Reciprocals of Fibonacci numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:59:28Z http://mathoverflow.net/feeds/question/51426 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51426/reciprocals-of-fibonacci-numbers Reciprocals of Fibonacci numbers vamsi krishna 2011-01-07T19:30:49Z 2011-01-09T02:57:26Z <p>Is the sum of the reciprocals of Fibonacci numbers a transcendental?</p> http://mathoverflow.net/questions/51426/reciprocals-of-fibonacci-numbers/51428#51428 Answer by David Speyer for Reciprocals of Fibonacci numbers David Speyer 2011-01-07T19:59:27Z 2011-01-07T20:15:22Z <p>Just to make the obvious comments:</p> <p>Let $q = (1 - \sqrt{5})/2$. Then <code>$F_n=((-q^{-1})^{n} - q^n)/\sqrt{5}$</code>. So $$\sum \frac{1}{F_n} = \sqrt{5} \sum \frac{q^n}{1-(-1)^n q^{2n}}.$$</p> <p>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 $q=e^{- \pi \sqrt{163}}$.</p> <p>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.</p> <hr> <p>Looks like I was too pessimistic. It is known to be irrational, see <a href="http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html" rel="nofollow">http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html</a> (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.</p> http://mathoverflow.net/questions/51426/reciprocals-of-fibonacci-numbers/51433#51433 Answer by Joseph O'Rourke for Reciprocals of Fibonacci numbers Joseph O'Rourke 2011-01-07T20:14:15Z 2011-01-09T00:59:58Z <p><a href="http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html" rel="nofollow">The sum</a> is well-known (e.g., its decimal expansion is <a href="http://oeis.org/A153386" rel="nofollow">OEIS A153386</a>), but it seems(?) that although it is known to be irrational, and several special sums of Fibonacci numbers are known to be transcendental, the <em>Reciprocal Fibonacci constant</em> itself (as it is known) remains unsettled.</p> <p>The paper "Irrationality results for reciprocal sums of certain Lucas numbers" by Paul-Georg Becker and Thomas Töpfer (<a href="http://www.springerlink.com/content/j041113q17261818/" rel="nofollow"><em>Archiv der Mathematik</em>, Volume 62, Number 4, 300-305, 1994</a>), says that Andre-Jeannin proved (in "A note on the irrationality of certain Lucas infinite series," <em>Fibonacci Quart.</em>, 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 <a href="http://archive.numdam.org/article/CM_1994__91_2_175_0.pdf" rel="nofollow">gave</a> "an <a href="http://mathworld.wolfram.com/IrrationalityMeasure.html" rel="nofollow">irrationality measure</a>" for this number.</p> <p>For special cases, see "Transcendence of reciprocal sums of binary recurrences" by Tomoaki Kanoko, Takeshi Kurosawa and Iekata Shiokawa (<a href="http://www.springerlink.com/content/l4601x2004603271/" rel="nofollow"><em>Monatshefte für Mathematik</em>, Volume 157, Number 4, 323-334, 2009</a>), and "Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers" by Daniel Duverney, Keiji Nishioka, Kumiko Nishioka, and Iekata Shiokawa (<a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.pja/1195509914" rel="nofollow"><em>Proc. Japan Acad. Ser. A Math. Sci.</em>, Volume 73, Number 7 (1997), 140-142</a>). For example, the latter paper establishes that $$\sum_{n=1}^\infty \frac{1}{F_n^{2s}}$$ is transcendental for any positive integer $s$.</p> http://mathoverflow.net/questions/51426/reciprocals-of-fibonacci-numbers/51456#51456 Answer by Wadim Zudilin for Reciprocals of Fibonacci numbers Wadim Zudilin 2011-01-08T03:23:24Z 2011-01-09T02:57:26Z <p>To supplement Joseph's answer, I add my review MR2354148 on [C. Elsner, S. Shimomura and I. Shiokawa, <em>Acta Arith.</em> <strong>130</strong>:1 (2007), 37--60].</p> <p>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, <em>Sb. Math.</em> <strong>187</strong>: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 <a href="http://arxiv.org/abs/math/0206179" rel="nofollow">[W. Zudilin, <em>Math. Notes</em> <strong>72</strong>:5-6 (2002), 858--862. MR1964151]</a> and <a href="http://arxiv.org/abs/math/0311033" rel="nofollow">[C. Krattenthaler, T. Rivoal, and W. Zudilin, <em>J. Inst. Math. Jussieu</em> <strong>5</strong>:1 (2006), 53--79. MR2195945]</a>. In particular, it is natural to expect a "Fibonacci" analogue of Rivoal's theorem [T. Rivoal, <em>C. R. Acad. Sci. Paris Ser. I Math.</em> <strong>331</strong>: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.</p> <p>Reviewed by Wadim Zudilin</p> <hr> <p>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.</p>