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This question already has an answer here:

Evaluate $$\sum_{k=0}^{\infty} \frac{1}{F_{2^n}} \;.$$

I'm thinking of using a relation from a term to another.

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marked as duplicate by Martin Sleziak, Workaholic, Watson, Ethan Bolker, Davide Giraudo Nov 6 '16 at 19:56

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    $\begingroup$ It's the Millin series mathworld.wolfram.com/MillinSeries.html. Here you can find a proof proofwiki.org/wiki/Millin_Series. $\endgroup$ – Marco Cantarini Mar 30 '15 at 5:29
  • $\begingroup$ Relevant (only gets relevant halfway down though). Specifically, the one by amwmath, with the time stamp "Posted on Sunday, 04 May, 2014 - 05:33 am:". $\endgroup$ – Akiva Weinberger Mar 31 '15 at 1:07
  • $\begingroup$ And the very next comment there is relevant, too: "All right, now that we have both proved it, in our own separate ways, let's look online and find eleven more." (By the way, a few comments up, Superbus finds a formula for the partial sum.) $\endgroup$ – Akiva Weinberger Mar 31 '15 at 1:10
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This series is known as the Millin series. One can show that, $$\sum_{n=0}^\infty\frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}$$ by first proving, using induction, that: $$\sum_{n=0}^\ell \frac{1}{F_{2^n}}=3-\frac{F_{2^\ell-1}}{F_{2^\ell}},$$ and then taking the limit as $\ell\to\infty$, knowing that $\lim\limits_{\ell\to\infty}\dfrac{F_{\ell+1}}{F_\ell}=\frac{1+\sqrt{5}}{2}$, $$\sum_{n=0}^\infty\frac{1}{F_{2^n}}=\lim_{\ell\to\infty}\left(3-\frac{F_{2^\ell-1}}{F_{2^\ell}}\right)=3-\dfrac{1}{(1+\sqrt{5})/2}=\dfrac{7-\sqrt{5}}{2}.$$

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