Rapidly Converging Series for 𝜁(2𝑛+1) from Fourier Series

Abstract

Ever since Euler first evaluated 𝜁(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.

1. Introduction and Preliminaries

The Riemann zeta function 𝜁(𝑠) is defined by (see, e.g., [1, p. 164]) 𝜁(𝑠)∶=^∞_𝑛=11𝑛^𝑠forℜ(𝑠)>1. (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: 𝜁(𝑠)=_𝑝1−𝑝^−𝑠−1forℜ(𝑠)>1, (2) where the product is taken over all primes.

The solution of the so-called Basler problem (cf., e.g., [2], [3, p. xxii], [4, p.66], [5, pp. 197-198], and [6, p. 364]) 𝜁(2)=^∞_𝑘=11𝑘²=𝜋²6 (3) 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,…}): 𝜁(2𝑛)=(−1)^𝑛+1(2𝜋)^2𝑛2()!𝐵2𝑛^2𝑛for𝑛∈ℕ₀∶=ℕ∪{0}, (4) where 𝐵_𝑛 (𝑛∈ℕ₀) are the 𝑛th Bernoulli numbers defined by the following generating function (see, e.g., [1, p. 81]): 𝑧𝑒^𝑧=−1^∞_𝑛=0𝐵_𝑛𝑧^𝑛𝑛!for|𝑧|<2𝜋. (5) The following recursion formula 𝐵_𝑛=^𝑛_𝑘=0𝑛𝑘𝐵_𝑘(𝑛∈ℕ⧵{1}),𝐵₀=1, (6) 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 for 𝜁(2𝑛) (see, e.g., [1, p. 167], [1, Section 4.1], and [14, Theorem I]): 𝜁()=22𝑛2𝑛+1^𝑛−1_𝑘=1𝜁(2𝑘)𝜁(2𝑛−2𝑘)for𝑛∈ℕ⧵{1}, (7) which can also be used to evaluate 𝜁(2𝑛)(𝑛∈ℕ⧵{1}).

The eta function or the alternating zeta function 𝜂(𝑠) is defined by 𝜂(𝑠)∶=^∞_𝑛=1(−1)^𝑛−1𝑛^𝑠forℜ(𝑠)>0. (8) Then it is easy to have the following relation between 𝜁(𝑠) and 𝜂(𝑠): 𝜂(𝑠)=1−2^1−𝑠𝜁(𝑠)forℜ(𝑠)>1. (9)

The 𝛽-function is defined by 𝛽(𝑠)∶=^∞_𝑗=0(−1)^𝑗2𝑗+1^𝑠forℜ(𝑠)>0. (10)

Remark 1. The 𝛽-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]), Ti_𝑛(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: ^𝑛_𝑘=1𝛽()𝛽()=12𝑘−12𝑛−2𝑘+1𝑛−21−2^−2𝑛𝜁(2𝑛), (11) 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 𝜁(2)=83𝛽(1)²=𝜋²6. (12)

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 of 𝛽(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 𝛽(2𝑚+1)=(−1)^𝑚𝐸_2𝑚2^2𝑚+2()!𝜋2𝑚^2𝑚+1for𝑚∈ℕ₀, (13) where 𝐸_2𝑚 are called Euler numbers (see, e.g., [1, pp. 86–89]) defined by 2𝑒^𝑧𝑒^2𝑧=+1sech𝑧=^∞_𝑚=0𝐸_𝑚𝑧^𝑚𝑚!for|𝑧|<𝜋2. (14)

Here we present a recurrence formula for evaluation of the 𝛽(2𝑚+1)(𝑚∈ℕ) given in (10) by using Fourier series. To do this, we choose the odd 2𝜋-periodic function 𝑔_𝑚 given by 𝑔_𝑚(𝑥)∶=𝜋²𝑥^2𝑚−1−𝑥^2𝑚+1for−𝜋≤𝑥≤𝜋,𝑚∈ℕ, (15) 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 𝑔_𝑚(𝑥): 𝑥^2𝑚−1𝜋²−𝑥²=^∞_𝑛=1𝑏_𝑛sin(𝑛𝑥)for𝑥∈ℝ, (16) where 𝑏_𝑛=2𝜋^𝜋₀𝜋²𝑥^2𝑚−1−𝑥^2𝑚+1sin(𝑛𝑥)𝑑𝑥for𝑛∈ℕ. (17)

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∈ℕ₀𝑥^𝑚sin(𝑎𝑥)𝑑𝑥=−^𝑚_𝑘=0𝑚𝑘𝑥𝑘!^𝑚−𝑘𝑎^𝑘+11cos𝑎𝑥+2𝑘𝜋 (18) to get the following two involved integral formulas: 𝜋^2𝜋₀𝑥^2𝑚−1sin(𝑛𝑥)𝑑𝑥=−^𝑚−1_𝑘=0(−1)^𝑛+𝑘()!𝜋2𝑘2𝑚−12𝑘^{2𝑚+1−2𝑘}𝑛^2𝑘+1(𝑚∈ℕ_𝜋0),𝑥^2𝑚+1sin(𝑛𝑥)𝑑𝑥=−^𝑚_𝑘=0(−1)^𝑛+𝑘()!𝜋2𝑘2𝑚+12𝑘^{2𝑚+1−2𝑘}𝑛^2𝑘+1𝑚∈ℕ₀. (19)

Now setting 𝑥=𝜋/2 in (16) and using the evaluation of 𝑏_𝑛 with (19) we get 𝑏_𝑛=2(−1)^𝑛+𝑚(2𝑚+1)!𝑛^2𝑚+1+2^𝑚−1_𝑘=0(−1)^𝑛+𝑘()−×𝜋2𝑘!⋅2𝑚+12𝑘2𝑚−12𝑘_{2(𝑚−𝑘)}𝑛^2𝑘+1=2^𝑚_𝑘=1(−1)^𝑛+𝑘()!−×𝜋2𝑘2𝑚+12𝑘2𝑚−12𝑘_{2(𝑚−𝑘)}𝑛^2𝑘+1(𝑛,𝑚∈ℕ), (20) and after some simplifications we finally obtain the result stated in Theorem 2.

Theorem 2. The following recurrence formula for evaluation of 𝛽(2𝑚+1)(𝑚∈ℕ) holds. For 𝑚∈ℕ, (−1)^𝑚()()32𝑚+1!𝛽2𝑚+1=−2^2𝑚+2𝜋^2𝑚+1+^𝑚−1_𝑘=1(−1)^𝑘+1(2𝑘)!−2𝑚+12𝑘2𝑚−12𝑘×𝜋^{2𝑚−2𝑘}𝛽(2𝑘+1), (21) where the empty sum is (as usual) understood to be nil throughout this paper.

For small values of 𝑚, we have 𝛽(3)=𝜋³(5)=532,𝛽𝜋1536⁵,𝛽(7)=61𝜋184320⁷,𝛽(9)=277𝜋8257536⁹,…. (22)

Remark 3. In order to get the evaluation of 𝛽(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 𝑓(𝑥)=𝜋²𝑥−𝑥³for−𝜋≤𝑥≤𝜋, (23) which is now continuous and piecewise differentiable. Here we use the function 𝑔_𝑚(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)=𝑥^2𝑘 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 for 𝜁(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: (1−𝑥)^𝑛=^𝑛_𝑗=0(−1)^𝑗𝑛𝑗𝑥^𝑗𝑛∈ℕ₀;𝑥∈ℂ, (24) where ℂ denotes the set of complex numbers, 𝑛𝑥(1−𝑥)^𝑛−1=^𝑛_𝑗=1(−1)^𝑗+1𝑗𝑛𝑗𝑥^𝑗(𝑛∈ℕ;𝑥∈ℂ),𝑛𝑥(1−𝑥)^𝑛−1−𝑛(𝑛−1)𝑥²(1−𝑥)^𝑛−2=^𝑛_𝑗=1(−1)^𝑗+1𝑗²𝑛𝑗𝑥^𝑗(𝑛∈ℕ;𝑥∈ℂ),𝑥𝑛=𝑥𝑛⟺𝑥−11𝑛−1𝑥𝑥𝑛=1𝑛,𝑥−1𝑛−1 (25)(26)(27) where (^𝑥_𝑛) is defined, for 𝑥∈ℂ, by 𝑥𝑛=𝑥(𝑥−1)⋯(𝑥−𝑛+1)(),1().𝑛!𝑛∈ℕ𝑛=0 (28)

Lemma 5. For each 𝑚∈ℕ and 𝑥∈ℂ, one has ℒ_𝑚(𝑥)=ℛ_𝑚(𝑥), where, for convenience, ℒ_𝑚(𝑥)∶=𝑥^𝑚2𝑚+1_𝑛=0212𝑛+1𝑥^2𝑛+1−𝑥^{2𝑚−1𝑚−1}_𝑛=0212𝑛+1𝑥^2𝑛+1,ℛ_𝑚(𝑥)∶=𝑥^2𝑚−1×^2𝑚−1_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚−1(𝑥+1^𝑗−(𝑥−1)^𝑗×1𝑥^𝑗−𝑥^2𝑚+1×^2𝑚+1_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗×1𝑥^𝑗. (29)

Proof. We proceed to prove by induction on 𝑚∈ℕ. We can give a direct evaluation to check the first three ones: ℒ₁(𝑥)4=−3+2𝑥²=ℛ₁(𝑥),ℒ₂(𝑥)4=−−4153𝑥²+2𝑥⁴=ℛ₂(𝑥),ℒ₃(𝑥)4=−−435𝑥15²−43𝑥⁴+2𝑥⁶=ℛ₃(𝑥). (30)
Now assume that ℒ_𝑚(𝑥)=ℛ_𝑚(𝑥) for some 𝑚∈ℕ. Then we have to show that ℒ_𝑚+1(𝑥)=ℛ_𝑚+1(𝑥). (31) By using the induction hypothesis, we find ℒ_𝑚+1(𝑥)4=−(2𝑚+1)(2𝑚+3)+𝑥²ℛ_𝑚(𝑥). (32) In view of (31) and (32), it is enough to show that 𝒫_𝑚(𝑥)∶=𝑥²ℛ_𝑚(𝑥)−ℛ_𝑚+1(𝑥)=4()(),2𝑚+1)(2𝑚+3𝑚∈ℕ (33) where, for convenience, 𝒫_𝑚(𝑥)=𝒫_𝑚,1(𝑥)+𝒫_𝑚,2(𝑥) with 𝒫_𝑚,1(𝑥)∶=^2𝑚+3_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚+3(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+3−𝑗}−^2𝑚+1_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+3−𝑗},𝒫_𝑚,2(𝑥)∶=^2𝑚−1_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚−1(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+1−𝑗}−^2𝑚+1_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+1−𝑗}. (34)
We first try to evaluate 𝒫_𝑚,1(𝑥). We find 𝒫_𝑚,1(𝑥)=^2𝑚+3_𝑗=1(−1)^𝑗𝑗𝑗−𝑗⋅)2𝑚+32𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+3−𝑗}=^2𝑚+3_𝑗=1(−1)^𝑗(4𝑚+5)−𝑗𝑗⋅)2𝑚+3−𝑗2𝑚+2−𝑗2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+3−𝑗}. (35) Let us consider the following partial fraction: (4𝑚+5)−𝑗=2𝑚+3−𝑗2𝑚+2−𝑗4𝑚+5−2𝑚+2−𝑗4𝑚+5−𝑗2𝑚+3−𝑗+𝑗2𝑚+2−𝑗.2𝑚+3−𝑗 (36) Using (27), we have 1𝑗=12𝑚+2−𝑗2𝑚+1=12𝑚+22𝑚+22𝑚+2−𝑗𝑗,12𝑚+22𝑚+2𝑗=2𝑚+3−𝑗2𝑚+12𝑚+2−𝑗()𝑗.2𝑚+2)(2𝑚+32𝑚+3 (37) Applying the last two identities to the last expression of 𝒫_𝑚,1(𝑥), we can separate 𝒫_𝑚,1(𝑥) into six polynomials as follows: 𝒫_𝑚,1(𝑥)=⁶_𝑘=1𝛼_𝑚,𝑘(𝑥), (38) where 𝛼_𝑚,1(𝑥)∶=4𝑚+52𝑚+2^2𝑚+2_𝑗=1(−1)^𝑗𝑗×)2𝑚+2(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+3−𝑗},𝛼_𝑚,2(𝑥)∶=4𝑚+52𝑚+3^2𝑚+3_𝑗=1(−1)^𝑗+1𝑗×)2𝑚+3(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+3−𝑗},𝛼_𝑚,3(𝑥)∶=−4𝑚+5(2𝑚+2)(2𝑚+3)^2𝑚+3_𝑗=1(−1)^𝑗+1𝑗𝑗⋅)2𝑚+3(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+3−𝑗},𝛼_𝑚,4(𝑥)1∶=2𝑚+2^2𝑚+2_𝑗=1(−1)^𝑗𝑗𝑗×)2𝑚+2(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+3−𝑗},𝛼_𝑚,5(𝑥)1∶=2𝑚+3^2𝑚+3_𝑗=1(−1)^𝑗𝑗𝑗×)2𝑚+3(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+3−𝑗},𝛼_𝑚,6(𝑥)1∶=(2𝑚+2)(2𝑚+3)^2𝑚+3_𝑗=1(−1)^𝑗+1𝑗²𝑗⋅)2𝑚+3(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+3−𝑗}. (39) Choosing to use some identities in Lemma 4, we can evaluate 𝛼_𝑚,𝑘(𝑥) as follows: 𝛼_𝑚,1(𝑥)=0,𝛼_𝑚,2(𝑥)2=4−,𝛼2𝑚+3_𝑚,3(𝑥)1=−4−𝑚+1,𝛼_𝑚,4(𝑥)=−2𝑥²,𝛼_𝑚,5(𝑥)=−2,𝛼_𝑚,6(𝑥)1=2+𝑚+1+2𝑥². (40) We thus have 𝒫_𝑚,1(𝑥)=⁶_𝑘=1𝛼_𝑚,𝑘(𝑥)2=−.2𝑚+3 (41)
Similarly we find 𝒫_𝑚,2(𝑥)=⁶_𝑘=1𝛽_𝑚,𝑘(𝑥), (42) where 𝛽_𝑚,1(𝑥)=12𝑚^2𝑚_𝑗=1(−1)^𝑗𝑗𝑗)2𝑚(𝑥+1^𝑗−(𝑥−1)^𝑗×𝑥^{2𝑚+1−𝑗},𝛽_𝑚,2(𝑥)=12𝑚+1^2𝑚+1_𝑗=1(−1)^𝑗+1𝑗𝑗×)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+1−𝑗},𝛽_𝑚,3(𝑥)=12𝑚(2𝑚+1)^2𝑚+1_𝑗=1(−1)^𝑗𝑗²𝑗⋅)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+1−𝑗},𝛽_𝑚,4(𝑥)=−4𝑚+12𝑚^2𝑚_𝑗=1(−1)^𝑗𝑗×)2𝑚(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+1−𝑗},𝛽_𝑚,5(𝑥)=4𝑚+12𝑚+1^2𝑚+1_𝑗=1(−1)^𝑗𝑗×)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+1−𝑗},𝛽_𝑚,6(𝑥)=4𝑚+12𝑚(2𝑚+1)^2𝑚+1_𝑗=1(−1)^𝑗+1𝑗𝑗⋅)2𝑚+1(𝑥+1^𝑗−(𝑥−1)^𝑗𝑥^{2𝑚+1−𝑗}. (43) Similarly as in evaluating 𝛼_𝑚,𝑘(𝑥), we have 𝛽_𝑚,1(𝑥)=2𝑥²,𝛽_𝑚,2(𝑥)𝛽=2,_𝑚,3(𝑥)1=−2−𝑚−2𝑥²,𝛽_𝑚,4(𝑥)𝛼=0,_𝑚,5(𝑥)2=−4+2𝑚+1,𝛽_𝑚,6(𝑥)1=4+𝑚. (44) We thus obtain 𝒫_𝑚,2(𝑥)=⁶_𝑘=1𝛽_𝑚,𝑘(𝑥)=2.2𝑚+1 (45) Now it is easy to see that 𝒫_𝑚(𝑥)=𝒫_𝑚,1(𝑥)+𝒫_𝑚,2(𝑥)=4().2𝑚+1)(2𝑚+3 (46) This completes the proof of (33) and so does Lemma 5.

Theorem 6. One has a recurrence formula for a rapidly converging series for 𝜁(2𝑚+1). For 𝑚∈ℕ, ^𝑚_𝑘=1(−1)^𝑘+1𝜋^{2(𝑚−𝑘)}()−×2𝑘!⋅2𝑚+12𝑘2𝑚−12𝑘1−2^−2𝑘𝜁()=2𝑘+12𝜋^2𝑚4𝑚²−1−4𝜋^2𝑚×^∞_𝑛=1𝜁(2𝑛)(2𝑛+2𝑚+1)(2𝑛+2𝑚−1)4^𝑛. (47)

Proof. We find from (15), (16), and (20) that, for 𝑥∈ℝ and 𝑚∈ℕ, 𝑔_𝑚(𝑥)=2^∞_𝑛=1^𝑚_𝑘=1(−1)^𝑘𝜋^{2(𝑚−𝑘)}()!⋅−()2𝑘2𝑚+12𝑘2𝑚−12𝑘−1^𝑛𝑛^2𝑘+1sin(𝑛𝑥). (48) Here we choose a method where the sine function disappears by using the following well-known result: sin(^∞₀𝑛𝑥)𝑥𝜋𝑑𝑥=2(𝑛∈ℕ). (49) The series for 𝑔_𝑚(𝑥)/𝑥 obtained by dividing (48) by 𝑥 converges uniformly on ℝ. Indeed, the value of 𝑔_𝑚(𝑥)/𝑥 at 𝑥=0 can be considered as lim_𝑥→0sin(𝑛𝑥)/𝑥=𝑛⋅lim_𝑥→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 ^∞₀𝑔_𝑚(𝑥)𝑥𝑑𝑥=2^∞_𝑛=1^𝑚_𝑘=1(−1)^𝑘𝜋^{2(𝑚−𝑘)}()!⋅−()2𝑘2𝑚+12𝑘2𝑚−12𝑘−1^𝑛𝑛^2𝑘+1×^∞₀sin(𝑛𝑥)𝑥=𝜋^𝑚_𝑘=1(−1)^𝑘+1𝜋^{2(𝑚−𝑘)}()!×−𝜂(),2𝑘2𝑚+12𝑘2𝑚−12𝑘2𝑘+1 (50) where 𝜂 is the eta function given in (8). Now using the relationship (9) with 𝑠=2𝑘+1 gives ^𝑚_𝑘=1(−1)^𝑘+1𝜋^{2(𝑚−𝑘)}()!×−⋅2𝑘2𝑚+12𝑘2𝑚−12𝑘1−2^−2𝑘𝜁()=12𝑘+1𝜋^∞₀𝑔_𝑚(𝑥)𝑥𝑑𝑥. (51)
Since 𝑔_𝑚 is 2𝜋-periodic, we have ^∞₀𝑔_𝑚(𝑥)𝑥𝑑𝑥=^𝜋₀𝜋²𝑥^2𝑚−2−𝑥^2𝑚+^∞_𝑘=1ℐ_𝑘(𝑚)=2𝜋^2𝑚+14𝑚²+−1^∞_𝑘=1ℐ_𝑘(𝑚), (52) where, for convenience, ℐ_𝑘(𝑚)∶=_()𝜋^{(2𝑘+1)𝜋}_2𝑘−1𝜋²(𝑥−2𝑘𝜋)^2𝑚−1−(𝑥−2𝑘𝜋)^2𝑚+1𝑥𝑑𝑥. (53)
By using the binomial formula and integrating the resulting identity, we get ℐ_𝑘(𝑚)=𝜋^2𝑚+1(2𝑘)^2𝑚+1−(2𝑘)^2𝑚−1ln1+1/2𝑘1−1/2𝑘+𝜋^2𝑚+1𝒞_𝑘(𝑚), (54) where 𝒞_𝑘(𝑚)∶=(2𝑘)^{2𝑚+12𝑚+1}_𝑗=1(−1)^𝑗𝑗𝑗×)2𝑚+1(2𝑘+1^𝑗−(2𝑘−1)^𝑗12𝑘^𝑗−(2𝑘)^{2𝑚−12𝑚−1}_𝑗=1(−1)^𝑗𝑗𝑗×)2𝑚−1(2𝑘+1^𝑗−(2𝑘−1)^𝑗12𝑘^𝑗. (55)
Applying the following Maclaurin series ln1+𝑥=1−𝑥^∞_𝑛=02𝑥2𝑛+1^2𝑛+1(−1<𝑥<1) (56) to ln(1+1/2𝑘)/(1−1/2𝑘), in view of Lemma 5, we can show that 𝒞_𝑘(𝑚)=(2𝑘)^{2𝑚−1𝑚−1}_𝑛=0212𝑛+12𝑘^2𝑛+1−(2𝑘)^𝑚2𝑚+1_𝑛=0212𝑛+12𝑘^2𝑛+1. (57) We therefore have ℐ_𝑘(𝑚)=2𝜋^2𝑚+1(2𝑘)^∞2𝑚+1_𝑛=𝑚+1112𝑛+12𝑘^2𝑛+1−(2𝑘)^∞2𝑚−1_𝑛=𝑚112𝑛+12𝑘^2𝑛+1=2𝜋^2𝑚+1∞_𝑛=1112𝑛+2𝑚+12𝑘^2𝑛−^∞_𝑛=1112𝑛+2𝑚−12𝑘^2𝑛=−4𝜋^∞2𝑚+1_𝑛=11(2𝑛+2𝑚+1)(2𝑛+2𝑚−1)4^𝑛1𝑘^2𝑛. (58)
Finally, setting the last expression of ℐ_𝑘(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 𝜁(3)=4𝜋²−278𝜋²9^∞_𝑛=1𝜁(2𝑛)(2𝑛+1)(2𝑛+3)4^𝑛, (59) which, upon using 𝜁(0)=−1/2 (see, e.g., [1, p. 165, (10)]), can be expressed in a more compact form: 𝜁(3)=−8𝜋²9^∞_𝑛=0𝜁(2𝑛)(2𝑛+1)(2𝑛+3)4^𝑛. (60) The formula (59) or (60) has already been presented (cf., e.g., [29, p. 31] and [30, p. 837]).

Remark 7. Since 1<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 𝑅_𝑁 satisfying 𝑅_𝑁<8𝜋²9𝜁(2𝑁+2)(2𝑁+3)(2𝑁+5)^∞_𝑛=𝑁+114^𝑛<16𝜋²127(2𝑁+3)(2𝑁+5)4^𝑁. (61) 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=11/𝑛³, we can compute 𝜁(3) with an error 𝑅^𝑜_𝑁 satisfying 𝑅^𝑜_𝑁∶=^∞_𝑛=𝑁+11𝑛³<^∞_𝑁1𝑥³1𝑑𝑥=2𝑁², (62) from which we get the error bound 𝑅^𝑜₄₅<3⋅10⁻⁴. Scheufens [29, p. 31] estimated the error bounds |𝑅₂₅|<9⋅10⁻¹⁹ and 𝑅^𝑜₂₅<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 of 𝜁(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.

For 𝑚∈ℕ, let 𝑓 be the even 2𝜋-periodic function given by 𝑓(𝑥)=𝑥^2𝑚, 𝑥∈[−𝜋,𝜋]. Since 𝑓 is continuous and piecewise differentiable, we have 𝑓(𝑥)=12𝑎₀+^∞_𝑛=1𝑎_𝑛cos(𝑛𝑥)(𝑥∈ℝ), (63) where 𝑎₀=2𝜋^𝜋₀𝑥^2𝑚𝑑𝑥=2𝜋^2𝑚,2𝑚+1𝑎_𝑛=2𝜋^𝜋₀𝑥^2𝑚cos(𝑛𝑥)𝑑𝑥(𝑛∈ℕ). (64)(65)Evaluation of 𝑎_𝑛 in (65). Use a known indefinite integral formula (see, e.g., [28, p. 211, Entry 2.633(2)]) to get the following equation: ^𝜋₀𝑥^2𝑚()=cos𝑛𝑥𝑑𝑥_2𝑚𝑘=0𝑘𝑥𝑘!2𝑚^2𝑚−𝑘𝑛^𝑘+11sin𝑛𝑥+2𝑘𝜋^𝜋₀. (66) For convenience, let the right-hand side of (66) be denoted by ℐ. Then we have ℐ=^2𝑚_𝑘=0𝑘𝜋𝑘!2𝑚^2𝑚−𝑘𝑛^𝑘+11sin𝑛𝜋+2𝑘𝜋. (67) By considering the following relation ^2𝑚_𝑘=0𝛼(𝑘)=^𝑚_𝑘=0𝛼(2𝑘)+^𝑚−1_𝑘=0𝛼()(),2𝑘+1𝑚∈ℕ (68) we have ℐ=^𝑚_𝑘=0()!𝜋2𝑘2𝑚2𝑘^{2𝑚−2𝑘}𝑛^2𝑘+1()+sin𝑛𝜋+𝑘𝜋_{𝑚−1𝑘=0}()!𝜋2𝑘+12𝑚2𝑘+1^{2𝑚−2𝑘−1}𝑛^2𝑘+2𝜋×sin𝑛𝜋+𝑘𝜋+2=^𝑚−1_𝑘=0(−1)^𝑛+𝑘()!𝜋2𝑘+12𝑚2𝑘+1^{2𝑚−2𝑘−1}𝑛^2𝑘+2. (69) We finally get 𝑎_𝑛=2(−1)^𝑛𝜋^2𝑚−1_𝑘=0(−1)^𝑘()!×𝜋2𝑘+12𝑚2𝑘+1_{2(𝑚−𝑘)}𝑛^2𝑘+2(𝑛,𝑚∈ℕ). (70)

Setting 𝑓(𝑥)=𝑥^2𝑚 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): 𝜁(2𝑚)=(−1)^𝑚−1𝑚()!𝜋2𝑚+1^2𝑚+(−1)^𝑚×^𝑚−1_𝑘=1(−1)^𝑘+1𝜋^{2(𝑚−𝑘)}()!𝜁().2𝑚−2𝑘+12𝑘)(𝑚∈ℕ (71)

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology of the Republic of Korea (2010-0011005).