# Recursion formulae for Riemann Zeta function and Dirchlet series.

Abstract In this paper, some recursion formulae of sums for the Riemann Zeta function and Dirchlet, series are obtained through expanding several simple function on [-[pi],[pi]] or [0, 2[pi]] by using the Dirichlet theorem in Fourier series theory.

Keywords Riemann zeta, function, Dirchlet series, recurrence formula. Fourier series.

[sections]1. Introduction

It is well-known that the Riemann Zeta function defined by

[zeta](s) = [[infinity].summation over (n=1)] 1/[n.sup.s], R(s)>1 (1)

and Dirchlet series

D(s) = [[infinity].summation over (n=1)] [(-1).sup.n-1]/[n.sup.s], R(s)>1 (2)

play very important roles in Analytic Number Theory, and so on.

In 1734, Euler gave sum of the following Bernoulli series

[zeta](2) = [[infinity].summation over (n=1)] 1/[n.sup.2] = [[pi].sup.2]/6. (3)

The formula (3) has been studied by many mathematicians and many proofs have been published, for example, see [2]. In 1748, Euler further gave the following general formula

[zeta](2k) = [[infinity].summation over (n=1)] 1[n.sup.2k] = [(-1).sup.k-1] [2.sup.2k-1][[pi].sup.2k]/(2k)! [B.sub.2k], (4)

where [B.sub.2k] (k = 1, 2, ...) denotes Bernoulli numbers, defined in [18,19] by

t/[e.sup.t] - 1 = [[infinity].summation over (k=0) [t.sub.k]/k! [B.sub.k], |t| < 2[pi].

For other proofs concerning formula (4), please refer to the references in this paper, for example, [18] and [21]. In 1999, the paper [9] gave an elementary expression for [zeta](2k): Let n [member of] N, then

[[infinity].summation over (n=1)] 1/[n.sup.2k] = [A.sub.k][[pi].sub.2k], (5)

where

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

It is still an open problem to prove irrationality of [zeta](2k + 1) for several centuries. Until 1978, R. Apery, a French mathematician, proved that the number [zeta](3) is irrational. But one can not generalize his proof to other cases. So, many mathematicians keep much interest in the evaluation of [zeta](s) and sums of related series. For some examples, see [10,20,22].

The following formulae involving [zeta](2k + 1) were given by Ramanujan, see [22], as follows:

1. If k > 1 and k [member of] N,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)

2. if k > 0 and k [member of] N,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (8)

where [B.sub.j] is the j-th Bernoulli number, [alpha] > 0 and [beta] > 0 satisfy [alpha][beta] = [[pi].sup.2], and [summation]' means that, when k is an odd number 2m - 1, the last term of the left hand side in (8) is taken as [(-1).sup.m][[pi].sup.2m][B.sup.2.sub.2m]/[(m!).sup.2].

In 1928, Hardy in [6] proved (7). In 1970, E. Grosswald in [3] proved (8). In 1970, E. Grosswald in [4] gave another expression of [zeta](2k + 1). In 1983, N.-Y. Zhang in [20] not only proved Ramanujan formulae (7) and (8), but also gave an explicit expression of [zeta](2k + 1) as follows:

1. If k is odd, then we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (9)

2. if k is even,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (10)

where [??]-k([alpha]) = [[infinity].summation over (n=1) 1/[n.sup.2k+1] ([e.sup.2n[alpha]-1], and [[??]'.sub.k] [alpha] is the derivative of [[??].sub.-k]([alpha]) with respect to [alpha].

There are a lot of literature on calculating of [zeta](s), for example, see [2,p.435] and [18,pp.144-145; p.149; pp.150-151]. As a matter of fact, many other recent investigations and important results on the subject of the Riemannian Zeta function [zeta](s) can be found in the papers [11,12,13,14,15,16,17] by H. M. Srivastava, and others. Furthermore, Chapter 4 entitled

"Evaluations and Series Representations" of the book [15] 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 Dirchlet series through expanding some simple function on [-[pi],[pi]] or [0; 2[pi]] by using the Dirchlet theorem in Fourier series theory.

[section]2. Main results and proofs

Theorem 1. Let [delta](s)[??][[infinity].summation over (n=1)] 1/[(2n - 1).sup.s] and sigma(s)[??][[infinity].summation over (n=1)] [(-1).sup.n-1]/[(2n - 1).sup.s] for s >1. Then we have for k member of N

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (11)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (12)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (14)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (15)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (16)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (17)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (18)

Proof. Define the function f by

f(x) = [pi] - x/2, x [member of] (0,2[pi]).

Easy Computation reveals the Fourier series of f on (0,2[pi]):

[pi] - x/2 = [[infinity].summation over (n=1)] sin nx/n, x [member of] (0, 2[pi]). (19)

Integration term-by-term yields

[x.sup.2]/2x2! + [pi]x/2 - [zeta](2) = - [[infinity].summation over (n=1) cos nx/[n.sup.2], x [member of] [0,2[pi]].

Clearly, if we integrate 2k - 1 tithes on each side of (19) from 0 to x, then we obtain

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (20)

Taking in (20) x = [pi]/2 and noticing that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

We Conclude that

D(2k) = [(-1).sup.k+1] (4k - 1)[[pi].sup.2k]/2 x (2k)! + [k.summation over (j=1)] [(1).sup.k+j+1][4.sup.j]][[pi].sup.2k-2j]/(2k - 2j)! [zeta](2j). (21)

Taking x = [pi] in (20) we conclude that

D(2k) = [(-1).sup.k+1][[pi].sup.2k]/4k x (2k - 2)! + [k.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/(2k - 2j)! [zeta](2j). (22)

Taking x = 3[pi]/2 in (20) and noticing that

[(-1).sup.k][[infinity].summation over (n=1) cos 3n[pi]/2/[n.sup.2k] = [(-1).sup.k+1] [[infinity].summation over (n=1)] [(-1).sup.n-1]/[(2n).sup.2k] = [(-1).sup.k+1]/[2.sup.2k] D(2k),

we conclude that,

D(2k) = [(-1).sup.k+1](4k - 3) [3.sup.2k-1][[pi].sup.2k]/2 x (2k)! + [k.summation over (j=1)] [(-1).sup.k+j+1][4.sup.j][(3[pi]).sup.2k-2j]/(2k - 2j)! [zeta](2j) (24)

Integrating on each side of (20) from 0 to 2[pi], we get

[zeta](2k) = [(-1).sup.k+1](2k - 1)[2.sup.2k-2][[pi].sup.2k]/(2k+1)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1] [(2[pi]).sup.2k-2j]/(2k - 2j + 1)! [zeta](2j). (24)

Integrating term-by-term on each side of (20) from 0 to x, we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (25)

Taking x = [pi]/2 a in (25) nd noticing that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

we conclude that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (26)

Taking x = 7 in (25) we conclude that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (27)

Taking x = 3[pi]/2, in (25) and noticing that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

we conclude that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (28)

Integrating on each side of (25) from 0 to 2[pi], we get

[zeta](2k) = [(-1).sup.k+1] k[(2[pi]).sup.2k]/(2k + 2)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1] [2.sup.2k-2j+1][[pi].sup.2k-2j]/(2k - 2j + 2)! [zeta](2j). (29)

The proof of Theorem 1 is complete.

Remark Form recurrence formula (14), or (16), or (18) we can obtain values of [zeta](2k), for examples,

[zeta](2) = [[pi].sup.2]/6, [zeta](4) = [[pi].sup.4]/90, [zeta](6) = [[pi].sup.6]/945, [zeta](8) = [[pi].sup.8]/9450, [zeta](10) = [[pi].sup.10]/93555.

By using values of ((2k), we can conclude values of D(2k) from (11), or (12), or (13), for examples,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

By using values of [zeta](2k), we can also conclude values of [sigma](2k) from (15), or (17), for examples,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Theorem 2. For k [member of] N, we have

D(2k) = 1/[4.sup.k] - 1 [[(-1).sup.k+1[[pi].sup.2k]/2 x (2k)! + + [k-1.summation over (j=1)] [(-1).sup.k+j+1] [4.sup.j][[pi].sup.2k-2j]/(2k - 2j)! D(2j)], (30)

D(2k) = [(-1).sup.k+1][[pi].sup.2k]/2 x (2k + 1)! | [k-1.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/(2k - 2j + 1)! D(2j), (31)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (32)

[zeta](2k) = [(-1).sup.k+1][[pi].sup.2k]/2 x (2k)! + [k.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/(2k - 2j)! D(2j). (33)

Proof. Define the function g by

g(x) = x, x [member of] (-[pi],[pi]).

Easy computation reveals the Fourier series of f on (-[pi], [pi]):

x = 2 [[infinity].summation over (n=1) [(-1).sup.n-1] sin nx/n, x [member of] (-[pi], [pi]). (34)

Integration term-bbl-term yields

[x.sup.2]/2 x 2! D(2) = -[[infinity].summation over (n=1) [(-1).sup.n-1 cos nx/[n.sup.2], x [member of] [-[pi], [pi]].

Clearly, if we integrate 2k - 1 times on each side of (34) from 0 to x, then we obtain

[x.sup.2k]/2 x (2k)! + [k.summation over (j=1)] [(-1).sup.j][x.sup.2k-2j]/(2k - 2j)! D(2j) = [(-1).sup.k] [[infinity].summation over (n=1)] [(-1).sup.n-1] cos nx/[n.sup.2k], x [member of] [-[pi], [pi]]. (35)

Taking x = [pi]/2 in (35) we conclude that

D(2k) = 1/[4.sup.k] - 1 [[(-1).sup.k+1][[pi].sup.2k]/2 x (2k)! + + [k-1.summation over (j=1)] [(-1).sup.k+j+1][4.sup.j][[pi].sup.2k-2j] (2k - 2j)! D(2j)]. (36)

Integrating on each side of (35) from 0 to [pi], we conclude that

D(2k) = [(-1).sup.k+1][[pi].sup.2k]/2 x (2k + 1)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/(2k - 2j + 1)! D(2j). (37)

Integrating term-by-term on each side of (35) from 0 to x, we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (38)

Taking x = [pi]2 in (38) we conclude that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (39)

Integrating on each side of (38) from 0 to [pi], we get

[zeta](2k) = [(-1).sup.k+1][[pi].sup.2k] 2 x (2k)! + [k.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/(2k - 2j)! D(2j). (40)

The proof of Theorem 2 is complete.

Theorem 3. For k [member of] N, we have

[sigma](2k + 1) = [(-1).sup.k]/2 x (2k)! [([pi]/2).sup.2k+1] [k.summation over (j=1)] [(-1).sup.k+j]/(2k - 2j + 1)! [([pi]/2).sup.2k-2j+1] [delta](2j), (41)

[delta](2k) = [(-1).sup.k+1][[pi].sup.2k]/8 x (2k - 1)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j] 2 x (2k - 2j)! [delta](2j), (42)

[delta](2k) = [(-1).sup.k]/2 x (2k - 1)! [([pi]/2).sup.2k] + [k-1.summation over (j=1)] [(-1).sup.k+j]/(2k - 2j)! [([pi]/2).sup.2k-2j] [delta](2j), (43)

[delta](2k) = [(-1).sup.k+1][[pi].sup.2k]/4 x (2k)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/ (2k - 2j + 1)! [delta](2j). (44)

Proof. Define the function h by

h(x) = (x) = |x| x [member of] [-[pi]; [pi]].

Easy computation reveals the Fourier series of f on [-[pi], [pi]].

|x| = [pi]/2 - 4/[pi] [[infinity].summation over (n=1)] cos(2n - 1)x/[(2n - 1).sup.2], x [member of] [-[pi], [pi]]. (45)

Taking in (45) x = 0, we get

[delta](2) = [[infinity].summation over (n=1)] 1/[(2n - 1).sup.2] = [[pi].sup.2]/8.

Thus we have from (45)

[pi]/4 x - [delta](2) = - [[infinity].summation over (n=1)] cos(2n - 1)x/[(2n - 1).sup.2], x [member of] [0, [pi]]. (46)

Integrating term-by-term, we have from (46)

[pi]/4 x [x.sup.2]2! - [delta](2)x = - [[infinity].summation over (n=1)] sin(2n - 1)x/ [(2n - 1).sup.3], x [member of] [0, [pi]].

Clearly, if we integrate 2k - 1 times on each side of (46) from 0 to x, then we obtain

[pi]/4 x [x.sup.2k]/(2k)! + [k.summation over (j=1)] [(-1).sup.j][x.sup.2k-2j+1]/(2k - 2j + 1)! [delta](2j) = [(-1).sup.k] [[infinity].summation over (n=1)] sin(2n - 1)x [(2n - 1).sup.2k+1], x [member of] [0, [pi]]. (47)

Taking in (47) x = [pi]/2 we conclude that

[delta](2k + 1) = [(-1).sup.k]/2 x (2k)! [([pi]/2).sup.2k+1] + [k.summation over (j=1)] [(-1).sup.k+j]/(2k - 2j + 1)! [([pi]/2).sup.2k-2j+1] [delta](2j). (48)

Integrating on each side of (47) from 0 to 1/4, we conclude that

[delta](2k) = [(-1).sup.k+1][[pi].sup.2k]/8 x (2k - 1)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1][[pi]2k-2j]/2 x (2k - 2j)! [delta](2j). (49)

Integrating term-by-term on each side of (47) from 0 to x, we obtain that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (50)

Taking x = [pi]/2 in (50), we conclude that

[delta](2k) = [(-1).sup.k]/2 x (2k - 1)! [([pi]/2).sup.2k] + [k-1.summation over (j=1)] [(-1).sup.k+j]/(2k - 2j)! [([pi]/2).sup.2k-2j] [delta](2j). (51)

Integrating on each side of (50) from 0 to [pi], we conclude that

[delta](2k) = [(-1).sup.k+1][[pi].sup.2k]/4 x (2k)! + [k-1.summation over (j=1)] [(-1).sup.k+j+1][[pi].sup.2k-2j]/(2k - 2j + 1)! [delta](2j). (52)

The proof of Theorem 3 is complete.

Remark. Form recurrence formula (42), (43) or (44), we can obtain values of [delta](2k), for examples,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

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Suling Zhang ([dagger]) and Chaoping Chen ([double dagger])

([dagger]). Department of Basic Courses, Jiaozuo University, Jiaozuo, Henan, China

([double dagger]). College of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, Henan, China

(1) The authors were supported in part by NNSF (#10001016) of CHINA, SF for the Prominent Youth of Henan Province (#0112000200), SF of Henan Innovation Talents at Universities, Doctor Fund of Jiaozuo Institute of Technology, CHINA