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The Zeta Quotient $\Zeta (3)/\Pi^ 3$ Is Irrational

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The Zeta Quotient $\Zeta (3)/\Pi^ 3$ Is Irrational The Zeta Quotient ζ(3)/π3 is Irrational N. A. Carella Abstract: This note proves that the first odd zeta value does not have a closed form formula ζ(3) = rπ3 for any rational number r Q. Furthermore, assuming the irrationality of the second odd zeta6 value ζ(5), it is shown that ζ∈(5) = rπ5 for any rational number r Q. 6 ∈ 1 Introduction The first even zeta value has a closed form formula ζ(2) = π2/6, and the first odd zeta value has a nearly closed formula given by the Lerch representation 7π3 1 ζ(3) = 2 , (1) 180 − n3(e2πn 1) nX≥1 − confer (45) for more general version. This representation is a special case of the Ramanujan formula for odd zeta values, see [6], and other generalized version proved in [11]. However, it is not known if there exists a closed form formula ζ(3) = rπ3 with r Q, see [27, p. 167], [18, p. 3] for related materials. ∈ Theorem 1.1. The odd zeta value ζ(3) = rπ3, (2) 6 for any rational number r Q. ∈ The proof of this result is simply a corollary of Theorem 4.4 in Section 4. The basic technique gen- eralizes to other odd zeta values. As an illustration of the versatility of this technique, conditional on the irrationality of the zeta value ζ(5), in Theorem 5.5 it is shown that ζ(5)/π5 is irrational. The preliminary Section 2 and Section 3 develop the required foundation, and the proofs of the various results are presented in Section 4 and Section 5. 2 Basic Foundation For 2n +1=3 The results for the nonvanishing of the sine function at certain real numbers can be derived by arXiv:1906.10618v2 [math.GM] 29 Jul 2019 several different methods such as Weil criterion for uniformly distributed sequences, evaluations of infinite products, and other techniques. Some of these ideas are considered here. The case is equivalent to the Weil criterion for uniformly distributed sequences, confer [16, p. 8, Theorem 2.1] for more details. Lemma 2.1. If k Z× is an integer, then ∈ (i) sin(ζ(3)k) = 0 (ii) sin(ζ(3)−1k) =0. 6 6 July 30, 2019 AMS MSC : Primary 11M06, 11J72, Secondary 11Y60. Keywords: Irrational number; Odd zeta value; Closed form formula. 1 The Zeta Quotient ζ(3)/π3 is Irrational 2 Proof. (i) Fix an integer k = 0, and consider the uniformly distributed sequence 2ζ(3)kn : n 1 . From the uniform distribution6 of this sequence, [24, Theorem 4.2], it immediately{ follows≥ that} ζ(3)k = mπ for all k Z. Applying Lemma 2.4 yields 6 ∈ 1 1 sin ((2x + 1)ζ(3)k) ei2ζ(3)kn = (3) x x sin (ζ(3)k) −xX≤n≤x = o(1). This implies that sin(ζ(3)k) = 0 as required. (ii) Since ζ(3) is irrational, see Theorem 4.1, statement (i) implies statement (ii). 6 Unfortunately, the validity of the opposite result sin(ζ(3)k) = 0 is false. If it was true, it would lead to a far more rewarding result, that is, ζ(3) = rπ for some r Q×. To see this, write ζ(3) = 7π3/180 v, where v R×, see (1). Then, ∈ − ∈ 0 = sin(ζ(3)k) (4) = sin(7π3k/180 vk) − = sin(7π3k/180) cos(vk) sin(vk) cos(7π3k/180). − (5) Equivalently tan(7π3k/180) = tan(vk). The tangent function is periodic of period π, one-to-one, and monotonically increasing over the interval [ π/2,π/2]. Hence, the two arguments satisfy the equation − 7π3 k = vk + mπ (6) 180 where m Z. Lastly, ζ(3) = (m/k)π. ∈ Lemma 2.2. If k Z× is an integer, then ∈ (i) sin(ζ(3)π−1k) =0, (ii) sin(ζ(3)−1πk) =0. 6 6 Proof. (ii) Evaluate the product representation of the sine function π πk (πk/ζ(3))2 sin k = 1 ζ(3) ζ(3) − π2n2 nY≥1 πk k2 = 1 (7) ζ(3) − ζ(3)2n2 nY≥1 = 0. 6 The last line is valid for any irrational value ζ(s), s 3, see Theorem 4.1. (i) The proof is similar. ≥ Lemma 2.3. If k Z× is an integer, then ∈ (i) sin(ζ(3)−1π2k) =0, (ii) sin(ζ(3)π−2k) =0. 6 6 Proof. (i) Fix an integer k 1, let un/vn : n 1 be the sequence of convergents of the irrational number π/ζ(3), refer to Theorem≥ 4.2,{ and let ≥ } π π k m < 1, (8) ζ(3) − where m N is an integer. Then, as n , ∈ → ∞ π2 π sin k = sin π k m ζ(3) ζ(3) − π sin π v u . (9) ≥ ζ(3) n − n The Zeta Quotient ζ(3)/π3 is Irrational 3 This follows from the Dirichlet approximation theorem π π v u k m (10) ζ(3) n − n ≤ ζ(3) − for any integer k v , see [24, Theorem 2.1] for the best rational approximations. Therefore, ≤ n π π sin sin π v u (11) ζ(3) ≥ ζ(3) n − n π v u ≫ ζ(3) n − n 1 ≫ vn+1 = 0. 6 The third line in (11) follows from the basic Diophantine inequality 1 1 αqn pn (12) 2qn+1 ≤ | − |≤ qn for any irrational number α R, confer [21, Theorem 3.8], [15, Theorem 13], and similar references. ∈ Lemma 2.4. For any real number t = kπ with k Z, and a large integer x 1, the finite sum 6 ∈ ≥ sin((2x + 1)t) ei2tn = . (13) sin(t) −xX≤n≤x Proof. Expand the complex exponential sum into two subsums: ei2tn = e−i2t e−i2tn + ei2tn. (14) −xX≤n≤x 0≤nX≤x−1 0≤Xn≤x Lastly, use the geometric series to determine the closed form. 3 Basic Foundation For 2n +1=5 The verification of sin α = 0, using a single method, but restricted to the zeta quotients α = ζ(5)/πa, a 3 is given below.6 ≤ Lemma 3.1. Let k Z× be an integer, and assume that ζ(5)π−5 = u/v Q× is rational, where v> 1, and a 1. Then∈ sin(ζ(5)π−ak) =0 for a 3. ∈ ≥ 6 ≤ Proof. The evaluation of the product representation of the sine function returns ζ(5) ζ(5) (ζ(5)π−ak)2 sin k = k 1 πa πa − π2n2 nY≥1 ζ(5) ζ(5)2k2 = k 1 . πa − π2+2an2 nY≥1 The hypothesis ζ(5)/π5 = u/v means that ζ(5)2k2 π8−2au2k2 1 =1 =0, (15) − π2+2an2 − v2n2 6 for a 3, and 0 = k,n,u,v Z. Hence, ≤ 6 ∈ ζ(5) ζ(5) (ζ(5)π−ak)2 sin k = k 1 = 0 (16) πa πa − π2n2 6 nY≥1 for a 3 as claimed. ≤ Lemma 3.2. If k Z× is an integer, then ∈ The Zeta Quotient ζ(3)/π3 is Irrational 4 (i) sin(ζ(5)−1π4k) =0, (ii) sin(ζ(5)π−4k) =0. 6 6 Proof. (i) Fix an integer k 1, let un/vn : n 1 be the sequence of convergents of the irrational number π3/ζ(5), refer to Theorem≥ { 4.2, and let≥ } π3 π k m < 1, (17) ζ(5) − where m N is an integer. Then, as n , ∈ → ∞ π4 π3 sin k = sin π k m ζ(5) ζ(5) − π3 sin π v u . (18) ≥ ζ(5) n − n This follows from the Dirichlet approximation theorem π3 π3 v u k m (19) ζ(5) n − n ≤ ζ(5) − for any integer k v , see [24, Theorem 2.1] for the best rational approximations. Therefore, ≤ n π4 π3 sin sin π v u (20) ζ(5) ≥ ζ(5) n − n π3 v u ≫ ζ(5) n − n 1 ≫ vn+1 = 0. 6 The third line in (20) follows from the basic Diophantine inequality 1 1 αqn pn (21) 2qn+1 ≤ | − |≤ qn for any irrational number α R, confer [21, Theorem 3.8], [15, Theorem 13], and similar references. ∈ 4 Basic Results For The Zeta Quotients ζ(3)/π3 The third odd zeta value has a few irrationality proofs, confer [1], [4], [25, Chapter 4], and [5]. Theorem 4.1. ([1]) The real numbers (i) ζ(3), (ii) 1/ζ(3), are irrationals. The decimal expansion ζ(3) = 1.2020569031595942 ..., appears as sequence A002117 in [22]. The irrationality proofs for various associated real numbers considered here are derived from the theory of equidistribution of sequences of real numbers, see [16] and similar references. Theorem 4.2. The real numbers (i) ζ(3)/π, (ii) π/ζ(3), are irrationals. The Zeta Quotient ζ(3)/π3 is Irrational 5 Proof. (i) On the contrary the number ζ(3)/π = A/B is a rational number, where A, B N are integers such that gcd(A, B) = 1. This implies that the two sequences ∈ 2Aπ n : n 1 = 2Bζ(3) n : n 1 (22) { · ≥ } { · ≥ } are equivalent. Likewise, the two limits 1 1 lim ei2Aπn = lim ei2Bζ(3)n (23) x→∞ 2x x→∞ 2x −xX≤n≤x −xX≤n≤x are equivalent. These limits are evaluated in two distinct ways. I. Based on the independent properties of the number π. Use the identity ei2Aπn = 1 to evaluate of the left side limit as 1 1 lim ei2Aπn = lim 1=1. (24) x→∞ 2x x→∞ 2x −xX≤n≤x −xX≤n≤x II. Based on the independent properties of the number ζ(3). By Lemma 2.1, sin(t) = sin(Bζ(3)) =0 for any integer B = 0.
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