Algebraic relations for reciprocal sums of Fibonacci numbers

ACTA ARITHMETICA 30. 2007 Algebraic relations for reciprocal sums of Fibonacci numbers by Carsten Elsner Hannover Shun Shimomura Yokohama and Iekata Shiokawa Yokohama. Introduction. Let {F n } n 0 and {L n } n 0 be Fibonacci numbers and Lucas numbers defined by F 0 = 0 F = F n+2 = F n+ + F n n 0 L 0 = 2 L = L n+2 = L n+ + L n n 0. Duverney Ke. Nishioka Ku. Nishioka and the last named author [9] see also [8] proved the transcendence of the numbers s = 2 3... n= F 2s n n= L 2s n F s n= 2n L s n= 2n by using Nesterenko s theorem on the Ramanujan functions Pq Qq and Rq see Section 3. In this paper we prove the algebraic independence of the numbers respectively F 2 n= n F 4 n= n and write each n= F 2s n F 6 n= n respectively n= L 2s n L 2 n= n L 4 n= n s = 4 5 6... L 6 n= n as a rational respectively algebraic function of these three numbers over Q see Theorems 3 and Example. Similar results are obtained for the alternating sums n+ n= F 2s n see Theorems 2 and 4. respectively n+ n= L 2s n s = 2 3... 2000 Mathematics Subject Classification: Primary J8; Secondary J9 33E05. Key words and phrases: Fibonacci numbers Lucas numbers. [37] c Instytut Matematyczny PAN 2007

38 C. Elsner et al. It is interesting to compare our results with some arithmetical properties of the values of the Riemann zeta function ζs = n= n s : a Apéry [3] proved the irrationality of ζ3 cf. [4] [6] [5]. It is still unknown whether ζ5 is irrational or not. Zudilin [20] showed that at least one of ζ5 ζ7 ζ9 ζ is irrational. b Euler s formula ζ2k = k 2 2k B 2k 2k k N 2k! where B 2k are the Bernoulli numbers implies the algebraic dependence of ζ2k on ζ2 = 2 /6 for any integer k 2. For Fibonacci numbers we consider the Fibonacci zeta function ζ F s = Res > 0 F s n= n which extends meromorphically to the whole complex plane cf. [4]. a André-Jeannin [2] proved the irrationality of ζ F see also [5] [7]. The arithmetical nature of ζ F 3 is unknown. b Our results in this paper imply that the values ζ F 2 ζ F 4 ζ F 6 are algebraically independent and that for any integer s 4 ζ F 2s 5 s 2 r s ζ F 4 Qu v u := ζ F 2 v := ζ F 6 with some r s Q r s = 0 if and only if s is odd where the rational function of u and v is explicit; for example ζ F 8 5 4 ζ F4 = 3784u + 5 2 256u6 3456u 5 + 2880u 4 + 792u 3 v 00u 3 + 2060u 2 v 025u 2 + 7560uv + 336v 2 050v see Theorem Example and Table. 2. Statement of results. Suppose that α β C satisfy β < and αβ =. We put U n = αn β n α β n 0 2 V n = α n + β n n 0. If α + β = a Z then {U n } n 0 and {V n } n 0 represent integer-valued linearly independent binary recurrences satisfying the relation X n+2 = ax n+ + X n n 0 with initial values X 0 X = 0 and 2 a respectively. In particular if β = 5/2 we have the Fibonacci and Lucas numbers: U n = F n V n = L n n 0.

Reciprocal sums of Fibonacci numbers 39 In what follows s always denotes a positive integer. Set σ 0 s = and for s 2 let σ s...σ s s be the elementary symmetric functions of the s numbers 2 2... s 2 defined by σ i s = i r 2 ri 2 i s. r < <r i s The coefficients of the expansions are given by cosec 2 x = x 2 + a j x 2j sec 2 x = a j = j 2j 2 2j B 2j 2j! b j x 2j b j = j 2j 2 2j 2 2j B 2j 2j! j where B 2 = /6 B 4 = /30 B 6 = /42... are the Bernoulli numbers. In this paper the symbol B n is used for the Bernoulli numbers only here. It will denote a q-series in Sections 3 and 4. Theorem. Let {U n } n be defined by with α β Q satisfying β < and αβ = and set Φ 2s := α β 2s n= Un 2s. Then the numbers Φ 2 Φ 4 Φ 6 are algebraically independent and for any integer s 4 the number Φ 2s is written as with Φ 2s = 2s! s σ s sµ s µ s = 3 ϕ = 4 3 Φ 2 j= j 2j! 2 2j+3 σ s j sϕ j s ψ j a j 4Φ 2 2 + 2Φ 2 8Φ 4 + ω 5 4 32Φ 2 2 5Φ 2 ω + 3 0 s odd s even

40 C. Elsner et al. where ϕ 2 = 4 63 24Φ 2 2Φ 2 2 2Φ 2 5ω + 77 2 j 2 3 ϕ j = ϕ i ϕ j i j 3 j 22j + 3 i= ψ = 4 6Φ 2 2 3Φ 2 5ω + 25 3 4 ψ 2 = 4 9 24Φ 2 6Φ 2 2 3Φ 2 5ω + 25 4 j 2 ψ j = 224Φ 2 ψ j 3 ψ i ψ j i j2j i= ω = 56Φ 6 + 5/4 4Φ 2 +. j 3 Remark. If s 4 then + 4Φ 2 [s/2] Φ 2s r s Φ 4 Q[Φ 2 Φ 6 ] and the total degree of this does not exceed s + [s/2] where r s Q r s = 0 if and only if s is odd. Table. Relations for Φ 2s in Theorem x = Φ 2 y = Φ 4 z = Φ 6 s = 4 Φ 8 = 3 70 y + 8904x + 2 280x6 3456x 5 + 576x 4 + 8960x 3 z 444x 3 + 2060x 2 z 8x 2 + 52xz + 5680z 2 42z s = 5 Φ 0 = 2974x + 2 52x7 704x 6 + 62x 5 600x 4 z 30x 4 + 2560x 3 z 5x 3 + 450x 2 z + 4760xz 2 + 75xz + 700z 2 + 5z s = 6 Φ 2 = 462 y + 62624x + 3 089536x9 77444x 8 68x 7 4558848x 6 z + 53424x 6 + 77444x 5 z 40608x 5 + 737792x 4 z + 8558592x 3 z 2 556x 4 + 3472x 3 z + 7547904x 2 z 2 3258x 3 + 772x 2 z + 202544xz 2 + 2458624z 3 35x 2 + 3780xz + 27260z 2 252z Theorem 2. Let {U n } n be defined as in Theorem and set Φ 2s := α β 2s n+. n= U 2s n

Reciprocal sums of Fibonacci numbers 4 Then the numbers Φ 2 Φ 4 Φ 6 are algebraically independent and for any integer s 4 the number Φ 2s is written as Φ 2s = 2s! s j 2j! σ s sµ s + 2 2j+3 σ s j sϕ j + s ψ j a j with j= { Φ 2 s odd µ s = 24 4ξ s even ϕ = 4 80Φ 4 0ξ 2 + 5ξ 45 8 ϕ 2 = 6 89 ξ 80Φ 4 6ξ 2 + 5ξ 8 j 2 3 ϕ j = ϕ i ϕ j i j 3 j 22j + 3 i= ψ = 4 80Φ 4 + 2ξ 2 + 5ξ 9 8 ψ 2 = 6 27 ξ 80Φ 4 + 2ξ 2 + 5ξ 8 j 2 ψ j = 8ξψ j + 3 ψ i ψ j i j2j i= j 3 where ξ = ξφ 2 Φ 4 Φ 6 is a number satisfying 3 8ξ 3 +5ξ 2 +440Φ 4 46ξ 252Φ 2 +260Φ 4 7560Φ 6 77 = 0. 6 Table 2. Relations for Φ 2s in Theorem 2 x = Φ 2 y = Φ 4 s = 4 Φ 8 = 5 3 y2 + 89 ξ2 378 ξ + 23 y 432 + 6804 ξ4 6804 ξ3 23 844 ξ2 + 7 08864 ξ 43 74824 s = 5 Φ 0 = 80 630 x 693 ξ 0 8 y 2 63 2079 ξ3 5 2079 ξ2 + 5 756 ξ 5 y 756 5633 ξ5 + 6452 ξ4 + 403 4490640 ξ3 5 40824 ξ2 + 007 3265920 ξ 40739 57480920

42 C. Elsner et al. Theorem 3. Let {V n } n be defined by 2 with α β Q satisfying β < and αβ = and set Ψ 2s := Vn 2s. n= Then the numbers Ψ 2 Ψ 4 Ψ 6 are algebraically independent and for any integer s 4 the number Ψ 2s is written as s j 2j! Ψ 2s = σ s sµ s + 2s! 2 2j+3 σ s j sψ j s ϕ j b j j= with { Ψ2 s odd µ s = 4Ψ2 2 + Ψ 2 6Ψ 4 s even ϕ = 2 8Ψ 2 + 8Ψ 2 + η + ϕ 2 = 2 8Ψ 2 + 8Ψ 2 + η + 24Ψ 2 + η + 3 j 2 ϕ j = 24Ψ 2 + η + 3ϕ j + 3 ϕ i ϕ j i j2j ψ = 2 8Ψ 2 + 8Ψ 2 η + i= j 3 ψ 2 = 2 8Ψ 2 + 8Ψ 2 η + 24Ψ 2 η + 3 j 2 ψ j = 24Ψ 2 η + 3ψ j + 3 ψ i ψ j i j2j i= j 3 Table 3. Relations for Ψ 2s in Theorem 3 x = Ψ 2 y = Ψ 4 s = 4 Ψ 8 = 3 70 y + 8 35 x4 + 4 35 x3 + 35 η2 + 45 η + 2 x 2 63 + 260 η2 + 80 η + x + 2520 2060 η2 + 2880 η 2520 s = 5 Ψ 0 = 2 5 x5 + 7 2 x4 + 5040 η2 + 504 η + 4 x 3 720 7 + 3440 η2 + 344 η + 3 x 2 920 + 5806080 η4 + 96768 η3 + 67 93536 η2 + 625 290304 η + 222 x 64520 + 46448640 η4 + 77444 η3 + 233 774440 η2 + 6 33776 η 560960

where η = ηψ 2 Ψ 6 is a number satisfying Reciprocal sums of Fibonacci numbers 43 η + 5 2 = 92Ψ2 2 48Ψ 2 + 6 + 3840Ψ 6 + 30. 8Ψ 2 + Theorem 4. Let {V n } n be defined as in Theorem 3 and set Ψ2s n+ :=. n= Then the numbers Ψ2 Ψ 4 Ψ 6 are algebraically independent and for any integer s 4 the number Ψ2s is written as Ψ2s s j 2j! = σ s sµ s + 2s! 2 2j+3 σ s j sψ j + s ϕ j b j with { Ψ 2 s odd µ s = 8 θ s even j= ϕ = 2 96Ψ 4 θ 2 + 2θ 3 V 2s n ϕ 2 = 2θ 96Ψ 4 θ 2 + 2θ 396Ψ4 3θ 2 + 2θ 3 ϕ j = 96Ψ4 3θ 2 + 2θ 3 ϕ j 2 j 3 ϕ i ϕ j i j2j θ ψ = 2 96Ψ 4 + θ 2 + 2θ 3 ψ 2 = 2θ 96Ψ 4 + θ 2 + 2θ 396Ψ4 + 3θ 2 + 2θ 3 ψ j = 96Ψ4 + 3θ 2 + 2θ 3 ψ j 2 j + 3 ψ i ψ j i j2j θ i= i= Table 4. Relations for Ψ 2s in Theorem 4 j 3 j 3 s = 4 Ψ 8 = x = Ψ 2 y = Ψ 4 35840θ 2 98304y3 02422θ + 9y 2 + 32θ 4 + 32θ 2 + 44θ + 9y 2θ 5 θ 4 + 38θ 2 22θ 3 s = 5 Ψ 0 = 630 x + 64520θ 786432y3 73728θ + y 2 + 576θ 4 + 3θ 2 + 8θ + 4y 3θ 5 8θ 4 6θ 3 54θ 2 + 5θ 24

44 C. Elsner et al. where θ = θψ2 Ψ 4 Ψ 6 is a number satisfying θ 2 92Ψ 4 6θ + 920Ψ 6 64Ψ 2 7 = 0. Remark 2. The quantities ξ η and θ in Theorems 2 3 and 4 respectively are algebraic functions of the corresponding sums for s 3. In applying these theorems we have to choose appropriate branches of them depending on the parameter a or β. This will be done in Section 5 Theorems 5 6 and 7. Example. The reciprocal sums of Fibonacci respectively Lucas numbers n= F n 2 n= F n 4 n= F n 6 respectively n= L 2 n n= L 4 n n= L 6 n are algebraically independent and for any integer s 4 the number 5 s n= F n 2s respectively n= L 2s n is written by the formula in Theorem respectively Theorem 3 with η in Theorem 6. The alternating sums n= n+ Fn 2s and n= n+ Ln 2s have similar properties mentioned in Theorems 2 and 4 with ξ and θ in Theorems 5 and 7 respectively. Example2. The Pell numbers 2 5 2 29... defined by P 0 = 0 P = P n+2 = 2P n+ + P n n 0 are expressible by with β = 2 and α = /β satisfying α β = 8 see [3]. To the numbers n= P n 2s and n= n+ Pn 2s we can apply Theorems and 2 with Theorem 5. 3. Preparation for the proofs. The reciprocal sums of {U n } n and {V n } n in our theorems are written as series of hyperbolic functions. In [9] Zucker gave a method of summing such series. He wrote them as q-series expressible in closed form in terms of K E and k where K and E are the complete elliptic integrals of the first and second kind with the modulus k 0 ± defined by dt k K = Kk = t 2 k 2 t 2 E = 2 t 2 t 2 dt 0 for k 2 C \ {0} {z z }. The branch of each integrand is chosen so that it tends to as t 0. The relation among q and these quantities is as follows: q = e c c = K /K K = Kk k 2 + k 2 =. We start with the following formulas. We choose c = cβ or q = qβ so that q = e c = β 2 β = e c/2. Then by [9 Tables i ii iv iii] 0

ν= Reciprocal sums of Fibonacci numbers 45 Σ := 2 2s cosech 2s s 4 νc = σ s j sa 2j+ β 2 2s! Σ 2 := 2 2s sech 2s νc = s s 5 σ s j sb 2j+ β 2 2s! ν= Σ 3 := 2 2s sech 2s2ν c = s s 6 σ s j sd 2j+ β 2 2 2s! 7 ν= Σ 4 := 2 2s ν= cosech 2s2ν c = 2 n= s 2s! note that σ i s denotes α i s in [9] where n 2j+ q 2n A 2j+ q = q 2n B 2j+q = C 2j+ q = n= n 2j+ q n q 2n D 2j+q = ν= n= n= σ s j sc 2j+ β 2 n+ n 2j+ q 2n q 2n n+ n 2j+ q n q 2n. Our reciprocal sums are expressible by these series of hyperbolic functions: Φ 2s = α β 2s 8 U2ν 2s + U2ν 2s = Σ 3 + Σ 9 0 Ψ 2s = ν= V2ν 2s + ν= Φ 2s = α β 2s Ψ 2s = ν= V2ν 2s ν= ν= V 2s 2ν U2ν 2s V 2s 2ν ν= = Σ 4 + Σ 2 ν= = Σ 4 Σ 2. U2ν 2s = Σ 3 Σ The q-series A 2j+ B 2j+ C 2j+ and D 2j+ are generated from Fourier expansions of the squares of Jacobian elliptic functions ns 2 z nc 2 z dn 2 z and nd 2 z respectively where ns z = snz ncz = sn 2 z dnz = k 2 sn 2 z ndz = dnz with w = snz defined by z = w 0 dw w 2 k 2 w 2. The power series expansions of these elliptic functions give the expressions

46 C. Elsner et al. of the corresponding q-series in terms of K E and k cf. [0]. For example we find in [7] the expressions 2K 2 3E Pq 2 := 24A q = K 2 + k2 2K 4 2 Qq 2 := + 240A 3 q = k 2 + k 4 2K 6 Rq 2 := 504A 5 q = 2 + k2 2k 2 2 k 2 with q = e c c = K /K. We recall here the theorem of Nesterenko and its corollary [6]. Nesterenko stheorem. If C with 0 < < then trans.deg Q Q P Q R 3. Corollary. If Q with 0 < < then P Q and R are algebraically independent. The corollary with 2 implies the following: Lemma. If q = e c Q with 0 < q < then K/ E/ and k are algebraically independent. To get the formulas stated in the theorems we use the recurrence relations satisfied by the coefficients of the power series expansions of Jacobian elliptic functions which are given by the following lemmas. Lemma 2. The coefficients of the expansion ns 2 z = z 2 + c j z 2j are given by c 0 = 3 + k2 c = 5 k2 + k 4 c 2 = 89 + k2 2k 2 2 k 2 j 2 j 22j + 3c j = 3 c i c j i j 3. i= Proof. Since w = snz is a solution of w 2 = w 2 k 2 w 2 such that w0 = 0 the function u = ns 2 z = w 2 satisfies u 2 = 4w 6 w 2 = 4w 2 w 2 w 2 k 2 that is 3 u 2 = 4uu u k 2 and differentiation of 3 leads us to 4 u = 6u 2 4 + k 2 u + 2k 2.

Reciprocal sums of Fibonacci numbers 47 Substituting u = z 2 + v v = c jz 2j into 4 we have 6z 4 + 2j2j c j z 2j 2 = 6 c j z 2j 2 + 2z 2 c j z 2j 4 + k 2 c j z 2j + 6z 4 4 + k 2 z 2 + 2k 2. Comparing the coefficients of z 2 and the constant terms on both sides we obtain c 0 = + k 2 /3 and c = k 2 + k 4 /5. For j 2 the coefficients of z 2j 2 on both sides satisfy j 2j2j c j = 6 c i c j i + 2c j 4 + k 2 c j. i=0 Since + k 2 = 3c 0 this can be written in the form j j 2 5 j 22j + 3c j = 3 c i c j i 6c 0 c j = 3 c i c j i. i=0 When j = 2 however both sides of 5 vanish and c 2 is not determined uniquely. To compute c 2 substitute u = z 2 + c 0 + c z 2 + c 2 z 4 + into equation 3 and compare the constant terms. Then we have c 2 = +k 2 2k 2 2 k 2 /89. Once c 0 c and c 2 are known the coefficients c j j 3 are uniquely determined by 5. It is also possible to derive the recurrence formula from the proof of Satz 3 in [ pp. 69 70] concerning the -function by choosing ns 2 z = z + +k 2 /3 g 2 = 4/3 k 2 +k 4 and g 3 = 4/27+k 2 2k 2 2 k 2. Lemma 3. The coefficients of the expansion k 2 nc 2 z = c j z 2j are given by j= c = k 2 c 2 = 3 k2 2 k 2 i= j 2 j2j c c j = 6c 2 c j + 3c c i c j i j 3. Proof. Since nc 2 z = sn 2 z the function u = k 2 nc 2 z is a solution of u 2 = 4uu + u + k 2 such that u0 = u 0 = 0 and hence u = 6u 2 + 42 k 2 u + 2 k 2. i=

48 C. Elsner et al. Substitution of u = j= c jz 2j yields the equality 2j2j c j z 2j 2 = 6 c j z 2j 2 + 42 k 2 c j z 2j + 2 k 2. j= j= Looking at the constant terms we have c = k 2. Furthermore for j 2 i= j= j 2 6 2j2j c j = 42 k 2 c j + 6 c i c j i in particular c 2 = 2 k 2 c /3 = k 2 2 k 2 /3. Multiplying both sides of 6 by c we obtain the desired recursive relation for j 3. Lemma 4. The coefficients of the expansion are given by k 2 nd 2 z = k 2 + c j z 2j j= c = k 2 k 2 c 2 = 3 k2 k 2 2k 2 j 2 j2j c c j = 6c 2 c j 3c c i c j i j 3. Proof. The function u = k 2 nd 2 z = k 2 k 2 sn 2 z which satisfies u 2 = 4u uu k 2 is a solution of i= u = 6u 2 + 42 k 2 u 2 k 2 with the initial condition u0 = k 2 u 0 = 0. Substitution of u = k 2 + v v = j= c jz 2j yields c = k 2 k 2 and for j 2 j 2 j2j c j = 2 2k 2 c j 3 c i c j i. Multiplying both sides by c and observing c 2 = 2k 2 c /3 we obtain the desired formula for j 3. Lemma 5. The coefficients of the expansion dn 2 z = + c j z 2j j= i=

Reciprocal sums of Fibonacci numbers 49 are given by c = k 2 c 2 = 3 k2 + k 2 j 2 j2j c c j = 6c 2 c j 3c c i c j i j 3. Proof. By definition u = dn 2 z satisfies u 2 = 4u uu k 2 u0 =. This lemma is verified by the same argument as in the proof of Lemma 4. 4. Proofs of the theorems 4.. Proof of Theorem. It follows from 4 6 and 8 that 7 Φ 2s = σ s sµ s + 2s! s j= where µ s = A s D. In particular 8 9 i= σ s j sa 2j+ s D 2j+ Φ 2 = A + D 6Φ 4 = A 3 D 3 A D 20Φ 6 = A 5 + D 5 5A 3 + D 3 + 4A + D. Recall the generating functions of A 2j+ and D 2j+ given by 2K 2 2Kx ns 2 4KK E = 2 + cosec 2 x 8 j 2x 2j A 2j+ 2j! 2K 2 2Kx k 2 nd 2 = 4KE 2 8 j 2x 2j D 2j+ 2j! cf. [9 Tables i iv] [2] [8 p. 535] [] which yield 2K 2j+2 20 ϕ j := c j = a j j 22j+3 2j! A 2j+ j 2K 2j+2 ψ j := c j+ 22j+3 2 j = 2j! D 2j+ j. Here c j and c j are the coefficients given by Lemmas 2 and 4 respectively. Note that 2K 2 E 8D = K + k2 2K 4 22 6D 3 = k 2 k 2 2K 6 6D 5 = k 2 k 2 2k 2

50 C. Elsner et al. cf. [9 Table iv] or 2 combined with Lemma 4. Then by 2 with q = β 2 and 22 A + D = 2K 2 23 2k 2 24 A D = 2K 2 6E 24 24 K 5 + 4k2 A 3 + D 3 = 2K 4 25 + 4k 2 4k 4 240 A 3 D 3 = 2K 4 26 6k 2 + 6k 4 240 A 5 + D 5 = 2K 6 27 2k 2 3k 2 + 3k 4 504 A 5 D 5 = 2K 6 28 2k 2 + 32k 2 32k 4. 504 Substituting these quantities into 8 and 9 we have expressions of Φ 2 Φ 4 and Φ 6 in terms of K E and k; and then we put 2K 2 29 X := 24Φ 2 = 2k 2 Let us set 30 x := Y := 440Φ 4 + 2K 2 6E = 0 K 5 + 4k2 Z := 20960Φ 6 + 4032Φ 2 + 23 2K 6 = 2 2k 2 3k 2 + 3k 4 2K 4 + 2 + 4k 2 4k 4. 2K 2 y := + 2K 2 E K z := 2K 4 6k 2 + 6k 4 2K 2 2k 2. Note that the same symbols x y and z denote different quantities in Section 2. Then the quantities above are written as follows: X = z Y = 06y 3x 2z + 4z 2 3x 2 Z = z 2 3z2 27x 2 + 2 2 9x2 7z 2. Solving these equations we obtain

Reciprocal sums of Fibonacci numbers 5 x 2 = Ω := 3X3 + 47X 2 + 2Z 27X + 7 302y x = Y 4X 2 20X + 3Ω z = X. Hence x y z are algebraic over QX Y Z = QΦ 2 Φ 4 Φ 6 and the algebraic independence of Φ 2 Φ 4 Φ 6 follows from Lemma. Substituting 20 and 2 into 7 we obtain the formula for Φ 2s stated in the theorem. Observe that Ω 6 = 24 3 Φ2 2 22 3 Φ 2 + 03 48 5 3 ω with ω = 56Φ 6 + 5/4 4Φ 2 +. The expression of µ s is obtained by writing 23 and 24 in terms of X Y and Ω namely Φ 2 Φ 4 and ω. Furthermore 20 and 2 for j = 2 together with 2 and 22 imply ϕ = 2K 4 k 2 + k 4 = 5 60 3x2 + z 2 ϕ 2 = 2K 6 2k 2 2 + k 2 k 4 = z 89 756 9x2 z 2 3 2K 4 ψ = k 2 k 4 = 4 x2 z 2 ψ 2 = 2K 2 2k 2 ψ = z 3 3 ψ with x 2 = Ω z = X from which the desired expressions follow. Multiplying both sides of the recurrence formula in Lemma 2 by 2K/ 2j+2 and using 20 we obtain the formula for ϕ j j 3. The recurrence formula for ψ j is obtained from Lemma 4 by using ψ 2 /ψ = X/3 = 24Φ 2 /3. 4.2. Proof of Remark. Denoting the denominator of ω by T :=4Φ 2 + we have the polynomials T ν ϕ 2ν T ν ϕ 2ν T ν ψ 2ν T ν ψ 2ν ν in Φ 2 Φ 6 of total degrees say δ 2ν δ 2ν δ 2ν δ 2ν respectively. Then the recurrence formulas concerning ϕ j and ψ j imply for ν δ 2ν δ 2ν δ 2ν δ 2ν max { + δ 2i + δ 2ν i δ 2i + δ 2ν i } i 2ν 3 max {δ 2i + δ 2ν i } i 2ν 2 max i 2ν 3 { + δ 2i + δ 2ν i δ 2i + δ 2ν i 2 + δ 2ν 2} max i 2ν 2 {δ 2i + δ 2ν i + δ 2ν }. Using these relations by induction on ν we can verify δ 2ν δ 2ν 3ν and δ 2ν δ 2ν 3ν + which yields the estimate for the total degree of the polynomial T [s/2] Φ 2s r s Φ 4.

52 C. Elsner et al. 32 4.3. Proof of Theorem 2. From 4 6 and 0 we have Φ 2s = 2s! s σ s sµ s σ s j s s D 2j+ β 2 + A 2j+ β 2 j= with µ s = s D β 2 + A β 2. In particular 33 34 Φ 2 = D A 6Φ 4 = D 3 + A 3 + D + A 20Φ 6 = D 5 A 5 5D 3 A 3 + 4D A. Substituting 23 26 and 28 into 33 and 34 we obtain 2K 2 2K 4 Y := 440Φ 4 = 0 2k 2 + 4k 2 4k 4 Z := 20960Φ 6 4032Φ 2 + 23 2K 4 2K 6 = 2 6k 2 + 6k 4 + 2 2k 2 + 32k 2 32k 4 which are written in the form Y = 0z 9x 2 7z 2 /2 Z = 24z 2 3x 2 + 2z9x 2 8z 2 with x and z given by 30. From these equations we obtain 8ξ 3 + 5ξ 2 + Y 35ξ 4Y Z/6 = 0 ξ = z/4 which is written as 3. It is easy to see that K/ E/ and k are algebraic functions of Φ 2 Φ 4 Φ 6 over Q. This fact combined with Lemma implies the algebraic independence of Φ 2 Φ 4 Φ 6. Formula 32 together with 20 and 2 implies the expression of Φ 2s in terms of ϕ j and ψ j. By 23 and 24 we can write µ s in terms of Φ 2 and ξ. Substituting z = 4ξ and 9x 2 = 67ξ 2 5ξ 2Y into 3 we obtain ϕ...ψ 2 expressed in terms of Φ 4 and ξ. The recurrence formula for ψ j follows from that of Lemma 4 combined with 2 and the relation ψ 2 /ψ = 4ξ/3. 4.4. Proof of Theorem 3. From 5 7 and 9 we have 35 Ψ 2s = σ s sµ s + 2s! s j= where µ s = C s B. In particular σ s j sc 2j+ s B 2j+ Ψ 2 = C + B 6Ψ 4 = C 3 B 3 C B 20Ψ 6 = C 5 + B 5 5C 3 + B 3 + 4C + B.

The formulas 2K 2 k 2 nc 2 2Kx 2K Reciprocal sums of Fibonacci numbers 53 = 4KE 2 + sec 2 x + 8 2 2Kx dn 2 = 4KE 2 + 8 j B 2j+ 2x 2j 2j! j C 2j+ 2x 2j 2j! cf. [9 Tables ii iii] imply the relations 2K 2j+2 36 ϕ j := c j = b j + j 22j+3 2j! B 2j+ j 2K 2j+2 37 ψ j := c j = j 22j+3 2j! C 2j+ j where c j and c j are the coefficients given in Lemmas 3 and 5 respectively. From [9 Tables ii iii] or Lemmas 3 and 5 we obtain + 8B = 4KE 2K 2 2 8C = E K 2K 4 2K 4 6B 3 = k 2 6C 3 = k 2 2K 6 2K 6 + 8B 5 = 2 k2 2 k 2 6C 5 = k 2 + k 2. Hence 38 39 40 4 42 43 C + B = 8 + 2K 2 8 C B = 8 + 2K 2 2E 8 K C 3 + B 3 = 6 2K 4 2k 2 6 C 3 B 3 = 6 + 2K 4 6 C 5 + B 5 = 8 + 2K 6 k 2 + k 4 8 C 5 B 5 = 8 2K 6 2k 2. 8

54 C. Elsner et al. So we can express Ψ 2 Ψ 4 and Ψ 6 in terms of K E and k. Put 2K 2 44 X := 8Ψ 2 + = 2K 4 2K 2 Y := 96Ψ 4 + 3 = 2 2E K Z := 920Ψ 6 + 5 2K 6 2K = 2 k 2 + k 4 + 5 4 2k 2 + 8 2K By the same argument as in Sections 4. and 4.3 with x y z given by 30 we have 45 x = X 4y = Y X 2 + 2X Xz 2 + 0Xz + 3X 3 + 6X 2Z = 0 implying that x y and z are algebraic over QX Y Z = QΨ 2 Ψ 4 Ψ 6. Substitution of 36 and 37 into 35 yields the desired formula for Ψ 2s in terms of ϕ j and ψ j. Furthermore using 36 and 37 for j = 2 together with Lemmas 3 and 5 respectively we have 2K 4 ϕ = k 2 = x z + x 2 ϕ 2 = 2K 2 2 k 2 ϕ = 3 6 z + 3xϕ 46 2K 4 ψ = k 2 = x z x ψ 2 = 3 2K 2 2 + k 2 ψ = 6 z 3xψ. The right hand sides of these equalities as well as µ s can be written in terms of X and z satisfying the quadratic equation 45. The recurrence formulas for ϕ j and ψ j are derived from those of Lemmas 3 and 5 combined with ϕ 2 /ϕ = 3X + z/6 and ψ 2 /ψ = 3X z/6 respectively. Replacing z by η we obtain the theorem. 47 4.5. Proof of Theorem 4. Relation together with 5 and 7 implies Ψ2s = 2s! s σ s sµ s + σ s j sc 2j+ β 2 + s B 2j+ β 2 j= 2.

Reciprocal sums of Fibonacci numbers 55 with µ s = C β 2 + s B β 2. In particular 48 49 Ψ 2 = C B 6Ψ 4 = C 3 + B 3 C + B 20Ψ 6 = C 5 B 5 5C 3 B 3 + 4C B. Substituting 38 4 and 43 into 48 and 49 we obtain 2K 4 2K 2 Y := 96Ψ4 3 = 2k 2 2 2K 6 2K 4 Z := 920Ψ6 64Ψ2 7 = 2 2k 2 5 which are written as Y = xz 2x Z = 2x 2 z 5x 2 with x and z given by 30. Then we have x 2 2Y x + Z = 0 z = 2 Y/x. The quantities K/ E/ and k are algebraic functions of Ψ2 Ψ 4 Ψ 6 over Q which implies the algebraic independence of Ψ2 Ψ 4 Ψ 6. The desired expression of Ψ2s is obtained by substituting 36 and 37 into 47. By 38 and 39 the quantity µ s is expressible in terms of Ψ2 and x. Using 46 we have the desired expressions of ϕ...ψ 2 in terms of Ψ4 and x. Observing the relations ϕ 2/ϕ = 3x + z/6 and ψ 2 /ψ = 3x z/6 we derive from Lemmas 3 and 5 the recurrence formulas for ϕ j and ψ j respectively. Replacing x by θ in these expressions we obtain the theorem. 5. Branches of the related algebraic functions. In what follows we put a = α + β C namely β = a/2 + 4a 2 where the branch is chosen so that βa = Oa as a. Then each reciprocal sum in the theorems is treated as a function of a or β. Theorem 5. Under the assumptions of Theorem 2 we have the following: i The function ξ = ξa is holomorphic in the domain a > 8.46 and is expressible in the form 50 ξa = 5 24 + 0 + γ / 0 0 = 0a := γ 3 2 + γ 3 γ2 2 i with γ = γ a γ 2 = γ 2 a expressed as 5 γ := 29 24 2 60Φ 4 γ 2 := 7963 24 3 + 63 4 Φ 2 + 95 2 Φ 4 945 2 Φ 6 satisfying γ a = 29 24 2 +Oa 2 and γ 2 a = 7963 24 3 + Oa 2 as a. Here the branches of the square and cube roots are taken so that γ 3 γ 2 2 = 7/92 4395 and that 52 0 = 48 4395i.

56 C. Elsner et al. ii For any real number a 2.6 the function ξa admits the expression 50 with the same branches as above where γ 3 γ2 2 > 0 γ 2 + γ 3 γ2 2 i 0. iii For a = respectively a = 2 corresponding to the Fibonacci respectively Pell numbers the quantity ξ = 2.6658... respectively ξ2 = 0.72378... is expressible as 50 with 0 =.22662... + 0.57598...i γ 3 γ 2 2 > 0 respectively 02 = 0.25772....00662...i γ 2 3 γ 2 2 2 > 0. Theorem 6. Under the assumptions of Theorem 3 we have the following: i The function η = ηa is holomorphic in the domain a > 5.43 and is expressible in the form with ηa = 5 + χa χa = 92Ψ 2 2 48Ψ 2 + 6 + 3840Ψ 6 + 30/8Ψ 2 + satisfying χa = 36 + Oa 2 as a. Here the branch is taken so that χ = 6. ii For a = corresponding to the Lucas numbers and for any integer a 2 η = 5 χ < 5 53 ηa = 5 + χa > 5. Furthermore the last equality holds for any real number a 2.4. Theorem 7. Under the assumptions of Theorem 4 we have the following: i The function θ = θa is holomorphic in the domain a > 5.89 and is expressible in the form with χ a = 96Ψ 4 3 θa = χ a + χ 2 a χ 2 a = 64Ψ 2 + 926Ψ 4 2 576Ψ 4 920Ψ 6 + 6 satisfying χ a = 3+Oa 4 and χ 2 a = 6+Oa 2 as a. Here the branch of the square root is taken so that χ 2 = 4. ii For a = corresponding to the Lucas numbers θ = χ χ 2 < χ

and for any integer a 2 Reciprocal sums of Fibonacci numbers 57 θa = χ a + χ 2 a > χ a. Furthermore the last equality holds for any real number a 2.5. 5.. Proof of Theorem 6. By definition we have V n = β n + β n β n β 2n so that V n β 2 β n. This implies for s 54 Ψ 2s n V n 2s β 2 2s n and hence Ψ 2s s N are holomorphic for β <. Set β 2sn β 2 2s β 2s 55 Ωβ = 8Ψ 2 + χa = 23 28Ψ 3 2 48Ψ 2 2 + 320Ψ 6. Under the assumption β < 0.3 we deduce from 54 that Ψ 2. 3 β 2 0.2 and that Ψ 6. 7 β 6 0.002. By these inequalities for β < 0.3 we get Ωβ /2 3 28 Ψ 2 3 48 Ψ 2 2 320 Ψ 6 > and 8Ψ 2 + 8 Ψ 2 > 0.03. Hence in the domain β < 0.3 the function χa 0 is holomorphic. Note that β = βa conformally maps the domain C\{a = yi 2 y 2} to the disc β <. To examine the corresponding domain around a = observe that for z < /7 z + z 2 3 + t3/2 dt 2 8 3/2 z 0.846 z. 3 7 0 Using this estimate we have 2β = a + 4a 2 0.846 4a = 3.2584 a provided that 4a 2 < /7. This fact implies that the image of the domain a > 5.43 is contained in the disc βa < 0.3. By 54 and 55 we see 56 χa = 36 + Oβ 2 = 36 + Oa 2 as a. To determine the branch of η recall that 57 η = 2K/ 2 2k 2 = 24Φ 2 cf. Section 4.4 and 29. Since α β U n β n β 2 we have Φ 2 = Oβ 2 = Oa 2 so that ηa = + Oa 2 as a. This fact combined with 56 implies the assertion i. To determine the sign of η we use the following numerical values for a = : Ψ 2 =.207299... Ψ 6 =.006249... Φ 2 = 0.485264... Then we have χ = 3.885... = 5.6463... 2. On the other hand η = 0.64633... by 57. Consideration of these values leads to the sign of the square root. Similarly for a = 2 using the values Ψ 2 = 0.2839243... Ψ 6 = 0.056465... Φ 2 = 0.622974...

58 C. Elsner et al. we get η2 = 2.8953... and χ2 = 2.048... 2 which yields the sign + in 53. Finally we consider the case a 2.4. Since V 2 = a 2 + 2 = aa + 2a 3.23a we have V n 3.23a n n 2 which yields Ψ 2 = n V n 2 a 2 + 3.23 2 a 2 j 0 a 2j <.6a 2 < 0.94 for a 2.4. Hence 55 implies χa > for a 2.4. Combining this fact with 56 and ηa = +Oa 2 a and taking the continuity of ηa into account we deduce formula 53 for a 2.4 which completes the proof. 5.2. Proof of Theorem 7. Since Ψ 2s n V n 2s β 2 2s β 2s we have Ψ 2s.2s+ β 2s for β < 0.3 so that χ 2 a 6 64 Ψ 2 926 Ψ 4 2 576 Ψ 4 920 Ψ 6 > 0.7 for β < 0.28. This implies that χ 2 a is holomorphic for β < 0.28. By the same argument as in Section 5. it is holomorphic for a > 5.89. Recall the relation 58 θ = 2K/ 2 = 8Ψ 2 + cf. 44. Observing the estimates Ψ 2 = Oa 2 χ a = 3 + Oa 4 and χ 2 a = 6 + Oa 2 as a we have θa = χ a + χ 2 a for a > 5.89 where the branch is taken so that χ 2 = 4. In this way the assertion i is verified. When a = by numerical computation we have Ψ 2 = 0.9370204... Ψ 4 = 0.992040... Ψ 6 = 0.9988644...; and relation 58 together with the numerical value of Ψ 2 cf. Section 5. yields θ = 0.65833... Similarly when a = 2 Ψ 2 = 0.226586... Ψ 4 = 0.067537... Ψ 6 = 0.056036... and θ2 = 3.2739... Using these numerical values we can check the equalities for a = 2. For a 2.5 observe that V n 3.3a n n 2. Then for a 2.5 we have Ψ6 m V 2m 6 a 6 + 3.3 6 a 2 j 0 a 2j.0000a 6 < 0.03a 2 and Ψ2 V 2 V2 2 a 2 +2 > 0.75a 2 so that χ 2 a 7 + 64Ψ2 920Ψ 6 > 7 9.6a 2 > 5 which implies the assertion. 5.3. Proof of Theorem 5. Put ξ = 5/24+λ. Then 3 is written in the form 59 λ 3 3γ λ 2γ 2 = 0 where γ and γ 2 are as given in the theorem which may be regarded as functions of a C or β. Since α β U n β n β 2 we infer for s that Φ 2s β 2 2s β 2s. Hence Φ 2s s N are holomorphic for β <. In particular for β < 0.2 we have Φ 2 < 0.046 Φ 4 < 0.002

Reciprocal sums of Fibonacci numbers 59 and Φ 6 < 0.000. Using 5 we obtain γ >.8 γ 2 < 2.3 for β < 0.2 and γ 29 24 2 γ 2 7963 24 3 as β 0 or a. To determine the branch of ξ we note that ξa = ηa/4 cf. 57 which implies ξ /4 as a. Let us find the solution of 59 corresponding to ξ. Such a solution may be expressed in the form λ 0 = γ / 0 + 0 0 = 3 γ 2 + γ 3 γ 2 2 i around β = 0 where the branches should be chosen so that 24 6 γ 3 γ2 2 29 3 7963 2 > 0 ξ /4 namely λ 0 /24 as β 0. The second condition is satisfied if 52 is valid. Furthermore provided β < 0.2 it is easy to see that γ 3 γ2 2 γ3 γ2 2 >.83 2.3 2 > 0 and that γ 2 + γ 3 γ2 2 i 0. Hence 0β is holomorphic for β < 0.2. By the same numerical argument as in Section 5. we can verify that the image of the domain a > 8.46 is contained in the disc β < 0.2. In this way the assertion i is proved. If a 2.6 then Φ 2s 0.76 s because α β 2 = a 2 + 4 0.76. Hence we have γ.44 and.68 γ 2.0 implying γ 3 γ2 2 > 0.6 for a 2.6. This fact combined with the continuity of ξ with respect to a implies the assertion ii. When a = by numerical computation we obtain Φ 2 = 0.0335078... Φ 4 = 0.002066... Φ 6 = 0.00045... Then we have γ =.83637... γ 2 = 0.624754... and ξ = η/4 = 2.6658... cf. Section 5.. Similarly when a = 2 we have Φ 2 = 0.0980088... Φ 4 = 0.046727... Φ 6 = 0.009227... implying γ 2 =.079707... γ 2 2 = 0.766330... and ξ2 = η2/4 = 0.72378... Using these numerical values we can determine the branches as in the assertion iii. References [] M. Abramowitz and I. A. Stegun Handbook of Mathematical Functions Dover New York 965. [2] R. André-Jeannin Irrationalité de la somme des inverses de certaines suites récurrentes C. R. Acad. Sci. Paris Sér. I Math. 308 989 539 54. [3] R. Apéry Irrationalité de ζ2 et ζ3 Astérisque 6 979 3. [4] F. Beukers A note on the irrationality of ζ2 and ζ3 Bull. London Math. Soc. 979 268 272. [5] P. Bundschuh and K. Väänänen Arithmetical investigations of a certain infinite product Compos. Math. 9 994 75 99. [6] H. Cohen Démonstration de l irrationalité de ζ3 d après R. Apéry in: Séminaire de Théorie des Nombres Grenoble 978 979 VI. VI.9.

60 C. Elsner et al. [7] D. Duverney Irrationalité de la somme des inverses de la suite de Fibonacci Elem. Math. 52 997 3 36. [8] D. Duverney Ke. Nishioka Ku. Nishioka and I. Shiokawa Transcendence of Jacobi s theta series Proc. Japan Acad. Ser. A Math. Sci. 72 996 202 203. [9] Transcendence of Rogers Ramanujan continued fraction and reciprocal sums of Fibonacci numbers ibid. 73 997 40 42. [0] H. Hancock Theory of Elliptic Functions Dover New York 958. [] A. Hurwitz und R. Courant Vorlesungen über allgemeine Funktionentheorie und elliptische Funktionen Springer Berlin 925. [2] C. G. J. Jacobi Fundamenta Nova Theoriae Functionum Ellipticarum Königsberg 829. [3] Th. Koshy Fibonacci and Lucas Numbers with Applications Wiley-Interscience 200. [4] L. Navas Analytic continuation of the Fibonacci Dirichlet series Fibonacci Quart. 39 200 409 48. [5] Yu. V. Nesterenko Some remarks on ζ3 Mat. Zametki 59 996 865 880 960 in Russian; English transl.: Math. Notes 59 996 625 636. [6] Modular functions and transcendence questions Mat. Sb. 87 996 65 96 in Russian; English transl.: Sb. Math. 87 996 39 348. [7] S. Ramanujan On certain arithmetical functions Trans. Cambridge Philos. Soc. 22 96 59 84. [8] E. T. Whittaker and G. N. Watson Modern Analysis 4th ed. Cambridge Univ. Press Cambridge 927. [9] I. J. Zucker The summation of series of hyperbolic functions SIAM J. Math. Anal. 0 979 92 206. [20] V. V. Zudilin One of the numbers ζ5 ζ7 ζ9 ζ is irrational Uspekhi Mat. Nauk 56 200 no. 4 49 50 in Russian; English transl.: Russian Math. Surveys 56 200 774 776. Fachhochschule für die Wirtschaft University of Applied Sciences Freundallee 5 D-3073 Hannover Germany E-mail: carsten.elsner@fhdw.de Department of Mathematics Keio University 3-4- Hiyoshi Kohoku-ku Yokohama 223-8522 Japan E-mail: shimomur@math.keio.ac.jp shiokawa@beige.ocn.ne.jp Received on 4.5.2006 599