ON EVALUATION OF RIEMANN ZETA FUNCTION ζ(s) 1. Introduction. It is well-known that the Riemann Zeta function defined by. ζ(s) = n s, R(s) > 1 (2)

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1 ON EVALUATION OF RIEMANN ZETA FUNCTION ζ(s) QIU-MING LUO, BAI-NI GUO, AND FENG QI Abstract. In this paper, by using Fourier series theory, several summing formulae for Riemann Zeta function ζ(s) and Dirichlet series are deduced.. Introduction It is well-known that the Riemann Zeta function defined by and Dirichlet series ζ(s) = D(s) =, R(s) > () ns ( ) n n s, R(s) > (2) are related to the gamma functions and have important applications in mathematics, especially in Analytic Number Theory. series In 734, Euler gave some remarkably elementary proofs of the following Bernoulli ζ(2) = n 2 = π2 6. (3) The formula (3) has been studied by many mathematicians and many proofs have been provided, for example, see 2]. In 748, Euler further gave the following general formula ζ(2k) = n 2k = ( )k 2 2k π 2k B 2k, (4) (2k)! where B 2k denotes Bernoulli numbers for k N Mathematics Subject Classification. R40, 42A6. Key words and phrases. Riemann Zeta function, Fourier series, recursion formula. The authors were supported in part by NNSF (#00006) of China, SF for the Prominent Youth of Henan Province (# ), SF of Henan Innovation Talents at Universities, NSF of Henan Province (# ), Doctor Fund of Jiaozuo Institute of Technology, CHINA. This paper was typeset using AMS-LATEX.

2 2 Q.-M. LUO, B.-N. GUO, AND F. QI The Bernoulli numbers B k and Euler numbers E k are defined in 22, 23] respectively by t e t = t k k! B k, t < 2π; (5) k=0 2e t e 2t + = t k k! E k, t π. (6) k=0 For other proofs concerning formula (4), please refer to the references in this paper, for example, 22] and 25]. In 999, the paper 0] gave the following elementary expression for ζ(2k): Let n N, then where n 2k = A kπ 2k, (7) A k = 3! A k 5! A k ( ) k 2 (2k )! A + ( ) k k (2k + )! k = ( ) k k (2k + )! + ( ) k i (2k 2i + )! A i. (8) i= For several centuries, the problem of proving the irrationality of ζ(2k + ) has remained unsolved. In 978, R. Apéry, a French mathematician, proved that the number ζ(3) is irrational. However, one cannot generalize his proof to other cases. Therefore, many mathematicians have much interest in the evaluation of ζ(s) and sums of related series. For some examples, see, 24, 26]. In 2], the lower and upper bounds for ζ(3) are given by using an integral expression ζ(3) = 8 7 (2i+) 3 = 2 7 π/2 0 x(π x) sin x dx in 9, p. 8] and refinements of the Jordan inequality x 6 x3 sin x x 6 x x5 in 3, 4]. The following formulae involving ζ(2k + ) were given by Ramanujan, see 24], as follows: () If k > and k N, α k 2 ζ( 2k) + ] n 2k e 2nα = ( β) k 2 ζ( 2k) + ] n 2k e 2nβ, (9)

3 ON EVALUATION OF RIEMANN ZETA FUNCTION ζ(s) 3 (2) if k > 0 and k N, 0 = (4α) k + ( 4β) k k+ j=0 2 ζ(2k + ) + 2 ζ(2k + ) + ] n 2k+ (e 2nα ) ( ) j π 2j B 2j B 2k 2j+2 (2j)!(2K 2J + 2)! n 2k+ (e 2nβ ) ] α k 2j+ + ( β) k 2j+], where B j is the j-th Bernoulli number, α > 0 and β > 0 satisfy αβ = π 2, and means that, when k is an odd number 2m, the last term of the left hand side in (0) is taken as ( )m π 2m B 2 2m (m!) 2. (0) In 928, Hardy in 6] proved (9). In 970, E. Grosswald in 3] proved (0). In 970, E. Grosswald in 4] gave another expression of ζ(2k + ). In 983, N.-Y. Zhang in 24] not only proved Ramanujan formulae (9) and (0), but also gave an explicit expression of ζ(2k + ) as follows: () If k is odd, then we have k+ ζ(2k + ) = 2ψ k (π) (2π) 2k+ (2) if k is even, j=0 ζ(2k + ) = 2ψ k (π) + 2π k ψ k(π) (2π)2k+ k ( ) j π 2j B 2j B 2k 2j+2 ; () (2j)!(2k 2j + 2)! k 2 j=0 ( ) j π 2j B 2j B 2k 2j+2, (2) (2j)!(2k 2j + 2)! where ψ k (α) = n 2k+ (e 2nα ), and ψ k (α) is the derivative of ψ k(α) with respect to α. There is much literature on calculating of ζ(s), for example, see 2, p. 435] and 22, pp ; p. 49; pp. 50 5]. As a matter of fact, many other recent investigations and important results on the subject of the Riemannian Zeta function ζ(s) can be found in the papers 5, 6, 7, 8, 20, 2] by H.M. Srivastava, and others. Furthermore, Chapter 4 entitled Evaluations and Series Representations of the book 9] contains a rather systematic presentation of much of these recent developments. The aim of this paper is to obtain recursion formulae of sums for the Riemann Zeta function and Dirichlet series through expanding the power function x n on

4 4 Q.-M. LUO, B.-N. GUO, AND F. QI π, π] by using the Dirichlet theorem in Fourier series theory. These recursion formulae are more beautiful than those from (4) to (2). To the best of our knowledge, these formulae are new. 2. Lemmas Lemma (Dirichlet Theorem 7, p. 28]). Let f(x) be a piecewise differentiable function on π, π]. () If f(x) is even on π, π], then the Fourier series expansion of f(x) on π, π] is where f(x + 0) + f(x 0) 2 a 0 = π 2 π 0 f(x) dx, a n = π 2 = a a n cos nx, (3) π 0 f(x) cos nx dx; (4) (2) if f(x) is odd on π, π], then we have f(x + 0) + f(x 0) = b n sin nx, (5) 2 where π b n = π f(x) sin nx dx. (6) 2 0 Lemma 2 (5, pp ]). Let n N and s R +, then s ( ) s ( ) x s i (2i)!x s 2i cos nx dx = sin nx 2i n 2i+ s + cos nx ( ) s ( ) i (2i + )!x s 2i 2i + n 2i+2, s ( ) s ( ) x s i+ (2i)!x s 2i sin nx dx = cos nx 2i n 2+ s + sin nx Lemma 3. For s >, let δ(s) ζ(s) = ζ(s) = ( ) s ( ) i (2i + )!x s 2i 2i + n 2i+2. (2n ) s (7) (8) and σ(s) ( ) n (2n ). Then s 2s 2 s δ(s), (9) 2s 2 s D(s), (20)

5 ON EVALUATION OF RIEMANN ZETA FUNCTION ζ(s) 5 δ(s) = σ(s) = (4n 3) s + (4n ) s, (2) (4n 3) s (4n ) s. (22) Proof. It is easy to see that, for s >, ζ(s), D(s), δ(s), and σ(s) converge absolutely. Since ζ(s) = n s = (2n ) s + the formula (9) follows from rewriting (23). Further, D(s) = ( ) n n s = (2n) s = δ(s) + ζ(s), (23) 2s (2n ) s (2n) s = δ(s) ζ(s), (24) 2s combining (9) with (24) yields (20). Lemma 4 (22, p. 5]). For k N, we have σ(2k + ) = ( ) n π2k+ = (2n ) 2k+ 2 2k+2 (2k)! E k. (25) 3. Main results and proofs We will use the usual convention that an empty sum is taken to be zero. For example, if k = 0 and k =, we take k i= Theorem. For k N, we have k + ζ(2k) = ( )k kπ 2k (2k + )! ζ(2k + ) = 22k+ 2 2k+ 2 i= = 0 in this paper. ( ) k+i+ π 2k 2i ζ(2i), (26) (2k 2i + )! ] + σ(2k + ) (4n ) 2k+ Proof. Let f(x) = x s, s N, then f(x) is differentiable on π, π].. (27) If s is even, then f(x) is an even function on π, π]. From (4) and (7), we obtain a 0 = 2πs s +, a n = 2 π s ( ) s ( ) n+i (2i + )!π s 2i 2i + n 2i+2. (28)

6 6 Q.-M. LUO, B.-N. GUO, AND F. QI Substituting (28) into (3) leads to a Fourier series expansion of f(x) = x s below s ( ) s 2i + ( ) i (2i + )!π s 2i ( ) n n 2i+2 cos nx = π ) (x s πs. (29) 2 s + If s is odd, then f(x) is an odd function on π, π]. Using (6) and (8) yields b n = 2 π s Substituting (30) into (5) give us the following s ( ) s ( ) i+ (2i)!π s 2i 2i ( ) s ( ) n+i+ (2i)!π s 2i 2i n 2i+. (30) Taking x = π and s = 2k in (29) produces k ( ) 2k ( ) i (2i )!π 2k 2i 2i i= Formula (26) follows from (32). ( ) n n 2i+ sin nx = π 2 xs. (3) Set s = 2k + in (9), (2), and (22), then we have n=0 kπ2k+ = n2i 2k +. (32) ζ(2k + ) = 22k+ 2 2k+ δ(2k + ), (33) δ(2k + ) = (2n ) 2k+ = (4n 3) 2k+ +, (4n ) 2k+ (34) σ(2k + ) = n=0 ( ) n (2n ) 2k+ = (4n 3) 2k+ Formula (27) follows from combining (33), (34), and (35). Remark. Using (26), we can obtain values of ζ(2k), for examples, ζ(2) = π2 6. (35) (4n ) 2k+ π4 π6 π8 π0, ζ(4) =, ζ(6) =, ζ(8) =, ζ(0) = Using (27), we also can approximate values of ζ(2k + ). Theorem 2. For k N, we have ζ(2k) = ( )k 2 2k π 2k B 2k, (36) (2k)! ζ(2k + ) = π 2k+ (2 2k+2 2)(2k)! E k + 22k+2 2 2k+. (37) (4n ) 2k+ where B 2k and E k denote Bernoulli numbers and Euler numbers, respectively. Proof. This follows from substituting (25) into (27).

7 ON EVALUATION OF RIEMANN ZETA FUNCTION ζ(s) 7 Theorem 3. For k N, we have Proof. Since D(2k) = ( )k π 2k 2(2k + )! k + i= σ(2k + ) = ( )k π 2k+ k 2 2k+2 (2k + )! + ( ) k+i+ π 2k 2i D(2i), (38) (2k 2i + )! sin nπ 2 = 0, for n = 2l, ( ) l, for n = 2l, In (29), taking x = 0 and s = 2k gives us k ( ) 2k ( ) i (2i + )!π 2k 2i 2i + From (4), formula (38) follows. ( ) k+i+ π 2k 2i σ(2i + ). (39) (2k 2i + )! (40) ( ) n n 2i+2 = π2k+ 2(2k + ). (4) In (3), taking x = π 2 and s = 2k + (k = 0,, 2... ) and using (40) yields k ( ) ] σ(2k + ) = ( )k+ 2k + ( π ) 2k+2 ( ) i (2i)!π 2k 2i+ σ(2i + ). (2k + )!π 2i 2 (42) From (42), we obtain (39). Remark 2. From (38), we can calculate values of Dirichlet series D(2k), for example, D(2) = D(4) = D(6) = ( ) n n 2 ( ) n n 4 ( ) n n 6 = π2 2, = 7π4 720, = 3π From (39), we can obtain values of σ(2k + ), for example, σ() = σ(3) = σ(5) = ( ) n 2n = π 4, ( ) n (2n ) 3 = π3 32, ( ) n (2n ) 5 = 5π5 536.

8 8 Q.-M. LUO, B.-N. GUO, AND F. QI Theorem 4. For k N, we have where B 2k denotes a Bernoulli number. D(2k) = ( )k (2 2k )π 2k B 2k, (43) (2k)! δ(2k) = ( )k (2 2k )π 2k B 2k. (44) 2(2k)! Proof. In (9) and (20), taking s = 2k and k N and using (36) leads to (43) and (44). Remark 3. From (44), we obtain δ(2) = δ(4) = δ(6) = (2n ) 2 = π2 8, (2n ) 4 = π4 96, (2n ) 6 = π Acknowledgments. The authors are indebted to the anonymous referees and the Editor, Professor H. M. Srivastava, for their fairly detailed comments and many valuable additions to the list of references. References ] Alfred van der Poorten, A proof missed by Euler, Shùxué Yìlín (Translations of Mathematics) (980), no. 2, (Chinese) 2] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers, McGraw-Hill, ] E. Grosswald, Comments on Some formulae of Ramnujan, Acta. Arith. 2 (972), ] E. Grosswald, Die Werte der Riemannschen Zeta Function an ungeraden Argumenstellen, Göttingen Nachrichten (970), ] Group of Compilation, Shùxué Shǒucè (Handbook of Mathematics), Higher Education Press, Beijing, China, 990. (Chinese) 6] G. H. Hardy, A formula of Ramanujan, J. London Math. Soc. 3 (928), ] L.-G. Hua, Gāoděng Shùxué Yǐnlùn (Introduction to Advanced Mathematics), Vol., Fascicule 2, Science Press, Beijing, China, 979. pp (Chinese) 8] R.-R. Jiang, Generalization and Applications of Fundamentional Residul Theorem, Shùxué de Shíjiàn yù Rènshī (Math. Practice Theory) 27 (997), no. 3, (Chinese)

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10 0 Q.-M. LUO, B.-N. GUO, AND F. QI 26] W.-P. Zhang, On identities of Riemann Zeta function, Kēxué Tōngbào (Chinese Sci. Bull.) 36 (99), no. 4, (Chinese) Department of Broadcast-Television-Teaching, Jiaozuo University, Jiaozuo City, Henan , CHINA address: Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan , CHINA address: Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan , CHINA address: