# Some Weighted Sum Formulas for Multiple Zeta, Hurwitz Zeta, and Alternating Multiple Zeta Values

Authors: Yuan He; Zhuoyu Chen
Journal: Journal of Mathematics
 (2021)

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## Abstract

We perform a further investigation for the multiple zeta values and their variations and generalizations in this paper. By making use of the method of the generating functions and some connections between the higher-order trigonometric functions and the Lerch zeta function, we explicitly evaluate some weighted sums of the multiple zeta, Hurwitz zeta, and alternating multiple zeta values in terms of the Bernoulli and Euler polynomials and numbers. It turns out that various known results are deduced as special cases.


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## Body

## 1. Introduction

Letα1,α2,…,αk be all positive integers with αk≥2. The multiple zeta values ζα1,…,αk of the depth k (sometimes called Euler sums) are usually defined by the following series:(1)ζα1,…,αk=∑1≤n1<n2<⋯<nk1n1α1n2α2⋯nkαk.These series appear in a variety of fields in mathematics, such as number theory, combinatorics, knot theory, the theory of mixed Tate motives, and quantum field theory; see, for example [1–4].It is well known that one of the central problems on the multiple zeta values is to determine all the possibleℚ-linear relations among them. Goncharov’s ([5], Conjecture 4.2) conjecture implies that it suffices to study the relations among the multiple zeta values of the same weight a=α1+α2+⋯+αk. Perhaps the earliest result in this direction is Euler’s [6] sum formula, namely,(2)∑α1+α2=nζα1,α2+1=ζn+1,n≥2,where ζ⋅ is the Riemann zeta function. In fact, there exists a general form of (2), as follows:(3)∑a=nζα1,α2,…,αk+1=ζn+1,n≥k,which is referred to as the “sum conjecture” in [7], and was proved by Granville [8] and Zagier [9] independently around the year 1995. Some new relations for the multiple zeta values and their different variations and generalizations have been found in recent years. For example, Ohno and Zudilin [10] in 2008 proved a weighted form of Euler’s sum formula:(4)∑α1+α2=n2α2+1ζα1,α2+1=n+2ζn+1,n≥2.It becomes obvious from (2) that Ohno and Zudilin’s weighed sum formula can be rewritten as(5)∑α1+α2=n2α2+1−1ζα1,α2+1=n+1ζn+1,n≥2.Based on the equivalence of (4) and (5), Guo and Xie [11] in 2009 extended the weighted sum formula of Ohno and Zudilin to arbitrary depth and discovered that for positive integers n,k with n≥k,(6)∑a=n2αk+2αk−1∑j=2k−12Sj−αk−j+2Sk−1−αk−k−1ζα1,α2,…,αk+1=n+1ζn+1,where Sj=αk−j+1+⋯+αk+1 for j=1,…,k−1. For another weighted sum formulas of the multiple zeta values, one is referred to [12], where the weight coefficients are given by (symmetric) polynomials of the arguments. On the contrary, let m,n,k be positive integers with m≥2 and n≥k, and let Emn,k be the sums of all multiple zeta values of the depth k and the weight mn given by(7)Emn,k=∑a=nζmα1,…,mαk.Gangl et al. [13] in 2004 proved that(8)E2n,2=34ζ2n.Shen and Cai [14] in 2012 obtained the formulas(9)E2n,3=58ζ2n−14ζ2ζ2n−2,E2n,4=3564ζ2n−516ζ2ζ2n−2.After that, Hoffman [15] used the theory of symmetric functions to establish the general sum formula(10)E2n,k=−1n−kπ2n2n+1!∑l=0n−kn−lk2n+12l2−22lB2l,where Bn is the n-th Bernoulli number. Furthermore, Zhao [16] used the ideas developed in [15] to evaluate the sums of all multiple Hurwitz zeta values of the depth k and the weight 2n in terms of the Euler numbers. Moreover, Zhao [17] used the theory of symmetric functions to consider the more complicated alternating multiple zeta values and depicted that the sums of all alternating multiple zeta values of the depth k and the weight 2n can be evaluated in terms of the Riemann zeta function and the Euler numbers. More recently, Chen et al. [18] used the method of the generating functions to express Emn,k by constructing a combinatorial identity of products of the multiple zeta values and the so-called multiple zeta-star values at the repetitions of m, and then used Muneta’s [19] and Nakamura’s [20] results to reobtain Hoffman’s sum formula (10) and confirm Genčev’s ([21], Conjecture 4.1) conjecture on the evaluation of E4n,k.(11)E4n,k=−1n−k22n+1π4n4n+2!∑l=0n−kn−lk4n+24l1−4l×∑r=02l4l2r−1r2−22rB2r2−24l−2rB4l−2r.Subsequently, Shen and Jia [22] extended the sums of the multiple Hurwitz zeta values previously considered in [23] and showed that the sums of all multiple Hurwitz zeta values of the depth k and the weight mn can be expressed by a combinatorial identity of products of the multiple Hurwitz zeta values and the so-called multiple Hurwitz zeta-star values at the repetitions of m. In particular, Shen and Jia [22] obtained Zhao’s [16] sum formula with a slight different notation and evaluated the sums of all multiple Hurwitz zeta values of the depth k and the weight 4n in terms of the Euler numbers.Motivated and inspired by the work of the above authors, we explicitly evaluate some weighted sums of the multiple zeta, Hurwitz zeta, and alternating multiple zeta values in terms of the Bernoulli and Euler polynomials and numbers by using the method of the generating functions and some connections between the higher-order trigonometric functions and the Lerch zeta function established by the first author [24]. The results presented here are the corresponding extensions of various known sum formulas.This paper is organized as follows: In Section2, we give several weighted sum formulas for the multiple zeta values, some of which generalize the sum formulas (10) and (11), and improve Eie and Ong’s [25] weighted sum formulas. In Section 3, we present some similar weighted sum formulas for the multiple Hurwitz zeta values and deduce Shen and Jia’s [22] sum formulas as special cases. Section 4 concentrates on the features that have contributed to the weighted sum formulas for the alternating multiple zeta values, and it then turns out that Zhao’s [17] sum formula is obtained in a rather simple way.


## 2. Sum Formulas for Multiple Zeta Values

For convenience, in the following, we always denote byi the square root of −1 such that i2=−1, sn,k the Stirling numbers of the first kind, Bnx the Bernoulli polynomials, and Enx the Euler polynomials. It is clear that taking x=0 and x=1/2 in the Bernoulli and Euler polynomials gives the Bernoulli numbers Bn=Bn0 and the Euler numbers En=2nEn1/2, respectively. We refer the reader to two standard books [26, 27] on basic properties for these special sequences and polynomials. We also write tnft as the coefficients of tn in ft for nonnegative integer n. We now state our first result as follows.Theorem 1.
Letm,n be positive integers. Then(12)∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=−1n22n+1π2nm−1!∑k=0m−1−1kB˜m,kUm,2n−m,k,where, and in what follows, B˜m,k is the linear combination of the Stirling numbers of the first kind satisfying that for positive integer m and nonnegative integer k,(13)B˜m,k=∑j=k+1mj−1ksm,j2j−k,and Um,n,k is the linear combination of the Bernoulli polynomials given for positive integer m and nonnegative integers n,k by(14)Um,n,k=1n!∑l=0nnlm−12n−lBl+k+11/2l+k+1.Proof.
Recall that Euler’s infinite product formula of the sine function is(15)sinπx=πx∏j=1∞1−x2j2,which holds true for arbitrary complex number x (see [26], p. 75 or [28], pp. 12–18). The binomial series asserts that for complex number α (see [27], p. 37),(16)1+tα=∑n=0∞αntn,t<1,where αn are the binomial coefficients given for nonnegative integer n by(17)α0=1,αn=αα−1⋯α−n+1n!,n≥1.
So from (15) and (16), we discover that for complex number x with 0<x<1,(18)sinπxπx−m=∏j=1∞∑k=0∞−1k−mkx2kj2k=∏j=1∞∑k=0∞m+k−1kx2kj2k.
Comparing the coefficients ofx2n on both sides of (18), it then follows that for complex number x with 0<x<1,(19)∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=x2nsinπxπx−m.
We now evaluate the right-hand side of (19) from another view. Let q,n be positive integers, and let θr be a real function defined on positive integer r. If θr≠0,±q,±2q,…, then (see [24], Theorem 3.2)(20)cscnπθrq=−in2n+1enπiθr/qn−1!∑k=0n−1−1kB˜n,kϕθrq,12,−k,where ϕa,x,s is the Lerch zeta function given for real number a, x≠ negative integer or zero, and complex number s by(21)ϕa,x,s=∑n=0∞e2πinan+xs,ℜs>1.
Note that the series is an entire function ofs when a is not an integer. Obviously, replacing n by m and θr/q by x in (20) gives that for real number x≠0,±1,±2,…,(22)cscmπx=−im2m+1emπixm−1!∑k=0m−1−1kB˜m,kϕx,12,−k.
It follows from (19) and (22) that(23)∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=x2nπmxm−im2m+1emπixm−1!∑k=0m−1−1kB˜m,kϕx,12,−k=−im2m+1πmm−1!∑k=0m−1−1kB˜m,kx2n−memπixϕx,12,−k.
It is easily seen from the Taylor series expansion for the complex exponential function and the familiar binomial theorem that(24)x2n−memπixϕx,12,−k=x2n−memπix∑j=0∞e2πijxj+1/2−k=x2n−m∑j=0∞1j+1/2−k∑l=0∞j+m/2l2πilxll!=2πi2n−m2n−m!∑j=0∞j+m/22n−mj+1/2−k=2πi2n−m2n−m!∑l=02n−m2n−mlm−122n−m−l∑j=0∞1j+1/2−l+k.
Since for nonnegative integern (see [29], Theorem 12.13),(25)ζ−n,x=−Bn+1xn+1,where ζs,x is the Hurwitz zeta function given for real number x>0 and complex number s by(26)ζs,x=∑n=0∞1n+xs,ℜs>1,so by applying (25) to (24), we arrive at(27)x2n−memπixϕx,12,−k=−2πi2n−mUm,2n−m,k.
Now (12) follows from (23) and (27). This completes the proof of Theorem 1.Corollary 1.
Letn be a positive integer. Then(28)∑k=1nE2n,k=−1nπ2n2−22nB2n2n!.Proof.
Since for nonnegative integern (see [26], p. 805),(29)Bn12=21−n−1Bn,so by taking m=1 in Theorem 1, in view of s1,1=1, we get the desired result.
Corollary1 is usually attributed to Hoffman ([15], Corollary 2) and was previously obtained by Aoki, Kombu, and Ohno ([30], Equation (4.6)), who stated it in the language of the multiple zeta-star values. We are in a good position to use Theorem 1 to yield the following result.Theorem 2.
Letm,n,k be positive integers with n≥k. Then(30)∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=−1n−k2π2nm−1!∑l=0n−kn−lk4l−1m−1n−lV1m,n−l×∑k=0m−1−1kB˜m,kUm,2l−m,k,where V1m,n is a rational number given for positive integer m and nonnegative integer n by(31)V1m,n=∑j1+⋯+jm=nj1,…,jm≥012j1+1!⋯12jm+1!.Proof.
Clearly, for real or complex parameterλ,(32)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=x2n∏j=1∞1+1+λm1x2j2+⋯+1+λm+k−1kx2kj2k+⋯.
Just as a polynomial function ofz in the order of ascending power is divided by another polynomial function of z in the order of ascending power, we discover that(33)λm1z2+⋯+λm+k−1kz2k+⋯1+m1z2+⋯+m+k−1kz2k+⋯=λm1z2−λm2z4+⋯+−1m−1λmmz2m=−λ1−z2m+λ.
Applying (33) to the right-hand side of (32), it then follows from (18) that(34)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=x2n∏j=1∞1+m1x2j2+⋯+m+k−1kx2kj2k+⋯×1−λ1−x2j2m+λ=∑l=0nx2lsinπxπx−mx2n−2l∏j=1∞1−λ1−x2j2m+λ.
Noticing that from (19) and Theorem 1, we have(35)x2lsinπxπx−m=−1l22l+1π2lm−1!∑k=0m−1−1kB˜m,kUm,2l−m,k.
We now evaluate the coefficients ofx2n−2l in the infinite product of the right-hand side of (34). Let μ1λ,…,μmλ be all complex numbers determined by the factorization of the polynomial function 1−λ1−zm+λ over the complex number field satisfying that(36)1−λ1−zm+λ=∏l=1m1+μlλz.
The famous Vieta’s theorem implies that(37)μ1λ⋯μmλ=−1m−1λ.
So from (36), (37), and the remarkable formula see [31], Equation (36), or [7], Corollary 2.3,(38)ζ2n=π2n2n+1!,n≥0,where an denotes the n repetitions of a, we get that(39)x2n−2l∏j=1∞1−λ1−x2j2m+λ=x2n−2l∏l=1m∏j=1∞1+μlλx2j2=x2n−2l∏l=1m∑j=0∞μlλjζ2jx2j=∑j1+⋯+jm=n−lj1,…,jm≥0μ1λj1ζ2j1⋯μmλjmζ2jm=−1m−1λn−lπ2n−2lV1m,n−l.
Inserting (35) and (39) into (34), it follows that(40)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk=2π2nm−1!∑l=0n−4l−1m−1λn−lV1m,n−l×∑k=0m−1−1kB˜m,kUm,2l−m,k.
Thus (30) follows immediately after making k-times derivative with respect to λ and then taking λ=−1 on both sides of (40). This concludes the proof of Theorem 2.
It is easy to check that takingm=1 in Theorem 2 and then applying (29) and s1,1=1 leads to Hoffman’s formula (10). It is worth noticing that the formula (40) can also be regarded as an extension of Theorem 1. In a similar consideration to Theorem 2, we have the following result.Theorem 3.
Letm,n,k be the positive integers with n≥k. Then(41)∑a=n∏j=1kαj+m−1αjζ4α1,…,4αk=−1n−k22n+m+1π4nm−1!∑l=0n−kn−lk−4l−1m−1n−lV2m,n−l×∑r=02l−1r4l−2r!B4l−2rmm2∑k=0m−1−1kB˜m,kUm,2r−m,k,where V2m,n is a rational number given for positive integer m and nonnegative integer n by(42)V2m,n=∑j1+⋯+jm=nj1,…,jm≥014j1+2!⋯14jm+2!,and Bnmx are the higher-order Bernoulli polynomials defined by the generating function (see [32]):(43)tet−1mext=∑n=0∞Bnmxtnn!,t<2π.Proof.
We know from (16) and (33) that(44)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ4α1,…,4αk=x4n∏j=1∞1+1+λm1x4j4+⋯+1+λm+k−1kx4kj4k+⋯=x4n∏j=1∞1+m1x4j4+⋯+m+k−1kx4kj4k+⋯×1−λ1−x4j4m+λ=∑l=0nx4l∏j=1∞1−x4j4−mx4n−4l∏j=1∞1−λ1−x4j4m+λ.
If we replacex by ix in (15), then we have(45)∏j=1∞1+x2j2−m=sinhπxπx−m=∑n=0∞Bnmm22πxnn!.
It follows from (15), (35), and (45) that(46)x4l∏j=1∞1−x4j4−m=∑r=02lx4l−2r∏j=1∞1+x2j2−mx2r∏j=1∞1−x2j2−m=∑r=02l2π4l−2r4l−2r!B4l−2rmm2x2rsinπxπx−m=24l+1π4lm−1!∑r=02l−1r4l−2r!B4l−2rmm2×∑k=0m−1−1kB˜m,kUm,2r−m,k.
On the contrary, from (36), (37), and the well known formula (see [31], Equation (37)),(47)ζ4n=22n+1π4n4n+2!,n≥0,we obtain that(48)x4n−4l∏j=1∞1−λ1−x4j4m+λ=x4n−4l∏l=1m∏j=1∞1+μlλx4j4=x4n−4l∏l=1m∑j=0∞μlλjζ4jx4j=∑j1+⋯+jm=n−lj1,…,jm≥0μ1λj1ζ4j1⋯μmλjmζ4jm=−1m−1λn−l22n−2l+mπ4n−4lV2m,n−l.
Inserting (46) and (48) into (44), we have(49)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ4α1,…,4αk=22n+m+1π4nm−1!∑l=0n4l−1m−1λn−lV2m,n−l×∑r=02l−1r4l−2r!B4l−2rmm2∑k=0m−1−1kB˜m,kUm,2r−m,k.
Therefore, we get (41) and finish the proof of Theorem 3 when making k-times derivative with respect to λ and then taking λ=−1 on both sides of (49).
It is trivial to check that Genčev’s conjecture (11) holds true when taking m=1 in Theorem 3. We here remark that Theorems 2 and 3, as well as the generalizations of the sum formulas (10) and (11), are the improvements of the results recently obtained by Eie and Ong ([25], Theorems 2.1 and 2.2), where they expressed the left-hand side of (30) by a combinatorial identity involving the higher derivative of one function and the sums of products of m Bernoulli polynomials, and the left-hand side of (41) by a combinatorial identity involving the higher derivative of another function and the sums of products of 2m Bernoulli polynomials.


## 3. Sum Formulas for Multiple Hurwitz Zeta Values

In this section, we shall study the multiple Hurwitz zeta values defined by the series (see [23, 33])(50)ζα1,…,αk;−12,…,−12=∑1≤n1<n2<⋯<nk1n1−1/2α1n2−1/2α2⋯nk−1/2αk,where α1,α2,…,αk are all positive integers with αk≥2 and present some weighted sum formulas for them. The results showed here are very analogous to the ones in Section 2.Theorem 4.
Letm,n be positive integers. Then(51)∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−12,…,−12=−1m+n−12m+2nπ2nm−1!∑k=0m−1−1kB˜m,kU^m,2n,k,where U^m,n,k is the linear combination of the Euler polynomials given for positive integer m and nonnegative integers n,k by(52)U^m,n,k=1n!∑l=0nnlm−12n−lEl+k12.Proof.
It is well known that Euler’s infinite product formula of the cosine function is(53)cosπx=∏j=1∞1−4x22j−12,which holds true for arbitrary complex number x (see [26], p. 75 or [28], pp. 12–18). Hence, we obtain from (16) and (53) that for complex number x with x<1/2,(54)x2ncosπx−m=x2n∏j=1∞1−x2j−1/22−m=x2n∏j=1∞∑k=0∞−1k−mkx2kj−1/22k=x2n∏j=1∞∑k=0∞m+k−1kx2kj−1/22k=∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−12,…,−12.
If we replacex by 1/2−x in (22), we get that for real number x≠±1/2,±3/2,…,(55)cosπx−m=−im2m+1emπi1/2−xm−1!∑k=0m−1−1kB˜m,kϕ12−x,12,−k,which can be converted by the expression of the higher-order secant function stated in ([24], Theorem 3.2). It then follows that(56)x2ncosπx−m=−1m−12m+1m−1!∑k=0m−1−1kB˜m,k×x2ne−mπixϕ12−x,12,−k.
Observe that(57)x2ne−mπixϕ12−x,12,−k=x2ne−mπix∑j=0∞−1je−2πijxj+1/2−k=x2n∑j=0∞−1j1j+1/2−k∑l=0∞j+m/2l−2πilxll!=2πi2n2n!∑j=0∞−1jj+m/22nj+1/2−k=2πi2n2n!∑l=02n2nlm−122n−l∑j=0∞−1j1j+1/2−l+k.
Since for nonnegative integern (see [34], Corollary 3),(58)η−n,x=12Enx,where ηs,x is the alternating Hurwitz zeta function (also called Hurwitz Euler-eta function) given for real number x>0 and complex number s by(59)ηs,x=∑n=0∞−1n1n+xs,ℜs>0,so by applying (58) to (57), we find that(60)x2ne−mπixϕ12−x,12,−k=2πi2n2U^m,2n,k.
Inserting (60) into (56), we have(61)x2ncosπx−m=−1m+n−12m+2nπ2nm−1!∑k=0m−1−1kB˜m,kU^m,2n,k.
Thus, we complete the proof of Theorem4 by equating (54) and (61).
In what follows, we denote byTmn,k the sums of all multiple Hurwitz zeta values of the depth k and the weight mn given for positive integers m,n,k with m≥2 and n≥k by(62)Tmn,k=∑a=nζmα1,…,mαk;−12,…,−12.
It follows that we state the following result.Corollary 2.
Letn be a positive integer. Then(63)∑k=1nT2n,k=−1nπ2nE2n2n!.Proof.
By takingm=1 in Theorem 4, in view of s1,1=1 and En=2nEn1/2 for nonnegative integer n, the desired result follows immediately.
We mention that Corollary2 was obtained by Shen and Jia ([22], p. 265) in the language of the multiple Hurwitz zeta-star values. We now give another weighted sum formula for the multiple Hurwitz zeta values as follows.Theorem 5.
Letm,n,k be positive integers with n≥k. Then(64)∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−12,…,−12=−1m+n−k−12mπ2nm−1!∑l=0n−kn−lk4l−1m−1n−lV3m,n−l×∑k=0m−1−1kB˜m,kU^m,2l,k,where V3m,n is a rational number given for positive integer m and nonnegative integer n by(65)V3m,n=∑j1+⋯+jm=nj1,…,jm≥012j1!⋯12jm!.Proof.
We obtain from (33) and (54) that(66)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−12,…,−12=x2n∏j=1∞1+1+λm1x2j−1/22+⋯+1+λm+k−1kx2kj−1/22k+⋯=x2n∏j=1∞1+m1x2j−1/22+⋯+m+k−1kx2kj−1/22k+⋯×1−λ1−x2j−1/22m+λ=∑l=0nx2lcosπx−mx2n−2l∏j=1∞1−λ1−x2j−1/22m+λ.
Noticing that from (61), we have(67)x2lcosπx−m=−1m+l−12m+2lπ2lm−1!∑k=0m−1−1kB˜m,kU^m,2l,k,and from (36), (37), and the formula (see [33] or [22], p. 265),(68)ζ2;−12n=π2n2n!,n≥0,we deduce that(69)x2n−2l∏j=1∞1−λ1−x2j−1/22m+λ=x2n−2l∏l=1m∏j=1∞1+μlλx2j−1/22=x2n−2l∏l=1m∑j=0∞μlλjζ2;−1/2jx2j=∑j1+⋯+jm=n−lj1,…,jm≥0μ1λj1ζ2;−1/2j1⋯μmλjmζ2;−1/2jm=−1m−1λn−lπ2n−2lV3m,n−l.
Inserting (67) and (69) into (66), we get that(70)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−12,…,−12=−1m−12mπ2nm−1!∑l=0n−4l−1m−1λn−lV3m,n−l×∑k=0m−1−1kB˜m,kU^m,2l,k.
Thus the desired result follows by makingk-times derivative with respect to λ and then taking λ=−1 on both sides of (70). This completes the proof of Theorem 5.Corollary 3.
Letn,k be positive integers with n≥k. Then(71)T2n,k=−1n−kπ2n2n!∑l=0n−kn−lk2n2lE2l.Proof.
Takingm=1 in Theorem 5, in light of s1,1=1 and En=2nEn1/2 for nonnegative integer n, we get the desired result.
Corollary3 is a general form of Shen and Cai’s [23] results for the cases 2≤k≤5 in T2n,k and was also found by Shen and Jia ([22], p. 265). For an equivalent version of Corollary 3, see ([16], Theorem 1.3) for details. Similarly, we state the following result.Theorem 6.
Letm,n,k be positive integers with n≥k. Then(72)∑a=n∏j=1kαj+m−1αjζ4α1,…,4αk;−12,…,−12=−1m+n−k−12m+2nπ4nm−1!∑l=0n−kn−lk−4l−1m−1n−lV4m,n−l×∑r=02l−1r4l−2r!E4l−2rmm2∑k=0m−1−1kB˜m,kU^m,2r,k,where V4m,n is a rational number given for positive integer m and nonnegative integer n by(73)V4m,n=∑j1+⋯+jm=nj1,…,jm≥014j1!⋯14jm!,and Enmx are the higher-order Euler polynomials defined by the generating function (see [35])(74)2et+1mext=∑n=0∞Enmxtnn!,t<π.Proof.
With the help of (16) and (33), we discover that(75)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ4α1,…,4αk;−12,…,−12=x4n∏j=1∞1+1+λm1x4j−1/24+⋯+1+λm+k−1kx4kj−1/24k+⋯=x4n∏j=1∞1+m1x4j−1/24+⋯+m+k−1kx4kj−1/24k+⋯×1−λ1−x4j−1/24m+λ=∑l=0nx4l∏j=1∞1−x4j−1/24−m×x4n−4l∏j=1∞1−λ1−x4j−1/24m+λ.
By replacingx by ix in (53), we find that(76)∏j=1∞1+x2j−1/22−m=coshπx−m=∑n=0∞Enmm22πxnn!.
It follows from (53), (67), and (76) that(77)x4l∏j=1∞1−x4j−1/24−m=∑r=02lx4l−2r∏j=1∞1+x2j−1/22−mx2r∏j=1∞1−x2j−1/22−m=∑r=02l2π4l−2r4l−2r!E4l−2rmm2x2rcosπx−m=−1m−12m+4lπ4lm−1!∑r=02l−1r4l−2r!E4l−2rmm2×∑k=0m−1−1kB˜m,kU^m,2r,k.
If we apply (36), (37), and the formula (see [22], p. 265),(78)ζ4;−12n=22nπ4n4n!n≥0,to the second infinite product in the right-hand side of (75), we have(79)x4n−4l∏j=1∞1−λ1−x4j−1/24m+λ=x4n−4l∏l=1m∏j=1∞1+μlλx4j−1/24=x4n−4l∏l=1m∑j=0∞μlλjζ4;−12jx4j=∑j1+⋯+jm=n−lj1,…,jm≥0μ1λj1ζ4;−12j1⋯μmλjmζ4;−12jm=−1m−1λn−l22n−2lπ4n−4lV4m,n−l.
Inserting (77) and (79) into (75), it then follows that(80)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ4α1,…,4αk;−12,…,−12=−1m−12m+2nπ4nm−1!∑l=0n4l−1m−1λn−lV4m,n−l×∑r=02l−1r4l−2r!E4l−2rmm2∑k=0m−1−1kB˜m,kU^m,2r,k.
Thus we prove (72) immediately by making k-times derivative with respect to λ and then taking λ=−1 on both sides of (80). This concludes the proof of Theorem 6.
In particular, we discover Shen and Jia’s ([22], p. 266) result as follows.Corollary 4.
Letn,k be positive integers with n≥k. Then(81)T4n,k=−1n−kπ4n4n!∑l=0n−kn−lk4n4l−1l4n−l×∑r=02l4l2r−1rE2rE4l−2r.Proof.
SinceEnx=En1x for nonnegative integer n, so by taking m=1 in Theorem 6, the desired result follows from s1,1=1 and En=2nEn1/2 for nonnegative integer n.


## 4. Sum Formulas for Alternating Multiple Zeta Values

As shown in [36, 37], the alternating multiple zeta values ζα1,…,αk;ε1,…,εk of depth k (also called alternating Euler sums) are defined by the series(82)ζα1,…,αk;ε1,…,εk=∑1≤n1<n2<⋯<nkε1n1ε2n2⋯εknn1α1n2α2⋯nkαk,where α1,α2,…,αk are all positive integers and ε1=±1,ε2=±1,…,εk=±1 with αk,εk≠1,1. We next use the methods and results described in the second and third sections to obtain some weighted sum formulas for the alternating multiple zeta values.Theorem 7.
Letm,n be positive integers. Then(83)∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−1α1,…,−1αk=2π2nm−1!∑r=0n−1r2n−2r!E2n−2rmm2×∑k=0m−1−1kB˜m,kUm,2r−m,k.Proof.
By substitutingx/2 for x in (15) and xi/2 for x in (53), we find that(84)sinπx/2πx/2coshπx2=∏j=1∞1−x22j2∏j=1∞1+x22j−12=∏j=1∞1−−1jx2j2.
It follows from (16) and (84) that(85)x2nsinπx/2πx/2coshπx2−m=x2n∏j=1∞∑k=0∞−mk−−1jkx2kj2k=x2n∏j=1∞∑k=0∞m+k−1k−1kjx2kj2k=∑k=1n∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−1α1,…,−1αk.
Trivially, the left-hand side of (85) can be written as(86)x2nsinπx/2πx/2coshπx2−m=∑r=0nx2n−2rcoshπx2−mx2rsinπx/2πx/2−m.
It is easy to see that from the generating function of the higher-order Euler polynomials, we have(87)x2n−2rcoshπx2−m=π2n−2r2n−2r!E2n−2rmm2,and from (22) and (27), we can write(88)x2rsinπx/2πx/2−m=−im2πmm−1!∑k=0m−1−1kB˜m,k×x2r−memπix/2ϕx2,12,−k=−1r2π2rm−1!∑k=0m−1−1kB˜m,kUm,2r−m,k.
Inserting (87) and (88) into (86), in view of (85), we get (83) immediately and thus finish the proof of Theorem 7.
If we denote byΞmn,k the sums of all alternating multiple zeta values of the depth k and the weight mn given for positive integers m,n,k with m≥2 and n≥k by(89)Ξmn,k=∑a=nζmα1,…,mαk;−1α1,…,−1αk,then we obtain Zhao’s ([17], Equation (7)) result as follows.Corollary 5.
Letn be a positive integer. Then(90)∑k=1nΞ2n,k=π2n4n2n!∑r=0n2n2r−1r2−22rB2rE2n−2r.Proof.
By takingm=1 in Theorem 7 and then using s1,1=1, the desired result follows from (29) and En=2nEn1/2 for nonnegative integer n.Theorem 8.
Letm,n,k be positive integers with n≥k. Then(91)∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−1α1,…,−1αk=−1n−k2π2nm−1!∑l=0n−kn−lk−1m−1n−l+l2n−lV5m,n−l×∑r=0l−1r2l−2r!E2l−2rmm2∑k=0m−1−1kB˜m,kUm,2r−m,k,where V5m,n is a rational number given for positive integer m and nonnegative integer n by(92)V5m,n=∑j1+⋯+jm=nj1,…,jm≥0−1j1+1/22j1+1!⋯−1jm+1/22jm+1!,and x denotes the greatest integer less than or equal to real number x.Proof.
It is easily seen from (33) and (85) that the following identities are complete:(93)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−1α1,…,−1αk=x2n∏j=1∞1+1+λm1−1jx2j2+⋯+1+λm+k−1k−1kjx2kj2k+⋯=x2n∏j=1∞1+m1−1jx2j2+⋯+m+k−1k−1kjx2kj2k+⋯×1−λ1−−1jx2j2m+λ=∑l=0nx2lsinπx/2πx/2coshπx2−m×x2n−2l∏j=1∞1−λ1−−1jx2j2m+λ.
Noticing that from (85) and Theorem 7, we have(94)x2lsinπx/2πx/2coshπx2−m=2π2lm−1!∑r=0l−1r2l−2r!E2l−2rmm2×∑k=0m−1−1kB˜m,kUm,2r−m,k,and from (36), (37), and the formula (see [17], Lemma 2.1),(95)ζ2;−1n=−1n+1/2π2n2n2n+1!,we obtain(96)x2n−2l∏j=1∞1−λ1−−1jx2j2m+λ=x2n−2l∏l=1m∏j=1∞1+μlλ−1jx2j2=x2n−2l∏l=1m∑j=0∞μlλjζ2;−1jx2j=∑j1+⋯+jm=n−lj1,…,jm≥0μ1λj1ζ2;−1j1⋯μmλjmζ2;−1jm=−1m−1λn−lπ2n−2l2n−lV5m,n−l.
Inserting (94) and (96) into (93), it then follows that(97)∑k=1n1+λk∑a=n∏j=1kαj+m−1αjζ2α1,…,2αk;−1α1,…,−1αk=2π2nm−1!∑l=0n−1m−1λn−l2n−lV5m,n−l∑r=0l−1r2l−2r!E2l−2rmm2×∑k=0m−1−1kB˜m,kUm,2r−m,k.
Therefore (91) follows immediately when making k-times derivative with respect to λ and then taking λ=−1 on both sides of (97). This concludes the proof of Theorem 8.Corollary 6.
Letn,k be positive integers with n≥k. Then(98)Ξ2n,k=−1n−kπ2n2n+1!∑l=0n−kn−lk2n+12l−1n−l+1/2+l2n+l×∑r=0l2l2r−1r2−22rB2rE2l−2r.Proof.
By takingm=1 in Theorem 8, in light of s1,1=1, we get the desired result after using (29) and En=2nEn1/2 for nonnegative integer n.
It is well known that the values of the Riemann zeta function at even positive integers can be computed by the Bernoulli numbers (see [29], Theorem 12.17):(99)ζ2n=−1n+122n−1π2nB2n2n!,n≥1.
Hence, by applying (99) to Corollary 6, one can easily get Zhao’s ([17], Theorem 1.3) sum formula:(100)Ξ2n,k=∑l=0n−kn−lk−1l−n−1/2+k2n+l−12n−2l+1!∑r=0lζ2l−2r¯π2n−2l+2rE2r2r!,where ζn¯=21−n−1ζn for nonnegative integer n.



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