Related to this question Find a solution for $f\left(\frac{1}{x}\right)+f(x+1)=x$, what is this sum: $$\sum_{n=1}^{\infty}(-1)^n\left(\frac{F_n}{F_{n+1}}-\frac1{\phi}\right)$$ where $F_n$ is the $n$th Fibonacci number and $\phi=\frac{1+\sqrt{5}}2$ is the golden ratio.
It is the (negative of the) sum of all the gaps, because the ratio alternates above and below the golden ratio.

  • 7
    $\begingroup$ Due to its alternating nature, the term $-\dfrac1\phi$ can be omitted. Also, $$S=-0.5779217972676187477988879656556639\ldots$$ $\endgroup$
    – Lucian
    Jan 9, 2015 at 14:11
  • 6
    $\begingroup$ With a bit of work one can show that the sum can be written $$-\sum_{n\ge 2}\frac1{F_n\varphi^n}\;.$$ $\endgroup$ Jan 10, 2015 at 1:58
  • 2
    $\begingroup$ @Lucian: Yeah, so far, I've gotten it to $$-\sum_{k=1}^\infty\frac1{F_{2k}\,F_{2k+1}}=-\sum_{k=1}^\infty\frac5{L_{4n+1}-1}$$ $\endgroup$
    – robjohn
    Jan 10, 2015 at 18:54
  • 2
    $\begingroup$ Alternative form: $$-\sum_{n=2}^{\infty}\dfrac{\sqrt{5}}{\varphi^{2n}+(-1)^{n-1}}.$$ $\endgroup$
    – Oleg567
    Jan 14, 2015 at 6:11
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    $\begingroup$ Another alternate form $$\frac{1}{\phi} - \sqrt{5}\left(L(\phi^{-2}) - 2( L(\phi^{-4}) - L(\phi^{-8}))\right)$$ where $L(\beta)$ is a Lambert series which can be computable using the q-polygamma function $\psi_\beta$ (QPolyGamma[] on WA): $$L(\beta) = \sum_{n=1}^\infty \frac{\beta^n}{1-\beta^n} = \frac{\psi_\beta(1) + \log(1-\beta)}{\log \beta}$$ $\endgroup$ Jan 15, 2015 at 9:54

1 Answer 1


Note: A lot of nice information is already given in the comment section to OPs question. Some aspects of my answer are stated there, so that this answer can be seen as supplement to the comment section.

First of all: I went through some papers regarding reciprocal series of Fibonacci sums and it seems there is no closed formula for OPs series. Arguments for this statement are given below.

We can write OPs series as:

\begin{align*} \sum_{n=1}^{\infty}(-1)^n\left(\frac{F_n}{F_{n+1}}-\frac{1}{\phi}\right) = -\sum_{n=1}^{\infty}\frac{1}{F_{2n}F_{2n+1}}\tag{1} \end{align*}


\begin{align*} \sum_{n=1}^{\infty}&(-1)^n\left(\frac{F_n}{F_{n+1}}-\frac{1}{\phi}\right)\\ &=\lim_{N\rightarrow\infty}\sum_{n=1}^{2N}(-1)^n\left(\frac{F_n}{F_{n+1}}-\frac{1}{\phi}\right)\\ &=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\left(\frac{F_{2n}}{F_{2n+1}}-\frac{F_{2n-1}}{F_{2n}}\right)\\ &=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{F_{2n}^2-F_{2n-1}F_{2n+1}}{F_{2n}F_{2n+1}}\\ &=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{-1}{F_{2n}F_{2n+1}}\tag{2}\\ &= -\sum_{n=1}^{\infty}\frac{1}{F_{2n}F_{2n+1}} \end{align*}

In (2) we use Cassini's identity $F_{n}^2-F_{n-1}F_{n+1}=(-1)^{n-1}$.

It's easy to see that the series

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_n} \end{align*} is absolutely convergent (see e.g. A.F. Horadam (1986)). So, convergence of OPs series follows due to $\sum_{n=1}^{\infty}\frac{1}{F_{2n}F_{2n+1}}<\sum_{n=1}^{\infty}\frac{1}{F_n}$.

Now we put the focus at the series:

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_{2n}F_{2n+1}}\tag{3} \end{align*}

and observe that according to S. Rabinowitz (1999) there is no known simple solution for the following series

\begin{align*} \mathbb{F}_N=\sum_{n=1}^{\infty}\frac{1}{F_n},\qquad \mathbb{G}_N=\sum_{n=1}^{\infty}\frac{(-1)^n}{F_n},\qquad \text{and} \qquad \mathbb{K}_N=\sum_{n=1}^{\infty}\frac{1}{F_nF_{n+1}} \end{align*}

The series (3) is non-alternating and structurally similar to $\mathbb{K}_N$. So, if there are no new insights since this paper was written, it's seems plausible that there is also no closed formula for (3) available.

What can we do? One way is to try to express other more complex reciprocal series of Fibonacci sums as expressions based upon the building blocks $\mathbb{F}_N,\mathbb{G}_N$ and $\mathbb{K}_N$. Another possibility is to express them via other standard functions:

The investigation of $\mathbb{F}_N$ by Catalan 1883 and earlier by Lucas 1878 was done by dividing $\mathbb{F}_n=\sum_{n=1}^{\infty}\frac{1}{F_n}$

  • into $\sum_{n=1}^{\infty}\frac{1}{F_{2n-1}}$ expressible in terms of Jacobian Elliptic Functions and

  • into$\sum_{n=1}^{\infty}\frac{1}{F_{2n}}$ expressible in terms of Lambert series

Landau elaborated based on Catalan's result presentation in terms of Theta Functions (see A.F. Horadam).

These are the typically used building blocks to express the reciprocal series of Fibonacci sums. Focussing on series of order two we find e.g. in A solution to a tantalizing problem by G. Almkvist (1984) \begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_{2n}^2}= \frac{5}{24}\left(1+\frac{1}{\pi^2}\frac{\vartheta_{1}^{\prime\prime}}{\vartheta_{1}^{\prime}}\right)\qquad\qquad \sum_{n=1}^{\infty}\frac{1}{F_{2n-1}^2}= -\frac{5}{8\pi^2}\frac{\vartheta_{3}^{\prime\prime}}{\vartheta_{3}} \end{align*}

a representations in terms of Theta functions or in R.S. Melham (1998) we find

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_n^2},\qquad\sum_{n=1}^{\infty}\frac{1}{F_{2n}^2},\qquad\text{and}\qquad\sum_{n=1}^{\infty}\frac{1}{F_{2n-1}^2} \end{align*} expressed in terms of Lambert series

\begin{align*} L(x)=\sum_{n=1}^{\infty}\frac{x^n}{1-x^n}\qquad |x|< 1 \end{align*}

and also in R.Andre-Jeannin (1988) we find

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_nF_{n+1}} =2\sqrt{5}\left[L\left(\frac{3-\sqrt{5}}{2}\right)-2L\left(\frac{7-3\sqrt{5}}{2}\right)+2L\left(\frac{47-21\sqrt{5}}{2}\right)\right]+\frac{1-\sqrt{5}}{2} \end{align*}

in terms of Lambert series. We follow this Ansatz imitate his proof of Lemma 2 and express the series (3) also in terms of Lambert series.

The following is valid

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_{2n}F_{2n+1}}&=\sqrt{5}\left[L(\psi^2)-2L(\psi^4)+2L(\psi^8)\right]+\psi\\ &=\sqrt{5}\left[L\left(\frac{3-\sqrt{5}}{2}\right)-2L\left(\frac{7-3\sqrt{5}}{2}\right) +2L\left(\frac{47-21\sqrt{5}}{2}\right)\right]+\frac{1-\sqrt{5}}{2}\\ &=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{F_nF_{n+1}}+\frac{1-\sqrt{5}}{4} \end{align*}

We use $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}=-\frac{1}{\phi}$ and observe

\begin{align*} \phi F_{2n+1}+F_{2n}&=\phi \frac{\phi^{2n+1}-\psi^{2n+1}}{\phi-\psi}+\frac{\phi^{2n}-\psi^{2n}}{\phi-\psi}\\ &=\frac{1}{\phi-\psi}\left[\phi\left(\phi^{2n+1}+\frac{1}{\phi^{2n+1}}\right)+\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)\right]\\ &=\frac{1}{\phi-\psi}\left(\phi^{2n+2}+\phi^{2n}\right)\\ &=\frac{\phi^{2n+1}}{\phi-\psi}\left(\phi+\frac{1}{\phi}\right)\\ &=\phi^{2n+1} \end{align*}

This implies

\begin{align*} \frac{1}{\phi^{2n}F_{2n}}+\frac{1}{\phi^{2n+1}F_{2n+1}}&=\frac{\phi F_{2n+1}+F_{2n}}{\phi F_{2n}F_{2n+1}} =\frac{1}{F_{2n}F_{2n+1}} \end{align*}

And we get

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{\phi^n F_n} &=\frac{1}{\phi}+\sum_{n=1}^{\infty}\left(\frac{1}{\phi^{2n}F_{2n}}+\frac{1}{\phi^{2n+1}F_{2n+1}}\right)\\ &=\frac{1}{\phi}+\sum_{n=1}^{\infty}\frac{1}{F_{2n}F_{2n+1}}\tag{4} \end{align*}

According to Lemma 3 in R.Andre-Jeannin (1988) the following identity is valid:

\begin{align*} \sum_{n=1}^{\infty}\frac{1}{\phi^n F_{n}}=\sqrt{5}\left[L(\psi^2)-2L(\psi^4)+2L(\psi^8)\right]\tag{5} \end{align*}

Comparing the RHS of (4) and (5) and observing that $\frac{1}{\phi}=-\psi$ the claim follows.

Note: Some additional information regarding closed formulae. In Summation of infinite Fibonacci Series by brother A. Brousseau (1969) we find some identities of reciprocal sums of order two. One of them is the alternating sum

\begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{F_nF_{n+k}}=\frac{1}{F_k}\left[k\phi^{-1}-\sum_{j=1}^k\frac{F_{j-1}}{F_j}\right]\qquad k\geq 1 \end{align*}

We further see that the non-alternating series \begin{align*} \sum_{n=1}^{\infty}\frac{1}{F_nF_{n+k}}\qquad k\geq 1 \end{align*} has a closed formula when $k$ is even and when $k$ is odd it is expressible as $a+b\sum_{n=1}^{\infty}\frac{1}{F_nF_{n+1}}$ with $a,b$ rational numbers, so $\sum_{n=1}^{\infty}\frac{1}{F_nF_{n+1}}$ is used as building block.

Some examples:

\begin{align*} &\sum_{n=1}^{\infty}\frac{1}{F_nF_{n+2}}=1\qquad&\qquad &\sum_{n=1}^{\infty}\frac{1}{F_{n}F_{n+3}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{F_{n}F_{n+1}}-\frac{1}{4}\\ \\ &\sum_{n=1}^{\infty}\frac{1}{F_nF_{n+4}}= \frac{7}{18}\qquad&\qquad &\sum_{n=1}^{\infty}\frac{1}{F_{n}F_{n+3}}=\frac{1}{5}\sum_{n=1}^{\infty}\frac{1}{F_{n}F_{n+1}}-\frac{17}{150} \end{align*}

  • $\begingroup$ @Michael: Thanks a lot for accepting my answer and granting the bounty. Best regards, $\endgroup$ Jan 19, 2015 at 17:48

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