Is the sum of the reciprocals of Fibonacci numbers a transcendental?


3 Answers 3


The sum is well-known (e.g., its decimal expansion is OEIS A153386), but it seems(?) that although it is known to be irrational, and several special sums of Fibonacci numbers are known to be transcendental, the Reciprocal Fibonacci constant itself (as it is known) remains unsettled.

The paper "Irrationality results for reciprocal sums of certain Lucas numbers" by Paul-Georg Becker and Thomas Töpfer (Archiv der Mathematik, Volume 62, Number 4, 300-305, 1994), says that Andre-Jeannin proved (in "A note on the irrationality of certain Lucas infinite series," Fibonacci Quart., 29, 132-136, 1991) that the sum $\sum_{n=0}^\infty 1/R_n$, where $R_{n+2}= a R_{n+1} + b R_n$, is irrational, and that Bundschuh and Väänänen gave "an irrationality measure" for this number.

For special cases, see "Transcendence of reciprocal sums of binary recurrences" by Tomoaki Kanoko, Takeshi Kurosawa and Iekata Shiokawa (Monatshefte für Mathematik, Volume 157, Number 4, 323-334, 2009), and "Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers" by Daniel Duverney, Keiji Nishioka, Kumiko Nishioka, and Iekata Shiokawa (Proc. Japan Acad. Ser. A Math. Sci., Volume 73, Number 7 (1997), 140-142). For example, the latter paper establishes that $$\sum_{n=1}^\infty \frac{1}{F_n^{2s}}$$ is transcendental for any positive integer $s$.


To supplement Joseph's answer, I add my review MR2354148 on [C. Elsner, S. Shimomura and I. Shiokawa, Acta Arith. 130:1 (2007), 37--60].

Let $\lbrace F_n\rbrace _{n\ge0}$ and $\lbrace L_n\rbrace _{n\ge0}$ be Fibonacci and Lucas numbers, respectively, $F_0=0$, $F_1=1$, $F_{n+2}=F_n+F_{n+1}$ for $n\ge0$, and $L_0=2$, $L_1=1$, $L_{n+2}=L_n+L_{n+1}$ for $n\ge0$. Using Nesterenko's theorem [Yu.V. Nesterenko, Sb. Math. 187:9 (1996), 1319--1348. MR1422383] and expressing the series $$ \zeta_F(2s)=\sum_{n=1}^\infty\frac1{F_n^{2s}} \quad\text{and}\quad \zeta_L(2s)=\sum_{n=1}^\infty\frac1{L_n^{2s}}, \qquad s=1,2,\dots, \qquad (1) $$ via the Eisenstein series $$ E_{2s}(q)=1-\frac{4s}{B_{2s}}\sum_{n=1}^\infty\sigma_{2s-1}(n)q^n, \qquad \sigma_k(n)=\sum_{d\mid n}d^k, $$ where $B_{2s}\in\mathbb Q$ are Bernoulli numbers, the authors prove the algebraic independence of the numbers in the collections $\zeta_F(2)$, $\zeta_F(4)$, $\zeta_F(6)$ and $\zeta_L(2)$, $\zeta_L(4)$, $\zeta_L(6)$ as well as express algebraically even "zeta values" $\zeta_F(2s)$ (and $\zeta_L(2s)$) for $s\ge4$ in terms of the three algebraically independent numbers in the corresponding collection. Similar algebraic independence results are shown for the alternating versions of (1). Known irrationality results for $\zeta_F(k)$ and $\zeta_L(k)$ with odd $k$ (when the series have no known relations with the modular world) are indicated. It is worth mentioning that these results go in a natural parallel with the ones for the so-called $q$-zeta values defined in [W. Zudilin, Math. Notes 72:5-6 (2002), 858--862. MR1964151] and [C. Krattenthaler, T. Rivoal, and W. Zudilin, J. Inst. Math. Jussieu 5:1 (2006), 53--79. MR2195945]. In particular, it is natural to expect a "Fibonacci" analogue of Rivoal's theorem [T. Rivoal, C. R. Acad. Sci. Paris Ser. I Math. 331:4 (2000), 267--270. MR1787183] on the infiniteness of irrational numbers in the set $\zeta_F(1),\zeta_F(3),\zeta_F(5),\dots$ (or $\zeta_L(1),\zeta_L(3),\zeta_L(5),\dots$), based on the techniques developed in the paper under review and in the joint paper of Krattenthaler, Rivoal and the reviewer cited above.

Reviewed by Wadim Zudilin

To summarize, the difficulty of proving the transcendence for odd $\zeta_F(s)$ with $s$ odd is similar to the one for odd zeta values. The irrationality of $\zeta_F(1)$ is known but already its non-quadraticity remains an open problem.


Just to make the obvious comments:

Let $q = (1 - \sqrt{5})/2$. Then $F_n=((-q^{-1})^{n} - q^n)/\sqrt{5}$. So $$\sum \frac{1}{F_n} = \sqrt{5} \sum \frac{q^n}{1-(-1)^n q^{2n}}.$$

This looks kind of like the logarithmic derivative of the modular form $\prod (1-q^n)$, but it's not exactly that. In any case, it is unlikely that someone has evaluated this sort-of-a-modular-form at this particular quadratic irrational. The natural thinkg to do is to evaluate modular forms when $\tau$ is a quadratic irrational, where $q=e^{2 \pi i \tau}$. For example, people might know what this sum equals at $q=e^{- \pi \sqrt{163}}$.

I'd be pleasantly surprised if there is a known answer. Of course, the safe money is always to bet on "transcendental" when you don't see a reason to expect anything else.

Looks like I was too pessimistic. It is known to be irrational, see https://mathworld.wolfram.com/ReciprocalFibonacciConstant.html (thanks Qiaochu!). They do indeed describe this quantity in terms of theta functions, which are a type of modular forms. To my surprise, they are able to prove things about these modular forms at $q$, although not to answer this question. Anyway, it looks like this reference has as much information as you can hope for.

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    $\begingroup$ It's known to be irrational, but neither Wikipedia nor Mathworld says anything about its transcendentality. $\endgroup$ Jan 7, 2011 at 20:10
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    $\begingroup$ Unless I'm missing something, it looks like the mathworld page only gives a theta function formula for the sum of reciprocals of odd-indexed Fibonacci numbers. $\endgroup$ Jan 7, 2011 at 21:35
  • $\begingroup$ Actually, Richard André-Jeannin proved something somewhat stronger, namely that this constant doesn't belong to $\mathbb{Q}(\sqrt{5})$. I was lucky enough to exchange with him on a French math forum, and I miss his kindness and delightful humor. $\endgroup$ Aug 12, 2022 at 11:41

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