A special constant and series with zeta values and harmonic numbers Khristo N. Boyadzhiev Department of Mathematics and Statistics, Ohio Northern University, Ada, OH 45810, USA [email protected] Abstract. In this paper we demonstrate the importance of a mathematical constant which is the value of several interesting numerical series involving harmonic numbers, zeta values, and logarithms. We also evaluate in closed form a number of numerical and power series. Key words and phrases harmonic numbers, skew-harmonic numbers, Riemann zeta function, Hurwitz zeta function, digamma function 2010 Mathematics Subject Classification. 11B34, 11M06, 33B15, 40C15 1. Introduction The purpose of this paper is to demonstrate the importance of the constant 1 ψγ(t ++ 1) M = ∫ dt ≈1.257746 0 t where ψ ()z=ΓΓ′ ()/() zz is the digamma function and γψ= − (1) is Euler’s constant. Theorem 1 in Section 2 shows that M is the numerical value of many interesting and important series involving logarithms, harmonic numbers Hn and zeta values ζ ()n . The harmonic numbers are defined by 11 Hn=ψγ( ++=+++ 1) 1 ... ,nH ≥ 1, = 0 n 2 n 0 1 with generating function ∞ ln(1− t ) n −=∑ Htn , | t | < 1. 1− t n=1 We also use the skew-harmonic numbers given by 1 1 (− 1) n−1 H−−=−+++1 ... ,nH ≥ 1. = 0 . n 23 n 0 Riemann’s zeta function is defined by ∞ 1 ζ (ss )= ∑ s , Re> 1. n=1 n − We consider series with terms Hnn (ζ ( )− 1) , Hnn (ζζ ( )−+ ( n 1)) , Hnnn (ζζ( )−+ ( 1) ) and − others similar. Related power series with coefficients Hnan ζ (+ 1, ) and Hn ζ ( na+ 1, ) , where ζ (,sa ) is the Hurwitz zeta function are also studied. Some series are evaluated in closed form in terms of known constants and special functions. Other interesting series will be presented in integral form. The main results of this paper are Theorem 1 in Section 2, Theorem 2 in Section 3, and Theorem 3 in Section 4. 2. A mosaic of series A classical theorem rooted in the works of Christian Goldbach (1690-1764) says that ∞ ∑{ζ (n )−= 1} 1 n=2 (see [1], [9], and [12, p. 142]; a short proof is given in the Appendix). The terms ζ ()n − 1 of the 1 above series decrease like the powers of . The following lemma was proved in [7, p. 51]. It 2 will be needed throughout the paper. (For convenience a proof is given in the Appendix.) Lemma 1. For every n ≥ 2 2 1 11n + <ζ ()n −< 1 . 2nn 21n − The lemma shows that the power series with general term {ζ (nx )− 1} n is convergent in the disk ||2x < . The following representation is true [12, p. 173, (139)]. ∞ (1) ∑{ζ (nx )− 1}n−1 =−− 1γψ ( 2 −x ) . n=2 Integration of this series yields [12, p. 173, (135)] ∞ ζ ()n − 1 (2) ∑ xxn =−(1γ ) +Γ− ln (2 x ) n=2 n and in particular, when x =1, this turns into a classical result published by Euler in 1776 (see [11]) ∞ ζ ()n − 1 (3) ∑ =1 −γ . n=2 n With x = −1 we find ∞ζζ()nn−− 1 ∞∞ () (1)n ∑(− 1) nn = ∑∑( − 1) − =γ −+1 ln Γ (3) n=2n nn= 22 nn= so that (as ln 2= ln Γ (3) ) ∞ ζ ()n (4) ∑(−= 1) n γ n=2 n which is another representation of γ often used by Euler. Cases when the series can be evaluated in explicit closed form are intertwined with similar cases when this is not possible (unless we accept integral form). Series of the form ∞ ζ ()n − 1 ∑ ,p = 1, 2 n=2 np+ 3 have been studied and evaluated in a number of papers - see, for instance, [9], [12, pp. 213-219, (469), (474), 517), (518)] and also [13], [14]. At the same time the series ∞∞ζζ(nn+− 1) 1 (+− 1) 1 ∑∑, (− 1) n−1 nn=22nn= resists evaluation in closed form in terms of recognized constants. However, they can be evaluated in integral form and we want to point out one simple integral which appears in several interesting cases as evidenced by Theorem 1 below. Namely, consider the constant 1 ψγ(t ++ 1) M = ∫ dt . 0 t Theorem 1. With M as defined above, ∞ (−+ 1)n−1ζ (n 1) (a) ∑ = M n=1 n ∞ 11 (b) ∑ ln 1+=M n=1 nn ∞ ln (n + 1) (c) ∑ = M n=1 nn(+ 1) ∞ (d) ∑ Hnn (ζ (+−= 1) 1) M n=1 ∞ (e) ∑ Hnn (ζγ()− 1) = M +− 1 n=2 ∞ 1 (f) ∑ (n−ζζ(2) − (3) −− ... ζ (nM )) = n=1 n (the first term in this series is just 1) ∞ − (g) ∑ Hnn (ζζ(+− 1) ( n + 2)) = M − ln 2 n=1 4 ∞ Ein(x ) (h) dx= M ∫ x 0 e −1 where Ein(x ) is the modified exponential integral, an entire function ∞ (− 1) nn−1 x Ein(x ) = ∑ n=1 nn! 1 (1−−uu ) ln(1 ) (i) ∫ du= M 0 uuln ∞ 1 n n −k −1ζ += (j) ∑∑n ( 1) (kM 1) . nk=11n2 = k Before proving the theorem we want to mention that the evaluations (a), (b), (c), and (i) are not new (for instance, they are listed on p. 142 in [3]). The series (c) appears in important applications in number theory – see [5] and [8]. The entire paper [4] is dedicated to that series. Proof. Starting from the well-known Taylor expansion for ψγ(1++x ) ∞ (5) ∑(− 1)nn−1ζ (nx + 1) = ψγ (1 ++ x ) n=1 we divide both sides by x and integrate from 0 to 1 to prove (a). Then (a) implies (b) n ∞(−+− 1)nn−−11ζ (n 1)∞ ( 1) ∞ 1 ∞∞ 1 (− 1)n−1 1∞ 1 1 = = = + ∑ ∑ ∑n+1 ∑∑ ∑ln 1 . n=1 nn=1 n k= 1 k kn= 11 k = nkk=1 k k Next we prove (b) → (d) n ∞ ∞ ∞11 ∞∞ 1 ζ +−= = ∑Hnnn( ( 1) 1) ∑ H ∑n+1 ∑∑H n n=1 n=1 k = 2kkk kn= 21 = and using the generating functions for the harmonic numbers we continue this way 5 ∞11−+ 1 ∞∞ 1 km1 1 =ln 1 −= ln =ln =M . ∑1 ∑∑−− k=2k1− k km= 21 k11 k= mm k Now starting from the first sum in the last equation we write ∞1111111−∞ ∞∞ M =ln 1 −= −−=ln 1 ∑1 ∑−− ∑∑m k=2k1− kkk=211 kkmkkm=21= k ∞∞11∞ 1 = = −ζζ − −− ζ ∑∑n ∑(nn(2) (3) ... ( )) nk=12n = kk(− 1) n=1 n by using the evaluation (see [7, p. 51]) ∞ 1 =−ζζ − −− ζ ∑ n nn(2) (3) ... ( ) . k =2 kk(− 1) Thus (f) is also proved. Equation (e) follows from (d) by writing ∞∞11∞ ∞ ∑∑Hnnn(ζ()1−=) H−1 +( ζζ()1 n −=) ∑ Hm m( ( +−+ 1)1) ∑( ζ ()1n −) nn=22= nnm=1 n= 2 and then applying (3). Next we prove (h). From the representation 1 ∞ xs−1 ζ ()s = dx, Re()1 s > ∫ x Γ−()se0 1 we find with sn= +1 ∞∞(−+ 1)n−−11ζ (n 1)∞∞ ( − 1) nnx dxEin( x ) = = dx ∑∑∫∫xx. nn=11n 00= nn! e−− 11 e The exponential integral has the representation 6 x 1− e−t Ein(x ) = ∫ dt 0 t and integrating by parts we find ∞∞ −x ∞ Ein()x Ein()(1 xd− e ) ∞ dx = =Ein(x )ln(1 −− e−−xx ) Ein′ (x )ln(1 − e ) dx ∫∫xx− 0 ∫ 00ee−−11 0 ∞ 1− e−x =−−∫ ln(1e−x ) dx . 0 x ∞ Here Ein(xe )ln(1−=−x ) 0 since Ein(x ) grows like ln x when x →∞ and also 0 − limxe ln(1− x )0= . In the last integral we make the substitution ue= −x to find x→0 ∞ 1 −e−x 1 (1−−uu ) ln(1 ) −∫∫ln(1 −=e−x ) dx du 00x uuln and (i) is proved. We shall prove now the implication (a) → (g) by using Abel’s lemma for transformation of series (see, for example, [10, Exercise 10, p.78]): Lemma 2. Let {}an and{bn }, np≥ , be two sequences of complex numbers and let Aaan= pp ++1 ++... a n. Then for every np> we have nn−1 ∑∑abkk=+− bA n n A k() b k b k+1 . kp= kp= (− 1) k −1 We take here p =1, a = , and bk=ζ ( +− 1) 1 . Then AH= − and we find k k k kk nnk −1 −1 (− 1) −− ∑∑{ζ (k+−= 1) 1} Hnnk { ζ ( +−+ 1) 1} Hk { ζζ ( +− 1) ( k + 2)} . kk=11k = 7 − Setting n →∞ and using the estimate from lemma1 and also the fact that ||HHnn≤ and Hnn ln at infinity we come to the equation ∞∞k −1 (− 1) − ∑∑{ζ (k+−= 1) 1} Hkk { ζζ ( +− 1) ( k + 2)} . kk=11k = According to (a) we have ∞ (− 1) k −1 ∑ {ζ (kM+−= 1) 1} − ln 2 k =1 k and (g) follows. The implication (b) → (c) also follows from Abel’s lemma. We take p =1 and 11 akk=ln 1 + = ln(k +− 1) ln( kb ), =. kk Here Ann =ln ( + 1) and lemma 2 provides the equation nn1 1 ln(nk++ 1)−1 ln ( 1) ∑∑ln 1+= + kk=11k k n= kk(+ 1) which clearly shows the relation between (b) and (c). Last we prove (j). The proof is based on Euler’s series transformation [2]. Given a power series f( x )=++ a01 ax ... we have for sufficiently small ||t ∞ n 1 t n n f= ∑∑ tak . 11−−ttnk=00 = k We take fx( )=ψγ ( x ++ 1) with the expansion (5), where af0 =(0) = 0 . Using the substitution t x = we compute 1− t 1 ψγ(x ++ 1) 1/2 1 t dt ∫∫dx= f 00x 11−−t tt 8 1/2 ∞∞nnnndt 1 =tknk( −+= 1)−−11ζζ ( 1) ( −+ 1)k (k 1) ∫ ∑∑ ∑n ∑ 0 nk=11 = kktnn=12 k = 1 after integrating term by term.