The sum of the reciprocals of the ${2^n}$th Fibonacci numbers is known to be $\dfrac{3-\sqrt{5}}{2}$. https://math.stackexchange.com/a/746678/134791

This may be written as the following closed form for an Egyptian fraction.

$$\varphi=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}$$

where $\varphi$ is the golden ratio $$\varphi = \frac{1+\sqrt{5}}{2}$$

and $F(n)$ are the Fibonacci numbers as described by $$F(n)=F(n-1)+F(n-2)$$ with $F(0)=0, F(1)=1$.

The result generalizes to other samplings.

$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$


Is there a similar formula for $\varphi- \dfrac{F(2n+2) }{F(2n+1)}$?

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  • $\begingroup$ $$\sum_{i=0}^\infty \frac{1}{F_{k 2^i}}+ \varphi = \frac{1}{F_k} + \frac{1}{F_{2k}}+\frac{F_{2k-1}}{F_{2k}} + 1$$ $\endgroup$ Jun 4, 2017 at 6:29
  • $\begingroup$ @BrevanEllefsen isn't this equivalent to what OP knows??? $\endgroup$ Jun 4, 2017 at 6:39
  • $\begingroup$ @achillehui I believe so. It's something I found on the Internet while searching and I thought it worth posting - hence its status merely as a comment. It's supposedly from a paper on the topic from the 70's, but I can't find the paper right now $\endgroup$ Jun 4, 2017 at 6:41
  • $\begingroup$ It is proved in this post $$ \frac{F_{n+1}}{F_n}-\varphi = \frac{\sqrt{5}(-\varphi)^{-n}}{\varphi^n - (-\varphi)^{-n}}\approx \frac{\sqrt{5}(-1)^n}{\varphi^{2n}}\approx \frac{(-1)^n}{\sqrt{5}F_n^2} $$ So, we can conclude that $$ \varphi -\frac{F_{2n+2}}{F_{2n+1}}\approx \frac{1}{\sqrt{5}F_{2n+1}^2} $$ Maybe helpful. $\endgroup$
    – Amin235
    Jun 4, 2017 at 7:07

2 Answers 2


Let $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2} = -\alpha^{-1}$. In terms of $\alpha, \beta$, we have the Binet's formula:

$$F(n) = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$

For any odd integer $m$, this leads to

$$\varphi - \frac{F(m+1)}{F(m)} = \alpha - \frac{\alpha^{m+1} - \beta^{m+1}}{\alpha^m - \beta^m} = \frac{\beta^{m-1} + \beta^{m+1}}{\alpha^m - \beta^m} = \frac{(\beta-\alpha)\beta^m}{\alpha^m - \beta^m} = \frac{\alpha-\beta}{\alpha^{2m} + 1} $$ Notice $\displaystyle\;\frac{1}{\alpha^{2m}+1}\;$ can be rewritten as $$\begin{align} & \left(\frac{1}{\alpha^{2m}+1} + \frac{1}{\alpha^{4m}-1}\right) -\left(\frac{1}{\alpha^{4m}-1} - \frac{1}{\alpha^{8m}-1}\right) -\left(\frac{1}{\alpha^{8m}-1} - \frac{1}{\alpha^{16m}-1}\right) -\cdots\\ = & \frac{\alpha^{2m}}{\alpha^{4m}-1} - \frac{\alpha^{4m}}{\alpha^{8m}-1} - \frac{\alpha^{8m}}{\alpha^{16m}-1} - \cdots\\ = & \frac{1}{\alpha^{2m} - \beta^{2m}} - \frac{1}{\alpha^{4m} - \beta^{4m}} - \frac{1}{\alpha^{8m} - \beta^{8m}} - \cdots \end{align} $$ Combine what is already known for even $m$, we obtain following formula for general $m$.

$$\varphi - \frac{F(m+1)}{F(m)} = \begin{cases} \displaystyle\;\frac{1}{F(2m)} - \sum_{k=2}^\infty \frac{1}{F(2^km)}, & m \equiv 1 \pmod 2\\ \displaystyle\;-\sum_{k=1}^\infty \frac{1}{F(2^km)}, & m \equiv 0 \pmod 2 \end{cases} $$

One may wonder what happens if we replace Fibonacci numbers by Lucas numbers.
Using a similar approach, one can show that $$\varphi - \frac{L(m+1)}{L(m)} = \begin{cases} \displaystyle\;-\sum_{k=1}^\infty \frac{1}{F(2^km)},& m \equiv 1 \pmod 2\\ \displaystyle\;\frac{1}{F(2m)} - \sum_{k=2}^\infty \frac{1}{F(2^km)}, & m \equiv 0\pmod 2 \end{cases} $$

  • $\begingroup$ I find your answer so inserting. Is your method can be applied to Fibonacci n-Step Number? $\endgroup$
    – Amin235
    Jun 4, 2017 at 7:17
  • $\begingroup$ Hmm, also decreasing... I was expecting an increasing series, maybe with Lucas numbers but couldn't find anything. $\endgroup$ Jun 4, 2017 at 7:23
  • 1
    $\begingroup$ @Amin235 I don't think this approach will work for the $n$-step numbers. This answer relies on the the functional form of $F(n)$ provided by Binet's formula. Since $\beta = -\alpha^{-1}$, we are dealing with rational functions in a single variable. For the other $n$-step numbers, I don't think there are similar relations among the roots which we can use. $\endgroup$ Jun 4, 2017 at 7:26
  • $\begingroup$ @achillehui Thanks for your comment. Your method applied on the case $\frac{F_{n+1}}{F_n}$. Is there a way such that we can find a closed form for the case ${F_{n+1}}^{F_n}$? Excuse me that I ask too question! $\endgroup$
    – Amin235
    Jun 4, 2017 at 7:37
  • 1
    $\begingroup$ @Amin235 I don't have any idea how to compute $F_{n+1}^{F_n}$. I also don't think Binet's formula has any use in computing this term. $\endgroup$ Jun 4, 2017 at 8:03

From a different expansion for the expression given by @achillehui $\frac{\alpha-\beta}{\alpha^{2m}+1}$, as a geometric series, particular cases are

$$\varphi = 1+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{2k}$$

$$\varphi = \frac{3}{2}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{3·2k}$$

$$\varphi = \frac{8}{5}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{5·2k}$$

$$\varphi = \frac{21}{13}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{7·2k}$$

so the general form seems to be

$$\varphi = \frac{F(2n+2)}{F(2n+1)} + \sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{(2n+1)2k}$$

This is alternating, an increasing version would be nice.

  • $\begingroup$ I suggest to add this new form as an edition of your question and not an answer to your question. $\endgroup$
    – Amin235
    Jun 4, 2017 at 8:15
  • $\begingroup$ @Amin235 This has $2k$ instead of $2^k$, maybe it should be deleted directly... $\endgroup$ Jun 4, 2017 at 19:21

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