Series Representations for the Apery Constant (3) Involving the Values

The Ramanujan Journal (2019) 48:477–494 https://doi.org/10.1007/s11139-018-0081-0

Series representations for the Apery constant (3) involving the values (2n)

Cezar Lupu1 · Derek Orr1

Received: 21 December 2016 / Accepted: 23 August 2018 / Published online: 17 January 2019 © Springer Science+Business Media, LLC, part of Springer Nature 2019

Abstract In this note, using the well-known series representation for the Clausen function, we also provide some new representations of Apery’s constant ζ(3). In addition, by an idea from De Amo et al. (Proc Am Math Soc 139:1441–1444, 2011) we derive some new rational series representations involving even zeta values and central binomial coefﬁcients. These formulas are expressed in terms of odd and even values of the Riemann zeta function and odd values of the Dirichlet beta function. In particular cases, we recover some well-known series representations of π.

Keywords Riemann zeta function · Clausen integral · Rational zeta series representations · Apery’s constant

Mathematics Subject Classiﬁcation Primary 41A58 · 41A60; Secondary 40B99

1 Introduction and preliminaries

In 1734, Leonard Euler produced a sensation when he proved the following formula:

∞ π 2 1 = . n2 6 n=1

Later, in 1740, Euler generalized the above formula for even positive integers:

B Cezar Lupu [email protected]; [email protected] Derek Orr [email protected]

1 Department of Mathematics, University of Pittsburgh, 301 Thackeray Hall, Pittsburgh, PA 15260, USA 123 478 C. Lupu, D. Orr

2n−1 2n + B 2 π ζ(2n) = (−1)n 1 2n , (1) (2n)!

∞ 1 where ζ(s) = is the celebrated Riemann zeta function. The coefﬁcients Bn n=1 ns are the so-called Bernoulli numbers and they are deﬁned in the following way:

∞ z B = n zn, |z| < 2π. ez − 1 n! n=0

An elementary proof of (1) can be found in [4]. In [19], De Amo et al. produced another proof of Euler’s formula (1) using the Taylor series expansion of the tangent function and Fubini’s theorem. In this paper, using similar ideas but for other functions, we provide a new proof for the Clausen acceleration formula that will serve as an application in displaying some fast representations for Apery’s constant ζ(3). Moreover, the Taylor series expansion for the secant and cosecant functions combined with some other integration techniques will give us some interesting rational series representations involving ζ(2n) and binomial coefﬁcients. Also, we display some particular cases of such series. Some of them are well-known representations of π. For the sake of completeness we display the Taylor series for the tangent, cotangent, secant and cosecant functions:

∞ n−1 2n 2n (−1) 2 (2 − 1)B − π tan x = 2n x2n 1, |x| < (2) (2n)! 2 n=1 ∞ n 2n (−1) 2 B − cot x = 2n x2n 1, |x| <π (3) (2n)! n=0 ∞ (−1)n E π sec x = 2n x2n, |x| < (4) (2n)! 2 n=0 ∞ n+1 2n−1 (−1) 2(2 − 1)B − csc x = 2n x2n 1, |x| <π, (5) (2n)! n=0 where En are the Euler numbers, and Bn the Bernoulli numbers. The Riemann zeta function ζ(s) and the Hurwitz (generalized) function ζ(s, a) are deﬁned by ⎧ ∞ ∞ ⎪ 1 1 1 ⎪ = Re(s)>1, ⎨ ns 1 − 2−s (2n − 1)s n=1 n=1 ζ(s) := ∞ (6) ⎪ n−1 ⎪ 1 (−1) ⎩⎪ Re(s)>0, s = 1, 1 − 21−s ns n=1 123 Series representations for the Apery constant... 479 and ∞ 1 ζ(s, a) := , Re(s)>1; a = 0, − 1, − 2,... (7) (n + a)s n=0 Both of them are analytic over the whole complex plane, except s = 1, where they have a simple pole. Also, from the two deﬁnitions above, one can observe that 1 1 ζ(s) = ζ(s, 1) = ζ s, = 1 + ζ(s, 2). 2s − 1 2

Clausen’s function (or Clausen’s integral, see [14]) Cl2(θ) is deﬁned by

∞ sin kθ θ x Cl (θ) := =− log 2sin dx. 2 k2 2 k=1 0

This integral was considered for the ﬁrst time by Clausen in 1832 [14], and it was investigated later by many authors [8–13,22,23,29,30]. The Clausen functions are very important in mathematical physics. Some well-known properties of the Clausen function include periodicity in the following sense:

Cl2(2kπ ± θ) = Cl2(± θ) =±Cl2(θ).

Moreover, it is quite clear from the deﬁnition that Cl2(kπ) = 0fork integer. For example, for k = 1 we deduce

π π x 2 π log 2sin dx = 0, log(sin x) dx =− log 2. 0 2 0 2 π 3π By periodicity we have Cl =−Cl = G, where G is the Catalan 2 2 2 2 constant deﬁned by

∞ (−1)n G := ≈ 0.9159 ... (2n + 1)2 n=0

More generally, one can express the above integral as the following:

θ 1 log(sin x)dx =− Cl (2θ)− θ log 2, 2 2 0 θ 1 log(cos x)dx =− Cl2(π − 2θ)− θ log 2, 0 2 θ log(1 + cos x)dx = 2Cl2(π − θ)− θ log 2, 0 123 480 C. Lupu, D. Orr and

θ π log(1 + sin x)dx = 2G − 2Cl2 + θ − θ log 2. 0 2

The Dirichlet beta function is deﬁned as

∞ (−1)n β(s) := . (2n + 1)s n=0

Alternatively, one can express the beta function in terms of Hurwitz zeta function by the following formula valid in the whole complex s-plane,

1 β(s) = (ζ(s, 1/4) − ζ(s, 3/4)) . 4s

β( ) = β( ) = π3 β( + ) = Clearly, 2 G (Catalan’s constant), 3 32 , and 2n 1 n 2n+1 (−1) E2nπ , where En are the Euler numbers mentioned above which are given 4n+1(2n)! ∞ 2 E by the Taylor series = n tn. et + e−t n! n=0

1.1 Organization of the paper

This note is organized as follows. In Sect. 2, we recall some well-known series representations for Apery constant ζ(3). In Sect. 3.1 of the next section, we give a new proof for the Clausen acceleration formula. This formula serves as an ingredient for Theo- rem 3.2 where we provide some “fast” converging series representations for Apery’s constant ζ(3). On the other hand, in Sect. 3.2, we produce some rational representations involving ζ(2n) and binomial coefﬁcients and we also provide some interesting series summations as corollaries.

2 Some known rational zeta series for Apery’s constant

ζ( + ) In [27], Strivastava et al. derive series representations for 2n 1 by evaluating π/ω s−1 ,ω ≥ the integral 0 t cot tdt, s 2 integers in two different ways. One of the ways involves the generalized Clausen functions. When they are evaluated in terms of ζ(2n + 1) one obtains the following formula for ζ(3), ∞ 2π 2 ζ(2n) ζ(3) = log 2 + 2 . (8) 9 (2n + 3)4n n=0 123 Series representations for the Apery constant... 481

Another remarkable result which led to Apery’s proof of the irrationality of ζ(3) is given by the rapidly convergent series

∞ 5 − 1 ζ(3) = (−1)n 1 . (9) 2 3 2n n=1 n n

Moreover, in [18], Cvijovic and Klinowski derive the following formula:

∞ π 2 (2n + 5)ζ(2n) ζ(3) =− . (10) 3 (2n + 1)(2n + 2)(2n + 3)22n n=0

This formula is related to the one found by Ewell [21],

∞ 4π 2 ζ(2n) ζ(3) =− . (11) 7 (2n + 1)(2n + 2)22n n=0

3 Main results

3.1 Clausen acceleration formula and some representations of (3)

We provide a new proof for the classical Clausen acceleration formula [11,20,24].

Proposition 3.1 We have the following representation for the Clausen function Cl2(θ),

∞ Cl (θ) ζ(2n) 2 = 1 − log |θ|+ θ 2n, |θ| < 2π. (12) θ (2π)2nn(2n + 1) n=1

Proof Integrating by parts the function xycot(xy) we have

π π 2 π π y 2 xycot(xy) dx = log sin − log(sin(xy)) dx 2 2 0 0 π π y 1 π y u = log sin − log sin du 2 2 2y 2 0 π π y π 1 = log sin + log 2 + Cl2(π y) 2 2 2 2y ∞ π π y y2 π 1 = − + + (π y) log 1 2 log 2 Cl2 2 2 = 4k 2 2y k 1 2 n ∞ ∞ y π π y π 4k2 π 1 = log − + log 2 + Cl (π y), 2 2 2 n 2 2y 2 k=1 n=1 123 482 C. Lupu, D. Orr

∞ 2 (πx) = πx − x where we have used the product formula for sine, sin k=1 1 k2 . ( ) θ θ = − On the other hand, using the formula 3 written in the form cot 1 ∞ ζ(2n) θ 2n 2 n=1 π2n ,wehave π ∞ π 2n+1 2 π ζ(2n) xycot(xy)dx = − 2 2 y2n. 2 π 2n(2n + 1) 0 n=1

Therefore, we obtain ∞ π 2n+1 ∞ π ζ(2n) π π ζ(2n) 1 − 2 2 y2n = log(π y) − y2n + Cl (π y). 2 π 2n(2n + 1) 2 2 n4n 2y 2 n=1 n=1

This implies that ∞ ∞ π ζ(2n) π π ζ(2n) Cl (π y) = 2y − π y2n − log(π y) + y2n , 2 2 (2n + 1)4n 2 2 n4n n=1 n=1 which, after some computations, is equivalent to

∞ ζ(2n)(π y)2n Cl (π y) = π y − π y log(π y) + π y . 2 4nn(2n + 1)π 2n n=1

Setting α = π y, we obtain our result.

Remark θ = π ( π ) = In particular case of 2 , using the fact that Cl2 2 G, we obtain

∞ ζ(2n) 2G π = − 1 + log . (13) n(2n + 1)16n π 2 n=1

In computations, the following accelerated peeled form [6]isused

∞ Cl (θ) θ 2 2π 2π + θ ζ(2n) − 1 θ 2n 2 = 3−log |θ| 1 − − log + . θ 4π 2 θ 2π − θ n(2n + 1) 2π n=1 (14) It is well-known that ζ(n) − 1 converges to zero rapidly for large values of n.In [30], Wu et al. derive the following representation for the Clausen function Cl2(θ),

∞ θ 2ζ(2n) + Cl (θ) = θ − θ log 2sin − θ 2n 1. (15) 2 2 (2n + 1)(2π)2n n=1 123 Series representations for the Apery constant... 483

It is interesting to see that integrating the above formula from 0 to π/2wehavethe following representation for ζ(3) (see [12]), ∞ 4π 2 1 2G ζ(2n) ζ(3) = + − . (16) 35 2 π (n + 1)(2n + 1)16n n=1

The following result will provide some new representations for Apery’s constant ζ(3). The main ingredients in the proof of this next result are Clausen acceleration formulae and Fubini’s theorem.

Theorem 3.2 We have the following series representations:

∞ 4π2 3 π ζ(2n) ζ(3) = − log + , ( + )( + ) n (17) 35 2 2 = n n 1 2n 1 16 n 1 ∞ 64 8π2 4 π ζ(2n) ζ(3) =− β(4) + − log + 3 , (18) 3π 9 3 2 n(2n + 1)(2n + 3)16n n=1 and ∞ 64 16π 2 1 3G ζ(2n) ζ(3) =− β(4) + + − 3 , (19) 3π 27 2 π (2n + 1)(2n + 3)16n n=1 where G is the Catalan constant, and β(s) is the Dirichlet beta function.

Proof For (17), integrating (12) from 0 to π/2 and using Fubini’s theorem, we have

π π π ∞ 2 2 2 ζ(2n) (y) y = (y − y y) y + y2n+1 y Cl2 d log d 2n d 0 0 0 (2π) n(2n + 1) n=1 π 2 2 π π π 1 2 = − log − y dy 8 8 2 2 0 ∞ π + ζ(2n) 2n 2 + 2 , (2π)2nn(2n + 1)(2n + 2) n=1 which is equivalent to π 2 ∞ 2 π 3 π ζ(2n) Cl (y) dy = − log + . 2 8 2 2 n(n + 1)(2n + 1)16n 0 n=1 123 484 C. Lupu, D. Orr

Alternatively, we can integrate the Clausen function using its deﬁnition given in the introduction and changing the order of integration as follows:

π π y 2 2 x Cl2(y) dy =− log 2sin dxdy 0 0 0 2 π π 2 2 x =− log 2sin dydx 0 x 2 π 2 π x =− log 2sin dx 0 2 2 π π 2 2 x + x log 2dx + x log sin dx. 0 0 2

= Using the deﬁnition of the Clausen function again and, after the substitution x 2u, π/4 35 πG π 2 using the identity u log(sin(u)) du = ζ(3) − − log 2 (see [12]), 0 128 8 32 we have

π 2 2 2 π π π 35 πG π Cl2(y) dy = Cl2 + log 2 + 4 ζ(3) − − log 2 0 2 2 8 128 8 32 35 = ζ(3). 32

Setting the two results equal to each other and solving for ζ(3) gives us our result. For (18), we proceed similarly to the previous method. We will begin with

π2/4 √ π2/4 √ √ √ Cl2( y) dy = y − y log( y) dy 0 0 2 ∞ π /4 ζ(2n)yn+1/2 + y 2n d 0 n(2n + 1)(2π) n=1 2 π 3 π 3 π 1 π /4 √ = − log − y dy 12 12 2 3 0 ∞ π2 + / ζ(2n) n 3 2 + 4 , n(2n + 1)(n + 3/2)(2π)2n n=1 which is equivalent to

2 ∞ π /4 √ π 3 4 π ζ(2n) Cl ( y) dy = − log + 3 . 2 12 3 2 n(2n + 1)(2n + 3)16n 0 n=1 123 Series representations for the Apery constant... 485

On the other hand, using the deﬁnition of the Clausen function and changing the order of integration, we see

π2/4 √ π 2 G 1 2 1 3 Cl2( y) dy = + 72πζ(3) − 192π G + ψ3 − ψ3 , 0 4 768 4 4

n+1 where ψ3 is the trigamma function. Using the identity ψn(z) = (−1) n!ζ(n + 1, z) where ζ(k, z) is the Hurwitz zeta function, and the relationship between the Hurwitz zeta function and the beta function, we have that

2 π /4 √ 3π Cl2( y) dy = ζ(3) + 2β(4). 0 32

Setting this result equal to the previous result of the integral and solving for ζ(3), we see (18) is indeed true. For (19), instead of integrating the Clausen function using (12), we will integrate it using (15). This gives us

π2 √ π 3 4 1 2 4 Cl2( y) dy = − 72πζ(3) − 192π G + 3!4 β(4) 0 12 384 π 3 ∞ ζ( ) − 2n . 2 (2n + 1)(2n + 3)16n n=1

Setting this result equal to the previous Clausen formula yields

3π 3π 9π + ζ(3) = ζ(3) 32 16 32 ∞ π 3 π 2G π 3 ζ(2n) = + − 6β(4) − , 12 2 2 (2n + 1)(2n + 3)16n n=1 and from here, (19) follows.

Remark To prove (17), one could solve for G in (13) and plug it into (16) and rear- π range. Also, to prove (19), one could solve for log 2 in (13) and plug that into (18) and rearrange. Further, if we integrate (14), we arrive at a rapidly converging series representation for ζ(3), that is 2π 2 ζ(3) = 9 + 138 log 2 − 18 log 3 − 50 log 5 − 2logπ 35 ∞ ζ(2n) − 1 + 2 . (20) n(2n + 1)(n + 1)16n n=1 123 486 C. Lupu, D. Orr

3.2 Rational series representations involving (2n) and binomial coefficients

As it has been already been highlighted in [6] one can relate the rational ζ -series with various Dirichlet L-series. A rational ζ -series can be accelerated for computational purposes provided that one solves the exact sum

∞ qn , an n=2 where a = 2, 3, 4,.... In fact, it has been already highlighted in [6] we shall call rational ζ -series of a real number x, the following representation:

∞ x = qnζ(n, m), n=2 where qn is a rational number and ζ(n, m) is the Hurwitz zeta function. For m > 1 integer, one has ⎛ ⎞ ∞ m−1 ⎝ −n⎠ x = qn ζ(n) − j . n=2 j=1

In the particular case m = 2, one has the following series representations:

∞ 1 = (ζ(n) − 1) n=2 ∞ 1 1 − γ = (ζ(n) − 1) n n=2 ∞ 1 log 2 = (ζ(2n) − 1), n n=2 where γ is the Euler–Mascheroni constant. For other rational zeta series representations we recommend [1–3,7,12,15–17,25–27]. In what follows we prove soem new rational zeta series representations involving binomial coefﬁcients.

Theorem 3.3 The following representation is true: ⎧ ⎪ 1 ∞ ⎨ m odd, ζ(2n) 2n = m (21) n ⎪ 1 1 = n4 m ⎩ 2ζ(m) 1 − − 1 m even . n 1 m 2m 123 Series representations for the Apery constant... 487

Proof We start by integrating xy csc(xy) two different ways. Applying integration by parts, L’Hospital’s rule, and properties of logarithm, we ﬁnd

π 2 π π y π y xy csc(xy) dx =− log 1 + cos − log sin 0 2 2 2 π π 2 2 + log(1 + cos(xy)) dx − log(sin(xy)) dx 0 0 π =− d (π − α) + 1 ( α) α 2Cl2 Cl2 2 2 d 2 1 1 + 2Cl (π − α) + Cl (2α) , y 2 2 2

π y where α = . Applying (8) to the result and simplifying, we get 2

π ∞ π 2n 2n ∞ 2n 2 ζ(2n)( ) (2 − y) π ζ(2n)(π y) π xy csc(xy) dx = π 2 − − n(2π)2n 2 n(2π)2n 2 0 n= n= 1 1 2π π + 1 − log − log(2 − y) y 2 ∞ π ζ(2n)( )2n+1(2 − y)2n+1 + 2 2 y n(2n + 1)(2π)2n n=1 ∞ ζ( )(π )2n+1 + 1 2n y . 2y (2π)2n n=1

On the other hand, we can apply Fubini’s theorem and integrate its power series term by term. Thus, we will have

π π ∞ n+1 n 2 2 (−1) (4 − 2)B xy csc(xy) dx = 2n (xy)2n dx (2n)! 0 0 n=0 ∞ n+1 n π 2n+1 (−1) (4 − 2)B2n( ) = 2 y2n. (2n + 1)(2n)! n=0

Setting the two results equal to each other and simplifying more, we see

∞ n ∞ ζ(2n) 2 2n (−1)k ζ(2k) yk − y2k n4n k 2k (2k + 1)4k n=1 k=0 k=1 ∞ 2 1 − + (1 − log π)+ 2 yk 1 y k2k k=1 123 488 C. Lupu, D. Orr ∞ 2n+1 k 1 ζ(2n) (−1) 2n + 1 − − + 2 yk 1 2 n(2n + 1)4n 2k k n=1 k=0 ∞ (−1)k+1(22k−1 − 1)B π 2k = 2k y2k. 4k(2k + 1)! k=0

Now we group the coefﬁcients on both sides. The odd powers of y (i.e., 2 j − 1for j = 1, 2, 3,...), we ﬁnd ∞ ζ( ) (− )2 j−1 2n 2n 1 + 2 n4n 2 j − 1 22 j−1 (2 j)22 j n=1 ∞ ζ(2n) (−1)2 j 2n + 1 +2 = 0. n(2n + 1)4n 22 j 2 j n=1

Multiplying by 22 j−1,wesee ∞ ζ( ) + 1 = 2n 2n − 1 2n 1 2 j n4n 2 j − 1 2n + 1 2 j n=1 − ∞ ζ( ) = 2 j 1 2n 2n . 2 j n4n 2 j − 1 n=1

Setting m = 2 j − 1, we arrive at the ﬁrst part of the theorem. For the even powers of y (i.e., 2 j for j = 1, 2, 3,...), we arrive at the following: ∞ ζ( ) (− )2 j ζ( ) 2n 1 2n − 2 j + 2 n4n 22 j 2 j (2 j + 1)4 j (2 j + 1)22 j+1 n=1 ∞ 2 j+1 j+1 2 j−1 2 j ζ(2n) (−1) 2n + 1 (−1) (2 − 1)B j π + 2 = 2 . n(2n + 1)4n 22 j+1 2 j + 1 4 j (2 j + 1)! n=1

Multiplying by 4 j and using (1) to replace the Bernoulli numbers by ζ(2 j),we have

∞ ζ(2n) 2n 1 2n + 1 ζ(2 j) 1 ζ(2 j) 2 − = − + 1 − . n4n 2 j 2n + 1 2 j + 1 2 j + 1 2 j + 1 (2 j + 1) 22 j n=1 2n 1 2n + 1 2 j 2n Using the binomial identity − = ,thissim- 2 j 2n + 1 2 j + 1 2 j + 1 2 j pliﬁes to

∞ ζ(2n) 2n 1 1 = 2ζ(2 j) 1 − . n4n 2 j 2 j 22 j n=1 123 Series representations for the Apery constant... 489

Letting m = 2 j gives the ﬁnal result of the theorem and thus, the proof is com- plete. Corollary 3.4 [5,28] We have

∞ ζ(2n) = log π − 1. (22) n(2n + 1)4n n=1

Proof This follows immediately by setting the coefﬁcients of y−1 equal to each other on both sides. Corollary 3.5 We have the following series representations:

∞ ζ(2n) 1 = , (23) 4n 2 n=1 ∞ ζ(2n)(2n − 1)(2n − 2) = 1, (24) 4n n=1 ∞ ζ(2n)(2n − 1) π 2 1 = − , (25) 4n 8 2 n=1 ∞ ζ(2n)n π 2 = , (26) 4n 16 n=1 and ∞ ζ(2n)n2 3π 2 = . (27) 4n 32 n=1 ∞ ζ(2n)nk Proof Deﬁne F(k) := . Letting m = 1, 3, 2, we obtain (23), (24), and 4n n=1 (25), respectively. Using (23), note that (25) can be rewritten as 2F(1) − F(0) = π2 − ( ) 8 F 0 , and so, (26) follows immediately. Finally, using (23) and (26), note that (24) can be rewritten as 4F(2) − 6F(1) + 2F(0) = 2F(0). From this, (27) follows immediately. Remark Letting m = 2k and m = 2k − 1 (k = 1, 2, 3,...)in the even and odd parts of (21) respectively, we can add the two results to ﬁnd ∞ ζ(2n) 2n 2n ζ(2k) 1 1 1 + = 1 − + − , n4n 2k − 1 2k k 4k 2k − 1 2k n=1 which is equivalent to

∞ ζ(2n) 2n + 1 ζ(2k) 1 1 = 1 − − . (28) n4n 2k k 4k 2k(2k − 1) n=1 123 490 C. Lupu, D. Orr

Theorem 3.6 We have the following series representation: ⎧ ⎪ 1 ∞ ⎨ (1 − β(m)) m odd, ζ(2n) 2n = m (29) n ⎪ 1 1 = n16 m ⎩ ζ(m) 1 − − 1 m even, n 1 m 2m

Proof Similar to the previous theorem’s proof, we integrate sec(xy) two different ways. First, integrating regularly and using properties of logarithm, we have

π 2 1 π y 1 π y sec(xy)dx = log 1 + sin − log cos 0 y 2 y 2 1 d π = − 2Cl + α − α log 2 y dα 2 2 ∞ 1 y 2 − log 1 − , y 2n − 1 n=1

π ∞ α = y (β) = − 2β 2 where and we have used the product formula cos 1 π(2k−1) . 2 k=1 Using Eq. (8) in the ﬁrst term and applying properties of logarithm and its power series to the second term, the integral becomes

π 2 1 π sec(xy)dx = 2log + 2log(1 + y) − log 2 0 y 2 ∞ π ∞ ∞ ζ(2n)( )2n(1 + y)2n 2k − 2 2 + y , y n(2π)2n k(2n − 1)2k n=1 n=1 k=1 and ∞ ∞ ∞ 1 1 1 1 = − = 1 − ζ(2k). (2n − 1)2k n2k (2n)2k 4k n=1 n=1 n=1

Alternatively, we can apply Fubini’s theorem once again and integrate the power series of the secant function term by term. Doing so will give us the following:

π π ∞ n ∞ n π 2n+1 2 2 (−1) E (−1) E2n( ) sec(xy)dx = 2n (xy)2ndx = 2 y2n. (2n)! (2n + 1)(2n)! 0 0 n=0 n=0

Setting the two results equal to each other and simplifying more, we see ∞ k+1 ∞ 2n 2 π (−1) − ζ(2n) 2n − log √ + 2 yk 1 − 2 yk 1 y k n16n k 2 2 k=1 n=1 k=0 123 Series representations for the Apery constant... 491 ∞ ∞ k π 2k+1 ζ(2k) 1 − (−1) E2k( ) + 1 − y2k 1 = 2 y2k. k 4k (2k + 1)! k=1 k=0

Now we can group coefﬁcients. For the even powers of y (i.e., 2 j for j = 1, 2, 3,...), we have 2 j+2 ∞ j 2 j+1 (−1) ζ(2n) 2n (−1) E j π 2β(2 j + 1) 2 − 2 = 2 = . 2 j + 1 n16n 2 j + 1 (2 j + 1)!22 j+1 2 j + 1 n=1

Setting m = 2 j + 1 and rearranging, we achieve the ﬁrst result of the theorem. For the odd powers of y (i.e., 2 j − 1for j = 1, 2, 3,...), we ﬁnd

∞ (−1)2 j ζ(2n) 2n ζ(2 j) 1 2 − 2 + 1 − = 0, 2 j n16n 2 j j 4 j n=1 and rearranging this gives the second part of the theorem. Corollary 3.7 We have the following representations:

∞ ζ(2n) π = log √ , (30) n16n n=1 2 2 and ∞ ζ(2n) 4 − π = . (31) 16n 8 n=1 Proof Formula (30) follows immediately by setting the coefﬁcients of y−1 equal to each other on both sides. For (31), we set the constant terms on both sides equal to ﬁnd

∞ ζ(2n) π 2 − 4 = . 16n 2 n=1

From here, (31) follows immediately. Remark In the previous theorem, if we used the Clausen identity for log(cos(x)) given in the introduction rather than the cosine product formula, we obtain the following relation:

∞ n ∞ π ζ(2n) 2 2n 2 yk log − yk − (−1)k + (1 − 2(−1)k) 4 n4n k 4n k n=1 k=0 k=1 ∞ 2β(2k + 1) = y2k+1. 2k + 1 k=0 123 492 C. Lupu, D. Orr

Gathering the coefﬁcients of y0, we ﬁnd

∞ π ζ(2n) 2 log − − 1 = 0. 4 n4n 4n n=1

Rearranging and using (30), we can see that

∞ ζ(2n) 1 π = log , (32) 2n22n 2 2 n=1 which is a very nice result. This series appears by setting the coefﬁcients of y0 equal to each other in the proof of Theorem 2.3; however, the series will cancel and you arrive at a truth statement. Here, we are able to recover that series.

Corollary 3.8 We have

∞ ζ(2n)(2n − 1) π 2 1 = − , (33) 16n 16 2 n=1 ∞ ζ(2n)(2n − 1)(2n − 2) π 3 = 1 − , (34) 16n 96 n=1 ∞ ζ(2n)n π π = − 1 , (35) 16n 16 2 n=1 and ∞ ζ(2n)n2 π 3π π 2 = − − 1 . (36) 16n 32 2 4 n=1

∞ ζ(2n)nk Proof Deﬁne G(k) := . Letting m = 2, 3in(29) gives us formulas 16n n=1 (33) and (34), respectively. Using (31), we can rewrite (33)as2G(1) − G(0) = π 2 π − G(0) − and from this, (35) follows immediately. Lastly, for (36), expanding 16 8 π π 3 (34)using(31), we have 4G(2) − 6G(1) + 2G(0) = + 2G(0) − .Using(35), 4 96 we achieve the desired result.

Corollary 3.9 We have the following series

∞ ζ(2n) 1 2n ζ(2k) 1 1 − = 1 − , (37) n4n 4n 2k 2k 4k n=1 123 Series representations for the Apery constant... 493 and ∞ ζ(2n) 1 2n β(2k + 1) 1 − = . (38) n4n 4n 2k + 1 2k + 1 n=1

Proof Denote (29.1) and (29.2)aswellas(21.1) and (21.2) as the formula for m odd and m even, respectively. Letting m = 2k for k = 1, 2, 3 ... and subtracting (29.2) from (21.2),formula(37) follows immediately. Similarly, letting m = 2k + 1for k = 0, 1, 2,...and subtracting (29.1) from (21.1), we obtain (38). 2n + 1 Remark You can add (37) and (38) together to get inside the series. You 2k + 1 2n + 1 can also change k to k − 1in(38) and add (37) and (38) again to give you 2k in the series.

Acknowledgements We would like to thank Piotr Hajlasz, Bogdan Ion, George Sparling and William C. Troy for some fruitful conversations which led to an improvement of the present paper.

References

1. Adamchik, V.S.:Contributions to the theory of the Barnes function. Int. J. Math. Comput. Sci. 9, 11–30 (2014) 2. Adamchik, V.S.,Strivastava, H.M.: Some series of the zeta and related functions. Analysis 18, 131–144 (1998) 3. Alzer, H., Karayannakis, D., Strivastava, H.M.: Series representations for some mathematical constants. J. Math. Anal. Appl. 320, 145–162 (2006) 4. Apostol, T.M.: Another elementary proof of Euler’s formula for ζ(2n).Am.Math.Mon.80, 425–431 (1973) 5. Boros, G., Moll, V.: Irresistible Integrals. Symbolics, Analysis and Experiments in the Evaluation of Integrals. Cambridge University Press, Cambridge (2004) 6. Borwein, J.M., Bradley, D.M., Crandall, R.E.: Computational strategies for the Riemann zeta function. J. Comput. Appl. Math. 121, 247–296 (2000) 7. Borwein, J.M., Broadhurst, D.J., Kamnitzer, J.: Central binomial sums, multiple Clausen values, and zeta values. Exp. Math. 10, 25–34 (2001) π 2 ( θ)n θ 8. Bowman, F.: Note on the integral 0 log sin d . J. Lond. Math. Soc. 22, 172–173 (1947) 9. Choi, J.: Some integral representations of the Clausen function Cl2(x) and the Catalan constant. East Asian Math. J. 32, 43–46 (2016) 10. Choi, J., Strivastava, H.: Certain classes of series involving the zeta function. J. Math. Anal. Appl. 231, 91–117 (1999) 11. Choi, J., Strivastava, H.M.: The Clausen function Cl2(x) and its related integrals. Thai J. Math. 12, 251–264 (2014) 12. Choi, J., Strivastava, H.M., Adamchik, V.S.: Multiple Gamma and related functions. Appl. Math. Comput. 134, 515–533 (2003) 13. Choi, J., Cho, Y.J., Strivastava, H.M.: Log-sine integrals involving series associated with the zeta function and polylogarithm function. Math. Scand. 105, 199–217 (2009) 14. Clausen, T.: Uber die function sin φ + 1 sin 2φ + 1 sin 3φ +etc.. J. Reine Angew. Math. 8, 298–300 22 32 (1832) 15. Coffey, M.W.: On one-dimensional digamma and polygamma series related to the evaluation of Feyn- man diagrams. J. Comput. Appl. Math. 183, 84–100 (2005) 16. Cvijovic, D.: New integral representations of the polylogarithm function. Proc. R. Soc. A 643, 897–905 (2007) 123 494 C. Lupu, D. Orr

17. Cvijovic, D.: Closed-form evaluation of some families of cotangent and cosecant integrals. Integral Transforms Spec. Func. 19, 147–155 (2008) 18. Cvijovic, D., Klinowski, J.: New rapidly convergent series representations for ζ(2n + 1).Proc.Am. Math. Soc. 125, 1263–1271 (1997) 19. De Amo, E., Diaz Carrillo, M., Hernandez-Sanchez, J.: Another proof of Euler’s formula for ζ(2k). Proc. Am. Math. Soc. 139, 1441–1444 (2011) 20. de Doeder, P.J.: On the Clausen integral Cl2(θ) and a related integral. J. Comput. Appl. Math. 11, 325–330 (1984) 21. Ewell, J.A.: A new series representation for ζ(3).Am.Math.Mon.97, 219–220 (1990) 22. Grosjean, C.C.: Formulae concerning the computation of the Clausen integral Cl2(θ). J. Comput. Appl. Math. 11, 331–342 (1984) 23. Koblig, K.S.: Chebyshev coefficients for the Clausen function Cl2(x). J. Comput. Appl. Math. 64, 295–297 (1995) 24. Lewin, L.: Polylogarithms and Associated Functions. Elsevier North Holland Inc., New York (1981) 25. Ogreid, O.M., Osland, P.: Some infinite series related to Feynman diagrams. J. Comput. Appl. Math. 140, 659–671 (2002) 26. Strivastava, H.M., Choi, J.: Series Associated with the Zeta and Related Functions. Kluwer Academic Publishers, Dordrecht (2001) 27. Strivastava, H.M., Glasser, M.L., Adamchik, V.S.:Some definite integrals associated with the Riemann zeta function. Z. Anal. Anwend. 19, 831–846 (2000) 28. Tyler, D., Chernoff, P.R.: An old sum reappears-elementary problem 3103. Am. Math. Mon. 92, 507 (1985) 29. Wood, V.E.: Efficient calculation of Clausen’s integral. Math. Comput. 22, 883–884 (1968) 30. Wu, J., Zhang, X., Liu, D.: An efficient calculation of the Clausen functions. BIT Numer. Math. 50, 193–206 (2010)

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