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