What is a better comparision series for $\sum \frac {1}{F_n}$ than $\sum_0^\infty2^{-k}$?

By better comparison series than $\sum_0^\infty2^{-k}$ we mean a series $\sum c_k$ s.t. $\frac{1}{F_k}<c_k<\frac{1}{2^k}$

If possible $\sum c_k$ is a geometric series, or some other well known type of series e.g. $\sum \frac{1}{\alpha2^k}$ for some $\alpha$

By ratio test the sum over Fibonacci numbers converges. Is there a well known series that can be used to bound the value of the sum from above? and/or tighten the upper bound? I am not after closed form for the sum over Fibonacci numbers Just looking for a well known series that can be used for a relatively good comparison (I.e. the series is not way too large). I am trying to prove $\sum_0^\infty2^{-k}$ can be used for comparison, and then find a better series for comparison.

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    $\begingroup$ See OEIS A$079586$ $\endgroup$ – Lucian Jan 20 '15 at 5:39
  • $\begingroup$ Thank you @Lucian, how did you find this? $\endgroup$ – Arjang Jan 20 '15 at 5:41
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    $\begingroup$ Google leads also to en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant, mathworld.wolfram.com/ReciprocalFibonacciConstant.html, mathoverflow.net/questions/51426/…. From these you'll find the OEIS reference. It was found by Googling sum of the reciprocals of the fibonacci numbers. $\endgroup$ – Jonas Meyer Jan 20 '15 at 5:43
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    $\begingroup$ Note that $F_{2n}<τ^{2n}/\sqrt5$ and $F_{2n+1}>τ^{2n+1}/\sqrt5$ for every $n$ hence, to compare the sums of the series $∑F_n^{-1}$ and $\sqrt{5}∑τ^{-n}$ requires other arguments than simply the equivalent $F_n\sim\tau^n/\sqrt5$ that Binet provides... $\endgroup$ – Did Jan 20 '15 at 11:35
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    $\begingroup$ Starting from $\sqrt5F_n=\tau^n-(-1)^n\tau^{-n}$ with $\tau=\frac12(\sqrt5+1)$, one gets $\sum1/F_n=\sqrt5\sum1/\tau^n+\sqrt5R=2+\tau+\sqrt5R$ where $R=\sum(-1)^n/(\tau^{2n}-(-1)^n)$. The series $R$ is alternated hence one gets explicit lower bounds and/or upper bounds of $\sum1/F_n$, keeping any odd and/or even number of terms in $R$. For example, one and two terms yield $3\lt\sum1/F_n\lt3+\sqrt5/\tau^2$ and $N$ terms yield an approximation of order $1/\tau^{2N}$. $\endgroup$ – Did Jan 20 '15 at 14:43

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