyle \frac{2({\tau}^{k-2}-{\mu}^{k-2})}{{\tau}^{k}-{\mu}^{k}}$

    But $\displaystyle \lim_{k\to\infty}{\mu}^{k}=0$

    So, we are left with $\displaystyle \frac{2{\tau}^{k-2}}{{\tau}^{k}}=\frac{2}{{\tau}^{2}}=3-\sqrt{5}$

    And we have $\displaystyle \sum_{n=1}^{\infty}\frac{1}{F_{2^{n}}}=4-{\tau}$
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