Explicit Evaluations of Sums of Sequence Tails

0.12 MB·

by  Ce Xu

Explicit Evaluations of Sums of Sequence Tails
Ce Xu∗ Xiaolian Zhou†
School of Mathematical Sciences, Xiamen University
7
Xiamen 361005,P.R. China
1
0
2

n Abstract In this paper, we use Abel’s summation formula to evaluate several quadratic sums
a of the form:
J
N

4 F (A,B;x) := (A−A )(B−B )xn, x ∈[−1,1],
N n n

] nX=1
T
where the sequences A ,B are defined by the sums
n n
N
n n
.
th An = ak,Bn = bk, ak,bk = o(n−p),ℜ(p) > 1
a k=1 k=1
m X X
[ and A = lim An,B = lim Bn,F (A,B;x) = lim Fn(A,B;x). We give an explicit formula
n→∞ n→∞ n→∞
1 of Fn(A,B;x). Furthermore, we also evaluate several other series involving multiple zeta star
v values.
5
2 Keywords Sequence; harmonic numbers; Abel’s summation formula; Riemann zeta function;
7
multiple zeta star values, tails.
3
0 AMS Subject Classifications (2010): 11M06; 11M32.
.
1
0
7 1 Introduction
1
:
v In recent years there has been significant interest in investigating sums of products of Riemann
i
X zeta tails and multiple zeta values. The subject of this paper is studying quadratic and cubic
r sums
a
∞ ∞
F (A,B;x) = (A−A )(B −B )xn,F (A,B,ζ(r))= (A−A )(B −B )(ζ(r)−ζ (r)),
n n n n n
n=1 n=1
X X
where
n n
1 1
H = ζ (1) = , ζ (r)= , r,n ∈ N= {1,2,3···}
n n j n jr
j=1 j=1
X X
are classical harmonic number and generalized harmonic numbers or partial sums of the series
∞
1
ζ(s)= , ℜ(s)> 1
js
j=1
X
∗Corresponding author. Email: [email protected]
†Email: [email protected](X. ZHOU)
1
defining the Riemann zeta function, respectively, where the empty sum ζ0(r) is conventionally
understood to be zero. In recent papers [9,10], O. Furdui, A. Sˆıntˇamaˇrian and C. V˘alean prove
some results on sums of products of the tails of the Riemann zeta function, i.e.,
∞
1
ζ(s)−ζ (s)= .
n js
j=n+1
X
In [9], O. Furdui and A. Sˆıntˇamaˇrian shown that quadratic series
∞
F(ζ(s),ζ(s)) = (ζ(s)−ζ (s))2, s ∈ N/{1}
n
n=1
X
can be expressed as a rational linear combination of zeta function values, binomial coefficients
and products of two zeta function values. In [10], O. Furdui and C. V˘alean proven that the
series
∞
F(ζ(s),ζ(s+1)) = (ζ(s)−ζ (s))(ζ(s+1)−ζ (s+1)), s∈ N/{1}
n n
n=1
X
can be expressed in terms of Riemann zeta function values and a special integral involving a
polylogarithm function. The polylogarithm function defined as follows
∞ xn
Li (x) = ,ℜ(p)> 1, |x|≤ 1.
p np
n=1
X
Furthermore, in [12], M. E. Hoffman shown a simple general result for the quadratic sum
∞
F(ζ(p),ζ(q)) = (ζ(p)−ζ (p))(ζ(q)−ζ (q)), p,q ∈ N/{1}
n n
n=1
X
in terms of the multiple zeta values ζ(s1,s2,...,sm) defined by
1
ζ(s1,s2,...,sm) = ks1···ksm,
k1>··X·>km≥1 1 m
wherefor convergence s1+s2+···+sj > j for j = 1,2,...,m. Moreover, Hoffman also provided
an explicit evaluation of sums of products of three or more Riemann zeta tails in a closed form
in terms of multiple zeta values (see Theorem 4.3 in the reference [12]). The multiple zeta value
ζ(s1,s2,...,sm)issaid tohavedepthmandweight s1+...+sm. Multiplezeta valuesofgeneral
depthwereintroducedindependentlyby Hoffman [11] andD.Zagier [19], butfordepthtwo they
were already studied by Euler. In [5], Borwein and Girgensohn gave some evaluation multiple
zeta values of depth three.
In this paper, we discuss the analytic representations of F(A,B;x) and F(A,B,ζ(r)). Using
certain integral representations, we can prove that the cubic series
∞
F (ζ(p),ζ(q),ζ(r)) = (ζ(p)−ζ (p))(ζ(q)−ζ (q))(ζ(r)−ζ (r)), p,q,r ∈ N/{1}
n n n
n=1
X
can be expressed in terms of multiple zeta star values defined by
1
ζ⋆(s1,s2,...,sm)= ks1···ksm, s1 ∈ N/{1},si ∈ N,(i = 2,3,...).
k1≥··X·≥km≥1 1 m
2
∞
ζ (p)
Obviously, ζ⋆(m,p) = n . Furthermore,wealsogive someevaluation ofsumsofproducts
nm
n=1
X
of alternating Riemann Zeta Tails. For example,
∞
(ζ(m)−ζ (m)) ζ¯(p)−ζ¯ (p)
n n
n=1
X (cid:0) (cid:1)
∞ ζ (m−1) ζ¯ (p−1)
= n (−1)n−1+ n −ζ(m)ζ¯(p)−ζ¯(m+p−1),
np nm
n=1(cid:26) (cid:27)
X
whereζ¯(p)andζ¯ (p)standforthealternatingRiemannZetafunctionandalternating harmonic
n
numbers defined by
∞ k−1 n k−1
(−1) (−1)
ζ¯(p)= ,ζ¯ (p)= ,n ∈N,ℜ(p) ≥ 1.
kp n kp
k=1 k=1
X X
∞ ζ (p) ∞ ζ¯ (p) ∞ ζ (p) ∞ ζ¯ (p)
n n n−1 n n−1 n
The evaluation of linear sums , (−1) , (−1) and in
nm nm nm nm
n=1 n=1 n=1 n=1
X X X X
terms of values of Riemann zeta function (or alternating Riemann Zeta function) and polylog-
arithm function at positive integers is known when (p,m) = (1,3),(2,2) or p+m is odd (see
[2,3,7,14,17,18]). For example,
∞ ζ¯ (1) 7 5
n
= ζ(3)ln2− ζ(4),
n3 4 16
n=1
X
∞
H 1 11 1 1 7
n n−1 2 4
n3 (−1) = −2Li4 2 + 4 ζ(4)+ 2ζ(2)ln 2− 12ln 2− 4ζ(3)ln2,
n=1 (cid:18) (cid:19)
X
∞ ζ¯ (1) 3 1 1 1
n n−1 2 4
n3 (−1) = 2ζ(4)+ 2ζ(2)ln 2− 12ln 2−2Li4 2 ,
n=1 (cid:18) (cid:19)
X
∞
ζ (2) 51 1 7 1
n n−1 2 4
n2 (−1) = −16ζ(4)+4Li4 2 + 2ζ(3)ln2−ζ(2)ln (2)+ 6ln 2.
n=1 (cid:18) (cid:19)
X
Furthermore, we consider the nested sum
∞ yn n xk
, x,y ∈ [−1,1), m,p ∈ N.
nm kp
!
n=1 k=1
X X
By takingthesumover complementary pairsofsummationindices, weobtain asimplereflection
formula
∞ yn n xk ∞ xn n yk
nm kp + np km = Lip(x)Lim(y)+Lip+m(xy).
! !
n=1 k=1 n=1 k=1
X X X X
Therefore, we can deduce that
∞ ζ¯ (2) 13
n n−1
(−1) = ζ(4),
n2 16
n=1
X
3
∞
ζ (3) 19 3
n n−1
(−1) = ζ(4)− ζ(3)ln2,
n 16 4
n=1
X
∞ ζ¯ (2) 85 1 1 7
n 2 4
n2 = 16ζ(4)−4Li4 2 +ζ(2)ln 2− 6ln 2− 2ζ(3)ln2,
n=1 (cid:18) (cid:19)
X
∞ ζ¯ (3) 1 3 1 1 1
n n−1 4 2
(−1) =2Li4 + ζ(3)ln2+ ln 2− ζ(4)− ζ(2)ln 2.
n 2 4 12 2 2
n=1 (cid:18) (cid:19)
X
Some results for sums of harmonic numbers or alternating harmonic numbers may be seen in
the works of [2,3,7,13-18] and references therein.
The main result of this paper is the following theorems.
Theorem 1.1 If a ,b = o(n−p),ℜ(p) > 1 and x ∈ [−1,1], then
k k
n
(1−x)F (A,B;x) = (1−x) (A−A )(B −B )xk
n k k
k=1
X
n n
= ABx−(A−A )(B −B )xn+1− a xk (B −B )− b xk (A−A )
n n k n k n
! !
k=1 k=1
X X
n k n k n
− a xi b − b xi a + a b xk. (1.1)
i k i k k k
! !
k=1 i=1 k=1 i=1 k=1
X X X X X
Theorem 1.2 If a ,b = o(n−p),ℜ(p) > 1 and r ∈ N/{1}, then
k k
∞
F (A,B,ζ(r))= (A−A )(B −B )(ζ(r)−ζ (r))
k k k
k=1
X
∞ n ∞ n ∞
=ζ(r) ka b + kb a − na b −AB
k n k n n n
( ! ! )
n=1 k=1 n=1 k=1 n=1
X X X X X
∞ n ∞ n
− a (kζ (r)−ζ (r−1)) b − b (kζ (r)−ζ (r−1)) a
k k k n k k k n
! !
n=1 k=1 n=1 k=1
X X X X
∞
+ a b (nζ (r)−ζ (r−1)). (1.2)
n n n n
n=1
X
In section 2, we will use Abels summation formula to prove Theorem 1.1 and 1.2. The Abels
summationbypartsformulastatesthatif(an)n≥1 and(bn)n≥1 aretwosequencesofrealnumbers
n n n
and An = ak, then akbk = Anbn+1+ Ak(bk −bk+1).
k=1 k=1 k=1
X X X
2 Proof of Theorem 1.1 and Theorem 1.2
We now prove our Theorems. Firstly, we are ready to prove Theorem 1.1.
Proof. From the definition F (A,B;x), we have
n
n
F (A,B;x) = (A−A )(B−B )xk
n k k
k=1
X
4
n−1
= (A−a1)(B −b1)x+ (A−Ak+1)(B −Bk+1)xk+1
k=1
X
n−1
= (A−a1)(B −b1)x+ (A−Ak −ak+1)(B −Bk −bk+1)xk+1
k=1
X
n−1
= (A−a1)(B −b1)x+ (A−Ak)(B −Bk)xk+1
k=1
X
n−1 n−1
− {ak+1(B−Bk)+bk+1(A−Ak)}xk+1+ ak+1bk+1xk+1
k=1 k=1
X X
n
= (A−a1)(B −b1)x+ (A−Ak)(B −Bk)xk+1−(A−An)(B−Bn)xn+1
k=1
X
n n
+ akbkxk −a1b1x− {ak(B−Bk +bk)+bk(A−Ak +ak)}xk +a1B +b1A
k=1 k=1
X X
= xF (A,B;x)+ABx−(A−A )(B −B )xn+1
n n n
n n
− {a (B −B )+b (A−A )}xk − a b xk. (2.1)
k k k k k k
k=1 k=1
X X
By Abel’s summation formula, we can deduce that
n n n k n
{a (B −B )}xk = a xk (B−B )+ a xi b − a b xk, (2.2)
k k k n i k k k
! !
k=1 k=1 k=1 i=1 k=1
X X X X X
n n n k n
{b (A−A )}xk = b xk (A−A )+ b xi a − a b xk. (2.3)
k k k n i k k k
! !
k=1 k=1 k=1 i=1 k=1
X X X X X
Substituting (2.2) (2.3) into (2.1) respectively, we can obtain (1.1). We have completed the
lnr−1x
proof of Theorem 1.1. Now, we are ready to prove Theorem 1.2. Multiplying (1.2) by
2
(1−x)
and integrating over the interval (0,1), and using the following elementary integral identities
1
xklnr−1x
dx = (−1)r−1(r−1)!(ζ(r)−ζ (r)), r ∈ N/{1},
k
1−x
Z0
1
xklnr−1x
dx = (−1)r−1(r−1)!(ζ(r−1)−ζ (r−1)−kζ(r)+kζ (r)),r ∈ N/{1,2},
2 k k
(1−x)
Z0
by a direct calculation, we obtain
F (A,B,ζ(r))=AB(ζ(r−1)−ζ(r))
∞
+ a b (ζ(r−1)−ζ (r−1)−kζ(r)+kζ (r))
k k k k
k=1
X
5
∞ n
− a (ζ(r−1)−ζ (r−1)−kζ(r)+kζ (r)) b
k k k n
!
n=1 k=1
X X
∞ n
− b (ζ(r−1)−ζ (r−1)−kζ(r)+kζ (r)) a . (2.4)
k k k n
!
n=1 k=1
X X
Taking x = 1 in (1.1), we get
n n n
A b + B a = a b +A B . (2.5)
k k k k k k n n
k=1 k=1 k=1
X X X
Letting n → ∞, then have
∞ ∞ ∞
A b + B a = a b +AB. (2.6)
k k k k k k
k=1 k=1 k=1
X X X
Substituting (2.6) into (2.4), we know that when r ∈ N/{1,2}, the equality (1.2) is holds.
lnx
Similarly, multiplying (1.2) by and integrating over the interval (0,t), then letting
2
(1−x)
t → 1, we can prove that if r = 2, the formula (1.2) is also true. The proof of Theorem 1.2 is
finished.
3 Some results
In this section, we will give some closed form of sums of sequences tails.
Firstly, multiplying (1−x)−1 in both sides of (1.1), then letting x → 1 and using L. Hopital’s
rule, we deduce that
n
(1−x)F (A,B;x)
n
(A−A )(B −B )= F (A,B;1) = lim
k k n
x→1 1−x
k=1
X
n k n k n n
= ia b + ib a + ka (B −B )+ kb (A−A )
i k i k k n k n
! ! ! !
k=1 i=1 k=1 i=1 k=1 k=1
X X X X X X
n
+(n+1)(A−A )(B −B )− ka b −AB. (3.1)
n n k k
!
k=1
X
Letting n → ∞ in (3.1), we have
∞ ∞ n ∞ n ∞
(A−A )(B −B )= ka b + kb a − na b −AB. (3.2)
n n k n k n n n
! !
n=1 n=1 k=1 n=1 k=1 n=1
X X X X X X
n−1 n−1 n−1
1 1 (−1) (−1) 1 (−1)
Putting (a ,b ) = , , , , , in (3.2), we can obtain
n n nm np nm np nm np
! !
(cid:18) (cid:19)
the following identities
∞
(ζ(m)−ζ (m))(ζ(p)−ζ (p))
n n
n=1
X
6
= ζ⋆(p,m−1)+ζ⋆(m,p−1)−ζ(m+p−1)−ζ(m)ζ(p), (3.3)
∞
ζ¯(m)−ζ¯ (m) ζ¯(p)−ζ¯ (p)
n n
n=1
X(cid:0) (cid:1)(cid:0) (cid:1)
∞ ζ¯ (p−1) ∞ ζ¯ (m−1)
= n (−1)n−1+ n (−1)n−1−ζ(m+p−1)−ζ¯(m)ζ¯(p), (3.4)
nm np
n=1 n=1
X X
∞
(ζ(m)−ζ (m)) ζ¯(p)−ζ¯ (p)
n n
n=1
X (cid:0) (cid:1)
∞ ζ (m−1) ζ¯ (p−1)
= n (−1)n−1+ n −ζ(m)ζ¯(p)−ζ¯(m+p−1). (3.5)
np nm
n=1(cid:26) (cid:27)
X
1 1
Similarly, taking (a ,b ) = , in (1.2), we arrive at the conclusion that
n n nm np
(cid:18) (cid:19)
∞
F (ζ(m),ζ(p),ζ(r)) = (ζ(m)−ζ (m))(ζ(p)−ζ (p))(ζ(r)−ζ (r))
n n n
n=1
X
=ζ(r){ζ⋆(p,m−1)+ζ⋆(m,p−1)−ζ(m+p−1)−ζ(m)ζ(p)}
+{ζ⋆(m+p−1,r)−ζ⋆(m+p,r−1)}
−{ζ⋆(p,m−1,r)−ζ⋆(p,m,r−1)}
−{ζ⋆(m,p−1,r)−ζ⋆(m,p,r−1)}. (3.6)
lnr−1x
In (1.1), letting n → ∞, then multiplying it by and integrating over the interval (0,1),
x(1−x)
we conclude that
∞ ∞ n ∞ n
(A−A )(B −B )
n n
nr = akζk−1(r) bn+ bkζk−1(r) an
! !
n=1 n=1 k=1 n=1 k=1
X X X X X
∞
− anbnζn−1(r), r ∈ N/{1}. (3.7)
n=1
X
In fact, it is easily shown that the formula (3.7) is also true when r = 1.
1 1
Taking (a ,b )= , in (3.7), we get
n n nm np
(cid:18) (cid:19)
∞
(ζ(m)−ζ (m))(ζ(p)−ζ (p))
n n
nr
n=1
X
= ζ(m+p+r)+ζ⋆(m,p,r)+ζ⋆(p,m,r)
−ζ⋆(m+p,r)−ζ⋆(m,p+r)−ζ⋆(p,m+r),m,p,r ∈ N. (3.8)
We define the multiple harmonic star number ζn⋆(s1,s2,...,sm) by
1
ζn⋆(s1,s2,...,sm) := ks1···ksm,si ∈ N, (i= 1,2,...,m).
1≤km≤X···≤k1≤n 1 m
7
From the definition multiple harmonic star number, we can deduce the following relation
∞
ζ (p)ζ (r)
ζ⋆(p,m,r)+ n n = ζ⋆(p+m,r)+ζ(p)ζ⋆(m,r), p,m ∈ N/{1},r ∈ N. (3.9)
nm
n=1
X
Substituting (3.9) into (3.8) respectively, we obtain the symmetry identity
∞
ζ (m)ζ (p) ζ (p)ζ (r) ζ (m)ζ (r)
n n n n n n
+ +
nr nm np
n=1(cid:26) (cid:27)
X
= ζ⋆(p+m,r)+ζ⋆(p+r,m)+ζ⋆(m+r,p)
+ζ(p)ζ(m)ζ(r)−ζ(p+m+r), p,m,r ∈ N/{1}. (3.10)
∞
ζ (m)ζ (p)
n n
From [7,17], we know that the quadratic sums can be evaluated in terms of
nr
n=1
zeta values and double zeta star values ζ⋆(s1,s2)Xin the following cases: m = p =r or m+p+r
even (m,p,r ∈N/{1}) and m = 1,p+r ≤ 8. From example
∞
H ζ (3) 83 1 11
n n
= ζ(7)+ ζ(3)ζ(4)− ζ(2)ζ(5),
n3 8 4 2
n=1
X
∞
H ζ (2) 343 5 3
n n = − ζ(8)+12ζ(3)ζ(5)− ζ(2)ζ2(3)− ζ⋆(6,2),
n5 48 2 4
n=1
X
∞ ζ2(2) 457
n = 11ζ⋆(6,2)+ ζ(8)+6ζ(2)ζ2(3)−40ζ(3)ζ(5).
n4 18
n=1
X
∞
ζ (2)
where ζ⋆(6,2) = n .
n6
n=1
X
Similarly, from (1.1), we have the following results
∞
n−1
(ζ(m)−ζ (m))(ζ(p)−ζ (p))(−1)
n n
n=1
X
1 ∞ ζ¯ (p) ∞ ζ¯ (m)
= ζ(m)ζ(p)+ζ¯(m+p)− n − n , (3.11)
2 nm np
( )
n=1 n=1
X X
∞
(ζ⋆(p+1,p)−ζ⋆(p+1,p))(ζ⋆(q+1,q)−ζ⋆(q+1,q))
n n
n=1
X
1 ∞ ζ2(p)ζ (q)+ζ (2p)ζ (q) ζ2(q)ζ (p)+ζ (2q)ζ (p)
= n n n n + n n n n
2 nq+1 np+1
n=1(cid:26) (cid:27)
X
∞ ∞ ∞
ζ (p)ζ (q) ζ (p) ζ (q)
n n n n
− − , (3.12)
np+q+1 np+1 nq+1
! !
n=1 n=1 n=1
X X X
∞
(ζ⋆(p+1,p)−ζ⋆(p+1,p))(ζ⋆(p+2,p)−ζ⋆(p+2,p))
n n
n=1
X
2
1 ∞ ζ3(p)+ζ (p)ζ (2p) ζ2(p) 1 ∞ ζ (p)
= n n n − n + n
2 np+2 n2p+2 2 np+1
!
n=1(cid:26) (cid:27) n=1
X X
8
∞ ∞
ζ (p) ζ (p)
n n
− . (3.13)
np+1 np+2
! !
n=1 n=1
X X
Next, we give a Theorem.
Theorem 3.1 Let m,p ≥ 2 be integers, then have
∞ ∞ ∞
ζ(m)ζ (p)−ζ(p)ζ (m) H H
n n n n
= ζ(p) −ζ(m) +ζ(m)ζ(p+1)−ζ(p)ζ(m+1).
n nm np
n=1 n=1 n=1
X X X
(3.14)
Proof. First, we construct the following power function
∞
y = {H ζ (m)−ζ (m+1)}xn−1, x ∈(−1,1).
n n n
n=1
X
By the definition of harmonic numbers, we can deduce that
∞
1 1 1
y = H + ζ (m)+ − ζ (m+1)+ xn
n n+1 n (n+1)m n (n+1)m+1
n=1(cid:26)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
X
∞
H ζ (m)
= H ζ (m)−ζ (m+1)+ n + n xn. (3.15)
n n n m
(n+1) n+1
n=1(cid:26) (cid:27)
X
Therefore, from (3.15), we obtain the following relation
∞ ∞ H ζ (m) xn
{H ζ (m)−ζ (m+1)}xn−1 = n + n . (3.16)
n n n m
(n+1) n+1 1−x
n=1 n=1(cid:26) (cid:27)
X X
Then multiplying (3.16) by lnp−1x and integrating over (0,1), we have
∞ ∞
H ζ (m)−ζ (m+1) H ζ (m)
n n n n n
= + {ζ(p)−ζ (p)}. (3.17)
np (n+1)m n+1 n
n=1 n=1(cid:26) (cid:27)
X X
By a direct calculation, we obtain that
∞ ∞ ∞ ∞
H ζ (m) H ζ (p) H H ζ (m)
n n n n n n n
+ = ζ(p) + +
np nm nm np+m np+1
n=1(cid:26) (cid:27) n=1 n=1 n=1
X X X X
∞ ∞
ζ (m+1) ζ (m)
n n
− + {ζ(p)−ζ (p)}. (3.18)
np n n
n=1 n=1
X X
Change m,p to p,m, we get
∞ ∞ ∞ ∞
H ζ (p) H ζ (m) H H ζ (p)
n n n n n n n
+ = ζ(m) + +
nm np np nm+p nm+1
n=1(cid:26) (cid:27) n=1 n=1 n=1
X X X X
∞ ∞
ζ (p+1) ζ (p)
n n
− + {ζ(m)−ζ (m)}. (3.19)
nm n n
n=1 n=1
X X
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Combining (3.18) and (3.19), we can obtain (3.14). (cid:3)
We note that
∞
ζ(m)ζ(p)−ζ (m)ζ (p)
n n
n
n=1
X
∞ ∞
ζ(m)−ζ (m) ζ (m)(ζ(p)−ζ (p))
= ζ(p) n + n n , m,p ∈ N/{1}. (3.20)
n n
n=1 n=1
X X
In [10], O. Furdui and C. Valean gave the following formula
∞
ζ(m)−ζ (m)
n = ζ⋆(m,1)−ζ(m+1). (3.21)
n
n=1
X
Hence, combining (3.18), (3.20) and (3.21), we have the following corollary.
Corollary 3.2 For positive integers m > 1,p > 1, then we have
∞
ζ(m)ζ(p)−ζ (m)ζ (p)
n n
n
n=1
X
∞
H ζ (m) H ζ (p)
= n n + n n +ζ⋆(p,m+1)
np nm
n=1(cid:26) (cid:27)
X
−ζ(p)ζ(m+1)−ζ⋆(p+m,1)−ζ⋆(p+1,m). (3.22)
Similarly, noting that
∞
ζ (m)ζ(p)ζ(q)−ζ(m)ζ (p)ζ (q)
n n n
n
n=1
X
∞ ∞
ζ(m)ζ(p)−ζ (m)ζ (p) ζ(m)−ζ (m)
= ζ(m) n n −ζ(p)ζ(q) n , m,p,q ∈ N/{1}.
n n
n=1 n=1
X X
Therefore, we also obtain the following corollary.
Corollary 3.3 For positive integers m,p,q > 1, then we have
∞
ζ (m)ζ(p)ζ(q)−ζ(m)ζ (p)ζ (q)
n n n
n
n=1
X
∞
H ζ (q) H ζ (p)
n n + n n +ζ⋆(p,q+1)
= ζ(m) np nq
n=1(cid:18) (cid:19) 
−Xζ⋆(p+q,1)−ζ⋆(p+1,q)−ζ(p)ζ(q+1)
−ζ(p)ζ(q){ζ⋆(m,1)−ζ(m+1)}. (3.23)
 
 
Furthermore, we consider the following power series function
∞
y = {H ζ (p)ζ (q)−ζ (p+q+1)}xn−1, x ∈ (−1,1),
n n n n
n=1
X
proceeding in a similar fashion to evaluation of the Theorem 3.1, we can give the following
Theorem.
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