R E S E A R C H
Open Access
Identities involving harmonic and
hyperharmonic numbers
Dae San Kim
1and Taekyun Kim
2**Correspondence: [email protected] 2Department of Mathematics,
Kwangwoon University, Seoul, 139-701, Republic of Korea Full list of author information is available at the end of the article
Abstract
In this paper, we give some new and interesting identities involving harmonic and hyperharmonic numbers which are derived from the transfer formula for the associated sequences.
1 Introduction
LetFbe the set of all formal power series in the variabletoverCwith
F=
f(t) =
∞
k=
ak
k!t
ka k∈C
. ()
Suppose thatPis the algebra of polynomials in the variablexoverCand thatP∗is the vector space of all linear functionals onP. The action of the linear functionalLon a poly-nomialp(x) is denoted byL|p(x).
Letf(t)∈F. Then we consider a linear functional onPby setting
f(t)|xn=an (n≥) (see [, ]). ()
From () and (), we note that
tk|xn=n!δn,k (n,k≥) (see [, –]), ()
whereδn,kis the Kronecker symbol.
LetfL(t) =
∞
k=L|x
n
k! tk. Then we see thatfL(t)|xn=L|xn. The mapL−→fL(t) is a
vector space isomorphism fromP∗ ontoF. Henceforth,F is thought of as both a for-mal power series and a linear functional. We callFthe umbral algebra. The umbral cal-culus is the study of umbral algebra. The orderO(f(t)) of the nonzero power seriesf(t) is the smallest integerk for which the coefficient oftk does not vanish. IfO(f(t)) = ,
thenf(t) is called an invertible series. IfO(f(t)) = , thenf(t) is called a delta series. Let O(f(t)) = andO(g(t)) = . Then there exists a unique sequencesn(x) of polynomials such
thatg(t)f(t)k|s
n(x)=n!δn,kforn,k≥. The sequencesn(x) is called the Sheffer sequence
for (g(t),f(t)) which is denoted bysn(x)∼(g(t),f(t)) (see [, , ]). Ifsn(x)∼(,f(t)), then
sn(x) is called the associated sequence forf(t). By (), we easily see thateyt|p(x)=p(y).
Letf(t)∈Fandp(x)∈P. Then we have
f(t) =
∞
k=
f(t)|xk
k! t
k, p(x) =
∞
k=
tk|p(x)
k! x
k (see [, , ]). ()
From (), we note that
p(k)() =tk|p(x), |p(k)(x)=p(k)(). ()
By (), we easily see that
tkp(x) =p(k)(x) =d
kp(x)
dxk (k≥) (see [, , , ]). ()
Letφn(x) be exponential polynomials which are given by
∞
k= φk(x)
k! t
k=ex(et–) (see [, , ]). ()
Thus, by (), we get
φn(x) = n
k=
S(n,k)xk∼ ,log( +t)
, ()
whereS(n,k) is the Stirling number of the second kind.
The Stirling number of the first kind is defined by
(x)n=x(x– )· · ·(x–n+ ) = n
k=
S(n,k)xk. ()
Thus, by (), we get
S(n,k) =
k!
tk|(x)n
(see [, ]). ()
Let pn(x)∼(,f(t)), qn(x)∼(,g(t)). Then the transfer formula for the associated
se-quences is given by
qn(x) =x
f(t) g(t)
n
x–pn(x) (see [, ]). ()
Thenth harmonic number isHn=
n
i=i (n≥) andH= .
In general, the hyperharmonic numberHn(r)of orderris defined by
Hn(r)=
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
ifn≤ orr< ,
n ifr= ,n≥,
n i=H
(r–)
i ifr,n≥
From (), we note thatHn()is the ordinary harmonic numberHn. It is known that
Hn(r)=
n+r– r–
(Hn+r––Hr–) (see [, ]). ()
The generating functions of the harmonic and hyperharmonic numbers are given by
∞
n=
Hntn= –
log( –t)
–t ()
and
∞
n=
Hn(r)tn= –log( –t)
( –t)r , respectively. ()
The purpose of this paper is to give some new and interesting identities involving har-monic and hyperharhar-monic numbers which are derived from the transfer formula for the associated sequences.
2 Identities involving harmonic and hyperharmonic numbers
From () and (), we note that
φn(x) = n
j=
S(n,j)xj∼ ,log( +t)
()
and
(–)nφn(–x)∼ , –log( –t)
. ()
Let us assume that
qn(x)∼ ,t( –t)r
. ()
From (), () andxn∼(,t), we note that
qn(x) =x
t t( –t)r
n
x–xn=x( –t)–rnxn–
=x
n–
k=
–rn k
(–t)kxn–=x
n–
k=
rn+k– k
tkxn–
=x
n–
k=
rn+k– k
(n– )kxn––k= n–
k=
rn+k– k
(n– )kxn–k
=
n
k=
rn+n–k– n–k
(n– )n–kxk. ()
Now, we use the following fact:
∞
n=
Hn(r)tn= –log( –t)
Forn≥, by (), () and (), we get
qn(x) =x
–log( –t) t( –t)r
n
x–(–)nφn(–x)
=x
∞
l=
Hl(+r)tl
n
x–(–)n
n
j=
S(n,j)(–x)j
= (–)n
n
j=
S(n,j)(–)jx ∞
l=
Hl(+r)tl
n
xj–
= (–)n
n
j=
S(n,j)(–)jx j–
l=
l+···+ln=l
Hl(+r) · · ·Hl(nr+)
tl
xj–
= (–)n
n
j=
j–
l=
l+···+ln=l
S(n,j)(–)jHl(+r) · · ·H (r)
ln+(j– )lx
j–l
= (–)n
n
j=
j
k=
l+···+ln=j–k
S(n,j)(–)jHl(+r) · · ·H (r)
ln+(j– )j–kx
k
= (–)n
n
k= n
j=k
l+···+ln=j–k
(–)jS(n,j)Hl(+r) · · ·H (r)
ln+(j– )j–k
xk. ()
Therefore, by comparing coefficients on both sides of () and (), we obtain the following theorem.
Theorem For n≥,r≥, ≤k≤n,we have
rn+n–k– n–k
(n– )n–k= (–)n n
j=k
l+···+ln=j–k
S(n,j)(–)jHl(+r) · · ·H (r)
ln+(j– )j–k.
We recall the following equation:
log( +t) t
n
=
∞
l=
n!
(l+n)!S(l+n,n)t
l. ()
Forn≥, from (), () and (), we have
qn(x) =x
–log( –t) t( –t)r
n
x–(–)nφn(–x)
=x
log( –t) –t
n
( –t)–rnx–(–)nφn(–x)
= (–)n
n
j=
S(n,j)(–)jx
log( –t) –t
n
( –t)–rnxj–
= (–)n
n
j=
S(n,j)(–)jx
log( –t) –t
n j–
l=
rn+l– l
= (–)n
n
j=
S(n,j)(–)j
j–
l=
rn+l– l
(j– )lx j––l
m=
n! (m+n)!
×S(m+n,n)(–t)mxj––l
= (–)n
n
j=
j–
l=
j––l m=
(–)j+m
rn+l– l
n! (m+n)!
(j– )! (j– –l–m)!
×S(m+n,n)S(n,j)xj–l–m
= (–)n
n
k= n
j=k j–k
l=
(–)k+l
rn+l– l
n! (j–l–k+n)!
(j– )! (k– )!
×S(j–l–k+n,n)S(n,j)
xk. ()
Therefore, by () and (), we obtain the following theorem.
Theorem For r,n≥, ≤k≤n,we have
rn+n–k– n–k
(n– )n–k
= (–)n
n
j=k j–k
l=
(–)k+l
rn+l– l
n! (j–l–k+n)!
(j– )! (k– )!
×S(j–l–k+n,n)S(n,j).
Here we invoke the following identity:
∞
n= n
m=
mHm(r)
tn=t( –rlog( –t))
( –t)r+ . ()
Let us consider the following associated sequence:
qn(x)∼ ,t( –t)r+
. ()
Forn≥, by () and (), we get
qn(x) = n
k=
(r+ )n–k– n–k
(n– )n–kxk. ()
Let us assume that
pn(x)∼ ,t –rlog( –t)
. ()
Forn≥, by (), () andxn∼(,t), we get
pn(x) = x
t t( –rlog( –t))
n
=x
Therefore, by () and (), we obtain the following theorem.
Here we use the following identity:
Let us consider the following associated sequence:
qn(x)∼ ,t( –t)r+
Let us assume that
pn(x)∼ ,t –rlog( –t)
Theorem For n,r≥, ≤k≤n,we have
(r+ )n–k– n–k
(n– )n–k
=
n
a=k n–a
l=
j+···+jn=a–k
(j+ )· · ·(jn+ )Hj(+r) · · ·H (r)
jn+
l!rl
×
n+l– l
n– a–
S(n–a,l)(a– )a–k.
Now, we utilize the following identity:
∞
n=
(n+ )Hntn=
t–log( –t)
( –t) . ()
Let us consider the following associated sequence:
qn(x)∼ ,t( –t)
. ()
Forn≥, from () and (), we have
qn(x) = n
k=
n–k– n–k
(n– )n–kxk. ()
Let us assume that
pn(x)∼ ,t–log( –t)
. ()
We observe that
t–log( –t) =t+
∞
n=
tn
n = t+
∞
n=
tn
n. ()
From (), (), () andxn∼(,t), we can derive the following equation:
pn(x) =x
t (t+∞n=tnn)
n
x–xn
= –nx
+
∞
n=
tn–
n
–n
xn–
= –nx
∞
l=
–n l
∞
n=
tn–
n
l
xn–
= –nx
n–
l=
(–)l
n+l– l
×
n––l m=
m+···+ml=m
l(m
+ )· · ·(ml+ )
= –n
Therefore, by () and (), we obtain the following theorem.
Now, we recall the following identity:
∞
n=
nHntn=
t{ + t– ( +t)log( –t)}
( –t) . ()
Let us consider the following associated sequence:
qn(x)∼ ,t( –t)
. ()
Forn≥, from () and (), we can derive the following equation:
qn(x) = n
k=
n–k– n–k
(n– )n–kxk. ()
Let us assume that
pn(x)∼ ,t
+ t– ( +t)log( –t). ()
We observe that
+ t– ( +t)log( –t) = + t+ ( +t)
∞
j=
tj j
= + t+t+
∞
j=
tj
j +
∞
j=
tj+ j
= + t+
∞
j=
tj+
j+ +
∞
j=
tj+
j+
= + t+
∞
j=
j+ (j+ )(j+ )t
j+. ()
Forn≥, by (), (), () andxn∼(,t), we get
pn(x) =x
t
t{ + t– ( +t)log( –t)}
n
x–xn
=x
+ t+
∞
j=
j+ (j+ )(j+ )t
j+ –n
xn–
=x
n–
l=
(–)l
n+l– l
+
∞
j=
j+ (j+ )(j+ )t
j+ l
tlxn–
=
n–
l=
n––l a=
n–a–l k=
j+···+ja=n–a–k–l
(–)l
n+l– l
l a
l–a(n– )n–k
×
a
i=(ji+ )
a
i=(ji+ )(ji+ )
=
Therefore, by () and (), we obtain the following theorem.
Theorem For n≥, ≤k≤n,we have
Here we invoke the following identity:
∞
Let us consider the following associated sequence:
qn(x)∼ ,t( –t)
From () and (), we note that
Let us assume that
pn(x)∼ ,t + t– ( +t)log( –t)
Theorem For n≥, ≤k≤n,we have
n–k– n–k
(n– )n–k
=
n
m=k n–m
l=
n–m–l a=
j+···+ja=n–a–m–l
b+···+bn=m–k (–)l
n+l– l
l a
l–a
×(n– )n–m(m– )m–k a
i=(ji+ )
a
i=(ji+ )(ji+ )
b+
c=
· · ·
bn+
cn=
n
i=
ciHci.
Here we use the following identity:
∞
n=
n(n+ )Hntn=
t{( +t) – (t+ )log( –t)}
( –t) . ()
Let us consider the following associated sequence:
qn(x)∼ ,t( –t)
. ()
By () and (), we get
qn(x) = n
k=
n–k– n–k
(n– )n–kxk (n≥). ()
Let us assume that
pn(x)∼ ,t
( +t) – (t+ )log( –t). ()
We see that
( +t) – (t+ )log( –t) = + t+
∞
n=
n+ n(n+ )t
n+. ()
Forn≥, from (), (), () andxn∼(,t), we have
pn(x) =x
t
t{( +t) – (t+ )log( –t)}
n
x–xn =x ( +t) – (t+ )log( –t)–nxn–
=x
+ t+
∞
j=
j+ j(j+ )t
j+ –n
xn–. ()
From (), by the same method of (), we get
pn(x) = –n n
k= n–k
l=
n–k–l a=
j+···+ja=n–a–l–k (–)l
n+l– l
l a
l–a
×(n– )n–k
a
i=
(ji+ )
(ji+ )(ji+ )
Forn≥, by (), (), (), () and (), we get
By the same method, we can derive the following identity from ():
qn(x) = –n
By comparing coefficients on both sides of () and (), we get
Remark Recently, several authors have studied theq-extension of harmonic and hyper-harmonic numbers (see [–]).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the manuscript and typed, read and approved the final manuscript.
Author details
1Department of Mathematics, Sogang University, Seoul, 121-742, Republic of Korea.2Department of Mathematics,
Kwangwoon University, Seoul, 139-701, Republic of Korea.
Acknowledgements
The authors express their sincere gratitude to the referees for their valuable suggestions and comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786).
Received: 26 April 2013 Accepted: 22 July 2013 Published: 7 August 2013
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doi:10.1007/s13370-012-0106-6
doi:10.1186/1687-1847-2013-235