# Rapidly Converging Series forζ(2n+1) from Fourier Series

Authors: Junesang Choi
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Ever since Euler first evaluatedζ(2) and ζ(2m), numerous interesting solutions of the problem of evaluating the ζ(2m)(m∈ℕ) have appeared in the mathematical literature. Until now no simple formula analogous to the evaluation of ζ(2m)  (m∈ℕ) is known for ζ(2m+1)(m∈ℕ) or even for any special case such as ζ(3). Instead, various rapidly converging series for ζ(2m+1) have been developed by many authors. Here, using Fourier series, we aim mainly at presenting a recurrence formula for rapidly converging series for ζ(2m+1). In addition, using Fourier series and recalling some indefinite integral formulas, we also give recurrence formulas for evaluations of β(2m+1) and ζ(2m)(m∈ℕ), which have been treated in earlier works.


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## Body

## 1. Introduction and Preliminaries

The Riemann zeta functionζ(s) is defined by (see, e.g., [1, p. 164])
(1)ζ(s)∶=∑n=1∞1nsforℜ(s)>1.
The Riemann zeta function ζ(s) in (1) plays a central role in the applications of complex analysis to number theory. The number-theoretic properties of ζ(s) are exhibited by the following result known as Euler's formula, which gives a relationship between the set of primes and the set of positive integers:
(2)ζ(s)=∏p(1-p-s)-1forℜ(s)>1,
where the product is taken over all primes.The solution of the so-calledBasler problem (cf., e.g., [2], [3, p. xxii], [4, p.66], [5, pp. 197-198], and [6, p. 364])
(3)ζ(2)=∑k=1∞‍1k2=π26
was first found in 1735 by Euler (1707–1783) [7], although Jakob Bernoulli (1654–1705) and Johann Bernoulli (1667–1748) did their utmost to sum the series in (3). The former of these Bernoulli brothers did not live to see the solution of the problem, and the solution became known to the latter soon after Euler found it (see, for details, Knopp [8, p.238]). Five years later in 1740, Euler (see [9]; see also [10, pp. 137–153]) succeeded in evaluating all of ζ(2n)(n∈ℕ:={1,2,3,…}):
(4)ζ(2n)=(-1)n+1(2π)2n2(2n)!B2nforn∈ℕ0∶=ℕ∪{0},
where Bn (n∈ℕ0) are the nth Bernoulli numbers defined by the following generating function (see, e.g., [1, p. 81]):
(5)zez-1=∑n=0∞Bnznn!for|z|<2π.
The following recursion formula
(6)Bn=∑k=0n‍(nk)Bk(n∈ℕ∖{1}),B0=1,
can be used for computing Bernoulli numbers. Ever since Euler first evaluated ζ(2) and ζ(2n), numerous interesting solutions of the problem of evaluating the ζ(2n)(n∈ℕ) have appeared in the mathematical literature. Even though there were certain earlier works which gave a rather long list of papers and books together with some useful comments on the methods of evaluation of ζ(2) and ζ(2n) (see, e.g., [5, 11, 12]), the reader may be referred to the very recent work [13] which contains an extensive literature of as many as more than 70 papers.We may recall here a known recursion formula forζ(2n) (see, e.g., [1, p. 167], [1, Section 4.1], and [14, Theorem I]):
(7)ζ(2n)=22n+1∑k=1n-1‍ζ(2k)ζ(2n-2k)forn∈ℕ∖{1},
which can also be used to evaluate ζ(2n)(n∈ℕ∖{1}).The eta function or the alternating zeta functionη(s) is defined by
(8)η(s)∶=∑n=1∞(-1)n-1nsforℜ(s)>0.
Then it is easy to have the following relation between ζ(s) and η(s):
(9)η(s)=(1-21-s)ζ(s)forℜ(s)>1.Theβ-function is defined by
(10)β(s)∶=∑j=0∞(-1)j(2j+1)sforℜ(s)>0.Remark 1.
Theβ-function (see [15, p. 329]) has been denoted in several ways, such as ξ(s) (see [14]), ℒ(s) (see [15, p. 329]; see also [1, p. 404]), L(s) (see [16, p. 125] and [17]), ψ(s) (see [18]), Tin(1) (see [19, p. 190]; see also [15, p. 332]), and S(s,1/4) (see [20, p. 375]). Williams [14, p. 22, Theorem II] gave an interesting companion of the result (7) in the following form:
(11)∑k=1nβ(2k-1)β(2n-2k+1)=(n-12)(1-2-2n)ζ(2n),
which appears erroneously in Hansen [21, p. 357, Entry (54.7.1)]. Since β(1) is the well-known Gregory series for π/4 (with β(2) being the familiar Catalan constant G), by setting n=1 in (11), we immediately obtain
(12)ζ(2)=83{β(1)}2=π26.Until now no simple formula analogous to (4) is known for ζ(2m+1) or even for any special case such as ζ(3). It is not even known whether ζ(2m+1) is rational or irrational, except that the irrationality of ζ(3) was proved by Apéry [22]. But it is known that there are infinitely many ζ(2n+1) which are irrational (see [23, 24]). On the other hand, various rapidly converging series for ζ(2m+1)(m∈ℕ) have been developed by many authors (see, e.g., [25, 26]; see also [1, Chapter 4] and the references cited in the chapter). Very recently, Choi and Chen [27] gave a double inequality approximating ζ(2m+1)(m∈ℕ) by a more rapidly convergent series. Here, using Fourier series, we aim mainly at presenting a recurrence formula for rapidly converging series for ζ(2m+1)(m∈ℕ). In addition, using Fourier series, we also give recurrence formulas for evaluations of β(2m+1) and ζ(2m)(m∈ℕ).


## 2. Evaluation ofβ(2m+1) from Fourier Series

Euler proved (see, e.g., [18, p. 1071], [16, p. 125], [15, p. 330], [17, p. 372], and [19, p. 196]) that
(13)β(2m+1)=(-1)mE2m22m+2(2m)!π2m+1form∈ℕ0,
where E2m are called Euler numbers (see, e.g., [1, pp. 86–89]) defined by
(14)2eze2z+1=sechz=∑m=0∞‍Emzmm!for|z|<π2.Here we present a recurrence formula for evaluation of theβ(2m+1)(m∈ℕ) given in (10) by using Fourier series. To do this, we choose the odd 2π-periodic function gm given by
(15)gm(x)∶=π2x2m-1-x2m+1for-π≤x≤π,m∈ℕ,
which is seen to be continuous and piecewise differentiable on the set of real numbers ℝ. Now we can get the following Fourier series expansion of gm(x):
(16)x2m-1(π2-x2)=∑n=1∞bnsin(nx)forx∈ℝ,
where
(17)bn=2π∫0π(π2x2m-1-x2m+1)sin(nx)dxforn∈ℕ.Evaluation of
b
n
in (17). We use a known indefinite integral formula (see, e.g., [28, p. 211, Entry 2.633(1)]) for m∈ℕ0(18)∫‍xmsin(ax)dx=-∑k=0m‍k!(mk)xm-kak+1cos(ax+12kπ)
to get the following two involved integral formulas:
(19)π2∫0πx2m-1sin(nx)dx=-∑k=0m-1(-1)n+k(2k)!(2m-12k)π2m+1-2kn2k+1(m∈ℕ),∫0π‍x2m+1sin(nx)dx=-∑k=0m‍(-1)n+k(2k)!(2m+12k)π2m+1-2kn2k+1(m∈ℕ0).Now settingx=π/2 in (16) and using the evaluation of bn with (19) we get
(20)bn=2(-1)n+m(2m+1)!n2m+1+2∑k=0m-1(-1)n+k(2k)!·{(2m+12k)-(2m-12k)}×π2(m-k)n2k+1=2∑k=1m(-1)n+k(2k)!{(2m+12k)-(2m-12k)}×π2(m-k)n2k+1(n,m∈ℕ),
and after some simplifications we finally obtain the result stated in Theorem 2.Theorem 2.
The following recurrence formula for evaluation ofβ(2m+1)(m∈ℕ) holds. For m∈ℕ,
(21)(-1)m(2m+1)!β(2m+1)=-322m+2π2m+1+∑k=1m-1(-1)k+1(2k)!{(2m+12k)-(2m-12k)}×π2m-2kβ(2k+1),
where the empty sum is (as usual) understood to be nil throughout this paper.For small values ofm, we have
(22)β(3)=π332,β(5)=51536π5,β(7)=61184320π7,β(9)=2778257536π9,….Remark 3.
In order to get the evaluation ofβ(3) and a rapidly converging series representation of ζ(3) from Fourier series, instead of using the periodic version of x3 which is not continuous, Scheufens [29] made a good choice of the odd 2π-periodic function f given by
(23)f(x)=π2x-x3for-π≤x≤π,
which is now continuous and piecewise differentiable. Here we use the function gm(x) in (15) which is a natural modification of the Scheufens chosen function (23) to give the results in this and the next sections.Chen [17] used the even 2π-periodic function f(x)=x2k on [-π,π] to get a recurrence formula for β(2m+1). Yue and Williams [20] used residue calculus to derive a recurrence formula for β(2m+1). Butzer and Hauss [15] presented diverse single and multiple integral representations of β(m).


## 3. A Recurrence Formula for a Rapidly Converging Series forζ(2m+1)

We begin by recalling some elementary known or easily derivable formulas for the binomial coefficients as in the lemma given below.Lemma 4.
Each of the following formulas holds:(24)(1-x)n=∑j=0n(-1)j(nj)xj(n∈ℕ0;x∈ℂ),
where ℂ denotes the set of complex numbers,
(25)nx(1-x)n-1=∑j=1n(-1)j+1j(nj)xj(n∈ℕ;x∈ℂ),(26)nx(1-x)n-1-n(n-1)x2(1-x)n-2=∑j=1n(-1)j+1j2(nj)xj(n∈ℕ;x∈ℂ),(27)(xn)=xn(x-1n-1)⟺1x(xn)=1n(x-1n-1),
where (xn) is defined, for x∈ℂ, by
(28)(xn)={x(x-1)⋯(x-n+1)n!(n∈ℕ),1(n=0).Lemma 5.
For eachm∈ℕ and x∈ℂ, one has ℒm(x)=ℛm(x), where, for convenience,
(29)ℒm(x)∶=x2m+1∑n=0m22n+1(1x)2n+1-x2m-1∑n=0m-122n+1(1x)2n+1,ℛm(x)∶=x2m-1×∑j=12m-1(-1)jj(2m-1j){(x+1)j-(x-1)j}×(1x)j-x2m+1×∑j=12m+1(-1)jj(2m+1j){(x+1)j-(x-1)j}×(1x)j.Proof.
We proceed to prove by induction onm∈ℕ. We can give a direct evaluation to check the first three ones:
(30)ℒ1(x)=-43+2x2=ℛ1(x),ℒ2(x)=-415-43x2+2x4=ℛ2(x),ℒ3(x)=-435-415x2-43x4+2x6=ℛ3(x).
Now assume thatℒm(x)=ℛm(x) for some m∈ℕ. Then we have to show that
(31)ℒm+1(x)=ℛm+1(x).
By using the induction hypothesis, we find
(32)ℒm+1(x)=-4(2m+1)(2m+3)+x2ℛm(x).
In view of (31) and (32), it is enough to show that
(33)𝒫m(x)∶=x2ℛm(x)-ℛm+1(x)=4(2m+1)(2m+3)(m∈ℕ),
where, for convenience, 𝒫m(x)=𝒫m,1(x)+𝒫m,2(x) with
(34)𝒫m,1(x)∶=∑j=12m+3(-1)jj(2m+3j){(x+1)j-(x-1)j}×x2m+3-j-∑j=12m+1(-1)jj(2m+1j){(x+1)j-(x-1)j}×x2m+3-j,𝒫m,2(x)∶=∑j=12m-1(-1)jj(2m-1j){(x+1)j-(x-1)j}×x2m+1-j-∑j=12m+1(-1)jj(2m+1j){(x+1)j-(x-1)j}×x2m+1-j.
We first try to evaluate𝒫m,1(x). We find
(35)𝒫m,1(x)=∑j=12m+3(-1)jj{(2m+3j)-(2m+1j)}·{(x+1)j-(x-1)j}x2m+3-j=∑j=12m+3(-1)j(4m+5)-j(2m+3-j)(2m+2-j)(2m+1j)·{(x+1)j-(x-1)j}x2m+3-j.
Let us consider the following partial fraction:
(36)(4m+5)-j(2m+3-j)(2m+2-j)=4m+52m+2-j-4m+52m+3-j-j2m+2-j+j2m+3-j.
Using (27), we have
(37)12m+2-j(2m+1j)=12m+2(2m+22m+2-j)=12m+2(2m+2j),12m+3-j(2m+1j)=2m+2-j(2m+2)(2m+3)(2m+3j).
Applying the last two identities to the last expression of 𝒫m,1(x), we can separate 𝒫m,1(x) into six polynomials as follows:
(38)𝒫m,1(x)=∑k=16‍αm,k(x),
where
(39)αm,1(x)∶=4m+52m+2∑j=12m+2(-1)j(2m+2j)×{(x+1)j-(x-1)j}x2m+3-j,αm,2(x)∶=4m+52m+3∑j=12m+3(-1)j+1(2m+3j)×{(x+1)j-(x-1)j}x2m+3-j,αm,3(x)∶=-4m+5(2m+2)(2m+3)∑j=12m+3(-1)j+1j(2m+3j)·{(x+1)j-(x-1)j}×x2m+3-j,αm,4(x)≔12m+2∑j=12m+2(-1)jj(2m+2j)×{(x+1)j-(x-1)j}x2m+3-j,αm,5(x)∶=12m+3∑j=12m+3(-1)jj(2m+3j)×{(x+1)j-(x-1)j}x2m+3-j,αm,6(x)≔1(2m+2)(2m+3)∑j=12m+3(-1)j+1j2(2m+3j)·{(x+1)j-(x-1)j}×x2m+3-j.
Choosing to use some identities in Lemma 4, we can evaluate αm,k(x) as follows:
(40)αm,1(x)=0,αm,2(x)=4-22m+3,αm,3(x)=-4-1m+1,αm,4(x)=-2x2,αm,5(x)=-2,αm,6(x)=2+1m+1+2x2.
We thus have
(41)𝒫m,1(x)=∑k=16‍αm,k(x)=-22m+3.
Similarly we find(42)𝒫m,2(x)=∑k=16‍βm,k(x),
where
(43)βm,1(x)=12m∑j=12m(-1)jj(2mj){(x+1)j-(x-1)j}×x2m+1-j,βm,2(x)=12m+1∑j=12m+1(-1)j+1j(2m+1j)×{(x+1)j-(x-1)j}x2m+1-j,βm,3(x)=12m(2m+1)∑j=12m+1(-1)jj2(2m+1j)·{(x+1)j-(x-1)j}x2m+1-j,βm,4(x)=-4m+12m∑j=12m(-1)j(2mj)×{(x+1)j-(x-1)j}x2m+1-j,βm,5(x)=4m+12m+1∑j=12m+1(-1)j(2m+1j)×{(x+1)j-(x-1)j}x2m+1-j,βm,6(x)=4m+12m(2m+1)∑j=12m+1(-1)j+1j(2m+1j)·{(x+1)j-(x-1)j}x2m+1-j.
Similarly as in evaluating αm,k(x), we have
(44)βm,1(x)=2x2,βm,2(x)=2,βm,3(x)=-2-1m-2x2,βm,4(x)=0,αm,5(x)=-4+22m+1,βm,6(x)=4+1m.
We thus obtain
(45)𝒫m,2(x)=∑k=16‍βm,k(x)=22m+1.
Now it is easy to see that
(46)𝒫m(x)=𝒫m,1(x)+𝒫m,2(x)=4(2m+1)(2m+3).
This completes the proof of (33) and so does Lemma 5.Theorem 6.
One has a recurrence formula for a rapidly converging series forζ(2m+1). For m∈ℕ,
(47)∑k=1m(-1)k+1π2(m-k)(2k)!·{(2m+12k)-(2m-12k)}×(1-2-2k)ζ(2k+1)=2π2m4m2-1-4π2m×∑n=1∞ζ(2n)(2n+2m+1)(2n+2m-1)4n.Proof.
We find from (15), (16), and (20) that, for x∈ℝ and m∈ℕ,
(48)gm(x)=2∑n=1∞‍[∑k=1m(-1)kπ2(m-k)(2k)!·{(2m+12k)-(2m-12k)}(-1)nn2k+1∑k=1m(-1)kπ2(m-k)(2k)!]sin(nx).
Here we choose a method where the sine function disappears by using the following well-known result:
(49)∫0∞‍sin(nx)xdx=π2(n∈ℕ).
The series for gm(x)/x obtained by dividing (48) by x converges uniformly on ℝ. Indeed, the value of gm(x)/x at x=0 can be considered as limx→0sin(nx)/x=n·limx→0sin(nx)/(nx)=n and then one may use the Weierstrass M-test. We therefore apply termwise integration to the resulting series and use (49) to get
(50)∫0∞‍gm(x)xdx=2∑n=1∞‍[(-1)nn2k+1∑k=1m‍(-1)kπ2(m-k)(2k)!·{(2m+12k)-(2m-12k)}(-1)nn2k+1∑k=1m‍]×∫0∞‍sin(nx)x=π∑k=1m‍(-1)k+1π2(m-k)(2k)!×{(2m+12k)-(2m-12k)}η(2k+1),
where η is the eta function given in (8). Now using the relationship (9) with s=2k+1 gives
(51)∑k=1m(-1)k+1π2(m-k)(2k)!×{(2m+12k)-(2m-12k)}·(1-2-2k)ζ(2k+1)=1π∫0∞gm(x)xdx.
Sincegm is 2π-periodic, we have
(52)∫0∞gm(x)xdx=∫0π(π2x2m-2-x2m)+∑k=1∞ℐk(m)=2π2m+14m2-1+∑k=1∞ℐk(m),
where, for convenience,
(53)ℐk(m)∶=∫(2k-1)π(2k+1)π‍π2(x-2kπ)2m-1-(x-2kπ)2m+1xdx.
By using the binomial formula and integrating the resulting identity, we get(54)ℐk(m)=π2m+1{(2k)2m+1-(2k)2m-1}ln1+1/2k1-1/2k+π2m+1𝒞k(m),
where
(55)𝒞k(m)∶=(2k)2m+1∑j=12m+1(-1)jj(2m+1j)×{(2k+1)j-(2k-1)j}(12k)j-(2k)2m-1∑j=12m-1(-1)jj(2m-1j)×{(2k+1)j-(2k-1)j}(12k)j.
Applying the following Maclaurin series(56)ln1+x1-x=∑n=0∞22n+1x2n+1(-1<x<1)
to ln(1+1/2k)/(1-1/2k), in view of Lemma 5, we can show that
(57)𝒞k(m)=(2k)2m-1∑n=0m-122n+1(12k)2n+1-(2k)2m+1∑n=0m22n+1(12k)2n+1.
We therefore have
(58)ℐk(m)=2π2m+1[(2k)2m+1∑n=m+1∞‍12n+1(12k)2n+1-(2k)2m-1∑n=m∞‍12n+1(12k)2n+1]=2π2m+1[∑n=1∞‍12n+2m+1(12k)2n-∑n=1∞‍12n+2m-1(12k)2n]=-4π2m+1∑n=1∞‍1(2n+2m+1)(2n+2m-1)4n1k2n.
Finally, setting the last expression ofℐk(m) in (52) and considering (51) yield our desired identity (47). This completes the proof of Theorem 6.The special case of (47) when m=1 yields
(59)ζ(3)=4π227-8π29∑n=1∞ζ(2n)(2n+1)(2n+3)4n,
which, upon using ζ(0)=-1/2 (see, e.g., [1, p. 165, (10)]), can be expressed in a more compact form:
(60)ζ(3)=-8π29∑n=0∞ζ(2n)(2n+1)(2n+3)4n.
The formula (59) or (60) has already been presented (cf., e.g., [29, p. 31] and [30, p. 837]).Remark 7.
Since1<ζ(2n)≤ζ(2)=π2/6<2(n∈ℕ), using the Nth partial sum of the infinite series in (59) or (60), we can compute ζ(3) with an error RN satisfying
(61)|RN|<8π29ζ(2N+2)(2N+3)(2N+5)∑n=N+1∞14n<16π2271(2N+3)(2N+5)4N.
Using the 45th partial sum in (60) we have an error bound |R45|<6·10-31, and approximately the value ζ(3)=1.20205690315959428539973816151. For comparison, using the Nth partial sum of ζ(3)=∑n=1∞1/n3, we can compute ζ(3) with an error RNo satisfying
(62)RNo∶=∑n=N+1∞‍1n3<∫N∞‍1x3dx=12N2,
from which we get the error bound R45o<3·10-4. Scheufens [29, p. 31] estimated the error bounds |R25|<9·10-19 and R25o<8·10-4. So it is easy to see that the original series of ζ(3) converges very slowly while the series representation (59) or (60) of ζ(3) converges very rapidly.
Srivastava and Choi [1, Chapter 3] presented a rather extensive collection of closed-form sums of series involving the zeta functions such as (59) or (60), together with an interesting historical introduction. In fact, the formula (47) may be obtained in a totally different way (see, e.g., [1, p. 259, (71)]).


## 4. Evaluation ofζ(2m) from Fourier Series

There have been earlier works (see, e.g., [17, 29, 31–34]) in which the authors evaluated ζ(2m) by using Fourier series. Here, for completeness, we also do the same thing. Yet we may very carefully emphasize that, when the involved coefficient in Fourier series is computed, its computation becomes a little easier by using a known indefinite integral formula.Form∈ℕ, let f be the even 2π-periodic function given by f(x)=x2m, x∈[-π,π]. Since f is continuous and piecewise differentiable, we have
(63)f(x)=12a0+∑n=1∞ancos(nx)(x∈ℝ),
where
(64)a0=2π∫0πx2mdx=2π2m2m+1,(65)an=2π∫0πx2mcos(nx)dx(n∈ℕ).Evaluation ofanin (65). Use a known indefinite integral formula (see, e.g., [28, p. 211, Entry 2.633(2)]) to get the following equation:
(66)∫0πx2mcos(nx)dx=∑k=02mk!(2mk)x2m-knk+1sin(nx+12kπ)|0π.
For convenience, let the right-hand side of (66) be denoted by ℐ. Then we have
(67)ℐ=∑k=02mk!(2mk)π2m-knk+1sin(nπ+12kπ).
By considering the following relation
(68)∑k=02mα(k)=∑k=0mα(2k)+∑k=0m-1α(2k+1)(m∈ℕ),
we have
(69)ℐ=∑k=0m(2k)!(2m2k)π2m-2kn2k+1sin(nπ+kπ)+∑k=0m-1(2k+1)!(2m2k+1)π2m-2k-1n2k+2×sin(nπ+kπ+π2)=∑k=0m-1(-1)n+k(2k+1)!(2m2k+1)π2m-2k-1n2k+2.
We finally get
(70)an=2(-1)nπ2∑k=0m-1(-1)k(2k+1)!(2m2k+1)×π2(m-k)n2k+2(n,m∈ℕ).Settingf(x)=x2m in (63) and setting x=π and applying (64) and (70) to the resulting identity, we finally get the following recurrence formula for evaluation of ζ(2m):
(71)ζ(2m)=(-1)m-1m(2m+1)!π2m+(-1)m×∑k=1m-1(-1)k+1π2(m-k)(2m-2k+1)!ζ(2k)(m∈ℕ).



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