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Evaluate the sum:

$$\sum_{n=0}^{\infty} \frac{1}{F_{(2^n)}}$$

where $F_{m}$ is the $m$-th term of the Fibonacci sequence. I need some support here. Thanks.

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    possible duplicate of [How can I show that $\sum\limits_{n=1}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}$ is algebraic?](http://math.stackexchange.com/questions/105412/how-can-i-show-that-sum-limits-n-1-infty-fracz2n1-z2n1-is)2012-07-24
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    It is true that an answer to 105412 is an answer to the current question. However, there are answers to this question that would not qualify as answers to the older question, so I don't think this one qualifies as a duplicate.2012-07-25

3 Answers 3

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As wikipedia claims the result follows from the identity $$ \sum\limits_{n=0}^N\frac{1}{F_{2^n}}=3-\frac{F_{2^N-1}}{F_{2^N}} $$ You can try to prove it by induction.

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    nice identity. I wonder what happens if one meets this problem in a competition because there you don't have access at wikipedia. :) Maybe there is a resonable approach for solving it. Thanks for information.2012-06-12
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See http://oeis.org/A079585 and references there.

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    @Chris: The paper by Rabinowitz is available as a .ps file [here](http://www.math-cs.ucmo.edu/~mjms/1998.3/stan.ps).2012-06-12
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    @Brian M. Scott: nice!!! Thanks for the paper. A good piece of information there!2012-06-12
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Typing Sum[1/Fibonacci[2^n],{n,0,infty}] in http://www.wolframalpha.com gives,

$$\sum_{n=0}^\infty \frac{1}{F_{(2^n)}} = \left(\frac{1-\sqrt{5}}{2}\right)^3+\left(\frac{1+\sqrt{5}}{2}\right)^2 = 2.381966\dots$$

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    an interesting result.2012-06-12
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    Your answer simplifies to $\frac{5-\sqrt5}2$.2018-01-22