Identities Involving Generalized Harmonic Numbers and Other Special Combinatorial

Sequences

Huyile Liang^{1} 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 H_{n,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 H_{n,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 H_{n,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; see, for instance, [1, 2, 3, 4, 5, 6, 7]. We recall the definition of harmonic numbers H_{n} = ∑_{k=1}^{n} \frac{1}{k} for n ≥ 0. The generating function of H_{n} is ∑_{n=1}^{∞} H_{n}t^{n} = −\frac{ln(1−t)}{1−t}. In this paper, we discuss a class of generalized harmonic numbers H_{n,k,r}(α, β). We refer to Zhao and Wuyungaowa [10] for this topic. The definition of H_{n,k,r}(α, β) is

1This work was supported by the Science Research Foundation of Inner Mongolia (2012MS0118) and the National Natural Science Foundation of China (NSFC Grant #11061020).

∑_{n=0}^{∞} H_{n,k,r}(α, β)t^{n} = \frac{(− 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 ≠ 0). From the generating function of H_{n,k,r}(α, β), we know that H_{n,1,1}(1, 1) = H_{n}(n ≥ 0).

From (1) we obtain

(− ln(1 − αt))^{r} = (1 − βt)^{k} ∑_{n=0}^{∞} H_{n,k,r}(α, β)t^{n}

= ∑_{i=0}^{k} \binom{k}{i}(−βt)^{i} ∑_{n=0}^{∞} H_{n,k,r}(α, β)t^{n}

= ∑_{n=0}^{∞} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{n−h,k,r}(α, β)t^{n}.  (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 second kind by S(n, k; r). Denote further C_{n}^{(k)}, Ĉ_{n}^{(k)}, B_{n}^{(r)}, G_{n}^{(x)}, G_{n}^{(k)} 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:

∑_{n=k}^{∞} s(n, k; r)\frac{(−1)^{n−k}t^{n}}{n!} = \frac{ln^{k}(1 + t)}{(1 + t)^{r}k!}, k = 0, 1, 2, …,  (3)

∑_{n=k}^{∞} |s(n, k; r)|\frac{t^{n}}{n!} = \frac{(− ln(1 − t))^{k}}{(1 − t)^{r}k!} , k = 0, 1, 2, …,  (4)

∑_{n=k}^{∞} S(n, k; r)\frac{t^{n}}{n!} = \frac{e^{rt}(e^{t} − 1)^{k}}{k!} , k = 0, 1, 2, …,  (5)

∑_{n=0}^{∞} C_{n}^{(k)} \frac{t^{n}}{n!} = (\frac{t}{ln(1 + t)})^{k},  (6)

∑_{n=k}^{∞} Ĉ_{n}^{(k)} \frac{t^{n}}{n!} = (\frac{t}{(1 + t) ln(1 + t)})^{k},  (7)

∑_{n=0}^{∞} G_{n}^{(k)} \frac{t^{n}}{n!} = (\frac{2t}{e^{t} + 1})^{k},  (8)

∑_{n=0}^{∞} \frac{G_{n}^{(x)}}{2^{n}} \frac{t^{n}}{n!} = (\frac{2}{e^{t} + 1})^{x},  (9)

∑_{n=0}^{∞} B_{n}^{(r)} \frac{t^{n}}{n!} = (\frac{t}{e^{t} − 1})^{r},  (10)

∑_{n=k}^{∞} L(n, k; r)\frac{t^{n}}{n!} = (1 + t)^{r} \frac{1}{k!}(\frac{−t}{1 + t})^{k},  (11)

∑_{n=0}^{∞} H_{n}^{(r)}(z)t^{n} = \frac{(− ln(1 − t))^{r+1}}{t(1 − t)^{1−z}} ,  (12)

∑_{n=1}^{∞} H_{n}^{(r)}t^{n} = \frac{− ln(1 − t)}{(1 − t)^{r}}.  (13)

Let [t^{n}]f(t) be the coefficient of t^{n} in the formal power series of f(t), where f(t) = ∑_{n=0}^{∞} f_{n}t^{n}. (See Merlini, Sprugnoli, and Verri [8] for related topics.) If f(t) and g(t) are formal power series, we get the following relations:

[t^{n}](αf(t) + βg(t)) = α[t^{n}]f(t) + β[t^{n}]g(t),  (14)
[t^{n}]f(t) = [t^{n−1}]f(t),  (15)
[t^{n}]f(t)g(t) = ∑_{j=0}^{n}[y^{j}]f(y)[t^{n−j}]g(t).  (16)

A Riordan array is a pair (d(t), h(t)) of formal power series with h_{0} = h(0) = 0. It defines an infinite lower triangular array (d_{n,k})_{n,k∈N} according to the rule

d_{n,k} = [t^{n}]d(t)(h(t))^{k} .

Hence we write {d_{n,k}} = (d(t), h(t)). Moreover, if (d(t), h(t)) is a Riordan array and f(t) is the generating function of the sequence {f_{k}}_{k∈N}, i.e., f(t) = ∑_{k=0}^{∞} f_{k}t^{k}, then we have

∑_{k=0}^{∞} d_{n,k}f_{k} = [t^{n}]d(t)f(h(t)) = [t^{n}]d(t)[f(y) | y = h(t)].  (17)

Furthermore, based on the generating function (1) we obtain the next three Riordan arrays:

\{H_{n,k,r}(α, β)\} = (\frac{1}{(1 − βt)^{k}} , \frac{− ln(1 − αt)}{t}),  (18)
\{H_{n,k,r+1}(α, β)\} = (\frac{− ln(1 − αt)}{(1 − βt)^{k}}, \frac{− ln(1 − αt)}{t}),  (19)
\{H_{n,k,r}(α, α)\} = (\frac{1}{(1 − αt)^{k}} , \frac{− ln(1 − αt)}{t}).  (20)

In this paper, we pay particular attention to the three Riordan arrays above.

2 Identities involving H_{n,k,r}(α, β), s(n, k; r), S(n, k; r), B_{n}^{(r)} and L(n, k; r)

Theorem 1. Let k, r, m ≥ 1, l ≥ 0 be integers. Then

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)|s(n − j, m; l)|\frac{α^{n−j}}{(n − j)!} = \frac{1}{m!}H_{n,l,m+r}(α, α). (21)

Proof. By applying (2), (4) and (16), we get

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)|s(n − j, m; l)|\frac{α^{n−j}}{(n − j)!}

= [t^{n}]\frac{(− ln(1 − αt))^{m+r}}{(1 − αt)^{l}m!} = \frac{1}{m!}H_{n,l,m+r}(α, α) .

Theorem 2. Let n, k, j ≥ 1, l ≥ 0 be integers. Then

∑_{j=m}^{n} H_{n,k,j}(α, β)S(j, m; l)\frac{m!}{j!} = ∑_{i=0}^{n−m} \binom{i + k − 1}{i}\binom{n − i + l − 1}{n − m − i}β^{i}α^{n−i}.  (22)

Proof. By using (5), (17) and (18), we obtain

∑_{j=m}^{n} H_{n,k,j}(α, β)S(j, m; l)\frac{m!}{j!} = [t^{n}]\frac{1}{(1 − βt)^{k}} [(e^{y} − 1)^{m}e^{yl} | y = − ln(1 − αt)]

= α^{m}[t^{n−m}]\frac{1}{(1 − βt)^{k}} \frac{1}{(1 − αt)^{l+m}} = ∑_{i=0}^{n−m} \binom{i + k − 1}{i}\binom{n − i + l − 1}{n − m − i}β^{i}α^{n−i} .

Corollary 3. The following relations hold:

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)|s(n − j, m)|\frac{α^{n−j}}{(n − j)!} = \frac{α^{n}}{m!}|s(n, m + r)|\frac{(m + r)!}{n!}, (23)

∑_{j=m}^{n} H_{n,k,j}(α, α)S(j, m; l)\frac{m!}{j!} = \binom{n + l + k − 1}{n − m}α^{n},  (24)

∑_{j=m}^{n} H(n, r − 1)S(j, m; l)\frac{m!}{j!} = \binom{n + l}{n − m}.  (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).

Theorem 4. Let n, k ≥ 1 and l, j ≥ 0 be integers. Then

∑_{j=m}^{n} H_{n,k,j+1}(α, β)S(j, m; l)\frac{m!}{j!} = ∑_{i=0}^{n−m} H_{i,k,1}(α, β)\binom{n − i + l − 1}{n − m − i}α^{n−i}.  (26)

Proof. By applying (5), (17) and (19), we have

∑_{j=m}^{n} H_{n,k,j+1}(α, β)S(j, m; l)\frac{m!}{j!} = [t^{n}]\frac{− ln(1 − αt)}{(1 − βt)^{k}}[(e^{y} − 1)^{m}e^{yl} | y = − ln(1 − αt)]

= α^{m}[t^{n−m}]\frac{− ln(1 − αt)}{(1 − βt)^{k}} \frac{1}{(1 − αt)^{l+m}} = ∑_{i=0}^{n−m} H_{i,k,1}(α, β)\binom{n − i + l − 1}{n − m − i}α^{n−i} .

Corollary 5. The following relations hold:

∑_{j=m}^{n} H_{n,k,j+1}(α, α)S(j, m; l)\frac{m!}{j!} = α^{n}H_{n−m−1}(1 − m − l − k),  (27)
∑_{j=m}^{n} H_{n,k,j+1}(α, α)S(j, m; l)\frac{m!}{j!} = α^{n}H_{n−m}^{(m+l+k)}.  (28)

Proof. Setting β = α in (26), we obtain (27) and (28). Theorem 6. Let k, r ≥ 1 and m, l ≥ 0 be integers. Then

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\frac{L(n − j, m; l)}{(n − j)!}(−α)^{n−j}

= \begin{cases}∑_{i=r}^{l−m} \binom{l−m}{i}|s(n − m − i, r)|\frac{(−1)^{i}α^{n}r!}{m!(n−m−i)!}, & if l > m; \\ \frac{α^{n}}{m!}|s(n − m, r)|\frac{r!}{(n−m)!}, & if l = m; \\ \frac{α^{m}}{m!} H_{n−m,m−l,r}(α, α), & if l < m. \end{cases}  (29)

Proof. By applying (2), (11) and (16), we get

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\frac{L(n − j, m; l)}{(n − j)!}(−α)^{n−j}

= [t^{n}](− ln(1 − αt))^{r} \frac{(1 − αt)^{l}}{m!} (\frac{αt}{1 − αt})^{m}

= \begin{cases}∑_{i=r}^{l−m} \binom{l−m}{i}|s(n − m − i, r)|\frac{(−1)^{i}α^{n}r!}{m!(n−m−i)!}, & if l > m; \\ \frac{α^{n}}{m!}|s(n − m, r)|\frac{r!}{(n−m)!}, & if l = m; \\ \frac{α^{m}}{m!} H_{n−m,m−l,r}(α, α), & if l < m. \end{cases}

Theorem 7. Let n, j, k, m ≥ 1 be integers. Then

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{B_{j}^{(m)}}{j!} = \begin{cases}\frac{1}{α^{m}} H_{n+m,k−m,m}(α, α), & if k > m; \\ α^{n}|s(n + m, m)|\frac{m!}{(n+m)!}, & if k = m; \\ ∑_{i=0}^{m−k} \binom{m−k}{i}|s(n + m − i, m)|\frac{(−1)^{i}α^{n}m!}{(n+m−i)!}, & if k < m. \end{cases}  (30)

Proof. By applying (10), (17) and (20), we get

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{B_{j}^{(m)}}{j!} = [t^{n}]\frac{1}{(1 − αt)^{k}} [(\frac{y}{e^{y} − 1})^{m} | y = − ln(1 − αt)]

= \begin{cases}\frac{1}{α^{m}} H_{n+m,k−m,m}(α, α), & if k > m; \\ α^{n}|s(n + m, m)|\frac{m!}{(n+m)!}, & if k = m; \\ ∑_{i=0}^{m−k} \binom{m−k}{i}|s(n + m − i, m)|\frac{(−1)^{i}α^{n}m!}{(n+m−i)!}, & if k < m. \end{cases}

3 Identities involving H_{n,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

∑_{i=1}^{n} H_{n,k,i}(α, β)\frac{G_{i}^{(x)}}{2^{i}i!} = ∑_{h=0}^{x} ∑_{j=0}^{n−h} \binom{n − h − j + k − 1}{k − 1}\binom{j + x − 1}{x − 1}\binom{x}{h}\frac{(−1)^{h}}{2^{j}} β^{n−j−h}α^{j+h}.

(31)

Proof. By using (9), (17) and (18), we get

∑_{i=1}^{n} H_{n,k,i}(α, β)\frac{G_{i}^{(x)}}{2^{i}i!} = [t^{n}]\frac{1}{(1 − βt)^{k}} [(\frac{2}{e^{y} + 1})^{x} | y = − ln(1 − αt)]

= [t^{n}]\frac{1}{(1 − βt)^{k}} \frac{1}{(1 − \frac{αt}{2})^{x}} (1 − αt)^{x}

= ∑_{h=0}^{x} ∑_{j=0}^{n−h} \binom{n − h − j + k − 1}{k − 1}\binom{j + x − 1}{x − 1}\binom{x}{h}\frac{(−1)^{h}}{2^{j}} β^{n−j−h}α^{j+h} .

Theorem 9. Let x be a real number, k, j ≥ 1 be integers. Then

∑_{j=1}^{∞} H_{n,k,j}(α, α)\frac{G_{j}^{(x)}}{2^{j}j!} = \begin{cases}∑_{i=0}^{n} \binom{i+k−x−1}{i}\binom{x+n−i−1}{n−i} \frac{α^{n}}{2^{n−i}} , & if k > x; \\ \binom{n+k−1}{n} \frac{α^{n}}{2^{n}} , & if k = x; \\ ∑_{i=0}^{x−k} \binom{x−k}{i}\binom{x+n−i−1}{n−i} \frac{(−1)^{i}α^{n}}{2^{n−i}} , & if k < x. \end{cases}  (32)

Proof. By using (9), (17) and (20), we have

∑_{j=1}^{∞} H_{n,k,j}(α, α)\frac{G_{j}^{(x)}}{2^{j}j!} = [t^{n}]\frac{1}{(1 − αt)^{k}} [(\frac{2}{e^{y} + 1})^{x} | y = − ln(1 − αt)]

= \begin{cases}∑_{i=0}^{n} \binom{i+k−x−1}{i}\binom{x+n−i−1}{n−i} \frac{α^{n}}{2^{n−i}} , & if k > x; \\ \binom{n+k−1}{n} \frac{α^{n}}{2^{n}} , & if k = x; \\ ∑_{i=0}^{x−k} \binom{x−k}{i}\binom{x+n−i−1}{n−i} \frac{(−1)^{i}α^{n}}{2^{n−i}} , & if k < x. \end{cases}

Corollary 10. The following relations hold:

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}}{2^{j}j!} = ∑_{i=0}^{n} \binom{i + k − 2}{i}\frac{α^{n}}{2^{n−i}} ,  (33)
∑_{j=1}^{n} H(n, j − 1)\frac{G_{j}}{2^{j}j!} = \frac{1}{2^{n}} .  (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

∑_{l=1}^{n} H_{n,k,l}(α, β)\frac{G_{l}^{(m)}}{l!} = ∑_{j=0}^{m} ∑_{i=0}^{n−j} H_{i,k,m}(α, β)\binom{m}{j}\binom{n − i − j + m − 1}{m − 1}\frac{(−1)^{j}α^{n−i}}{2^{n−i−j}} . (35)

Proof. By applying (8), (17) and (18), we have

∑_{l=1}^{n} H_{n,k,l}(α, β)\frac{G_{l}^{(m)}}{l!} = [t^{n}]\frac{1}{(1 − βt)^{k}} [(\frac{2y}{e^{y} + 1})^{m} | y = − ln(1 − αt)]

= [t^{n}]\frac{(− ln(1 − αt))^{m}}{(1 − βt)^{k}} \frac{1}{(1 − \frac{αt}{2})^{m}} (1 − αt)^{m}

= ∑_{j=0}^{m} ∑_{i=0}^{n−j} H_{i,k,m}(α, β)\binom{m}{j}\binom{n − i − j + m − 1}{m − 1}\frac{(−1)^{j}α^{n−i}}{2^{n−i−j}} .

Theorem 12. Let n, k, j, m ≥ 1 be integers. Then

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(m)}}{j!} = \begin{cases}∑_{i=0}^{n} H_{n−i,k−m,m}(α, α)\binom{i+m−1}{m−1} \frac{α^{i}}{2^{i}} , & if k > m; \\ ∑_{i=m}^{n} |s(i, m)|\frac{m!}{i!} \binom{n−i+m−1}{m−1} \frac{α^{n}}{2^{n−i}} , & if k = m; \\ ∑_{i=0}^{m−k} ∑_{j=0}^{n−i} \binom{m−k}{i}\binom{j+m−1}{m−1}|s(n − i − j, m)|\frac{(−1)^{i}α^{n}m!}{2^{j}(n−i−j)!}, & if k < m. \end{cases} (36)

Proof. By using (8), (17) and (20), we have

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(m)}}{j!} = [t^{n}]\frac{1}{(1 − αt)^{k}} [(\frac{2y}{e^{y} + 1})^{m} | y = − ln(1 − αt)]

= \begin{cases}∑_{i=0}^{n} H_{n−i,k−m,m}(α, α)\binom{i+m−1}{m−1} \frac{α^{i}}{2^{i}} , & if k > m; \\ ∑_{i=m}^{n} |s(i, m)|\frac{m!}{i!} \binom{n−i+m−1}{m−1} \frac{α^{n}}{2^{n−i}} , & if k = m; \\ ∑_{i=0}^{m−k} ∑_{j=0}^{n−i} \binom{m−k}{i}\binom{j+m−1}{m−1}|s(n − i − j, m)|\frac{(−1)^{i}α^{n}m!}{2^{j}(n−i−j)!}, & if k < m. \end{cases}

Theorem 13. Let n, k, r, m ≥ 1 be integers. Then

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\frac{(−α)^{n−j}C_{n−j}^{(m)}}{(n − j)!} = \begin{cases}α^{n}|s(n − m, r − m)|\frac{(r−m)!}{(n−m)!}, & if r > m; \\ α^{m}δ_{n,m}, & if r = m; \\ \frac{(−1)^{n−r}α^{n}}{(n−r)!} C_{n−r}^{(m−r)}, & if r < m. \end{cases} (37)

Proof. By using (2), (6) and (16), we have

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\frac{(−α)^{n−j}C_{n−j}^{(m)}}{(n − j)!} = [t^{n}](− ln(1 − αt))^{r}(\frac{−αt}{ln(1 − αt)})^{m}

= \begin{cases}α^{n}|s(n − m, r − m)|\frac{(r−m)!}{(n−m)!}, & if r > m; \\ α^{m}δ_{n,m}, & if r = m; \\ \frac{(−1)^{n−r}α^{n}}{(n−r)!} C_{n−r}^{(m−r)}, & if r < m. \end{cases}

Theorem 14. Let n, k, r, m ≥ 1 be integers. Then

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\frac{(−α)^{n−j}Ĉ_{n−j}^{(m)}}{(n − j)!}

= \begin{cases}α^{m}H_{n−m,m,r−m}(α, α), & if r > m; \\ α^{n}\binom{n−1}{n−m}, & if r = m; \\ ∑_{i=m−r}^{n−r} Ĉ_{n−r−i}^{(m−r)} \frac{(−1)^{n−r−i}α^{n}}{(n−r−i)!} \binom{i+r−1}{r−1}, & if r < m. \end{cases}  (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

∑_{j=0}^{n} H_{j,k,r}(α, β)\frac{(−α)^{n−j}C_{n−j}}{(n − j)!} = αH_{n−1,k,r−1}(α, β),  (39)
∑_{j=0}^{n} H_{j,k,r}(α, β)\frac{(−α)^{n−j}Ĉ_{n−j}}{(n − j)!} = ∑_{i=0}^{n−1} H_{i,k,r−1}(α, β)α^{n−i}.  (40)

Proof. By applying (1), (6) and (16), we get

∑_{j=0}^{n} H_{j,k,r}(α, β)\frac{(−α)^{n−j}C_{n−j}}{(n − j)!} = [t^{n}]\frac{(− ln(1 − αt))^{r}}{(1 − βt)^{k}} \frac{αt}{− ln(1 − αt)} = αH_{n−1,k,r−1}(α, β) .

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

∑_{j=r}^{n} \frac{H(j, r − 1)(−α)^{n−j}C_{n−j}}{(n − j)!} = α^{n}H(n − 1, r − 2),  (41)

∑_{j=r}^{n} \frac{H(j, r − 1)(−α)^{n−j}Ĉ_{n−j}}{(n − j)!} = α^{n} ∑_{i=r}^{n−1} H(i − 1, r − 2).  (42)

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 H_{n,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 − \frac{z}{w})^{α}g(z),  (43)

where g(z) is analytic near w, g(w) ≠ 0 and α ∉ {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 w_{j}, α_{j} and g_{j}(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) = ∑_{j} \frac{g_{j}(w_{j})n^{−α_{j}−1}}{Γ(−α_{j})w_{j}^{n}} + o(R^{−n}n^{−a−1}).

Lemma 18. (see [11]) Let α be a real number and

L(z) = ln(\frac{1}{1 − z}) .

When n → ∞,

[z^{n}](1 − z)^{α}L^{k}(z) ∼ \frac{1}{Γ(−α)}n^{−α−1} ln^{k} n , (α ∉ \{0, 1, 2, …\})

[z^{n}](1 − z)^{m}L^{k}(z) ∼ (−1)^{m}km!n^{−m−1} ln^{k−1} n , (m ∈ ℤ_{≥0}, k ∈ ℤ_{≥1}).

Lemma 19. (see [11]) Suppose that a(z) = ∑ a_{n}z^{n} and b(z) = ∑ b_{n}z^{n} are power series with radii of convergence α > β ≥ 0, respectively. Suppose that \frac{b_{n−1}}{b_{n}} → β as n → ∞. If a(β) ≠ 0 and ∑ c_{n}z^{n} = a(z)b(z), then

c_{n} ∼ a(β)b_{n} as n → ∞.

Theorem 20. Let k, r ≥ 1. As n → ∞, we get

∑_{h=0}^{n} \binom{k}{h}(−β)^{h}H_{n−h,k,r}(α, β) ∼ \frac{α^{n}r}{n} ln^{r−1} n .

Proof. By Eq. (2) and Lemma 18, we get

∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{n−h,k,r}(α, β) = [t^{n}](− ln(1 − αt))^{r} ∼ \frac{α^{n}r}{n} ln^{r−1} n .

Theorem 21. Let k, r, m ≥ 1. As n → ∞, we have

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)|s(n − j, m)|\frac{α^{n−j}}{(n − j)!} ∼ \frac{α^{n}(m + r)}{m!n}(ln n)^{m+r−1} .

Proof. By Eq. (23) and Lemma 18, we obtain

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)|s(n − j, m)|\frac{α^{n−j}}{(n − j)!} = \frac{1}{m!}[t^{n}](− ln(1 − αt))^{m+r}

∼ \frac{α^{n}(m + r)}{m!n}(ln n)^{m+r−1} .

Theorem 22. Let k, j, m ≥ 1 and l ≥ 0. As n → ∞, we have

∑_{j=1}^{n} H_{n,k,j}(α, α)S(j, m; l)\frac{m!}{j!} ∼ \frac{α^{n}n^{k+m+l−1}}{Γ(k + l + m)} .

Proof. By Eq. (24) and Lemma 17, we obtain

∑_{j=1}^{n} H_{n,k,j}(α, α)S(j, m; l)\frac{m!}{j!} = α^{m}[t^{n}]\frac{t^{m}}{(1 − αt)^{k+l+m}} ∼ \frac{α^{n}n^{k+m+l−1}}{Γ(k + l + m)} .

Theorem 23. Let k, r ≥ 1. As n → ∞, we get

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\binom{n − j + l}{n − j}β^{n−j} ∼ \frac{α^{n}n^{l}}{l!} ln^{r} n .

Proof. It is well known that

∑_{n=0}^{∞} \binom{n + l}{n}t^{n} = \frac{1}{(1 − t)^{l+1}} .  (44)

By Lemma 18 and Eq. (44), we get

∑_{j=0}^{n} ∑_{h=0}^{k} \binom{k}{h}(−β)^{h}H_{j−h,k,r}(α, β)\binom{n − j + l}{n − j}α^{n−j} = [t^{n}]\frac{(− ln(1 − αt))^{r}}{(1 − αt)^{l+1}} ∼ \frac{α^{n}n^{l}}{l!} ln^{r} n .

Theorem 24. Let j, k, m ≥ 1. As n → ∞, we have

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{B_{j}^{(m)}}{j!} ∼ \begin{cases}(−1)^{m−k}α^{n}m(m − k)!(n + m)^{k−m−1}(ln(n + m))^{m−1}, & if m − k ∈ ℤ_{≥0}; \\ \frac{α^{n}(n+m)^{k−m−1}}{Γ(m−k)} (ln(n + m))^{m}, & if m − k ∉ ℤ_{≥0}. \end{cases}

Proof. By the proof of Eq. (28) and Lemma 18, we obtain

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{B_{j}^{(m)}}{j!} = [t^{n}]\frac{1}{(1 − αt)^{k}} [(\frac{y}{e^{y} − 1})^{m} | y = − ln(1 − αt)]

∼ \begin{cases}(−1)^{m−k}α^{n}m(m − k)!(n + m)^{k−m−1}(ln(n + m))^{m−1}, & if m − k ∈ ℤ_{≥0}; \\ \frac{α^{n}(n+m)^{k−m−1}}{Γ(m−k)} (ln(n + m))^{m}, & if m − k ∉ ℤ_{≥0}. \end{cases}

Theorem 25. Let j be fixed and x be a real number. As n → ∞, we have

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(x)}}{2^{j}j!} ∼ \begin{cases}\frac{2^{x}α^{n}n^{k−x−1}}{Γ(k−x)} , & if k > x; \\ \frac{α^{n}n^{k−1}}{2^{n}Γ(k)} , & if k = x; \\ \frac{(−1)^{x−k}α^{n}n^{x−1}}{2^{n}Γ(x)} , & if k < x. \end{cases}

Proof. By the proof of (32) and Lemma 17, we obtain

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(x)}}{2^{j}j!} = [t^{n}] \begin{cases}\frac{1}{(1−αt)^{k−x}} \frac{1}{(1− \frac{αt}{2} )^{x}} , & if k > x; \\ (1 − \frac{αt}{2})^{x}, & if k = x; \\ \frac{(1−αt)^{x−k}}{(1− \frac{αt}{2} )^{x}} , & if k < x. \end{cases} ∼ \begin{cases}\frac{2^{x}α^{n}n^{k−x−1}}{Γ(k−x)} , & if k > x; \\ \frac{α^{n}n^{k−1}}{2^{n}Γ(k)} , & if k = x; \\ \frac{(−1)^{x−k}α^{n}n^{x−1}}{2^{n}Γ(x)} , & if k < x. \end{cases}

Theorem 26. Let j, k, m ≥ 1, j be fixed. As n → ∞, we have

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(m)}}{j!} ∼ \begin{cases}\frac{α^{n}2^{m}n^{k−m−1} ln^{m+1} n}{Γ(k−m)}, & if k > m; \\ \frac{2^{m}α^{n}m}{n} ln^{m−1} n, & if k = m; \\ \frac{(−1)^{m−k}2^{m}α^{n}(m−k)!}{n^{m+1−k}} ln^{m} n, & if k < m. \end{cases}

Proof. By (36), we see that

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(m)}}{j!} = [t^{n}] \begin{cases}\frac{1}{(1−αt)^{k−m}} \frac{1}{(1− \frac{αt}{2} )^{m}} (− ln(1 − αt))^{m}, & if k > m; \\ \frac{1}{(1− \frac{αt}{2} )^{m}} (− ln(1 − αt))^{m}, & if k = m; \\ (1 − αt)^{m−k} \frac{1}{(1− \frac{αt}{2} )^{m}} (− ln(1 − αt))^{m}, & if k < m. \end{cases}

Let k = m, and set

a(t) = \frac{1}{(1 − \frac{αt}{2})^{m}} = ∑_{n=0}^{∞} \binom{n + m − 1}{m − 1}(\frac{αt}{2})^{n} and b(t) = (− ln(1−αt))^{m} = ∑_{n=m}^{∞} |s(n, m)|\frac{m!}{n!}α^{n}t^{n} .

in Lemma 19. Since \frac{|s(n−1,m)| \frac{m!}{(n−1)!}}{|s(n,m)| \frac{m!}{n!}} → \frac{1}{α} as n → ∞, a(\frac{1}{α}) = 2^{m} ≠ 0. Then we have

∑_{j=1}^{n} H_{n,k,j}(α, α)\frac{G_{j}^{(m)}}{j!} ∼ \begin{cases}\frac{α^{n}2^{m}n^{k−m−1} ln^{m+1} n}{Γ(k−m)}, & if k > m; \\ \frac{2^{m}α^{n}m}{n} ln^{m−1} n, & if k = m; \\ \frac{(−1)^{m−k}2^{m}α^{n}(m−k)!}{n^{m+1−k}} ln^{m} n, & if k < m. \end{cases}

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 H_{n,k,r}(α, β) by Laplace’s method.

Theorem 27. Let k, r ≥ 1, 0 < β, (b + 1)^{b+1} ≠ βb^{b}, (b + 1)^{b+1} > αb^{b} and b be a positive integer. As k → ∞, we get

∑_{n=0}^{∞} \frac{H_{n,k,r}(α, β)}{[(b + 1)n + 1]\binom{(b+1)n}{n}}

∼ (\frac{(b + 1)^{b+1}}{(b + 1)^{b+1} − βb^{b}} )^{k− \frac{1}{2}} ( ln \frac{(b + 1)^{b+1}}{(b + 1)^{b+1} − αb^{b}} )^{r}\sqrt{\frac{2π(b + 1)^{b−2}}{βkb^{b−1}}} .  (45)

Proof. From Tiberiu [9], we know that the inverse of a binomial coefficient is related to an integral as follows:

\binom{n}{m}^{−1} = (n + 1) ∫_{0}^{1} t^{m}(1 − t)^{n−m}dt.  (46)

Owing to (46), we have

∑_{n=0}^{∞} \frac{H_{n,k,r}(α, β)}{[(b + 1)n + 1]\binom{(b+1)n}{n}} = ∑_{n=0}^{∞} H_{n,k,r}(α, β) ∫_{0}^{1} t^{n}(1 − t)^{nb}dt

= ∫_{0}^{1} ∑_{n=0}^{∞} H_{n,k,r}(α, β)(t(1 − t)^{b})^{n}dt = ∫_{0}^{1} \frac{(− ln(1 − αt(1 − t)^{b}))^{r}}{(1 − βt(1 − t)^{b})^{k}} dt

= ∫_{0}^{1}(− ln(1 − αt(1 − t)^{b})^{r}e^{−k ln(1−βt(1−t)^{b})}dt .

Let φ(t) = (− ln(1 − αt(1 − t)^{b})^{r} and h(t) = − ln(1 − βt(1 − t)^{b}). Then h(t) reaches a maximum at t = \frac{1}{b+1}, h′(\frac{1}{b+1}) = 0 and h′′(\frac{1}{b+1}) < 0. By applying Laplace’s method, we have

∑_{n=0}^{∞} \frac{H_{n,k,r}(α, β)}{[(b + 1)n + 1]\binom{(b+1)n}{n}} = ∫_{0}^{1}(− ln(1 − αt(1 − t)^{b})^{r}e^{−k ln(1−βt(1−t)^{b}}dt

∼ φ(\frac{1}{b + 1})e^{kh( \frac{1}{b+1} )}\sqrt{\frac{−2π}{kh′′(\frac{1}{b+1})}} .

Corollary 28. Let k, r ≥ 1, α < 4, 0 < β, β ≠ 4 and b be a positive integer. As k → ∞, we get

∑_{n=0}^{∞} \frac{H_{n,k,r}(α, β)}{(2n + 1)\binom{2n}{n}} ∼ (\frac{4}{4 − β} )^{k− \frac{1}{2}} ( ln \frac{4}{4 − α})^{r}\sqrt{\frac{π}{kβ}} .  (47)

Proof. Setting b = 1 in (45), we obtain (47).

Theorem 29. Let k, r ≥ 1, 0 < α, (b + 1)^{b+1} ≠ −βb^{b} and b be a positive integer. As r → ∞, we get

∑_{n=0}^{∞} \frac{(−1)^{n}H_{n,k,r}(α, β)}{[(b + 1)n + 1]\binom{(b+1)n}{n}}

∼ (−1)^{r}(\frac{(b + 1)^{b+1}}{(b + 1)^{b+1} + βb^{b}} )^{k}( ln \frac{(b + 1)^{b+1} + αb^{b}}{(b + 1)^{b+1}} )^{r+ \frac{1}{2}} \sqrt{\frac{2π(b + 1)^{b−2}((b + 1)^{b+1} + αb^{b})}{αrb^{b−1}(b + 1)^{b+1}}}. (48)

Proof. Owing to (46), we get

∑_{n=0}^{∞} \frac{(−1)^{n}H_{n,k,r}(α, β)}{[(b + 1)n + 1]\binom{(b+1)n}{n}} = ∑_{n=0}^{∞}(−1)^{n}H_{n,k,r}(α, β) ∫_{0}^{1} t^{n}(1 − t)^{nb}dt

= ∫_{0}^{1} ∑_{n=0}^{∞} H_{n,k,r}(α, β)(−t(1 − t)^{b})^{n}dt = ∫_{0}^{1} \frac{(− ln(1 + αt(1 − t)^{b}))^{r}}{(1 + βt(1 − t)^{b})^{k}} dt

= (−1)^{r} ∫_{0}^{1} \frac{e^{r ln ln(1+αt(1−t)^{b}}}{(1 + βt(1 − t)^{b})^{k}} dt .

Let φ(t) = \frac{1}{(1+βt(1−t)^{b})^{k}} and h(t) = ln ln(1 + αt(1 − t)^{b}). Then h(t) reaches the maximum at t = \frac{1}{b+1}, h′(\frac{1}{b+1}) = 0 and h′′(\frac{1}{b+1}) < 0. By applying Laplace’s method, we have

∑_{n=0}^{∞} \frac{(−1)^{n}H_{n,k,r}(α, β)}{[(b + 1)n + 1]\binom{(b+1)n}{n}} = ∫_{0}^{1}(− ln(1 − αt(1 − t)^{b})^{r}e^{−k ln(1−βt(1−t)^{b}}dt

∼ φ(\frac{1}{b + 1})e^{rh( \frac{1}{b+1} )}\sqrt{\frac{−2π}{rh′′(\frac{1}{b+1})}} .

Corollary 30. Let k, r ≥ 1, 0 < α, β ≠ −4. As r → ∞, we get

∑_{n=0}^{∞} \frac{(−1)^{n}H_{n,k,r}(α, β)}{(2n + 1)\binom{2n}{n}} ∼ (−1)^{r}(\frac{4}{4 + β} )^{k}( ln \frac{4 + α}{4})^{r+ \frac{1}{2}} \sqrt{\frac{π(4 + α)}{4αr}}.  (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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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.

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