Hyperharmonic series involving Hurwitz zeta function

HYPERHARMONIC SERIES INVOLVING HURWITZ ZETA FUNCTION ISTV´ AN MEZ˝ O AND AYHAN DIL Abs ac . We show ha he sum o he se ies o med by he so-called hype ha monic numbe s can be exp essed in e ms o he Riemann ze a unc ion. These esul s enable us o e o mula e Eule ’s o mula in ol ing he Hu wi z ze a unc ion. In addi on, we imp o e Conway and Guy’s o mula o hype ha monic numbe s. 1. Hype ha monic numbe s In oduc ion. In 1996, J. H. Conway and R. K. Guy in [CG] ha e de ined he no ion o hype ha monic numbe s. The n- h ha monic numbe o o de 0 is H(0) n=1 n(n > 0), and o all > 0 le H( ) n= n X k=1 H( −1) k be he n- h hype ha monic numbe o o de . In special, he hype - ha monic numbe s o o de 1 a e simply called ha monic numbe s: H(1) n=Hn= n X k=1 1 k. I u ned ou ha hese numbe s ha e many combina o ial connec- ions. Ge ing deepe insigh , see [BGG] and he e e ences gi en he e. H( ) ncan be exp essed by binomial coe icien s and o dina y ha monic numbe s [CG]: H( ) n=n+ −1 −1(Hn+ −1−H −1).(1) In his pape we gi e a new p oo o (1) and, in addi ion, we p o ide a mo e gene al o mula. 2000 Ma hema ics Subjec Classi ica ion. 11B83. Key wo ds and ph ases. hype ha monic numbe s, Eule sums, Riemann ze a unc ion, Hu wi z ze a unc ion, Hype geome ic se ies. 1 2 ISTV ´ AN MEZ ˝ O AND AYHAN DIL Mo eo e , we in es iga e sums o se ies in ol ing hype ha monic numbe s. Among o he s, we p o e ha o he Hu wi z ze a unc ion ∞ X n=1 H( ) n nm= ∞ X n=1 H( −1) nζ(m, n). As a special case, we ge he nex cu ious iden i ies o Riemann ze a and Hu wi z ze a unc ion ∞ X k=1 ζ(2, k) k= 2ζ(3),(2) ∞ X k=1 ζ(3, k) k=5 4ζ(4).(3) We close ou in oduc ion wi h he ela ion be ween hype ha monics and he so-called -S i ling numbe s. -S i ling numbe s. The -S i ling numbe o he i s kind wi h pa ame e s nand k, deno ed by n k , is he numbe o pe mu a ions o he se {1, . . . , n}ha ing kdisjoin , non-emp y cycles, in which he elemen s 1 h ough a e es ic ed o appea in di e en cycles (n≥k≥ ). The ollowing iden i y in eg a es he hype ha monic- and he - S i ling numbe s o he i s kind. n+ +1 n!=H( ) n. This equali y will be used in he special case = 1 [GKP]: (4) n+1 2 n!:= n+1 21 n!=Hn. 2. A ela ion be ween hype ha monic numbe s Gene a ing unc ions. Le (an)n∈Nbe a eal sequence. Then he unc ion (z) := ∞ X n=0 anzn is called he gene a ing unc ion o (an)n∈N. I an=Hnwe ge ha (see [GKP, BGG]) (5) ∞ X n=0 Hnzn=−ln(1 −z) 1−z, and in gene al (c . [D]) (6) ∞ X n=0 H( ) nzn=−ln(1 −z) (1 −z) . HYPERHARMONIC SERIES INVOLVING HURWITZ ZETA FUNCTION 3 Beside hese we also need New on’s binomial o mula 1 (1 −z)k= ∞ X n=0 n+k−1 nzn. We p o ide a mo e gene al o m o (1). Theo em 1. We ha e k+ −1 kH(k+ ) n=n+k nH( ) n+k−n+k+ −1 nH( ) k P oo . Le us conside he gene a ing unc ion o he hype ha monic numbe s in (6). I we di e en ia e he le hand side k- imes and make some ea agemen , we ob ain dk dzk−ln (1 −z) (1 −z) =(k+ −1)! ( −1)! (Hk+ −1−H −1−ln (1 −z)) (1 −z)k+ . We can w i e his in e ms o New on’s binomial se ies and gene a ing unc ion o hype ha monic numbe s as ollows: dk dkz−ln(1 −z) (1 −z) = k! ∞ X n=0 n+k+ −1 nH( ) k+k+ −1 kH(k+ ) nzn. On he o he hand, i we iew he gene a ing unc ion as a powe se ies, we ob ain dk dzk(∞ X n=1 H( ) nzn)=k! ∞ X n=0 n+k kH( ) n+kzn. Compa ing he coe icien s o bo h sides gi es he s a emen .  I we subs i u e = 1 and k= −1 in he o mula abo e, we ge back Conway and Guy’s esul (1). 3. Asymp o ic app oxima ion To ha e he exac asymp o ic beha iou o hype ha monic numbe s we need he ollowing inequali y om [CG]. (7) 1 2(n+ 1) + ln(n) + γ < Hn<1 2n+ ln(n) + γ(n∈N), whe e γ= 0.5772 . . . is he Eule -Masche oni cons an . Lemma 2. Fo all n∈Nand o a ixed o de ≥2we ha e H( ) n∼1 ( −1)! n −1ln(n), ha is, he quo ien o he le and igh hand side ends o 1 as n→ ∞. 4 ISTV ´ AN MEZ ˝ O AND AYHAN DIL P oo . Fo he binomial coe icien in (1), n+ −1 −1∼1 ( −1)!n −1. Fo he con enience le us in oduce he a iable := −1. We would like o es ima e he ac o Hn+ −H in (1). Acco ding o (7), we ge ha ln(n+ )−ln( √e)< Hn+ −H <ln(n+ ), whence 1−ln( √e) ln(n+ )<Hn+ −H ln(n+ )<1. The limi o he le -hand side is 1 as n ends o in ini y. The e o e ( emembe ha = −1) Hn+ −1−H −1∼ln(n+ −1) ∼ln(n). Collec ing he esul s abo e we ge he s a emen o he Lemma.  Co olla y 3. We ha e ∞ X n=1 H( ) n nm<+∞, whene e m> . 4. A connecion wi h he Hu wi z ze a unc ion The Hu wi z ze a unc ion is de ined as ζ(m, n) = ∞ X p=0 1 (n+p)m. We poin ou ha he sums in ol ing hype ha monic numbe s can be ans o med in o he o m as in he nex heo em. Theo em 4. I ≥1and m≥ + 1, hen ∞ X n=1 H( ) n nm= ∞ X n=1 H( −1) nζ(m, n). P oo . We ans o m he le hand side as H( ) 1 1m+H( ) 2 2m+H( ) 3 3m+··· =H( −1) 1 1m+H( −1) 1+H( −1) 2 2m+H( −1) 1+H( −1) 2+H( −1) 3 3m+··· =H( −1) 1 ∞ X p=1 1 pm+H( −1) 2 ∞ X p=1 1 (p+ 1)m+H( −1) 3 ∞ X p=1 1 (p+ 2)m+··· = ∞ X n=1 H( −1) n ∞ X p=1 1 (p+n−1)m, HYPERHARMONIC SERIES INVOLVING HURWITZ ZETA FUNCTION 5 and he esul comes.  In he case = 1, his heo em and Eule ’s summa ion o mula gi e he iden i ies (2) and (3). And, in gene al, ∞ X n=1 Hn nm= ∞ X k=1 ζ(m, k) k. 5. Gene a ing unc ions, Eule sums and Hype geome ic Se ies In his sec ion we in oduce he no ions needed in wha ollows. The gene a ing unc ion (8) 1 m!(−ln(1 −z))m= ∞ X n=1 n mzn n! can be ound in [GKP, B]. The well known polyloga i hm unc ions can also be conside ed as gene a ing unc ions belong o an=1 nk( o a ixed k). Lik(z) = ∞ X n=1 zn nk(k= 1,2, . . . ). Eule sums. The gene al Eule sum is an in ini e sum whose gene al e m is a p oduc o ha monic numbe s di ided by some powe o n, see he comp ehensi e pape [FS]. The sum ∞ X n=1 Hn nm=1 2(m+ 2)ζ(m+ 1) − m−2 X k=1 ζ(m−k)ζ(k+ 1) was de i ed by Eule (see [BB] and he e e ences gi en he e). Rela ed se ies we e s udied by De Doelde in [dD] and Shen [S], o ins ance. Hype geome ic se ies. The Pochhamme symbol is de ined by (9) (x)n=x(x+ 1) ···(x+n−1), wi h special cases (1)n=n! and (x)1=x. The de ini ion o he hype geome ic unc ion (o hype geome ic se ies) is he ollowing: nFma1, a2, . . . , an b1, b2,. . . , bm z= ∞ X k=0 (a1)k(a2)k···(an)k (b1)k(b2)k···(bm)k zk k!. This unc ion will appea in he sum o he hype ha monic numbe s. We shall need one mo e s a emen . Lemma 5. We ha e Zln(z) (1 −z)zdz = Li2(1 −z) + 1 2ln2(z), 6 ISTV ´ AN MEZ ˝ O AND AYHAN DIL and o all 2≤ ∈N (10) Zln(z) (1 −z)z dz =Zln(z) (1 −z)z −1dz −ln(z) ( −1)z −1−1 ( −1)2z −1, o , equi alen ly, Zln(z) (1 −z)z dz = Li2(1 −z) + 1 2ln2(z)− −1 X k=1 ln(z) kzk+1 k2zk. up o addi i e cons an s P oo . The de ini ion o Li2(z) eadily gi es ha Li0 2(1 −z) = ln(z) 1−z. Mo eo e , 1 2ln2(z)0 =ln(z) z, whence Li0 2(1 −z) + 1 2ln2(z)0 =ln(z) (1 −z)z. The i s s a emen is p o ed. The second one also can be deduced by di e en ia ion. The de i a i e o he igh -hand side o (10) has he o m ln(z) (1 −z)z −1−( −1)z −2−( −1)2ln(z)z −2 ( −1)2(z −1)2−−( −1) ( −1)2z =ln(z) z (1 −z), as we wan .  6. The summa ion o mula Fo he sake o simplici y, we in oduce he no a ions S( , m) := ∞ X n=1 H( ) n nm, and B(k, m) := m+1Fm1,1, . . . , 1, k + 1 2,2, . . . , 2 1. A e hese in oduc o y s eps we a e eady o deduce a ecu sion o mula o S( , m). Theo em 6. I ≥2and m≥ + 1, hen S( , m) = S(1, m) + −1 X k=1 1 k[S(k, m −1) −B(k, m)] . HYPERHARMONIC SERIES INVOLVING HURWITZ ZETA FUNCTION 7 P oo . We begin wi h he gene a ing unc ion (6): ∞ X n=1 H( ) n nzn=−Zln(1 −z) z(1 −z) dz. F om he p e ious lemma −Zln(1 −z) z(1 −z) dz =Zln(z) (1 −z)z dz = = Li2(z) + 1 2ln2(1 −z)− −1 X k=1 ln(1 −z) k(1 −z)k+1 k2(1 −z)k. Acco ding o (6) and (8) one can w i e ∞ X n=1 H( ) n nzn= Li2(z) + ∞ X n=1 n 2zn n!− −1 X k=1 1 k(−1) ∞ X n=0 H(k) nzn+1 k2 ∞ X n=0 n+k−1 nzn!. Le us deal wi h he second e m. The S i ling numbe s o he i s kind sa is y he ecu ence ela ion [GKP] n k= (n−1)n−1 k+n−1 k−1(n > 0). F om his n+ 1 2=nn 2+n 1=nn 2+ (n−1)! (n > 0). Now, (4) can be ew i en as ollows Hn=1 n!n+ 1 2=1 (n−1)!n 2+1 n, whence 1 n!n 2=Hn n−1 n2. The e o e he second sum is ∞ X n=1 n 2zn n!= ∞ X n=1 Hn nzn− ∞ X n=1 zn n2. Since he las membe equals o Li2(z), i cancels he i s membe o he sum abo e. Hence ∞ X n=1 H( ) n nzn= ∞ X n=1 Hn nzn+ −1 X k=1 1 k ∞ X n=0 H( ) nzn−1 k2 ∞ X n=0 n+k−1 nzn!. 8 ISTV ´ AN MEZ ˝ O AND AYHAN DIL An easy induc ion shows ha (a e di iding wi h z, in eg a ing, and epea ing hese s eps (m−1)- imes and inally subs i u ing z= 1) (11) ∞ X n=1 H( ) n nm=S(1, m)+ −1 X k=1 1 kS( , m −1) −1 k2 ∞ X n=1 n+k−1 n1 nm−1!. The las s ep is he ans o ma ion o he las membe . ∞ X n=1 n+k−1 n1 nm−1=1 (k−1)! ∞ X n=1 (n)k nm, because o he de ini ion o he Pochhamme symbol (9). On he o he hand, he de ini ion o B(k, m) yields ha B(k, m) = ∞ X n=0 (n!)m (n+ 1)!m (k+ 1)n n!= ∞ X n=0 1 (n+ 1)m (k+ 1)n n!. The nex con e sion should be applied: k!(k+ 1)n n!=(k+n)! n!= (n+ 1)(n+ 2) ···(n+k) = (n+ 1)k. I means ha he equali y (12) B(k, m) = 1 k! ∞ X n=0 (n+ 1)k (n+ 1)m=1 k! ∞ X n=1 (n)k nm holds. Tha is, ∞ X n=1 n+k−1 n1 nm−1=kB(k, m). Conside ing his and (11) he esul ollows.  7. Tables o he low-o de sums In he ollowing ables we collec he low-o de esul s o he Sum- ma ion Theo em. We used he ollowing iden i ies which can be easily de i ed om (9) and (12). B(1, m) = ζ(m−1), B(2, m) = 1 2(ζ(m−1) + ζ(m−2)) , B(3, m) = 1 6ζ(m−3) + 1 2ζ(m−2) + 1 3ζ(m−1). S(2, m) Powe o nClosed o m App ox. alue m= 3 π4 72 −π2 6+ 2ζ(3) 2.112083781 m= 4 π4 72 + 3ζ(5) −ζ(3) 1 + π2 61.284326055 HYPERHARMONIC SERIES INVOLVING HURWITZ ZETA FUNCTION 9 m= 5 π6 540 −π4 90 −1 2ζ(3)2+ 3ζ(5) −π2 6ζ(3) 1.109035642 m= 6 π6 540 + 4ζ(7) −π4 90 ζ(3) −1 2ζ(3)2− ζ(5) 1 + π2 6 1.047657410 m= 7 π8 4200 −π6 945 −ζ(5)ζ(3) + 4ζ(7) − π2 6ζ(5) −π4 90 ζ(3) 1.022090029 m= 8 π8 4200 + 5ζ(9) −π6 945 ζ(3) −π4 90 ζ(5) − ζ(5)ζ(3) −ζ(7) 1 + π2 6 1.010557246 m= 9 π10 34020 −π8 9450 −ζ(7)ζ(3) −1 2ζ(5)2+ 5ζ(9) −π2 6ζ(7) −π6 945 ζ(3) −π4 90 ζ(5) 1.005133570 m= 10 π10 34020 +6ζ(11)−π8 9450 ζ(3)−π6 945 ζ(5)− 1 2ζ(5)2−π4 90 ζ(7) −ζ(7)ζ(3) − ζ(9) 1 + π2 6 1.002522063 S(3, m) Powe o nClosed o m App ox. alue m= 4 π4 48 −π2 8−π2 6ζ(3) −1 4ζ(3) + 3ζ(5) 1.628620203 m= 5 π6 540 −π4 144 −π2 4ζ(3)−3 4ζ(3)−1 2ζ(3)2+ 9 2ζ(5) 1.180103635 m= 6 π6 360 −π4 120 +4ζ(7)−π2 6ζ(5)−π4 90 ζ(3)− 3 4ζ(3)2+1 4ζ(5) −π2 12 ζ(3) 1.072362484 m= 7 π8 4200 −π6 2520 −π4 60 ζ(3) −1 4ζ(3)2− ζ(5)ζ(3) −ζ(5) π2 4+3 4+ 6ζ(7) 1.032351029 m= 8 π8 2800 −π6 1260 −ζ(3) π4 180 +π6 945 − ζ(5) π2 12 +π4 90 −3 2ζ(5)ζ(3) + ζ(7) 3 4−π2 6+ 5ζ(9) 1.015179175 S(4, m) Powe o nClosed o m App ox. alue m= 5 π6 540 −π4 810 −11π2 216 −ζ(3)−11π2 36 ζ(3)− 1 2ζ(3)2+11 2ζ(5) 1.310990854 m= 6 11π6 3240 −π4 80 −ζ(3) π4 90 +1π2 6+11 36 + +11 12 ζ(3)2+ζ(5) 59 36 −π2 6+4ζ(7) 1.103348021