Automorphic Representations and L-Functions Editors: D. Prasad, C.S. Rajan, A. Sankaranarayanan, J. Sengupta Copyright c 2013 Tata Institute of Fundamental Research Publisher: Hindustan Book Agency, New Delhi, India The Fibonacci Zeta Function M. Ram Murty1 Abstract We consider the lacunary Dirichlet series obtained by taking the re- ciprocals of the s-th powers of the Fibonacci numbers. This series admits an analytic continuation to the entire complex plane. Its spe- cial values at integral arguments are then studied. If the argument is a negative integer, the value is algebraic. If the argument is a positive even integer, the value is transcendental by Nesterenko’s work. This is a result of Duverney, Ke. Nishioka, Ku. Nishioka and Shiokawa. We present a simplified proof of their result. If the argument is 1, the value has been shown to be irrational by Andr´e-Jeannin. We present a slight modification of Duverney’s proof of this fact. At the same time, we highlight the “modular connection” of these questions as well as signal some new results in the theory of special values of q-analogues of classical Dirichlet L-functions. 1 Introduction The sequence of Fibonacci numbers is defined by the recurrence relation fn+1 = fn + fn 1, n 1 − ≥ with initial values f0 = 0 and f1 = 1. The Fibonacci zeta function is the series ∞ s F (s) := fn− . n=1 Since the n-th Fibonacci number has exponential growth, it is easy to see that the series converges for (s) > 0. Its analytic continuation is easily derived using a technique theℜ author and Sinha [12] used to derive the analytic continuation for the Riemann, Hurwitz and multiple Hurwitz zeta functions. Indeed, if α = 1+√5 and β = 1/α, then, 2 − αn βn f = − . n α β − 1Research of the author was partially supported by an NSERC Discovery grant. 1 2 M. Ram Murty Thus, using the binomial theorem, s ∞ s ∞ ns n s (α β)− 1/f = α− (1 (β/α) )− − n − n=1 n=1 ∞ ns ∞ s j nj = α− − ( 1) (β/α) j − n=1 j=0 ∞ s βj /αs+j = − ( 1)j . j − 1 (βj /αs+j ) j=0 − The right hand side is easily seen to converge for all complex values of s. From this derivation, we immediately see that the special values of F (s) are all algebraic (and in fact lying in the quadratic field Q(√5)) when s is either zero or a negative integer, since in these cases the sum is a finite sum. In this paper, we will be interested in the special values F (k) when k is a natural number. What kind of numbers are these? Are they transcen- dental? Are they algebraic? It turns out that much like the case with the Riemann zeta function, the special values F (2k) are all transcendental and the nature of F (2k +1) is still unknown. Analogous questions can be asked for any second order recurrence sequence. The discussion below extends to some of these sequences but not all. There are some subtle issues that enter into the derivations that seem at present to be insurmountable. We discuss these issues in the final section of the paper. For the sake of elegance of exposition, we focus on the Fibonacci sequence first. The problem of evaluating F (1) seems to go back to Laisant in 1899 (see [11]). Later that year, Edmund Landau [11] addressed the question and was unable to give a definite answer. However, he noted that the sum ∞ 1 f n=1 2n+1 can be expressed as special values of classical theta functions, giving us the first hint of a “modular connection.” More precisely, Landau proved that ∞ 1 √5 3 √5 = θ2 − , f 4 2 2 n=1 2n+1 where 2 ∞ (n+1/2) θ2(q)= q . n= −∞ The Fibonacci Zeta Function 3 Beyond this, Landau was unable to say if this number is transcendental but this is now known due to recent advances in transcendental number theory. Our current state of knowledge on this problem is recorded in the fol- lowing two theorems: Theorem 1.1 (Andr´e-Jeannin, 1989) F (1) is irrational. Theorem 1.2 (Duverney, Ke. Nishioka, Ku. Nishioka, Shiokawa, 1998, [6]) F (2k) is transcendental for all k 1. ≥ As will be explained below, the essential ingredient in the proof of The- orem 1.2 is a deep theorem of Nesterenko regarding the transcendence of special values of Eisenstein series, proved in 1996. We will attempt to ex- plain why the case of even exponents can be solved and where the difficulty lies for the odd exponents. The proof of Theorem 1.2 is somewhat “easier” than Theorem 1.1. But both proofs require our entry into the world of q-series. Theorem 1.1 re- quires the q-exponential and q-logarithm functions and the identities seem to suggest a connection to Ramanujan’s “mock-theta” world. For the most part, this paper is a survey of known results. However, the presentation and arrangement of ideas is new. In particular, the emphasis on the use of the q-exponential and q-logarithm in the proof of Theorem 1.1 following [5] is with a view to give conceptual unity to these isolated results. We also introduce q-analogues of Dirichlet L-series L(s,χ) and report on the transcendence of some of their special values. These results will appear in a forthcoming paper [3]. 2 Nesterenko’s Theorem To facilitate the proof of Theorem 1.2, we review Nesterenko’s theorem regarding transcendental values of Eisenstein series. Recall that the Eisen- stein series of weight 2, 4, and 6 for the full modular group are given by ∞ E (q) = 1 24 σ (n)qn, 2 − 1 n=1 ∞ n E4(q) = 1 + 240 σ3(n)q , n=1 ∞ E (q) = 1 504 σ (n)qn, 6 − 5 n=1 j where σj(n)= d n d . We will make fundamental use of: | 4 M. Ram Murty Theorem 2.1 (Nesterenko, 1996) For any q with q < 1, the transcen- dence degree of the field | | Q(q,E2(q),E4(q),E6(q)) is at least 3. Thus, for q algebraic, E2(q),E4(q), and E6(q) are algebraically independent. Let us also recall that the general Eisenstein series 4k ∞ n E2k(q) = 1 σ2k 1(n)q , − B − 2k n=1 where B2k is the 2k-th Bernoulli number, is a polynomial in E4 and E6 (see for example, [15]). We also record here the following theorems. Theorem 2.2 For k 1 and m 1, we have that E (qm) is algebraic ≥ ≥ 2k over the field generated by E2,E4,E6. Theorem 2.3 Let K be the field generated by E2,E4,E6 over the rational numbers. Suppose that f is a non-constant function which is algebraic over the function field K. If α is an algebraic number with 0 < α < 1 and f(α) is defined, then f(α) is transcendental. | | For proofs, we refer the reader to Lemmas 2 and 3 of [6]. 3 Proof of Theorem 1.2 Let α = 1+√5 and β = 1/α. Then, 2 − αn βn f = − , n α β − as is easily verified by induction. Thus, the study of the series F (k) reduces to the study of the series ∞ 1 , (αn βn)k n=1 − which is ∞ 1 ( 1)k . (3.1) − (βn ( 1)nβ n)k n=1 − − − The Fibonacci Zeta Function 5 Breaking up the sum into n even and n odd leads us to consider the two series ∞ 1 A (q)= , k (qn q n)k n=1 − − and ∞ 1 B (q)= . k (qn + q n)k n=1 − For then, we can re-write (3.1) as ( 1)k A (β2)+ B (β) B (β2) , − k k − k We now look at Ak(q) and Bk(q) separately. Clearly, ∞ qnk ( 1)kA (q)= . (3.2) − k (1 q2n)k n=1 − Now for q < 1, we have | | 1 ∞ = qm. 1 q m=0 − Differentiating this (k 1) times with respect to q leads to − (k 1)! ∞ m k+1 − = m(m 1) (m (k 2))q − . (1 q)k − ··· − − m=k 1 − − Thus, ( 1)kA (q)(k 1)! = − k − ∞ nk ∞ 2n(m k+1) q m(m 1) (m (k 2))q − − ··· − − n=1 m=k 1 − which is equal to n(2m k+2) q − m(m 1) (m (k 2)). − ··· − − n 1;m k 1 ≥ ≥ − Now suppose k = 2j is even. Then, 2n(m j+1) A (q)(2j 1)! = q − m(m 1) (m 2j + 2). 2j − − ··· − n 1;m 2j 1 ≥ ≥ − 6 M. Ram Murty For j = 1, we have 2nm ∞ 2r A2(q)= q m = σ1(r)q . n,m 1 r=1 ≥ We immediately recognize that this is, apart from the constant term, the 2 weight 2 Eisenstein series, E2(q ) (upto a constant). For j = 2, we have 2n(m 1) A (q)3! = q − m(m 1)(m 2). 4 − − n 1;m 3 ≥≥ Putting m 1 as d, we see that m(m 1)(m 2) is (d + 1)d(d 1) so that setting r =−nd, the above can be re-written− as− − ∞ q2r d(d2 1) − r=1 d r | 2 2 and again we see that this is a linear combination of E2(q ) and E4(q ) (minus the constant terms). A similar calculation shows that A6(q) is a lin- 2 2 2 ear combination of E2(q ), E4(q ) and E6(q ) (minus the constant terms).