Explicit evaluation of harmonic sums

0.16 MB·

by  Ce Xu

Explicit evaluation of harmonic sums
Ce Xu∗
School of Mathematical Sciences, Xiamen University
7
Xiamen 361005,P.R. China
1
0
2
Abstract In this paper, we obtain some formulae for harmonic sums, alternating harmonic
n
a sums and Stirling number sums by using the method of integral representations of series. As
J
applications of these formulae, we give explicit formula of several quadratic and cubic Euler
2
sums through zeta values and linear sums. Furthermore, some relationships between harmonic

] numbers and Stirling numbers of the first kind are established.
T
N Keywords Harmonic number; Euler sum; Riemann zeta function; Stirling number.
.
h AMS Subject Classifications (2010): 11M06; 11M32; 33B15
t
a
m
1 Introduction
[

1 In this paper, the generalized harmonic numbers and alternating harmonic numbers of order k
v
are defined respectively as
8
8 n 1 n (−1)j−1
3 ζ (k) := , L (k) := , 1 ≤ n,k ∈ Z, (1.1)
0 n jk n jk
0 Xj=1 Xj=1
.
1 n 1
0 where H ≡ ζ (1) = denotes the classical harmonic number.
n n
7 j
j=1
1 X
From [12,27], we know that the classical Euler sums are the infinite sums whose general term is
:
v product of 1 or (−1)n−1, harmonic numbers and alternating harmonic numbers of index n and
i
X a power of n−1. Therefore, the Euler sums of index
r
a
π := k ,··· ,k ,··· ,k ,··· ,k ,π := h ,··· ,h ,··· ,h ,··· ,h ,p
1  1 1 m1 m1 2  1 1 m2 m2
 q1 qm1   l1 lm2 
   
are defined by | {z } | {z } | {z } | {z }
m1 ζqi (k ) m2 Llj (h )
∞ n i n j
i=1 j=1
S ≡ S := ,
π1,π2,p mQ1kqi,mQ2hlj,p Q npQ
i=1 i j=1 j nX=1
m1 ζqi (k ) m2 Llj (h )
∞ n i n j
S¯ ≡ S¯ := i=1 j=1 (−1)n−1,
π1,π2,p mQ1kqi,mQ2hlj,p Q npQ
i=1 i j=1 j nX=1
∗Corresponding author. Email: [email protected]
1
m1 m2
where the quantity ω = π +π +p = (k q )+ (h l )+p being called the weight and the
1 2 i i j j
i=1 j=1
X X
quantities m ,m being thedegree. Here p(p > 1),m ,m ,q ,k ,h ,l are non-negative integers,
1 2 1 2 i i j j
1 ≤ k < k < ··· < k ∈Z, 1≤ h < h <··· < h ∈Z. For example,
1 2 m1 1 2 m2
∞ H3ζ5(2)ζ2(3)L2 (1)L3 (2) ∞ L (p )L (p )
S132532,1223,4 = n n n n4 n n ,S¯0,p1p2,p = n 1np n 2 (−1)n−1
n=1 n=1
X X
∞ H2ζ4(2)ζ (3)L2 (1)L3 (2) ∞ ζ (p )ζ (p )
S¯12243,1223,5 = n n n n5 n n (−1)n−1,Sp1p2,0,p = n 1npn 2 .
n=1 n=1
X X
From the definition of Euler sums, there are altogether four types of linear sums:
∞ ∞ ∞ ∞
ζ (p) ζ (p) L (p) L (p)
S = n ,S¯ = n (−1)n−1,S = n ,S¯ = n (−1)n−1.
p,0,q nq p,0,q nq 0,p,q nq 0,p,q nq
n=1 n=1 n=1 n=1
X X X X
The evaluation of linear sums in terms of values of Riemann zeta function and polylogarithm
function at positive integers is known when (p,q) = (1,3),(2,2), or p + q is odd [6,13]. For
instance, we have
∞
H 1 11 1 1 7
S¯ = n(−1)n−1 = −2Li + ζ(4)+ ζ(2)ln22− ln42− ζ(3)ln2,
1,0,3 n3 4 2 4 2 12 4
n=1 (cid:18) (cid:19)
X
∞
L (1) 3 1 1 1
S¯ = n (−1)n−1 = ζ(4)+ ζ(2)ln22− ln42−2Li .
0,1,3 n3 2 2 12 4 2
n=1 (cid:18) (cid:19)
X
In [12], Philippe Flajolet and Bruno Salvy gave explicit reductions to zeta values for all linear
sums S ,S¯ ,S ,S¯ when p+q is an odd weight. The relationship between the values
p,0,q p,0,q 0,p,q 0,p,q
oftheRiemannzetafunctionandEulersumshasbeenstudiedbymany authors,forexamplesee
[2-3,6-14,16-27]. The Riemann zeta function and alternating Riemann zeta function are defined
respectively by [1,4,5]
∞
1
ζ(s):= ,ℜ(s) > 1,
ns
n=1
X
and
∞ (−1)n−1
ζ¯(s):= , ℜ(s) ≥ 1.
ns
n=1
X
The general multiple zeta functions is defined as
1
ζ(s ,s ,··· ,s ):= ,
1 2 m ns1ns2···nsm
n1>n2>X···>nm>0 1 2 m
where s +···+s is called the weight and m is the multiplicity. In this paper, we show that
1 m
the Euler-type sums with harmonic numbers:
∞ ∞
ζ (m) L (m)
n , n ,1 ≤ m,k ∈ Z
n(n+k) n(n+k)
n=1 n=1
X X
2
can be expressed in terms of series of Riemann zeta values and harmonic numbers. We also
provide an explicit evaluation of
∞
S(n+1,p)
(p−1)! ,2 ≤ p ∈ Z,1 ≤ k ∈ Z
n!n(n+k)
n=p−1
X
in a closed form in terms of zeta values and the complete exponential Bell polynomial Y (n),
k
S(n,k) stands for the Stirling number of the first kind. Specifically, we investigate closed-form
representations for sums of the following form:
∞ H2 ∞ H L (1) ∞ H L (1)
S12,0,p = npn,S1,1,p = nnnp ,S¯1,1,p = nnnp (−1)n−1,
n=1 n=1 n=1
X X X
∞ ∞ ∞
H2 L2 (1) L2 (1)
S¯12,0,p = npn(−1)n−1,S0,12,p = nnp ,S¯0,12,p = nnp (−1)n−1.
n=1 n=1 n=1
X X X
Furthermore, we evaluate several other series involving harmonic numbers. For example
H3+3H ζ (2)+2ζ (3)
3ζ(3)+ k k k k
∞ H2 1 3
nX=1n(n+n k) = k −Hk2+kζk(2) −k−1 Hi2i +ζ(2)Hk−1. (1.2)
i=1
X
 
 
 
 
2 Main Theorems and Proof
In this section, we usecertain integral of polylogarithm function representations to evaluate sev-
eral series with alternating (or non-alternating) harmonic numbers. The polylogarithm function
defined as follows
∞ xn
Li (x) := ,ℜ(p) > 1, |x| ≤ 1.
p np
n=1
X
when x takes 1 and −1, then the function Li (x) are reducible to Riemann zeta function and
p
alternating Riemann zeta function, respectively.
Theorem 2.1 For positive integers m and k, then
∞ m−1 k−1
ζ (m) 1 H
n = ζ(m+1)+ (−1)j−1ζ(m+1−j)ζ (j)+(−1)m−1 i , (2.1)
n(n+k) k  k−1 im
nX=1  Xj=1 Xi=1 
 m−1 
ζ¯(m+1)+ (−1)j−1ζ¯(m+1−j)ζ (j)
∞ k−1
nX=1nL(nn(+mk)) = k1 +(−1)m−1lnXj2=(1ζk−1(m)+Lk−1(m))+(−1)mk−1 (−i1m)i−1Li(1). (2.2)
i=1
 X 
 
Proof. By the definition of polylogarithm function and Cauchy product formula, we can verify
that
∞
Li (x)
m = ζ (m)xn, x ∈ (−1,1), (2.3)
n
1−x
n=1
X
3
∞
Li (−x)
− m = L (m)xn, x ∈ (−1,1). (2.4)
n
1−x
n=1
X
Multiplying (2.3) and (2.4) by xr−1−xk−1 and integrating over (0,1), we obtain
1 ∞
Li (x) ζ (m)
xr−1−xk−1 m dx = (k−r) n (0 ≤ r < k, r,k ∈ Z), (2.5)
1−x (n+r)(n+k)
Z0 (cid:16) (cid:17) nX=1
1 ∞
Li (−x) L (m)
xk−1−xr−1 m dx = (k−r) n (0 ≤ r < k, r,k ∈ Z). (2.6)
1−x (n+r)(n+k)
Z0 (cid:16) (cid:17) Xn=1
We now evaluate the integral on the left side of (2.5) and (2.6). Noting that
1 k−r 1
Li (x)
xr−1−xk−1 m dx = xr+i−2Li (x)dx, (2.7)
m
1−x
Z0 (cid:16) (cid:17) Xi=1Z0
1 k−r −1
Li (−x)
xk−1−xr−1 m dx = (−1)r+i xr+i−2Li (x)dx. (2.8)
m
1−x
Z0 (cid:16) (cid:17) Xi=1 Z0
Using integration by parts we have
x q−1 xn (−1)q (−1)q n xk
tn−1Li (t)dt = (−1)i−1 Li (x)+ ln(1−x)(xn −1)− .
q ni q+1−i nq nq k
!
Z i=1 k=1
0 X X
(2.9)
Taking r = 0 in (2.5)-(2.8), substituting (2.9) into (2.5)-(2.8), we can obtain (2.1) and (2.2). (cid:3)
If taking r ≥ 1 in (2.5)-(2.8), using (2.9), we have
m−1
(−1)j−1ζ(m+1−j)(ζ (j)−ζ (j))
∞ k−1 r−1
(n+ζrn)((mn)+k) = k−1 r Xj=1 k−1 H r−1 H , (2.10)
nX=1 +(−1)m−1 i − i 
im im
!
i=1 i=1
 X X 
 
 
m−1 
(−1)j−1ζ¯(m+1−j)(ζ (j)−ζ (j))
k−1 r−1
 
∞ Ln(m) = 1 +Xj=(1−1)m−1ln2(ζ (m)−ζ (m)+L (m)−L (m)).
nX=1(n+r)(n+k) k−r +(−1)mk−1 (−1)ki−−11L (1) r−1 k−1 r−1 
im i
 
 Xi=r (2.11)
 
 
Letting m =1 in (2.1) and (2.2), we conclude that 
∞
H 1 1 1 H
n = H2+ ζ (2)+ζ(2)− k , (2.12)
n(n+k) k 2 k 2 k k
n=1 (cid:18) (cid:19)
X
4
1+(−1)k−1
∞ Ln(1) = 1 ζ¯(2)+ln2(Hk +Lk(1))−ln2 k . (2.13)
Xn=1n(n+k) k −12 L2k(1)+ζk(2) + Lkk(1)(−1)k−1 
 (cid:0) (cid:1) 
 
Theorem 2.2 For integers n ≥1,k ≥ 0, we have
1
Y (n)
tn−1lnk(1−t)dt = (−1)k k , (2.14)
n
Z
0
where Y (n)= Y (ζ (1),1!ζ (2),2!ζ (3),··· ,(r−1)!ζ (r),···), Y (x ,x ,···) stands for the
k k n n n n k 1 2
complete exponential Bell polynomial is defined by (see [15])
tm tk
exp x = 1+ Y (x ,x ,···) . (2.15)
m k 1 2
 m! k!
m≥1 k≥1
X X
 
Proof. Using the definition of the complete exponential Bell polynomial, it is easily shown that
tk 1
1+ Y (n) =
k
k! t t
k≥1 (1−t) 1− ··· 1−
X 2 n
(cid:18) (cid:19) (cid:18) (cid:19)
and
n
1 t
= 1+
t t t t
(1−t) 1− ··· 1− m=1m(1−t) 1− ··· 1−
2 n X 2 m
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Therefore we obtain
n
Y (m)
Y (n)= k k−1 , Y (n)= 1. (2.16)
k 0
m
m=1
X
It is easily shown using integration by parts that
x 1 n xj
tn−1ln(1−t)dt = xnln(1−x)− −ln(1−x) ,−1≤ x < 1.
n j 
Z0  Xj=1 
Letting x → 1−, we have  
x H Y (n)
lim tn−1ln(1−t)dt = − n = − 1 . (2.17)
x→1−Z0 n n
Using integration by parts and (2.17) , we can find that
x 1 2 n 1 k xj
tn−1ln2(1−t)dt = (xn −1)ln2(1−x)− xkln(1−x)− −ln(1−x)
n n k  j 
Z0 Xk=1  Xj=1 
(2.18)
 
5
Letting x tend to 1− in (2.18), we deduce that
x 2 n 1 k 1 Y (n)
lim tn−1ln2(1−t)dt = = 2 . (2.19)
x→1−Z0 n k=1 k j=1 j n
X X
Further, it is easily verify that
x (−1)m n 1 k1 1 km−1 1 Y (n)
lim tn−1lnm(1−t)dt = m! ··· = (−1)m m ,1 ≤ m ∈ Z.
x→1−Z0 n kX1=1 k1 kX2=1k2 kXm=1km n
(2.20)
We complete the proof of (2.14). (cid:3)
From the definition of the complete exponential Bell polynomial, we have
Y (n)= H ,Y (n)= H2+ζ (2),Y (n)= H3+3H ζ (2)+2ζ (3),
1 n 2 n n 3 n n n n
Y (n)= H4+8H ζ (3)+6H2ζ (2)+3ζ2(2)+6ζ (4).
4 n n n n n n n
Theorem 2.3 For integer p ≥ 2,k ≥ 1, we have
∞
S(n+1,p) 1 Y (k) Y (k)
p p−1
(p−1)! = (p−1)!ζ(p)+ − . (2.21)
n!n(n+k) k p k
n=p−1 (cid:26) (cid:27)
X
where S(n,k) are the Stirling number of the first kind defined by
n
x x
n!(1+x) 1+ ··· 1+ = S(n+1,k+1)xk.
2 n
(cid:16) (cid:17) (cid:16) (cid:17) Xk=0
From the definition S(n,k), we can rewrite it as
n n
x
S(n+1,k+1)xk = n!exp ln 1+
 j 
Xk=0 Xj=1 (cid:18) (cid:19)
 n ∞ xk
= n!exp (−1)k−1
 kjk
Xj=1Xk=1 
∞ ζ (k)xk
= n!exp (−1)k−1 n  .
k
( )
k=1
X
Therefore, we know that S(n,k) is a rational linear combination of products of harmonic num-
bers. The following identities is easily derived
(n−1)!
S(n,1)= (n−1)!,S(n,2)= (n−1)!H ,S(n,3) = H2 −ζ (2) ,
n−1 2 n−1 n−1
(n−1)! (cid:2) (cid:3)
S(n,4)= H3 −3H ζ (2)+2ζ (3) ,
6 n−1 n−1 n−1 n−1
(n−1)!(cid:2) (cid:3)
S(n,5)= H4 −6ζ (4)−6H2 ζ (2)+3ζ2 (2)+8H ζ (3) .
24 n−1 n−1 n−1 n−1 n−1 n−1 n−1
(cid:2) (cid:3)
6
Proof. From [15], we have
∞ xn
lnp(1−x)= (−1)pp! S(n,p) ,1 ≤ p ∈ Z,−1≤ x < 1. (2.22)
n!
n=p
X
Differentiating this equality, we obtain
lnp−1(1−x) ∞ xn
= (−1)p−1(p−1)! S(n+1,p) ,p ≥ 2. (2.23)
1−x n!
n=p−1
X
Let k,r be integers with k > r ≥ 0, then multiplying (2.23) by xr−1−xk−1 and integrating over
(0,1), we get
1 lnp−1(1−x) ∞ (k−r)S(n+1,p)
xr−1−xk−1 dx = (−1)p−1(p−1)! ,p ≥ 2. (2.24)
1−x n!(n+r)(n+k)
Z0 (cid:16) (cid:17) n=Xp−1
Noting that
1 lnp−1(1−x) k−r 1
xr−1−xk−1 dx = xr+i−2lnp−1(1−x)dx (2.25)
1−x
Z0 (cid:16) (cid:17) Xi=1Z0
and
1
lnp−1(1−x)
dx = (−1)p−1(p−1)!ζ(p), 2 ≤ p ∈ Z. (2.26)
x
Z
0
Taking r = 0,k ≥ 1 in (2.24), using (2.14)(2.16)(2.25) and (2.26), we can obtain (2.21). (cid:3)
If taking r ≥ 1 in (2.24), then we have
∞
S(n+1,p) 1
(p−1)!(k−r) = {Y (k−1)−Y (r−1)}. (2.27)
p p
n!(n+r)(n+k) p
n=p−1
X
Letting p = 3,4 in (2.8), we can give the following Corollary.
Corollary 2.4 If 1≤ k ∈ Z, then we have
∞ H2−ζ (2) 1 H3+3H ζ (2)+2ζ (3) H2+ζ (2)
n n = 2ζ(3)+ k k k k − k k , (2.28)
n(n+k) k 3 k
n=1 (cid:26) (cid:27)
X
H4+8H ζ (3)+6H2ζ (2)+3ζ2(2)+6ζ (4)
∞ H3−3H ζ (2)+2ζ (3) 1 k k k k k k k
n n n n = 4 .
nX=1 n(n+k) k −Hk3+3Hkζkk(2)+2ζk(3) +6ζ(4) 
(2.29)
 
 
From (2.1), taking m = 2, we have
∞ k−1
ζ (2) 1 H
n i
= ζ(3)+ζ(2)H − . (2.30)
n(n+k) k k−1 i2
( )
n=1 i=1
X X
Substituting (2.30) into (2.28), we obtain (1.2). In the same manner, we obtain the following
Theorem:
7
Theorem 2.5 For integer m,k > 0, then
∞
ζ (m)
n (−1)n−1
n+k
n=1
X
∞
ζ (m)
= (−1)k n (−1)n−1−ζ¯(m+1) +(−1)k+m+1ln2(ζ (m)+L (m))
n k−1 k−1
!
n=1
X
m−1 k−1
L (1)
+(−1)k (−1)j−1ζ¯(m+1−j)L (j)+(−1)m+k i , (2.31)
k−1 im
j=1 i=1
X X
∞
L (m)
n (−1)n−1
n+k
n=1
X
∞ k−1
L (m) H
= (−1)k−1 ζ(m+1)− n (−1)n−1+(−1)m i(−1)i−1
n im
!
n=1 i=1
X X
m−1
+(−1)k (−1)j−1ζ(m+1−j)L (j) (2.32)
k−1
j=1
X
From (2.31) and (2.32), we obtain
∞ k−1
H 1 L (1)
n (−1)n−1 = (−1)k−1 ln22−ln2(H +L (1))+ i , (2.33)
n+k 2 k−1 k−1 i
!
n=1 i=1
X X
∞ L (1) ζ(2)−ln22 k−1 H
n (−1)n−1 = (−1)k−1 − i(−1)i−1 . (2.34)
n+k 2 i
!
n=1 i=1
X X
Theorem 2.6 For 1≤ l ,l ,m ∈ Z and x,y,z ∈ [−1,1), we have the following relation
1 2
∞ ∞ ∞
ζ (l ,x)ζ (l ,y) ζ (l ,x)ζ (m,z) ζ (l ,y)ζ (m,z)
n 1 n 2 zn+ n 1 n yn+ n 2 n xn
nm nl2 nl1
n=1 n=1 n=1
X X X
∞ ∞ ∞
ζ (m,z) ζ (l ,x) ζ (l ,y)
= n (xy)n+ n 1 (yz)n+ n 2 (xz)n
nl1+l2 nm+l2 nl1+m
n=1 n=1 n=1
X X X
+Li (z)Li (x)Li (y)−Li (xyz) (2.35)
m l1 l2 l1+l2+m
n xk
where the partial sum ζ (l,x) of polylogarithm function is defined by ζ (l,x) := .
n n kl
k=1
X
∞
Proof. We construct the function F (x,y,z) := {ζ (l ,x)ζ (l ,y)−ζ (l +l ,xy)}zn−1,
n 1 n 2 n 1 2
n=1
X
z ∈(−1,1). By the definition of ζ (l,x), we have
n
∞
ζ (l ,x) ζ (l ,y)
F (x,y,z)= zF (x,y,z)+ n 1 yn+1+ n 2 xn+1 zn. (2.36)
((n+1)l2 (n+1)l1 )
n=1
X
8
Moving zF(x,y,z) from right to left and then multiplying (1−z)−1 to the equation (2.36) and
integrating over the interval (0,z), we obtain
∞
ζ (l ,x)ζ (l ,y)−ζ (l +l ,xy)
n 1 n 2 n 1 2 zn
n
n=1
X
∞
ζ (l ,x) ζ (l ,y)
= n 1 yn+1+ n 2 xn+1 {Li (z)−ζ (1,z)}. (2.37)
((n+1)l2 (n+1)l1 ) 1 n
n=1
X
Furthermore, using integration and the following formula
∞
ζ (l ,x) ζ (l ,y)
n 1 yn+1+ n 2 xn+1 = Li (x)Li (y)−Li (xy),
((n+1)l2 (n+1)l1 ) l1 l2 l1+l2
n=1
X
we deduce (2.35) to complete the proof of Theorem 2.6. (cid:3)
Letting (x,y,z) = (−1,−1,1), (l ,l ,m) = (1,1,m) and (x,y,z) = (−1,−1,−1), (l ,l ,m) =
1 2 1 2
(1,1,m) in (2.35) gives the following
Corollary 2.7 For integer m > 1 we have that
∞ ∞
L2 (1) L (1)ζ (m)
n +2 n n (−1)n−1
nm n
n=1 n=1
X X
∞ ∞
ζ (m) L (1)
= n +2 n (−1)n−1+ln22ζ(m)−ζ(m+2), (m > 1) (2.38)
n2 nm+1
n=1 n=1
X X
∞ L2 (1) ∞ L (1)L (m)
n (−1)n−1+2 n n (−1)n−1
nm n
n=1 n=1
X X
∞ ∞
L (m) L (1)
= n +2 n +ln22ζ¯(m)−ζ¯(m+2), (m > 0). (2.39)
n2 nm+1
n=1 n=1
X X
In fact, multiplying (2.35) by (1−z)−1 and integrating over (0,z) (z ∈ (−1,1)), we can obtain
the following Corollary.
Corollary 2.8 For positive integers l > 0,l > 0,m > 1 and x,y,z ∈ [−1,1), then we have
1 2
n n
ζ (1,z) ζ (1,z)
k k
ζ (l ,x) ζ (l ,y)
∞ ∞ n 1 km ∞ n 2 km
ζ (l ,x)ζ (l ,x)ζ (1,z) ! !
n 1 n 2 n + Xk=1 yn+ Xk=1 xn
nm nl2 nl1
n=1 n=1 n=1
X X X
n
ζ (1,z)
k
∞ km ∞ ∞
! ζ (l ,x)ζ (1,z) ζ (l ,y)ζ (1,z)
= Xk=1 xnyn+ n 1 n yn+ n 2 n xn
nl1+l2 nm+l2 nm+l1
n=1 n=1 n=1
X X X
∞ ∞
ζ (1,z) ζ (1,z)
+Li (x)Li (y) n − n xnyn . (2.40)
l1 l2 nm nm+l1+l2
! !
n=1 n=1
X X
9
Letting x,y,z → 1 in (2.40), we arrive at the conclusion that, for integers m,l ,l > 1
1 2
n n
H H
k k
ζ (l ) ζ (l ,y)
∞ ∞ n 1 km ∞ n 2 km
H ζ (l )ζ (l ) ! !
n n 1 n 2 k=1 k=1
+ X + X
nm nl2 nl1
n=1 n=1 n=1
X X X
n
H
k
∞ km ∞ ∞ ∞ ∞
! H ζ (l ) H ζ (l ) H H
= Xk=1 + n n 1 + n n 2 +ζ(l )ζ(l ) n − n .
nl1+l2 nm+l2 nm+l1 1 2 nm nm+l1+l2
! !
n=1 n=1 n=1 n=1 n=1
X X X X X
(2.41)
From [13], we derive the following identity
∞ n ∞ ∞ ∞
1 H H H H ζ (p+1)
k n n n n
= +ζ(p+1) − . (2.42)
np+1 km nm+p+1 nm nm
!
n=1 k=1 n=1 n=1 n=1
X X X X X
Therefore, we may rewrite (2.41) as
n n
H H
k k
ζ (l ) ζ (l ,y)
∞ ∞ n 1 km ∞ n 2 km
H ζ (l )ζ (l ) ! !
n n 1 n 2 k=1 k=1
+ X + X
nm nl2 nl1
n=1 n=1 n=1
X X X
∞
H ζ (l ) H ζ (l ) H ζ (l +l )
n n 1 n n 2 n n 1 2
= + −
nm+l2 nm+l1 nm
n=1(cid:26) (cid:27)
X
∞ ∞
H H
n n
+ζ(l +l ) +ζ(l )ζ(l ) . (2.43)
1 2 nm 1 2 nm
! !
n=1 n=1
X X
Taking l = l = m = 2l+1 (l is a positive integer) in (2.43), we conclude that
1 2
n
H
k
ζ (2l+1)
∞ H ζ2(2l+1) ∞ n k2l+1!
n n +2 Xk=1
n2l+1 n2l+1
n=1 n=1
X X
∞ ∞
H ζ (2l+1) H ζ (4l+2) H
= 2 n n − n n + ζ(4l+2)+ζ2(2l+1) n . (2.44)
n4l+2 n2l+1 n2l+1
!
n=1(cid:26) (cid:27) n=1
X (cid:0) (cid:1) X
3 Closed form of quadratic Euler sums
In this section we evaluate some quadratic Euler sums involving harmonic numbers and alter-
nating harmonic numbers.
Theorem 3.1 For integers l > 0,m > 0,p > 1, we have
∞ n ∞ n
ζ (l) H ζ (m) H
(−1)m−1 n k −(−1)p+l n k
np+1 km np+1 kl
! !
n=1 k=1 n=1 k=1
X X X X
10