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: xuce1242063253@163.com 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 several 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 numbers. 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 alternating 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

Explicit evaluation of harmonic sums

Similar Explicit evaluation of harmonic sums