1
2
3
47
6
Journal of Integer Sequences, Vol. 15 (2012),
Article 12.9.6
23 11
Identities Involving Generalized Harmonic
Numbers and Other Special Combinatorial
Sequences
Huyile Liang1 and Wuyungaowa
Department of Mathematics
College of Sciences and Technology
Inner Mongolia University
Hohhot 010021
P. R. China
lianghuyile@gmail.com
wuyungw@163.com
Abstract
In this paper, we study the properties of the generalized harmonic numbers Hn,k,r (α, β).
In particular, by means of the method of coefficients, generating functions and Riordan arrays, we establish some identities involving the numbers Hn,k,r (α, β), Cauchy
numbers, generalized Stirling numbers, Genocchi numbers and higher order Bernoulli
numbers. Furthermore, we obtain the asymptotic values of some summations associated with the numbers Hn,k,r (α, β) by Darboux’s method and Laplace’s method.
1
Introduction
Harmonic numbers are important in various branches of combinatorics and number theory,
and they also frequently appear in the analysis of algorithms and expressions for special
functions. Recently, many papers have been devoted to the study of harmonic number
identities by various methods;
P see, for instance, [1, 2, 3, 4, 5, 6, 7]. We recall thePdefinitionn of
harmonic numbers Hn = nk=1 k1 for n ≥ 0. The generating function of Hn is ∞
n=1 Hn t =
1
This work was supported by the Science Research Foundation of Inner Mongolia (2012MS0118) and the
National Natural Science Foundation of China (NSFC Grant #11061020).
1
− ln(1−t)
. In this paper, we discuss a class of generalized harmonic numbers Hn,k,r (α, β). We
1−t
refer to Zhao and Wuyungaowa [10] for this topic. The definition of Hn,k,r (α, β) is
∞
X
Hn,k,r (α, β)tn =
n=0
(− ln(1 − αt))r
,
(1 − βt)k
(1)
where k ≥1 and r ≥1 are integers. Let (α, β) be a pair of real numbers and (αβ 6= 0). From
the generating function of Hn,k,r (α, β), we know that Hn,1,1 (1, 1) = Hn (n ≥ 0).
From (1) we obtain
(− ln(1 − αt))r = (1 − βt)k
∞
X
Hn,k,r (α, β)tn
n=0
k X
∞
X
k
(−βt)i
Hn,k,r (α, β)tn
i
n=0
i=0
∞
k
XX k
(−β)h Hn−h,k,r (α, β)tn .
=
h
n=0 h=0
=
(2)
For convenience, let us recall some definitions and notations. Denote the generalized
Stirling numbers of the first kind by s(n, k; r), and the generalized Stirling numbers of the
(k)
(k)
(r)
(x)
(k)
second kind by S(n, k; r). Denote further Cn , Ĉn , Bn , Gn , Gn be the higher order
Cauchy numbers of both kinds, higher order Bernoulli numbers, the generalized Genocchi
numbers, the higher order Genocchi numbers and the generalized Lah numbers. These
numbers satisfy the following generating functions respectively:
∞
X
n=k
∞
X
n=k
∞
X
s(n, k; r)
(−1)n−k tn
lnk (1 + t)
=
,
n!
(1 + t)r k!
|s(n, k; r)|
(− ln(1 − t))k
tn
=
,
n!
(1 − t)r k!
k = 0, 1, 2, . . . ,
k = 0, 1, 2, . . . ,
tn
ert (et − 1)k
=
, k = 0, 1, 2, . . . ,
n!
k!
n=k
k
∞
n
X
t
(k) t
=
,
Cn
n!
ln(1
+
t)
n=0
k
∞
n
X
t
(k) t
Ĉn
,
=
n!
(1 + t) ln(1 + t)
n=k
k
∞
n
X
2t
(k) t
=
,
Gn
n!
et + 1
n=0
x
∞
(x)
X
2
Gn t n
=
,
n n!
t+1
2
e
n=0
S(n, k; r)
2
(3)
(4)
(5)
(6)
(7)
(8)
(9)
∞
X
n=0
t
Bn(r)
n
n!
=
∞
X
t
et − 1
r
,
k
−t
tn
r 1
,
L(n, k; r) = (1 + t)
n!
k!
1
+
t
n=k
∞
X
n=0
∞
X
n=1
Hn(r) (z)tn =
Hn(r) tn =
(− ln(1 − t))r+1
,
t(1 − t)1−z
− ln(1 − t)
.
(1 − t)r
(10)
(11)
(12)
(13)
n
n
P∞Let [tn ]f (t) be the coefficient of t in the formal power series of f (t), where f (t) =
n=0 fn t . (See Merlini, Sprugnoli, and Verri [8] for related topics.) If f (t) and g(t) are
formal power series, we get the following relations:
[tn ](αf (t) + βg(t)) = α[tn ]f (t) + β[tn ]g(t),
[tn ]f (t) = [tn−1 ]f (t),
n
X
[tn ]f (t)g(t) =
[y j ]f (y)[tn−j ]g(t).
(14)
(15)
(16)
j=0
A Riordan array is a pair (d(t), h(t)) of formal power series with h0 = h(0) = 0. It defines
an infinite lower triangular array (dn,k )n,k∈N according to the rule
dn,k = [tn ]d(t)(h(t))k .
Hence we write {dn,k } = (d(t), h(t)). Moreover, if (d(t), h(t)) P
is a Riordan array and f (t) is
k
the generating function of the sequence {fk }k∈N , i.e., f (t) = ∞
k=0 fk t , then we have
∞
X
dn,k fk = [tn ]d(t)f (h(t)) = [tn ]d(t)[f (y) | y = h(t)].
(17)
k=0
Furthermore, based on the generating function (1) we obtain the next three Riordan
arrays:
− ln(1 − αt)
1
,
,
(18)
{Hn,k,r (α, β)} =
(1 − βt)k
t
− ln(1 − αt) − ln(1 − αt)
{Hn,k,r+1 (α, β)} =
,
(19)
,
(1 − βt)k
t
1
− ln(1 − αt)
{Hn,k,r (α, α)} =
.
(20)
,
(1 − αt)k
t
In this paper, we pay particular attention to the three Riordan arrays above.
3
2
(r)
Identities involving Hn,k,r (α, β), s(n, k; r), S(n, k; r), Bn
and L(n, k; r)
Theorem 1. Let k, r, m ≥ 1, l ≥ 0 be integers. Then
n X
k X
k
j=0 h=0
h
(−β)h Hj−h,k,r (α, β)|s(n − j, m; l)|
1
αn−j
=
Hn,l,m+r (α, α).
(n − j)!
m!
(21)
Proof. By applying (2), (4) and (16), we get
n X
k X
k
j=0 h=0
h
(−β)h Hj−h,k,r (α, β)|s(n − j, m; l)|
αn−j
(n − j)!
1
(− ln(1 − αt))m+r
=
Hn,l,m+r (α, α) .
= [t ]
l
(1 − αt) m!
m!
n
Theorem 2. Let n, k, j ≥ 1, l ≥ 0 be integers. Then
n−m
n
X i + k − 1n − i + l − 1
X
m!
β i αn−i .
=
Hn,k,j (α, β)S(j, m; l)
n−m−i
j!
i
i=0
j=m
(22)
Proof. By using (5), (17) and (18), we obtain
n
X
m!
1
= [tn ]
[(ey − 1)m eyl | y = − ln(1 − αt)]
k
j!
(1
−
βt)
j=m
n−m
X i + k − 1n − i + l − 1
1
1
m n−m
β i αn−i .
=
= α [t
]
n
−
m
−
i
(1 − βt)k (1 − αt)l+m
i
i=0
Hn,k,j (α, β)S(j, m; l)
Corollary 3. The following relations hold:
n X
k X
k
j=0 h=0
n
X
j=m
n
X
j=m
h
αn
(m + r)!
αn−j
=
|s(n, m + r)|
,
(n − j)!
m!
n!
n+l+k−1 n
=
α ,
n−m
n+l
.
=
n−m
(−β)h Hj−h,k,r (α, β)|s(n − j, m)|
Hn,k,j (α, α)S(j, m; l)
m!
j!
H(n, r − 1)S(j, m; l)
m!
j!
(23)
(24)
(25)
Proof. Setting l = 0 in (21), we get (23). Setting β = α in (22), we have (24). Setting
β = α = k = 1 in (22), we obtain (25).
4
Theorem 4. Let n, k ≥ 1 and l, j ≥ 0 be integers. Then
n
X
n−m
X
m!
n − i + l − 1 n−i
Hn,k,j+1 (α, β)S(j, m; l)
α .
=
Hi,k,1 (α, β)
j!
n
−
m
−
i
j=m
i=0
(26)
Proof. By applying (5), (17) and (19), we have
n
X
m!
− ln(1 − αt) y
= [tn ]
[(e − 1)m eyl | y = − ln(1 − αt)]
k
j!
(1 − βt)
j=m
n−m
X
1
n − i + l − 1 n−i
m n−m − ln(1 − αt)
=
Hi,k,1 (α, β)
α .
= α [t
]
(1 − βt)k (1 − αt)l+m
n
−
m
−
i
i=0
Hn,k,j+1 (α, β)S(j, m; l)
Corollary 5. The following relations hold:
n
X
j=m
n
X
Hn,k,j+1 (α, α)S(j, m; l)
m!
= αn Hn−m−1 (1 − m − l − k),
j!
(27)
Hn,k,j+1 (α, α)S(j, m; l)
m!
(m+l+k)
= αn Hn−m .
j!
(28)
j=m
Proof. Setting β = α in (26), we obtain (27) and (28).
Theorem 6. Let k, r ≥ 1 and m, l ≥ 0 be integers. Then
n X
k X
k
j=0 h=0
h
(−β)h Hj−h,k,r (α, β)
L(n − j, m; l)
(−α)n−j
(n − j)!
Pl−m l−m
(−1)i αn r!
,
|s(n
−
m
−
i,
r)|

 i=r
m!(n−m−i)!
i
αn
r!
= m! |s(n − m, r)| (n−m)! ,

 αm
Hn−m,m−l,r (α, α),
m!
if l > m;
if l = m;
if l < m.
Proof. By applying (2), (11) and (16), we get
n X
k X
k
L(n − j, m; l)
(−α)n−j
(n
−
j)!
h
j=0 h=0
m
l
αt
n
r (1 − αt)
= [t ](− ln(1 − αt))
m!
1 − αt
Pl−m l−m
(−1)i αn r!
|s(n
−
m
−
i,
r)|
, if l > m;

 i=r
i
m!(n−m−i)!
αn
r!
= m! |s(n − m, r)| (n−m)! ,
if l = m;

 αm
Hn−m,m−l,r (α, α),
if l < m.
m!
(−β)h Hj−h,k,r (α, β)
5
(29)
Theorem 7. Let n, j, k, m ≥ 1 be integers. Then
 1

(m)
n
 αm Hn+m,k−m,m (α, α),
X
Bj
m!
,
= αn |s(n + m, m)| (n+m)!
Hn,k,j (α, α)

j!
P

m−k
(−1)i αn m!
m−k
j=1
,
|s(n
+
m
−
i,
m)|
i=0
(n+m−i)!
i
if k > m;
if k = m;
(30)
if k < m.
Proof. By applying (10), (17) and (20), we get
m
(m)
n
X
Bj
1
y
n
| y = − ln(1 − αt)
= [t ]
Hn,k,j (α, α)
k
y −1
j!
(1
−
αt)
e
j=1
 1
if k > m;

 αm Hn+m,k−m,m (α, α),
m!
n
if k = m;
= α |s(n + m, m)| (n+m)! ,

Pm−k m−k
(−1)i αn m!
|s(n + m − i, m)| (n+m−i)! , if k < m.
i=0
i
3
Identities involving Hn,k,r (α, β), Genocchi numbers and
Cauchy numbers
Theorem 8. Let x be a real number, k, i ≥ 1, n ≥ 0 be integers. Then
n−h x X
n
(x)
X
X
n − h − j + k − 1 j + x − 1 x (−1)h n−j−h j+h
Gi
β
α .
Hn,k,i (α, β) i =
k−1
x−1
2 i!
h
2j
i=1
h=0 j=0
(31)
Proof. By using (9), (17) and (18), we get
x n
(x)
X
1
2
Gi
n
Hn,k,i (α, β) i = [t ]
y = − ln(1 − αt)
k
y +1
2
i!
(1
−
βt)
e
i=1
1
1
(1 − αt)x
k
x
(1 − βt) (1 − αt
)
2
n−h
x
X X n − h − j + k − 1j + x − 1x (−1)h
β n−j−h αj+h .
=
j
2
k−1
x−1
h
h=0 j=0
= [tn ]
Theorem 9. Let x be a real number, k, j ≥ 1 be integers. Then
Pn i+k−x−1 x+n−i−1 n
α

(x)
n−i ,
∞

i=0
2
i
n−i
X
Gj
n+k−1 αn
Hn,k,j (α, α) j =
,
n
2n

2 j!
Px−k x−k x+n−i−1 (−1)i αn
j=1
,
i=0
n−i
i
2n−i
6
if k > x;
if k = x;
if k < x.
(32)
Proof. By using (9), (17) and (20), we have
x (x)
Gj
1
2
n
Hn,k,j (α, α) j = [t ]
y = − ln(1 − αt)
k
y +1
2
j!
(1
−
αt)
e
j=1
Pn i+k−x−1 x+n−i−1 n
α

, if k > x;
 i=0 i
n−i
2n−i
n+k−1 αn
=
,
if k = x;
n
2n

Px−k x−k x+n−i−1 (−1)i αn
, if k < x.
i=0
2n−i
n−i
i
∞
X
Corollary 10. The following relations hold:
n
X
n X
Gj
i + k − 2 αn
Hn,k,j (α, α) j =
,
2 j!
2n−i
i
j=1
i=0
n
X
H(n, j − 1)
j=1
(33)
Gj
1
= n.
j
2 j!
2
(34)
Proof. Setting x = 1 in (32), we have (33). Setting x = k = 1 in (32), we obtain (34).
Theorem 11. Let k, l, m ≥ 1 be integers. Then
n
X
l=1
(m)
G
Hn,k,l (α, β) l
l!
=
n−j
m X
X
j=0 i=0
m n − i − j + m − 1 (−1)j αn−i
Hi,k,m (α, β)
.
m−1
j
2n−i−j
(35)
Proof. By applying (8), (17) and (18), we have
m n
(m)
X
Gl
1
2y
n
Hn,k,l (α, β)
= [t ]
y = − ln(1 − αt)
l!
(1 − βt)k
ey + 1
l=1
1
(− ln(1 − αt))m
(1 − αt)m
k
m
(1 − βt)
(1 − αt
)
2
n−j
m
XX
m n − i − j + m − 1 (−1)j αn−i
.
=
Hi,k,m (α, β)
n−i−j
2
j
m
−
1
j=0 i=0
= [tn ]
Theorem 12. Let n, k, j, m ≥ 1 be integers. Then
 Pn
αi
Hn−i,k−m,m (α, α) i+m−1
,

(m)
n
i=0

m−1
2i
X
Gj
Pn
m! n−i+m−1 αn
,
=
Hn,k,j (α, α)
i=m |s(i, m)| i!
m−1
2n−i

j!
i αn m!
Pm−k Pn−i m−k j+m−1
j=1
|s(n − i − j, m)| 2(−1)
j (n−i−j)! ,
j=0
i=0
m−1
i
7
if k > m;
if k = m;
if k < m.
(36)
Proof. By using (8), (17) and (20), we have
m (m)
n
X
Gj
2y
1
n
Hn,k,j (α, α)
= [t ]
y = − ln(1 − αt)
j!
(1 − αt)k
ey + 1
j=1
 Pn
i+m−1 αi
,
if k > m;

Pi=0 Hn−i,k−m,m (α, α) m−1
2i
n
m! n−i+m−1 αn
,
if k = m;
=
i=m |s(i, m)| i!
m−1
2n−i

Pm−k Pn−i m−k j+m−1
(−1)i αn m!
|s(n − i − j, m)| 2j (n−i−j)! , if k < m.
j=0
i=0
m−1
i
Theorem 13. Let n, k, r, m ≥ 1 be integers. Then
 n
(r−m)!

n X
k
α |s(n − m, r − m)| (n−m)! ,
n−j (m)
X
(−α)
C
k
n−j
= αm δn,m ,
(−β)h Hj−h,k,r (α, β)

(n
−
j)!
h
 (−1)n−r αn (m−r)
j=0 h=0
Cn−r ,
(n−r)!
if r > m;
if r = m;
if r < m.
(37)
Proof. By using (2), (6) and (16), we have
m
(m)
n X
k X
(−α)n−j Cn−j
k
−αt
h
n
r
(−β) Hj−h,k,r (α, β)
= [t ](− ln(1 − αt))
h
(n
−
j)!
ln(1 − αt)
j=0 h=0
 n
(r−m)!

α |s(n − m, r − m)| (n−m)! , if r > m;
= αm δn,m ,
if r = m;

 (−1)n−r αn (m−r)
if r < m.
Cn−r ,
(n−r)!
Theorem 14. Let n, k, r, m ≥ 1 be integers. Then
(m)
n X
k X
(−α)n−j Ĉn−j
k
h
(−β) Hj−h,k,r (α, β)
(n − j)!
h
j=0 h=0

m
(α, α),
if r > m;

α Hn−m,m,r−m
n n−1
if r = m;
= α n−m ,

Pn−r
(m−r) (−1)n−r−i αn i+r−1
, if r < m.
i=m−r Ĉn−r−i (n−r−i)!
r−1
(38)
Proof. The proof of (38) is similar to that of (37), and it is omitted here.
Theorem 15. Let k, r ≥ 1 be integers. Then
n
X
j=0
Hj,k,r (α, β)
(−α)n−j Cn−j
= αHn−1,k,r−1 (α, β),
(n − j)!
n
X
(39)
n−1
(−α)n−j Ĉn−j X
Hj,k,r (α, β)
=
Hi,k,r−1 (α, β)αn−i .
(n − j)!
j=0
i=0
8
(40)
Proof. By applying (1), (6) and (16), we get
n
X
Hj,k,r (α, β)
j=0
αt
(− ln(1 − αt))r
(−α)n−j Cn−j
= [tn ]
= αHn−1,k,r−1 (α, β) .
k
(n − j)!
(1 − βt)
− ln(1 − αt)
Hence (39) holds. The proof of (40) is similar to that of (39), and it is omitted here.
Corollary 16. Let n, r ≥ 2. Then
n
X
H(j, r − 1)(−α)n−j Cn−j
(n − j)!
j=r
n
X
H(j, r − 1)(−α)n−j Ĉn−j
(n − j)!
j=r
= αn H(n − 1, r − 2),
=α
n
n−1
X
H(i − 1, r − 2).
(41)
(42)
i=r
Proof. Setting β = α and k = 1 in (39) and (40), we obtain (41) and (42).
4
Asymptotics
In this section, we give the asymptotic expansion of certain sums involving Hn,k,r (α, β). We
first recall three lemmas.
A singularity of f (z) at |z| = w is called algebraic if f (z) can be written as the sum of a
function analytic in a neighborhood of w and a finite number of terms of the form
(1 −
z α
) g(z),
w
(43)
where g(z) is analytic near w, g(w) 6= 0 and α 6∈ {0, 1, 2, . . .}.
Lemma 17. (See [8].) Suppose that f (z) is analytic for |z| < R, R > 0 and has only
algebraic singularities on |z| = R. Let a be the minimum of ℜ(α) for the terms of the form
at the singularity of f (z) on |z| = R, and let wj , αj and gj (z) be the w, α and g(z) for those
terms of the form (43) for which Re(α) = a. Then, as n → ∞,
[z n ]f (z) =
X gj (wj )n−αj −1
j
Γ(−αj )wjn
+ o(R−n n−a−1 ).
Lemma 18. (see [11]) Let α be a real number and
L(z) = ln(
1
).
1−z
When n → ∞,
1
n−α−1 lnk n , (α 6∈ {0, 1, 2, . . .})
Γ(−α)
[z n ](1 − z)m Lk (z) ∼ (−1)m km!n−m−1 lnk−1 n , (m ∈ Z≥0 , k ∈ Z≥1 ).
[z n ](1 − z)α Lk (z) ∼
9
P
P
Lemma 19. (see [11]) Suppose that a(z) =
an z n and b(z) =
bn z n are power series
→ β as n → ∞. If
with radii of convergence
α > β ≥ 0, respectively. Suppose that bn−1
bn
P
n
a(β) 6= 0 and
cn z = a(z)b(z), then
cn ∼ a(β)bn
as n → ∞.
Theorem 20. Let k, r ≥ 1. As n → ∞, we get
n X
αn r r−1
k
(−β)h Hn−h,k,r (α, β) ∼
ln n .
h
n
h=0
Proof. By Eq. (2) and Lemma 18, we get
k X
αn r r−1
k
ln n .
(−β)h Hn−h,k,r (α, β) = [tn ](− ln(1 − αt))r ∼
n
h
h=0
Theorem 21. Let k, r, m ≥ 1. As n → ∞, we have
n X
k X
αn−j
k
αn (m + r)
(−β)h Hj−h,k,r (α, β)|s(n − j, m)|
∼
(ln n)m+r−1 .
h
(n
−
j)!
m!n
j=0 h=0
Proof. By Eq. (23) and Lemma 18, we obtain
n X
k X
αn−j
k
1 n
h
(−β) Hj−h,k,r (α, β)|s(n − j, m)|
=
[t ](− ln(1 − αt))m+r
h
(n − j)!
m!
j=0 h=0
∼
αn (m + r)
(ln n)m+r−1 .
m!n
Theorem 22. Let k, j, m ≥ 1 and l ≥ 0. As n → ∞, we have
n
X
Hn,k,j (α, α)S(j, m; l)
j=1
m!
αn nk+m+l−1
∼
.
j!
Γ(k + l + m)
Proof. By Eq. (24) and Lemma 17, we obtain
n
X
j=1
Hn,k,j (α, α)S(j, m; l)
tm
αn nk+m+l−1
m!
= αm [tn ]
∼
.
j!
(1 − αt)k+l+m
Γ(k + l + m)
Theorem 23. Let k, r ≥ 1. As n → ∞, we get
n X
k X
α n nl r
k
n − j + l n−j
h
ln n .
(−β) Hj−h,k,r (α, β)
β
∼
l!
h
n−j
j=0 h=0
10
Proof. It is well known that
∞ X
n+l
n
n=0
tn =
1
.
(1 − t)l+1
(44)
By Lemma 18 and Eq. (44), we get
n X
k X
k
(− ln(1 − αt))r
n − j + l n−j
α n nl r
(−β) Hj−h,k,r (α, β)
α
= [tn ]
ln n .
∼
h
n−j
(1 − αt)l+1
l!
j=0 h=0
h
Theorem 24. Let j, k, m ≥ 1. As n → ∞, we have
(
(m)
n
X
Bj
(−1)m−k αn m(m − k)!(n + m)k−m−1 (ln(n + m))m−1 ,
Hn,k,j (α, α)
∼ αn (n+m)k−m−1
j!
(ln(n + m))m ,
Γ(m−k)
j=1
if m − k ∈ Z≥0 ;
if m − k 6∈ Z≥0 .
Proof. By the proof of Eq. (28) and Lemma 18, we obtain
n
X
(m)
Bj
Hn,k,j (α, α)
j!
j=1
∼
(
1
= [t ]
(1 − αt)k
n
y
ey − 1
m
| y = − ln(1 − αt)
(−1)m−k αn m(m − k)!(n + m)k−m−1 (ln(n + m))m−1 , if m − k ∈ Z≥0 ;
αn (n+m)k−m−1
(ln(n + m))m ,
if m − k 6∈ Z≥0 .
Γ(m−k)
Theorem 25. Let j be fixed and x be a real number. As n → ∞, we have
 x n k−x−1
n

,
if k > x;
 2 αΓ(k−x)
(x)
n

X
Gj
n
k−1
if k = x;
Hn,k,j (α, α) j ∼ α2nnΓ(k) ,

2 j!
x−k αn nx−1

(−1)
j=1

, if k < x.
n
2 Γ(x)
Proof. By the proof of (32) and Lemma 17, we

1
1

,

(1−αt)k−x (1− αt
(x)
)x
n

2
X
Gj
n
αt x
Hn,k,j (α, α) j = [t ] (1 − 2 ) ,

2 j!
x−k

j=1
 (1−αt)
,
(1− αt )x
2
obtain
 x n k−x−1
2 α n

if k > x;

 Γ(k−x) ,
αn nk−1
if k = x;
if k = x; ∼ 2n Γ(k) ,

x−k
n
x−1

 (−1) n α n , if k < x.
if k < x.
2 Γ(x)
if k > x;
Theorem 26. Let j, k, m ≥ 1, j be fixed. As n → ∞, we have
 αn 2m nk−m−1 lnm+1 n
,
if k > m;

(m)
n

Γ(k−m)
X
Gj
m
n
Hn,k,j (α, α)
∼ 2 αn m lnm−1 n,
if k = m;

j!
 (−1)m−k 2m αn (m−k)! m
j=1
ln n, if k < m.
nm+1−k
11
Proof. By (36), we see that
(m)
n
X
Gj
Hn,k,j (α, α)
j!
j=1

1
1

(− ln(1 − αt))m ,
if k > m;

 (1−αt)k−m (1− αt2 )m
if k = m;
= [tn ] (1− 1αt )m (− ln(1 − αt))m ,
2


(1 − αt)m−k 1αt (− ln(1 − αt))m , if k < m.
(1− )m
2
Let k = m, and set
∞ ∞
X
X
1
n + m − 1 αt n
m!
m
( ) and b(t) = (− ln(1−αt)) =
a(t) =
|s(n, m)| αn tn .
αt m =
m−1
2
n!
(1 − 2 )
n=0
n=m
in Lemma 19. Since
n
X
m!
|s(n−1,m)| (n−1)!
|s(n,m)| m!
n!
Hn,k,j (α, α)
j=1
→
(m)
Gj
j!
1
α
as n → ∞, a( α1 ) = 2m 6= 0. Then we have
 αn 2m nk−m−1 lnm+1 n
,
if k > m;


Γ(k−m)
m−1
2m αn m
∼
ln
n,
if k = m;
n

 (−1)m−k 2m αn (m−k)! m
ln n, if k < m.
nm+1−k
This gives the result for the case k = m. The proofs of the cases k > m and k < m are
similar to that of k = m.
In the final result of this section, we give the asymptotic expansion of certain sums for
binomial coefficients and Hn,k,r (α, β) by Laplace’s method.
Theorem 27. Let k, r ≥ 1, 0 < β, (b + 1)b+1 6= βbb , (b + 1)b+1 > αbb and b be a positive
integer. As k → ∞, we get
∞
X
Hn,k,r (α, β)
[(b + 1)n + 1] (b+1)n
n
k− 21 r s
(b + 1)b+1
(b + 1)b+1
2π(b + 1)b−2
.
∼
ln
(b + 1)b+1 − βbb
(b + 1)b+1 − αbb
βkbb−1
n=0
(45)
Proof. From Tiberiu [9], we know that the inverse of a binomial coefficient is related to an
integral as follows:
−1
Z 1
n
tm (1 − t)n−m dt.
(46)
= (n + 1)
m
0
Owing to (46), we have
∞
X
Hn,k,r (α, β)
=
(b+1)n
∞
X
Hn,k,r (α, β)
[(b + 1)n + 1] n
n=0
Z
∞
1X
b n
=
Hn,k,r (α, β)(t(1 − t) ) dt =
n=0
Z
=
Z
0
n=0
1
1
0
b
Z
1
tn (1 − t)nb dt
0
(− ln(1 − αt(1 − t)b ))r
dt
(1 − βt(1 − t)b )k
(− ln(1 − αt(1 − t)b )r e−k ln(1−βt(1−t) ) dt .
0
12
Let ϕ(t) = (− ln(1 − αt(1 − t)b )r and h(t) = − ln(1 − βt(1 − t)b ). Then h(t) reaches a
1
1
1
, h′ ( b+1
) = 0 and h′′ ( b+1
) < 0. By applying Laplace’s method, we have
maximum at t = b+1
Z 1
∞
X
Hn,k,r (α, β)
b
(− ln(1 − αt(1 − t)b )r e−k ln(1−βt(1−t) dt
=
(b+1)n
0
n=0 [(b + 1)n + 1]
sn
1
−2π
1
)ekh( b+1 )
∼ ϕ(
.
1
b+1
kh′′ ( b+1
)
Corollary 28. Let k, r ≥ 1, α < 4, 0 < β, β 6= 4 and b be a positive integer. As k → ∞,
we get
k− 21 r r
∞
X
4
4
π
Hn,k,r (α, β)
.
(47)
ln
2n ∼
4−β
4−α
kβ
(2n + 1) n
n=0
Proof. Setting b = 1 in (45), we obtain (47).
Theorem 29. Let k, r ≥ 1, 0 < α, (b + 1)b+1 6= −βbb and b be a positive integer. As
r → ∞, we get
∞
X
(−1)n Hn,k,r (α, β)
(b+1)n
[(b
+
1)n
+
1]
n=0
n
k 1s
b+1
b+1
b r+ 2
(b
+
1)
(b
+
1)
+
αb
2π(b + 1)b−2 ((b + 1)b+1 + αbb )
∼ (−1)r
ln
.
(b + 1)b+1 + βbb
(b + 1)b+1
αrbb−1 (b + 1)b+1
(48)
Proof. Owing to (46), we get
∞
X
∞
X
(−1)n Hn,k,r (α, β)
=
(b+1)n
n
(−1) Hn,k,r (α, β)
n=0
n=0 [(b + 1)n + 1]
n
Z 1X
Z
∞
b n
=
Hn,k,r (α, β)(−t(1 − t) ) dt =
0
n=0
= (−1)r
at
Z
1
0
1
0
Z
1
tn (1 − t)nb dt
0
(− ln(1 + αt(1 − t)b ))r
dt
(1 + βt(1 − t)b )k
r ln ln(1+αt(1−t)b
e
dt .
(1 + βt(1 − t)b )k
1
b
Let ϕ(t) = (1+βt(1−t)
b )k and h(t) = ln ln(1 + αt(1 − t) ). Then h(t) reaches the maximum
1
1
1
, h′ ( b+1
) = 0 and h′′ ( b+1
) < 0. By applying Laplace’s method, we have
t = b+1
∞
X
(−1)n Hn,k,r (α, β)
n=0
[(b + 1)n + 1]
∼ ϕ(
=
(b+1)n
1
1
)erh( b+1 )
b+1
sn
Z
1
b
(− ln(1 − αt(1 − t)b )r e−k ln(1−βt(1−t) dt
0
−2π
.
1
)
rh′′ ( b+1
13
Corollary 30. Let k, r ≥ 1, 0 < α, β 6= −4. As r → ∞, we get
k r+ 12 r
∞
X
4+α
4
(−1)n Hn,k,r (α, β)
π(4 + α)
r
∼ (−1)
ln
.
2n
4
+
β
4
4αr
(2n
+
1)
n
n=0
(49)
Proof. Setting b = 1 in (48), we derive (49).
5
Acknowledgment
The authors are grateful to the anonymous referee for his/her helpful comments.
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14
2010 Mathematics Subject Classification: Primary 05A16, 05A19; Secondary 05A15.
Keywords: generalized harmonic number, Genocchi number, Stirling number, Cauchy number, Riordan array, method of coefficients.
Received October 25 2012; revised version received November 20 2012. Published in Journal
of Integer Sequences, December 27 2012.
Return to Journal of Integer Sequences home page.
15