# Algorithms for Some Euler-Type Identities for Multiple Zeta Values

Authors: Shifeng Ding; Weijun Liu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

Multiple zeta values are the numbers defined by the convergent seriesζ(s1,s2,…,sk)=∑n1>n2>⋯>nk>0(1/n1s1n2s2⋯nksk), where s1, s2, …, sk are positive integers with s1>1. For k≤n, let E(2n,k) be the sum of all multiple zeta values with even arguments whose weight is 2n and whose depth is k. The well-known result E(2n,2)=3ζ(2n)/4 was extended to E(2n,3) and E(2n,4) by Z. Shen and T. Cai. Applying the theory of symmetric functions, Hoffman gave an explicit generating function for the numbers E(2n,k) and then gave a direct formula for E(2n,k) for arbitrary k≤n. In this paper we apply a technique introduced by Granville to present an algorithm to calculate E(2n,k) and prove that the direct formula can also be deduced from Eisenstein's double product.


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## Body

## 1. Introduction

The multiple zeta sums,(1)ζ(s1,s2,…,sm)=∑n1>n2>⋯>nk>01n1s1n2s2⋯nmsm,
are also called Euler-Zagier sums, where s1,s2,…,sm are positive integers with s1≥2. Clearly, the Riemann zeta function ζ(s),
(2)ζ(s)=∑n=1∞1ns,s>1,
is the case m=1 in (1). The multiple zeta functions have attracted considerable interest in recent years.For Riemann’s zeta functionζ(s), Euler proved the following identity:
(3)∑j1+j2=nj1,j2≥1ζ(2j1)ζ(2j2)=2n+12ζ(2n).
Recently, some identities similar to (3) have also been established. Given two positive integers n and k (suppose n≥k), define a number Z(n,k) by
(4)Z(n,k)=∑j1+j2+⋯+jk=nj1,j2,…,jk≥1ζ(2j1)ζ(2j2)⋯ζ(2jk).
Then, for k∈{3,4,…,9}, the value of Z(n,k) is known [1–5].Following [6], for k≤n, let E(2n,k) be the sum of all multiple zeta values with even arguments whose weight is 2n and whose depth is k; that is,
(5)E(2n,k)=∑j1+⋯+jk=nj1,j2,…,jk≥1ζ(2j1,2j2,…,2jk).In [7], Gangl et al. proved the following identities:
(6)E(2n,2)=34ζ(2n),forn≥2,(7)∑r=1n-1ζ(2r+1,2n-2r-1)=14ζ(2n),forn≥2.Recently, using harmonic shuffle relations, Shen and Cai proved the following results in [8]:
(8)E(2n,3)=58ζ(2n)-14ζ(2)ζ(2n-2),forn≥3,E(2n,4)=3564ζ(2n)-516ζ(2)ζ(2n-2),forn≥4.In [6], applying the theory of symmetric functions, Hoffman established the generating function for the numbers E(2n,k). He proved that
(9)1+∑n≥k≥1E(2n,k)tnsk=sin(π1-st)1-ssin(πt).
Based on this generating function, some formulas for E(2n,k) for arbitrary n≥k are given. For example, Hoffman obtained that
(10)E(2n,k)=ζ(2n)22(k-1)(2k-1k)-∑j=1[(k-1)/2]ζ(2j)ζ(2n-2j)22k-3(2j+1)B2j(2k-2j-1k),
where B2j is the 2jth Bernoulli number.In this paper we use a technique introduced by Granville [9] to present an elementary recursion algorithm to calculate E(2n,k), we also give some direct formula for E(n,k) for arbitrary n≥k. Our algorithm may be of some interest if we note that it is obtained through an elementary analytic method and that the statement of the algorithm is fairly simple.


## 2. Statements of the Theorems

Theorem 1.
LetN denote a positive integer. Let a0(N),a1(N),a2(N),…, be a series of numbers defined by
(11)∏r=1r≠N∞(1+xr2-N2)=a0(N)+a1(N)x+⋯+ak-1(N)xk-1+⋯.
Then, for any two positive integers n and k with n≥k, one has
(12)E(2n,k)=∑N=1∞ak-1(N)N2n-2k+2.Theorem 2.
Given a positive integerN, we have
(13)∏r=1r≠N∞(1+xr2-N2)=2N2(-1)N+1sin(πN2-x)πxN2-x.Whenk is not large, we may use the following recursion algorithm to calculate ak(N) then use Theorem 1 to get the formula for E(2n,k).Theorem 3.
The coefficientsa0(N),a1(N),…,ak(N),…, can be calculated recursively by the following formulas:
(14)a0(N)=1;kak(N)=-∑j=0k-1bk-jaj(N),fork≥1,
where b1,b2,…, are the numbers defined by
(15)bj≜∑r=1r≠N∞1(N2-r2)j,∀j=1,2,3,….In [6], Hoffman established an interesting result [6, Lemma 1.3] to obtain his formula (10) for E(2n,k). This lemma might be deduced from the theory of Bessel functions. Using the expressions for the Bessel functions of the first kind with a half integer index, we may deduce from the generating function (13) a direct formula for ak(N).Theorem 4.
Fork≥1, one has
(16)ak-1(N)=πk-1k!(2N)k-1{sin(kπ2)∑j=0[k/2]‍(-1)j(k+2j)!(2j)!(k-2j)!·1(2Nπ)2j-cos(kπ2)×∑j=0[(k-1)/2](-1)j(k+2j+1)!(2j+1)!(k-2j-1)!·1(2Nπ)2j+1∑j=0[k/2]‍},(17)ak-1(N)=122k-2k!∑j=0[(k-1)/2]‍(-1)j(2π)2j(2k-1-2j)!(k-1-2j)!(2j+1)!·1N2k-2-2j.To deduce (17) from (16), we only need to write the expression of ak-1(N), respectively, according to whether k is odd or even, and use [(k-1)/2]-j (if k is odd) or [k/2]-j (if k is even) to replace j. In the two cases, we will get the expression (17) for ak-1(N). By Theorem 1, we have
(18)E(2n,k)=∑j=0[(k-1)/2](-1)jπ2j(2k-1-2j)!22k-2-2jk!(k-1-2j)!(2j+1)!ζ(2n-2j),
which reproduces Hoffman’s formula (10).


## 3. Proofs of the Theorems

Proof of Theorem1.
The left side of (12) is
(19)∑j1+⋯+jk=nj1,…,jk≥1ζ(2j1,2j2,…,2jk)=∑j1+⋯+jk=nj1,…,jk≥1∑n1>n2>⋯>nk>01n12j1n22j2⋯nk2jk=∑n1>n2>⋯>nk>0∑j1+⋯+jk=nj1,…,jk≥11n12j1n22j2⋯nk2jk.
The second sum in (19) is the coefficient of x2n in the formal power series
(20)∑j=1∞(x2n12)j∑j=1∞(x2n22)j⋯∑j=1∞(x2nk2)j=x2k(n12-x2)(n22-x2)⋯(nk2-x2)=∑j=1k(x2knj2-x2∏1≤m≤km≠j1nm2-nj2).
It follows that the coefficient of x2n earlier is
(21)∑j=1k‍(1nj2n-2k+2∏1≤m≤km≠j1nm2-nj2).
Hence, the sum (19) is
(22)∑n1>n2>⋯>nk>0∑j=1k1nj2n-2k+2∏1≤m≤km≠j1nm2-nj2.
Now, consider the function(23)fN(x)=∏r=1r≠N(1+xr2-N2).
We partition fN(x) into two parts. Let
(24)∏r>N(1+xr2-N2)=P0(N)+P1(N)x+P2(N)x2+⋯,∏1≤r<N(1+xr2-N2)=Q0(N)+Q1(N)x+Q2(N)x2+⋯.
Then, we have P0(N)=Q0(N)≜1, Qm(N)=0, for allm≥N, and
(25)Pj-1(N)=∑n1>n2>⋯>nj-1>N1(n12-N2)⋯(nj-12-N2),∀1<j≤k,Qk-j(N)=∑N>nj+1>nj+2>⋯>nk≥11(nj+12-N2)⋯(nk2-N2),∀1≤j<k.
Consider the sum (22). For j∈{1,2,…,k}, we treat each sum in (22) with respect to nj as follows:
(26)∑n1>n2>⋯>nk>0‍1nj2n-2k+2∏1≤m≤km≠j1nm2-nj2=∑nj=k-j+1∞‍[1nj2n-2k+2{(∑nj>nj+1>⋯>nk≥1‍∏j<m≤k‍1nm2-nj2)}×(∑n1>n2>⋯>nj-1>nj‍∏1≤m<j‍1nm2-nj2)×(∑nj>nj+1>⋯>nk≥1‍∏j<m≤k‍1nm2-nj2)]=∑nj=k-j+1∞‍1nj2n-2k+2Pj-1(nj)Qk-j(nj)=∑N=k-j+1∞‍1N2n-2k+2Pj-1(N)Qk-j(N)=∑N=1∞‍1N2n-2k+2Pj-1(N)Qk-j(N).
In the last step, N begins with 1 since Qk-j(N)=0 for 1≤N<k-j.
It follows that the sum (22) becomes that
(27)∑N=1∞‍P0(N)Qk-1(N)+P1(N)Qk-2(N)+⋯+Pk-1(N)Q0(N)N2n-2k+2.
Clearly, the sum P0(N)Qk-1(N)+P1(N)Qk-2(N)+⋯+Pk-1(N)Q0(N) in (27) is the coefficient of xk-1 in the Cauchy product of
(28)[P0(N)+P1(N)x+P2(N)x2+⋯]·[Q0(N)+Q1(N)x+Q2(N)x2+⋯];
that is, it is the coefficient of xk-1 in the power series
(29)fN(x)=a0(N)+a1(N)x+⋯+ak-1(N)xk-1+⋯.
Therefore, the sum (27) is
(30)∑N=1∞ak-1(N)N2n-2k+2.
The proof is completed.Remark 5.
If we takex to be a complex variable, then the series
(31)∑r=1r≠N∞xr2-N2
is absolutely and uniformly convergent for x in any compact set in the complex plane; thus, the function
(32)fN(x)=∏r=1r≠N∞(1+xr2-N2)
is analytic in the complex plane. Hence, it may be expanded as a Taylor series.Proof of Theorem2.
First we recall Euler’s classical formula(33)sin(πz)=πz∏r=1∞(1-z2r2),z∈ℂ.
Similar to Euler’s formula, Eisenstein studied a product of two variables and proved that for (ω,z)∈(ℂ∖ℤ)×ℂ the following formula holds (see [10, page 17]):
(34)sin[π(ω-z)]sin(πω)=∏r=-∞∞(1-zr+ω)≜limn→∞∏r=-nn(1-zr+ω).
Let N≥1 be temporarily fixed. By (34), for ω∉ℤ we have
(35)∏r=1r≠N∞(1-z2-2ωzr2-ω2)=sin[π(ω-z)]sin(πω)(1-zω)-1(1-z2-2ωzN2-ω2)-1.
Now, let ω→N. We get
(36)∏r=1r≠N∞(1-z2-2Nzr2-N2)=2N2(-1)Nsin[π(N-z)]π(N-z)(z2-2Nz).
We write z2-2Nz=-x. Or equivalently, let z=N±N2-x. Then, we get
(37)∏r=1r≠N∞(1+xr2-N2)=2N2(-1)N+1sin(πN2-x)πxN2-x.Proof of Theorem3.
Taking logarithms of both sides of (32), we get that
(38)log(fN(x))=∑r=1r≠N∞log(1-xN2-r2).
By Remark 5, the series may be differentiated term-by-term; hence, we have
(39)fN′(x)fN(x)=∑r=1r≠N∞-1N2-r2-x=-∑r=1r≠N∞∑j=0∞xj(N2-r2)j+1‍=-∑j=0∞‍(xj∑r=1r≠N∞1(N2-r2)j+1)=-∑j=0∞‍bj+1xj,
where we denote
(40)bj+1=∑r=1r≠N∞1(N2-r2)j+1,forj=0,1,2,….
The order of the summation can be changed since the series ∑r=1r≠N∞(-1/(N2-r2-x)) is dominated by ∑n=1∞(L/n2) for some positive constant L. From (39), we get that
(41)fN′(x)=-fN(x)∑j=0∞‍bj+1xj,
or
(42)∑k=1∞kak(N)xk-1=-(∑i=0∞ai(N)xi)(∑j=0∞bj+1xj).
Write out the Cauchy product in the right side of (42), then compare the coefficient of xk-1 on both sides. We get that
(43)kak(N)=-∑i+j=k-1i,j≥0bj+1ai(N)=-∑j=0k-1bk-jaj(N).Proof of Theorem4.
We now study the the generating function(44)fN(x)=2N2(-1)N+1sin(πN2-x)πxN2-x.
We may use L’Hospital’s rule to verify that
(45)limx→02N2(-1)N+1sin(πN2-x)πxN2-x=1.
Now we expand out fN(x). We have
(46)fN(x)=2N2(-1)N+1x∑m=0∞‍(-1)mπ2m(N2-x)m(2m+1)!=2N2(-1)N+1x∑m=0∞‍(-1)mπ2m(2m+1)!×∑k=0m‍(mk)N2m-2k(-1)kxk=2N2(-1)N+1∑k=0∞‍(-1)kN2k×∑m=k∞‍(-1)m(Nπ)2m(2m+1)!(mk)xk-1.
By (11) and (13), we have
(47)ak-1(N)=2N2-2k(-1)N+k+1×∑m=k∞(-1)m(Nπ)2m(2m+1)!(mk),∀k=1,2,3,….
Consider the function(48)sin(Nπx)Nπx=∑m=0∞‍(-1)m(Nπ)2m(2m+1)!xm.
Clearly, the sum in (47) can be rewritten as
(49)∑m=k∞‍(-1)m(Nπ)2m(2m+1)!(mk)=1k!(sin(Nπx)Nπx)x(k)|x=1,
where ()x(k) means the kth derivative of a function with respect to x.
We denoteϕ(x)=sinx/x. Then, we have
(50)ϕ(N2π2t)=sin(Nπt)Nπt,
and, hence,
(51)dkϕ(N2π2t)dtk=(Nπ)2kdkϕ(x)dxk|x=N2π2t,∀k=0,1,2,…,
which implies that
(52)(sin(Nπt)Nπt)t(k)|t=1=(Nπ)2k(sinxx)x(k)|x=(Nπ)2.
Finally, from (47) (49) we get that
(53)ak-1(N)=2N2π2k(-1)N+k+1k!(sinxx)x(k)|x=(Nπ)2,k≥1.
We may apply Hoffman’s result [6, Lemma 1.3] to get the direct formula for
(54)(sinxx)x(k)|x=(Nπ)2.
Here, we use some simple properties of the Bessel functions of the first kind to give its direct expression.
Lemma 6.Letk≥0be an integer and letx>0. Then one has(55)dkdxk(sinxx)=π2(-1)k2-kx-(2k+1)/4Jk+1/2(x),whereJk+1/2(x)denotes the Bessel function of the first kind of indexk+1/2.
The Bessel functions with a half-integer index can be represented by elementary functions. The following lemma is well known.
Lemma 7.Letk≥0be an integer, and letx>0. Then, one has(56)Jk+1/2(x)=2π1x{sin(x-kπ2)∑j=0[k/2]‍(-1)j(k+2j)!(2j)!(k-2j)!·1(2x)2j+cos(x-kπ2)×∑j=0[(k-1)/2]‍(-1)j(k+2j+1)!(2j+1)!(k-2j-1)!·1(2x)2j+1}.
From Lemmas 6 and 7, and (53), we get that
(57)ak-1(N)=πk-1k!(2N)k-1{sin(kπ2)∑j=0[k/2]‍(-1)j(k+2j)!(2j)!(k-2j)!·1(2Nπ)2j-cos(kπ2)×∑j=0[(k-1)/2]‍(-1)j(k+2j+1)!(2j+1)!(k-2j-1)!·1(2Nπ)2j+1∑j=0[k/2]‍}.
This completes the proof of Theorem 4.


## 4. Examples

The direct formula forak(N) can be found from Theorem 4. However, we would like to use Theorem 3 to present some concrete examples to show how to calculate ak(N) for small k. The difficult part of the recursion formula (14) is for j≥1 to calculate the sum
(58)bj=∑r=1r≠N∞1(N2-r2)j=(12N)j∑r=1r≠N∞[1N-r+1N+r]j=λj∑r=1r≠N∞[Ar+Br]j,
where we denote Ar=1/(N-r),Br=1/(N+r), and λ=1/2N.It follows fromArBr=λ[Ar+Br] that
(59)[Ar+Br]2=[Ar2+Br2]+2λ[Ar+Br],[Ar+Br]3=[Ar3+Br3]+3λ[Ar2+Br2]+6λ2[Ar+Br].
Generally, we can use induction on j to prove that if for j≥2 we have gotten some positive integers c1,c2,…,cj-1 such that
(60)[Ar+Br]j=[Arj+Brj]+c1λ[Arj-1+Brj-1]+⋯+cj-1λj-1[Ar+Br],
then the expression for [Ar+Br]j+1 is
(61)[Ar+Br]j+1=[Arj+1+Brj+1]+(1+c1)λ[Arj+Brj]+(1+c1+c2)λ2[Arj-1+Brj-1]+⋯+(1+c1+c2+⋯+cj-1)λj-1[Ar2+Br2]+2(1+c1+c2+⋯+cj-1)λj[Ar+Br].Note that ifi≥2 is an even integer, then we have
(62)∑r=1r≠N∞[Ari+Bri]=∑1≤r<N[1(N-r)i+1(N+r)i]+∑r>N[1(N-r)i+1(N+r)i]=2ζ(i)-1+2i(2N)i=2ζ(i)-(1+2i)λi.
Similarly, if i≥1 is an odd integer, then we have
(63)∑r=1r≠N∞[Ari+Bri]=-1+2i(2N)i=-(1+2i)λi.From (58)–(63), we get that(64)b1=-a1(N)=λ∑r=1r≠N∞[Ar+Br]=-3λ2,b2=λ2∑r=1r≠N∞{[Ar2+Br2]+2λ[Ar+Br]}=λ2·{[2ζ(2)-5λ2]-6λ2}=-11λ4+2ζ(2)λ2,b3=λ3∑r=1r≠N∞{[Ar3+Br3]+3λ[Ar2+Br2]+6λ2[Ar+Br]}=λ3·{-9λ3+3λ[2ζ(2)-5λ2]-18λ3}=-42λ6+6ζ(2)λ4,b4=λ4∑r=1r≠N∞{[Ar4+Br4]+4λ[Ar3+Br3]+10λ2[Ar2+Br2]+20λ3[Ar+Br]}=-163λ8+20ζ(2)λ6+2ζ(4)λ4.From formula (14), we get that
(65)a2(N)=-12{a0(N)b2+a1(N)b1}=-12{[-11λ4+2ζ(2)λ2]-(3λ2)2}=10λ4-ζ(2)λ2=58N4-ζ(2)4N2,a3(N)=-13{a0(N)b3+a1(N)b2+a2(N)b1}=-13{[-42λ6+6ζ(2)λ4]+3λ2[-11λ4+2ζ(2)λ2]-3λ2[10λ4-ζ(2)λ2]}=35λ6-5ζ(2)λ4=3564N6-5ζ(2)16N4,a4(N)=-14{a0(N)b4+a1(N)b3+a2(N)b2+a3(N)b1}=126λ8-21ζ(2)λ6-12ζ(4)λ4+12ζ2(2)λ4=63128N8-21ζ(2)64N6+3ζ(4)64N4.Fork=2,3,4, using a1(N), a2(N), and a3(N) in formula (12), respectively, we will get identities (6) and (8). Moreover, for n≥5, we have
(66)E(2n,5)=63128ζ(2n)-2164ζ(2)ζ(2n-2)+364ζ(4)ζ(2n-4).



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