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Math Help - 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:

    \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: \sum_{n=0}^{\infty}\frac{1}{F_{2^{n}}}=4-{\tau}

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

    golden ratio = \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:

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

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

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

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

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

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

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

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

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

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

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