Incomplete Finite Binomial Sums of Harmonic Numbers

Richard Kollár Faculty of Mathematics, Physics and Informatics Comenius University Mlynská dolina 842 45 Bratislava Slovakia

             richard.kollar@fmph.uniba.sk

Abstract

We explore a straightforward recursive relation for an incomplete binomial series. Through this approach, we establish novel identities for the incomplete finite binomial sum of harmonic numbers. Additionally, we introduce a new proof for an identity related to the incomplete finite alternating binomial sum of harmonic numbers. These identities act as analogues to their respective well-established formulas for complete series and enable the characterization of the asymptotic behavior of the incomplete binomial series of harmonic numbers.

1 Introduction

Let N be a positive integer, and let m be an integer satisfying 0 ≤ m ≤ N. We define the incomplete finite binomial sum of harmonic numbers, and the incomplete finite alternating binomial sum of harmonic numbers, respectively, as follows:

S_{N,m} = ∑_{k=1}^{m} \binom{N}{k}H_{k}, and I_{N,m} = ∑_{k=1}^{m} \binom{N}{k}(−1)^{k+1}H_{k}.

Here, H_{k} represents the harmonic number, defined by

H_{k} = 1 + \frac{1}{2} + ⋯ + \frac{1}{k}, for k ≥ 1,  and H_{0} = 0.  (1)

Furthermore, for m = ⌊N/2⌋, we introduce the following notation:

S_{N} = S_{N,⌊N/2⌋}, and I_{N} = I_{N,⌊N/2⌋}.

For n ≥ 1, we prove the following new identities:

S_{2n−1} = 2^{2n−2} (H_{2n−1} − 2C_{2n−1} + D_{n−1}) ,  (2)
S_{2n} = 2^{2n−1} (H_{2n} − 2C_{2n} + D_{n}) + \frac{1}{2}\binom{2n}{n}H_{n}.  (3)

In addition, for all positive integers N, and for all integers m satisfying 0 ≤ m ≤ N, we present a new proof of the identity originally proved by Batir [2]:

I_{N,m} = \frac{1}{N} + (−1)^{m+1}\binom{N − 1}{m} (H_{m} + \frac{1}{N} ) .  (4)

Identities (2)–(3) are incomplete binomial sum analogues of the well-known identity for the complete binomial sum of harmonic numbers:

∑_{k=0}^{N} \binom{N}{k}H_{k} = 2^{N} (H_{N} − C_{N}) ,  (5)

proved by various techniques by Spivey [16], Paule and Schneider [14], Boyadzhiev [4], Gonzalez [9], Batir [1, 2], and Batir and Sofo [3]. Identity (4) is the incomplete analogue of the complete alternating binomial sum of harmonic numbers:

∑_{k=0}^{N} \binom{N}{k}(−1)^{k+1}H_{k} = \frac{1}{N} ,

proved by various techniques by Euler [8], Jin and Du [11], Wang [17], Paule and Schneider [14], Gonzalez [9], and Batir [1, 2]. Here we introduce the following notation:

C_{n} = ∑_{k=1}^{n} \frac{1}{k2^{k}} , and D_{n} = ∑_{k=1}^{n} \frac{1}{(2k)2^{2k}} \binom{2k}{k},  (6)

for n ≥ 1, with C_{0} = D_{0} = 0.

Combinatorial identities involving harmonic numbers, including their binomial sums, have been extensively studied. In addition to the references mentioned previously, significant contributions have been made by Spiess [15], Liu and Wang [13], Chu [6], Choi [5], and Wei and Wang [18]. Batir [2] provides a comprehensive survey of the literature.

The term incomplete finite sum refers to a finite sum where the summation is taken over only a part of the full index range, not over all integer indices k such that 0 ≤ k ≤ N.

Incomplete binomial sums related to those presented in this work have been studied by Batir [2, Corollaries 3–4], who proved the following identities for all m ∈ ℕ and s ∈ ℂ \ ℤ^{−}:

∑_{k=1}^{m} \binom{m + s}{k}H_{k} = 2^{m} ∑_{k=1}^{m} \binom{s + k − 1}{k}\frac{1}{2^{k}} H_{k} + 2^{m} ∑_{k=1}^{m} \frac{1}{2^{k}} ∑_{j=0}^{k−1} \binom{s + k − 1}{j}\frac{1}{m + 1},

∑_{k=1}^{m} \binom{m + s}{k}(−1)^{k}H_{k} = (−1)^{m}\binom{s + m − 1}{m}H_{m} + \frac{(−1)^{m}}{s + m} \binom{s + m − 1}{m} − \frac{1}{s + m}.

By setting s = N − m and N ≥ m, the second identity reduces to (4), while the first one transforms to

S_{N,m} = 2^{m} ∑_{k=1}^{m} \binom{N − m + k − 1}{k}\frac{1}{2^{k}} H_{k} + 2^{m} ∑_{k=1}^{m} \frac{1}{2^{k}} ∑_{j=0}^{k−1} \binom{N − m + k − 1}{j}\frac{1}{j + 1}.

Incomplete binomial sums were also studied by Worsch [19], Dence [7], and Katsuura [12]. Gould’s eight-volume online book [10] features a comprehensive list of various types of both complete and incomplete finite binomial sums but lacks detailed proofs and references.

Our research is motivated by applications in which the binomial coefficient encodes the probability distribution of states in a system, and the system measure exhibits a symmetry k ↔ (N −k). Specifically, we focus on problems related to the mean time of self-organization in abstract model systems, where the measured quantity is proportional to the harmonic numbers H_{k}.

For proofs, we adapted techniques from Gonzalez [9], modifying them for incomplete finite sums. Our method is based on a general recursive formula for finite binomial series of an arbitrary sequence of real numbers, yielding a second-order linear recurrence relation for S_{N} and an explicit expression for I_{N,m}.

The derived formulae enable the asymptotic analysis of S_{N} and I_{N} as N → ∞. Note that such approximations cannot be simply deduced from the asymptotic behavior of H_{k} for k ≫ 1, as the sums S_{N} and I_{N} include terms H_{k} for small values of k. We show that as N → ∞,

\frac{S_{N}}{2^{N−1}} = ln (\frac{N}{2} ) + γ + o(1),

and

\frac{I_{N}}{2^{N−1}} \sqrt{\frac{πN}{2}} = (−1)^{⌊^{\frac{N}{2}} ⌋+1} (ln (\frac{N}{2} ) + γ + o(1)) .

Here, γ denotes the Euler-Mascheroni constant, γ ≐ 0.577.

2 Recursive combinatorial identity

Analogously to Gonzalez [9], we first derive a recursive combinatorial identity.

Lemma 1. Let m and N be positive integers, with m ≤ N, and let (a_{k})_{k=0}^{m} be a sequence of real numbers. Then, the following equation holds:

∑_{k=0}^{m} \binom{N}{k}a_{k} = ∑_{k=0}^{m−1} \binom{N − 1}{k}(a_{k} + a_{k+1}) + \binom{N − 1}{m}a_{m}.  (7)

Here, we define \binom{n}{k} = 0 for all integers n and k, where n < k.

Proof. The claim is valid for all N ≥ 1 and is demonstrated by the following calculation:

∑_{k=0}^{m} \binom{N}{k}a_{k} = \binom{N}{0}a_{0} + ∑_{k=1}^{m} \binom{N}{k}a_{k}

= \binom{N − 1}{0}a_{0} + ∑_{k=1}^{m} (\binom{N − 1}{k} + \binom{N − 1}{k − 1}) a_{k}

= ∑_{k=0}^{m} \binom{N − 1}{k}a_{k} + ∑_{k=0}^{m−1} \binom{N − 1}{k}a_{k+1}

= ∑_{k=0}^{m−1} \binom{N − 1}{k}(a_{k} + a_{k+1}) + \binom{N − 1}{m}a_{m}.

Note that Lemma 1 generalizes the identity derived in [9, Theorem 2.1] for the full sum:

∑_{k=0}^{N} \binom{N}{k}a_{k} = ∑_{k=0}^{N−1} \binom{N − 1}{k}(a_{k} + a_{k+1}),

which holds for all N ≥ 1 and for all sequences (a_{k})_{k=0}^{N}. This result follows from Lemma 1 by setting m = N.

In part of our work, we utilize a slight modification of the identity (7).

Corollary 2. Let n be a positive integer and let m be a nonnegative integer with m ≤ n. Suppose (a_{k})_{k=0}^{m+1} is a sequence of real numbers. Then,

∑_{k=0}^{m} \binom{2n − 1}{k}a_{k} = ∑_{k=0}^{m} \binom{2n − 2}{k}(a_{k} + a_{k+1}) − \binom{2n − 2}{m}a_{m+1}.  (8)

Similarly, let m and n be positive integers with m ≤ n, and let (a_{k})_{k=0}^{m} be a sequence of real numbers. Then,

∑_{k=0}^{m} \binom{2n}{k}a_{k} = ∑_{k=0}^{m−1} \binom{2n − 1}{k}(a_{k} + a_{k+1}) + \binom{2n − 1}{m}a_{m}.  (9)

Proof. To prove equation (8) for all m and n where n ≥ 1 and 1 ≤ m ≤ n, we set N = 2n−1 in Lemma 1:

∑_{k=0}^{m} \binom{2n − 1}{k}a_{k} = ∑_{k=0}^{m−1} \binom{2n − 2}{k}(a_{k} + a_{k+1}) + \binom{2n − 2}{m}a_{m}

= ∑_{k=0}^{m} \binom{2n − 2}{k}(a_{k} + a_{k+1}) − \binom{2n − 2}{m}a_{m+1}.

For m = 0, equation (8) simplifies to a_{0} = (a_{0} + a_{1}) − a_{1}. Equation (9) follows directly from Lemma 1 by setting N = 2n.

3 Incomplete binomial sum of harmonic numbers

In this section, we use the recurrence relations (8)–(9) to derive formulae for the incomplete binomial sum of harmonic numbers, where the summation is over half of the full index range.

Theorem 3. Let n be a positive integer. Then,

S_{2n−1} = ∑_{i=0}^{n−1} \binom{2n − 1}{i}H_{i} = 2^{2n−2} (H_{2n−1} − 2C_{2n−1} + D_{n−1}) ,  (10)

S_{2n} = ∑_{i=0}^{n} \binom{2n}{i}H_{i} = 2^{2n−1} (H_{2n} − 2C_{2n} + D_{n}) + \frac{1}{2}\binom{2n}{n}H_{n}.  (11)

Proof. First, we define A_{n} = S_{2n−1} and B_{n} = S_{2n}. Then, using equation (8) for m = n − 1 and a_{k} = H_{k} for k = 0, …, n, we obtain

A_{n} = ∑_{i=0}^{n−1} \binom{2n − 2}{i} (H_{i} + H_{i+1}) − \binom{2n − 2}{n − 1}H_{n}

= ∑_{i=0}^{n−1} \binom{2n − 2}{i} (2H_{i} + \frac{1}{i + 1}) − \binom{2n − 2}{n − 1}H_{n}.

Therefore,

A_{n} = 2 ∑_{i=0}^{n−1} \binom{2n − 2}{i}H_{i} + ∑_{i=0}^{n−1} \binom{2n − 2}{i}\frac{1}{i + 1} − \binom{2n − 2}{n − 1}H_{n}

= 2 ∑_{i=0}^{n−1} \binom{2n − 2}{i}H_{i} + \frac{1}{2n − 1} ∑_{i=0}^{n−1} \binom{2n − 1}{i + 1} − \binom{2n − 2}{n − 1}H_{n}.

Continuing, we find

A_{n} = 2 ∑_{i=0}^{n−1} \binom{2n − 2}{i}H_{i} − \binom{2n − 2}{n − 1}H_{n} + \binom{2n − 2}{n − 1}\frac{1}{n}

+ \frac{1}{2n − 1} (−1 + ∑_{i=0}^{n−1} \binom{2n − 1}{i}) .

Hence,

A_{n} = 2B_{n−1} − \binom{2n − 2}{n − 1}H_{n−1} + \frac{2^{2n−2} − 1}{2n − 1}.  (12)

Here, we used the identity

∑_{i=0}^{n−1} \binom{2n − 1}{i} = ∑_{i=n}^{2n−1} \binom{2n − 1}{i} = \frac{1}{2} ∑_{i=0}^{2n−1} \binom{2n − 1}{i} = 2^{2n−2}.

Similarly, using the identity (9) for m = n and the same choice of (a_{k})_{k=1}^{n} as in the first part, we obtain the following expression for B_{n}:

B_{n} = ∑_{i=0}^{n−1} \binom{2n − 1}{i} (H_{i} + H_{i+1}) + \binom{2n − 1}{n}H_{n}

= ∑_{i=0}^{n−1} \binom{2n − 1}{i} (2H_{i} + \frac{1}{i + 1}) + \binom{2n − 1}{n}H_{n}.

Consequently,

B_{n} = 2 ∑_{i=0}^{n−1} \binom{2n − 1}{i}H_{i} + ∑_{i=0}^{n−1} \binom{2n − 1}{i}\frac{1}{i + 1} + \binom{2n − 1}{n}H_{n}

= 2 ∑_{i=0}^{n−1} \binom{2n − 1}{i}H_{i} + \frac{1}{2n} ∑_{i=0}^{n−1} \binom{2n}{i + 1} + \binom{2n − 1}{n}H_{n}

= 2 ∑_{i=0}^{n−1} \binom{2n − 1}{i}H_{i} + \binom{2n − 1}{n}H_{n} + \frac{1}{2n} (∑_{i=0}^{n} \binom{2n}{i} − 1) .

Thus,

B_{n} = 2A_{n} + \binom{2n − 1}{n}H_{n} + \frac{2^{2n−1} − 1}{2n} + \frac{1}{4n}\binom{2n}{n}.  (13)

Here, we used the identity

∑_{i=0}^{n} \binom{2n}{i} = ∑_{i=n}^{2n} \binom{2n}{i} = \frac{1}{2} (∑_{i=0}^{2n} \binom{2n}{i} + \binom{2n}{n}) = 2^{2n−1} + \frac{1}{2}\binom{2n}{n}.

Furthermore, let α_{n} and β_{n}, n ≥ 1, denote the terms

α_{n} = \frac{2^{2n} − 1}{2n + 1} − \binom{2n}{n}H_{n},
β_{n} = \frac{2^{2n−1} − 1}{2n} + \binom{2n − 1}{n}H_{n} + \frac{1}{4n}\binom{2n}{n}.

The pair of equations (12)–(13) reduces to

A_{n} = 2B_{n−1} + α_{n−1},  B_{n} = 2A_{n} + β_{n}.  (14)

By combining these two recurrence relations, we obtain

A_{n} = 4A_{n−1} + α_{n−1} + 2β_{n−1},  (15)
B_{n} = 4B_{n−1} + β_{n} + 2α_{n−1}.  (16)

We set ω_{k} = α_{k} + 2β_{k} for 1 ≤ k ≤ n − 1. Then, for all k, 1 ≤ k ≤ n − 1, we can express ω_{k} as follows:

ω_{k} = \frac{2^{2k} − 1}{2k + 1} − \binom{2k}{k}H_{k} + \frac{2(2^{2k−1} − 1)}{2k} + 2\binom{2k − 1}{k}H_{k} + \frac{1}{2k} \binom{2k}{k}

= \frac{2^{2k} − 1}{2k + 1} − \binom{2k}{k}H_{k} + \frac{2^{2k} − 2}{2k} + (\binom{2k − 1}{k} + \binom{2k − 1}{k − 1}) H_{k} + \frac{1}{2k} \binom{2k}{k}

= \frac{2^{2k}}{2k + 1} − \frac{1}{2k + 1} − \binom{2k}{k}H_{k} + \frac{2^{2k}}{2k} − \frac{2}{2k} + \binom{2k}{k}H_{k} + \frac{1}{2k} \binom{2k}{k}

= 2^{2k} (\frac{1}{2k + 1} + \frac{1}{2k} ) − (\frac{1}{2k + 1} + \frac{2}{2k} ) + \frac{1}{2k} \binom{2k}{k}.  (17)

By a recursive application of equation (15), we derive

A_{n} = 2^{2}A_{n−1} + ω_{n−1} = 2^{4}A_{n−2} + ω_{n−1} + 4ω_{n−2} = ⋯

= 2^{2n−2}A_{1} + ∑_{k=1}^{n−1} 2^{2k−2}ω_{n−k} = ∑_{k=1}^{n−1} 2^{2k−2}ω_{n−k}.

Using the identity A_{1} = 0 in the last equality, we then have

A_{n} = (2^{2n−2} (\frac{1}{2n − 1} + \frac{1}{2n − 2}) − (\frac{1}{2n − 1} + \frac{2}{2n − 2}) + \frac{1}{2n − 2}\binom{2n − 2}{n − 1})

+ 2^{2} (2^{2n−4} (\frac{1}{2n − 3} + \frac{1}{2n − 4}) − (\frac{1}{2n − 3} + \frac{2}{2n − 4}) + \frac{1}{2n − 4}\binom{2n − 4}{n − 2})

+ ⋯ + 2^{2n−4} (2^{2} (\frac{1}{3} + \frac{1}{2}) − (\frac{1}{3} + \frac{2}{2}) + \frac{1}{2}\binom{2}{1})

= 2^{2n−2} ∑_{k=2}^{2n−1} \frac{1}{k} − 2^{2n−1} ∑_{k=2}^{2n−1} \frac{1}{k2^{k}} + 2^{2n−2} ∑_{k=1}^{n−1} \frac{1}{(2k)2^{2k}} \binom{2k}{k}.

Using the notation H_{n}, C_{n}, D_{n} introduced in equations (1) and (6), the last identity yields the claim (10) of the Theorem:

A_{n} = 2^{2n−2} (H_{2n−1} − 1) − 2^{2n−1} (C_{2n−1} − \frac{1}{2}) + 2^{2n−2}D_{n−1}
= 2^{2n−2}H_{2n−1} − 2^{2n−1}C_{2n−1} + 2^{2n−2}D_{n−1}.  (18)

Combining equations (14) and (18), we obtain the identity (11):

\begin{aligned}B_{n} & = 2A_{n} + β_{n} = 2 (2^{2n−2}H_{2n−1} − 2^{2n−1}C_{2n−1} + 2^{2n−2}D_{n−1}) \\ & + \frac{2^{2n−1} − 1}{2n} + \binom{2n − 1}{n}H_{n} + \frac{1}{4n}\binom{2n}{n} \end{aligned}

\begin{matrix}= 2^{2n−1}H_{2n−1} − 2^{2n} ∑_{k=1}^{2n−1} \frac{1}{k2^{k}} + 2^{2n−1} ∑_{k=1}^{n−1} \frac{1}{(2k)2^{2k}} \binom{2k}{k} \\ + 2^{2n−1} \frac{1}{2n} − \frac{1}{2n} + \binom{2n − 1}{n}H_{n} + \frac{1}{4n}\binom{2n}{n} \end{matrix}

\begin{matrix}= 2^{2n−1}H_{2n} − \frac{1}{2n} − 2^{2n} (∑_{k=1}^{2n} \frac{1}{k2^{k}} − \frac{1}{(2n)2^{2n}} ) \\ + 2^{2n−1} (∑_{k=1}^{n} \frac{1}{(2k)2^{2k}} − \frac{1}{(2n)2^{2n}} \binom{2n}{n}) + \binom{2n − 1}{n}H_{n} + \frac{1}{4n}\binom{2n}{n} \end{matrix}

= 2^{2n−1}H_{2n} − 2^{2n}C_{2n} + 2^{2n−1}D_{n} + \binom{2n − 1}{n}H_{n}

= 2^{2n−1}H_{2n} − 2^{2n}C_{2n} + 2^{2n−1}D_{n} + \frac{1}{2}\binom{2n}{n}H_{n}.

Let δ_{N}, for N ≥ 1, denote

δ_{N} = \begin{cases}0, & if N odd; \\ \frac{1}{2}\binom{N}{N/2}H_{N/2}, & if N even. \end{cases}

A combination of the identities (2)–(3) and (5) implies

∑_{k=⌊N/2⌋+1}^{N} \binom{N}{k}H_{k} = 2^{N−1}(H_{N} − C_{N}) − 2^{N−1} (C_{N} − D_{⌊N/2⌋}) − δ_{N}.

The last identity allows us to quantify an asymmetry between the first and the second half of the binomial sum of harmonic series

∑_{k=0}^{⌊N/2⌋} \binom{N}{k}H_{k} − ∑_{k=⌊N/2⌋+1}^{N} \binom{N}{k}H_{k} = 2^{N} (C_{N} − D_{⌊N/2⌋}) + δ_{N}.

Note that an analogous calculation can be used to prove a generalization of the recurrence relations (15)–(16) for the incomplete binomial sums of harmonic numbers S_{2n−1,m} and S_{2n,m}. However, the cancellation of the terms \binom{2n}{m}H_{m} and 2\binom{2n−1}{m}H_{m} in equation (17) only occurs for m = n. In the general case, where m ≠ n, these terms do not cancel, and the explicit formula (18) contains a binomial sum of harmonic numbers on the right hand side, meaning the double sums are not eliminated in the calculation.

4 Incomplete binomial sum of alternating harmonic numbers

Using Lemma 1, we provide a proof of (4) that differs from the original proof by Batir [2].

Theorem 4. Consider a positive integer N and a non-negative integer m. Then

I_{N,m} = \frac{1}{N} + (−1)^{m+1}\binom{N − 1}{m} (H_{m} + \frac{1}{N} ) .  (19)

Proof. The claim of the theorem reduces to the equality 0 = 0 for m = 0. Therefore, we assume that m ≥ 1. The identity (7) for a_{k} = (−1)^{k+1}H_{k}, with 0 ≤ k ≤ m, implies

I_{N,m} = ∑_{i=0}^{m−1} \binom{N − 1}{i} ((−1)^{i+1}H_{i} + (−1)^{i+2}H_{i+1}) + \binom{N − 1}{m}(−1)^{m+1}H_{m}

= ∑_{i=0}^{m−1} \binom{N − 1}{i}(−1)^{i} (−H_{i} + H_{i+1}) + (−1)^{m+1}\binom{N − 1}{m}H_{m}

= ∑_{i=0}^{m−1}(−1)^{i}\binom{N − 1}{i}\frac{1}{i + 1} + (−1)^{m+1}\binom{N − 1}{m}H_{m}.

Thus,

I_{N,m} = \frac{1}{N} ∑_{i=0}^{m−1}(−1)^{i}\binom{N}{i + 1} + (−1)^{m+1}\binom{N − 1}{m}H_{m}

= \frac{1}{N} ∑_{i=0}^{m−1}(−1)^{i} (\binom{N − 1}{i} + \binom{N − 1}{i + 1}) + (−1)^{m+1}\binom{N − 1}{m}H_{m}.

We conclude the proof of (19) by observing that the telescopic sum on the right-hand side of the previous equation reduces to

∑_{i=0}^{m−1}(−1)^{i} (\binom{N − 1}{i} + \binom{N − 1}{i + 1}) = \binom{N − 1}{0} + (−1)^{m−1}\binom{N − 1}{m}.

5 Asymptotic approximation

While there is no explicit formula for the finite sum C_{n}, the infinite sum can be evaluated as

∑_{k=1}^{∞} \frac{1}{k2^{k}} = ln 2.

This identity directly follows from the Taylor expansion of ln(1−x) = − ∑_{k=1}^{∞} \frac{x^{k}}{k}, for |x| < 1, evaluated at x = \frac{1}{2}.

Another straightforward calculation yields the Taylor expansion for the generating function, valid for |x| ≤ \frac{1}{2}:

∑_{k=1}^{∞} \binom{2k}{k}\frac{x^{2k}}{2k} = ln \frac{2}{\sqrt{1 − 4x^{2}} + 1}.

For x = \frac{1}{2}, it implies

∑_{k=1}^{∞} \binom{2k}{k}\frac{1}{(2k)2^{2k}} = ln 2.

The calculation of the asymptotic approximation of A_{n} as n → ∞ is then as follows:

A_{n} = 2^{2n−2} (ln(2n − 1) + γ) − 2^{2n−1} ln 2 + 2^{2n−2} ln 2 + 2^{2n−2}o(1)
= 2^{2n−2} (ln n + γ + o(1)) .

Similarly, for B_{n}, as n → ∞, we have B_{n} = 2^{2n−1} (ln n + γ + o(1)). This implies

S_{N} = 2^{N−1} (ln N − ln 2 + γ + o(1)) .

for N → ∞. A similar calculation yields an asymptotic approximation of I_{N}:

I_{N} = \frac{(−1)^{⌊N/2⌋+1}}{\sqrt{πN/2}} 2^{N−1} (ln N − ln 2 + γ + o(1)) .

6 Acknowledgment

Funding: this work has been supported by the Scientific Grant Agency of the Slovak Republic under the grants no. 1/0521/20 and 1/0755/19, and by the Slovak Research and Development Agency under the contract no. APVV-18-0308.

References

[1] N. Batir, Combinatorial identities involving harmonic numbers, Integers 20 (2020), #A25.

[2] N. Batir, Finite binomial sum identities with harmonic numbers, J. Integer Sequences 24 (2021), Article 21.4.3.

[3] N. Batir and A. Sofo, A unified treatment of certain classes of combinatorial identities, J. Integer Sequences 24 (2021), Article 21.3.2.

[4] K. N. Boyadzhiev, Harmonic number identities via Euler’s transform, J. Integer Sequences 12 (2009), Article 09.6.1.

[5] J. Choi, Finite summation formulas involving binomial coefficients, harmonic numbers and generalized harmonic numbers, J. Inequal. Appl. 2013 (2013), 49.

[6] W. Chu, Summation formulae involving harmonic numbers, Filomat 26 (2012), 143–152.

[7] T. P. Dence, Some half-row sums from Pascal’s triangle via Laplace transforms, College Math. J. 38 (2007), 205–209.

[8] L. Euler, Demonstratio insignis theorematis numerici circa uncias potestatum binomialium, Nova Acta Acad. Sci. Petropol. 15 (1799/1802), 33–43; reprinted in Opera Omnia, Ser. I, 16 (1935), 104–116.

[9] L. Gonzalez, A new approach for proving or generating combinatorial identities, Internat. J. Math. Ed. Sci. Tech. 41 (2010), 359–372.

[10] H. W. Gould, Combinatorial Identities, Revised edition published by the author, 1972.

Available at https://math.wvu.edu/~hgould/.

[11] H. T. Jin and D. K. Du, Abel’s lemma and identities on harmonic numbers, Integers 15 (2015), #A22.

[12] H. Katsuura, Summations involving binomial coefficients, College Math. J. 40 (2009), 275–278.

[13] H. Liu and W. Wang, Harmonic number identities via hypergeometric series and Bell polynomials, Integral Transform Spec. Funct. 23 (2012), 49–68.

[14] P. Paule and C. Schneider, Computer proofs of a new family of harmonic number identities, Adv. Appl. Math. 31 (2003), 359–378.

[15] J. Spiess, Some identities involving harmonic numbers, Math. Comput. 55 (1990), 839– 863.

[16] M. Z. Spivey, Combinatorial sums and finite differences, Discrete Math. 307 (2007), 3130–3146.

[17] W. Wang, Riordan arrays and harmonic number identities, Comput. Math. Appl. 60 (2010), 1494–1509.

[18] C. Wei and X. Wang, Whipple-type 3F 2- series and summation formulae involving generalized harmonic numbers, Int. J. Number Theory 14 (2018), 2385–2407.

[19] T. Worsch, Lower and upper bounds for (sums of) binomial coefficients, Interner Bericht

    94, Univ. Karlsruhe, 1992. Available at https://publikationen.bibliothek.kit.
    edu/181894.

2020 Mathematics Subject Classification: Primary 05A10; Secondary 05A19. Keywords: binomial coefficient, incomplete finite sum, combinatorial identity, harmonic number, harmonic number sum.

Received November 9 2023; revised versions received December 14 2023; January 15 2024. Published in Journal of Integer Sequences, January 15 2024.

Return to Journal of Integer Sequences home page.