Finite Binomial Sum Identities with Harmonic Numbers

Necdet Batır Department of Mathematics Nevşehir Hacı Bektaş Veli University 50300 Nevşehir Turkey

                      nbatir@hotmail.com

Abstract We prove some general combinatorial identities involving a variable, a parameter, and partial sums of arbitrary sequences. Applying these formulas, we deduce many finite binomial and central binomial sum identities involving the harmonic numbers. Most of our results are new and some known formulas are particular cases of those obtained here.

1 Introduction

Let s be a complex number and n ∈ ℕ. We recall that the generalized harmonic numbers H_{n}^{(s)} of order s are defined by ,

H_{n}^{(s)} = ∑_{k=1}^{n} \frac{1}{k^{s}}.

H_{0}^{(s)} = 0 and H_{n}^{(1)} is the familiar harmonic number H_{n} = 1 + \frac{1}{2} + ⋯ + \frac{1}{n}. See [18, Section 6.3]. Alternating or skew-harmonic numbers H_{n}^{−} are defined by

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

which is the n th partial sum of log 2 = ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{k}, see [4, 18]. The harmonic numbers have been studied since antiquity, and they have involved in a wide range of diverse areas of science and mathematics such as number theory, combinatorics, analysis, computer science, and differential equations. It is a well-known fact that the Riemann zeta function ζ has many series representations involving harmonic numbers; see, for example, the excellent book by Vǎlean [31], and [2]. Lagarias [21] proved that the Riemann hypothesis is equivalent to the statement σ(n) ≤ H_{n} + (log H_{n})e^{H_{n}}, where σ(n) is the sum of positive divisors of n. In a very recent paper, Elliot [12] discovered many expressions involving the harmonic numbers, which are equivalent to the Riemann hypothesis. Throughout this paper, we let ℕ_{0} = ℕ ∪ {0}, ℤ^{−} = {−1, −2, −3, ⋯ }, and ℤ_{0}^{−} = ℤ^{−} ∪ {0}. We let C_{n} denote the central binomial coefficients, that is,

C_{n} = \binom{2n}{n}.

Let Γ be the classical gamma function of Euler, and ψ(x) = Γ′(x)/Γ(x) (x > 0) be the digamma function. We may recall some basic properties of these important functions, which will be used extensively in this paper. The gamma function satisfies the reflection formula

Γ(s)Γ(1 − s) = \frac{π}{sin(πs)} (s ∈ ℂ\ℤ)  (1)

(see [23, p. 344] and [11, p. 253]) and the duplication formula

Γ (s + \frac{1}{2}) = \frac{Γ(2s)Γ(1/2)}{2^{2s−1}Γ(s)}  (2)

(see [23, p. 349] and [11, p. 252]). The binomial coefficients \binom{s}{t} (s, t ∈ ℂ\ℤ^{−}) are defined by

\binom{s}{t} = \frac{Γ(s + 1)}{Γ(t + 1)Γ(s − t + 1)}.  (3)

Making use of formulas (1) and (2) with s = 1/2 − k, we obtain

Γ (\frac{1}{2} − k) = \frac{(−1)^{k}Γ(1/2)k!2^{2k}}{(2k)!} .  (4)

We shall extensively use the following form of the binomial coefficients, which can be shown using the expressions (1), (2) and (3).

\binom{−\frac{1}{2}}{k} = \frac{(−1)^{k}C_{k}}{4^{k}} and \binom{−\frac{3}{2}}{k} = \frac{(−1)^{k}(2k + 1)C_{k}}{4^{k}} .  (5)

The digamma function ψ and harmonic numbers H_{n} are connected with

ψ(n + 1) = −γ + H_{n} (n ∈ ℕ_{0});  (6)

see [28, p. 31], where γ = 0.57721 ⋯ is the Euler-Mascheroni constant. The digamma function ψ possesses the following properties:

ψ(s) − ψ(1 − s) = −π cot(πs) (s ∈ ℂ\ℤ),  (7)

and

ψ (s + \frac{1}{2}) = 2ψ(2s) − ψ(s) − 2 log 2 (s ∈ ℂ\ℤ^{−});  (8)

see [28, p. 25]. Using (6), (7) and (8) we get

ψ (\frac{1}{2} − k) = ψ (\frac{1}{2} + k) = 2H_{2k} − H_{k} + ψ(1/2).  (9)

The binomial coefficients satisfy the following useful identities for n, k ∈ ℕ with k ≤ n

\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k − 1} and \frac{n + 1}{k + 1}\binom{n + 1}{k + 1} = \binom{n}{k}. (10)

In the literature, there exist many interesting identities for finite sums involving the harmonic numbers and the binomial coefficients. As examples,

∑_{k=0}^{n} \binom{n}{k}^{2}\binom{2n + k}{k}(H_{k} − H_{n−k}) = C_{n}^{2}(H_{2n} − H_{n});

see [9],

∑_{k=0}^{n}(−1)^{k}\binom{n}{k}H_{n+k}^{2} = \frac{1}{nC_{n}} (H_{n} − H_{2n} − \frac{2}{n});

see [32],

∑_{k=1}^{n} \frac{H_{k}C_{k}}{4^{k}} = 2 + \frac{(H_{n} − 2)(2n + 1)C_{n}}{4^{n}},

see [7], and finally

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

which is a slightly simplified form of [3, Identity 15)]. Recently, many other remarkable finite and infinite sum identities involving the harmonic numbers have been developed by many authors in different forms using a variety of methods. In [9] and [29] the authors used some identities for classical hypergeometric functions. By making use of the residue theorem in complex analysis, Flajolet and Sedgewick [14] established many elegant harmonic sum identities. Wang [32] obtained many nice harmonic number identities using the method of Riordan arrays. In [27] the author used generating function method to evaluate many finite and infinite harmonic sums in closed form. In [7] and [20] the authors evaluated many finite harmonic sums in closed form by using Abel-Gosper algorithm. By using finite difference method, Spivey [26] presented many summation formulas involving the binomial coefficients and harmonic numbers. We refer the interested readers to the papers given in [2, 5, 8, 10, 15, 16, 17, 19, 22, 24, 25, 29, 30] and the references therein for more identities. As we have seen in the brief review, harmonic numbers appear in a variety of useful identities. In a very recent paper, Batır and Sofo [1, Theorem 2.2] proved the following identity: Let x ∈ ℂ and (a_{n})_{n≥1} be any sequence of complex numbers, and A_{n} = a_{1} + a_{2} + ⋯ + a_{n}. Then

∑_{k=1}^{n} \binom{n}{k}A_{k}x^{k} = (1 + x)^{n−1} ∑_{k=0}^{n−1} ( ∑_{j=0}^{k} \binom{k}{j}a_{j+1}x^{j+1})\frac{1}{(1 + x)^{k}} .  (11)

Using this identity, they deduced many interesting binomial sum identities involving the harmonic numbers, and Fibonacci and Lucas sequences. This article can be regarded as a continuation of the papers [3] and [1]. Our aim in this paper is to generalize formula (11), and to prove the formula (12) noted below. This formula allows us to establish further interesting finite binomial and central binomial sum identities involving harmonic numbers. To demonstrate the usefulness of our formula we present many finite binomial and central binomial sum identities involving harmonic numbers by choosing particular values for s, x and the sequence (a_{n}). Most of our results are new and many known formulas are special cases of those obtained here.

Now we are ready to present our main results.

2 Main Results

Theorem 1. Let (a_{n}) be a sequence in ℂ, x ∈ ℂ, and s ∈ ℂ\ℤ^{−}. Let A_{n} = a_{1} + a_{2} + ⋯ + a_{n}. Then

∑_{k=0}^{n} \binom{n + s}{k}A_{k}x^{k} = (1 + x)^{n}( ∑_{k=0}^{n−1} \binom{s + k}{1 + k}A_{k+1}(\frac{x}{x + 1})^{k+1}

+ ∑_{k=0}^{n−1} ( ∑_{j=0}^{k} \binom{s + k}{j}a_{j+1}x^{j+1})\frac{1}{(x + 1)^{k+1}}).  (12)

Proof. The proof is by mathematical induction on n. The theorem is obviously true for n = 1. Assume that it is also true for n. Let us define

f_{n}(x) = ∑_{k=1}^{n} \binom{n + s}{k}A_{k}x^{k}.

Then we have by the first identity in (10)

f_{n+1}(x) = ∑_{k=1}^{n+1} \binom{n + s + 1}{k}A_{k}x^{k}

= \binom{n + s + 1}{n + 1}A_{n+1}x^{n+1} + ∑_{k=1}^{n} (\binom{n + s}{k} + \binom{n + s}{k − 1})A_{k}x^{k}

= \binom{n + s + 1}{n + 1}A_{n+1}x^{n+1} + f_{n}(x) + ∑_{k=0}^{n−1} \binom{n + s}{k}A_{k+1}x^{k+1}.

Using A_{k+1} = A_{k} + a_{k+1}, we get after a simple computation

f_{n+1}(x) = f_{n}(x) + \binom{n + s + 1}{n + 1}A_{n+1}x^{n+1} + ∑_{k=0}^{n−1} \binom{n + s}{k}A_{k}x^{k+1}

+ ∑_{k=0}^{n−1} \binom{n + s}{k}a_{k+1}x^{k+1}.

Making some algebraic manipulations, this becomes

f_{n+1}(x) = (1 + x)f_{n}(x) + (\binom{n + s + 1}{n + 1} − \binom{n + s}{n}) A_{n}x^{n+1}

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

By the first identity in (10), we have

f_{n+1}(x) = (1 + x)f_{n}(x) + \binom{n + s}{n + 1}A_{n+1}x^{n+1} + ∑_{k=0}^{n} \binom{n + s}{k}a_{k+1}x^{k+1}. (13)

Now using the inductive hypothesis and (13), we obtain

f_{n+1}(x) = (1 + x)^{n+1} ∑_{k=0}^{n−1} \binom{s + k}{1 + k}A_{k+1}(\frac{x}{x + 1})^{k+1}

+ (1 + x)^{n+1} ∑_{k=0}^{n−1} ( ∑_{j=0}^{k} \binom{s + k}{j}a_{j+1}x^{j+1})\frac{1}{(x + 1)^{k+1}}

+ \binom{n + s}{n + 1}A_{n+1}x^{n+1} + ∑_{k=0}^{n} \binom{n + s}{k}a_{k+1}x^{k+1}.

This can be simplified to

f_{n+1}(x) = (1 + x)^{n+1}( ∑_{k=0}^{n} \binom{s + k}{1 + k}A_{k+1}(\frac{x}{x + 1})^{k+1}

+ ∑_{k=0}^{n} ( ∑_{j=0}^{k} \binom{s + k}{j}a_{j+1}x^{j+1})\frac{1}{(x + 1)^{k+1}}).

So, identity (12) is also valid for n + 1 and the proof is completed.

Corollary 2. Letting x = −1 in (12) yields for n ∈ ℕ, and s ∈ ℂ\ℤ^{−}

∑_{k=1}^{n}(−1)^{k}\binom{n + s}{k}A_{k}

= (−1)^{n}\binom{s + n − 1}{n}A_{n} + ∑_{k=0}^{n−1}(−1)^{k+1}\binom{s + n − 1}{k}a_{k+1}.  (14)

Corollary 3. For s ∈ ℂ\ℤ^{−}, and n ∈ ℕ the following identity holds.

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

= (−1)^{n}\binom{s + n − 1}{n}H_{n} + \frac{(−1)^{n}}{s + n}\binom{s + n − 1}{n} − \frac{1}{s + n}.

Proof. If we put a_{k} = \frac{1}{k} in (14) we get

∑_{k=0}^{n}(−1)^{k}\binom{n + s}{k}H_{k}

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

Letting k + 1 = k′ and then deleting the prime, we get by the second identity in (10)

∑_{k=0}^{n}(−1)^{k}\binom{n + s}{k}H_{k}

= (−1)^{n}\binom{s + n − 1}{n}H_{n} + \frac{1}{s + n} ∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}.  (15)

In [3, Eq. 33] it was proved that

∑_{k=0}^{n}(−1)^{k}\binom{s + n}{k} = (−1)^{n}\binom{s + n − 1}{n}.  (16)

Upon using this identity in (15), one easily obtains the desired conclusion.

Letting x = 1 in (12) we obtain Corollary 4. Let n ∈ ℕ and s ∈ ℂ\ℤ^{−}. Then we have

∑_{k=1}^{n} \binom{n + s}{k}A_{k}

= 2^{n} ∑_{k=1}^{n} \binom{s + k − 1}{k}\frac{A_{k}}{2^{k}} + 2^{n} ∑_{k=1}^{n} \frac{1}{2^{k}} ∑_{j=0}^{k−1} \binom{s + k − 1}{j}a_{j+1}.  (17)

Differentiating both sides of the equation given in Corollary 3 with respect to s, and then setting a_{k} = \frac{1}{k} we get, by (10), the following corollary. Corollary 5. For s ∈ ℂ\ℤ^{−} and n ∈ ℕ the following identity holds.

∑_{k=1}^{n}(−1)^{k}\binom{n + s}{k}(ψ(s + n + 1) − ψ(s + n − k + 1))H_{k}

= (−1)^{n}H_{n}\binom{s + n − 1}{n}(ψ(s + n) − ψ(s)) − \frac{(−1)^{n}}{(s + n)^{2}} \binom{s + n − 1}{n}

+ \frac{(−1)^{n}}{s + n}\binom{s + n − 1}{n}(ψ(s + n) − ψ(s)) + \frac{1}{(s + n)^{2}}.  (18)

Proposition 6. For s ∈ ℂ\ℤ^{−} and n ∈ ℕ the following identity is valid.

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}H_{k}^{(2)}

= (−1)^{n}\binom{s + n − 1}{n}H_{n}^{(2)} − \frac{1}{s + n}(H_{n} − s ∑_{k=1}^{n} \frac{(−1)^{k}\binom{s+k−1}{k−1}}{k^{2}\binom{n}{k}}).  (19)

Proof. Putting a_{k} = \frac{1}{k^{2}} in (14), we get

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}H_{k}^{(2)}

= (−1)^{n}\binom{s + n − 1}{n}H_{n}^{(2)} + ∑_{k=0}^{n−1} \frac{(−1)^{k+1}}{(k + 1)^{2}} \binom{s + n − 1}{k}.  (20)

Using (10), we arrive at

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

If we let k + 1 = k′ (and then k′ = k) we get

∑_{k=0}^{n−1} \frac{(−1)^{k+1}}{(k + 1)^{2}} \binom{s + n − 1}{k} = \frac{1}{s + n} ∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}\frac{1}{k}.  (21)

The following identity comes from [3, Theorem 4].

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}\frac{1}{k} = s ∑_{k=1}^{n} \frac{(−1)^{k}\binom{s+k−1}{k−1}}{k^{2}\binom{n}{k}} − H_{n}.  (22)

Combining (20), (21) and (22), we see that (19) is valid. Theorem 7. For all s ∈ ℂ, which is not a negative integer, and n ∈ ℕ we have

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

+ \frac{H_{n}}{s + n} − \frac{2}{(s + n)^{2}} − \frac{s}{s + n} ∑_{k=1}^{n} \frac{(−1)^{k}\binom{s+k−1}{k−1}}{k^{2}\binom{n}{k}}.  (23)

Proof. Letting a_{k} = H_{k}^{2} − H_{k−1}^{2} in Corollary 2, we get

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}H_{k}^{2}

= (−1)^{n}\binom{s + n − 1}{n}H_{n}^{2} + ∑_{k=0}^{n−1}(−1)^{k+1}\binom{s + n − 1}{k} (H_{k+1}^{2} − H_{k}^{2}) . (24)

Using (10) and H_{k+1}^{2} − H_{k}^{2} = \frac{2H_{k+1}}{k+1} − \frac{1}{(k+1)^{2}}, (24) can be easily simplified to

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}H_{k}^{2} = (−1)^{n}\binom{s + n − 1}{n}H_{n}^{2}

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

If we apply Corollary 3 and (22) to this identity, we complete the proof.

Differentiating both sides of (23) with respect to s, we arrive at the following corollary. Corollary 8. Let s ∈ ℂ\ℤ^{−} and n ∈ ℕ. Then we have

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (ψ(s + n + 1) − ψ(s + n − k + 1)) H_{k}^{2}

= (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{2} + \frac{2H_{n}}{s + n} + \frac{2}{(s + n)^{2}}) (ψ(s + n) − ψ(s))

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

− \frac{H_{n}}{(s + n)^{2}} + \frac{4}{(s + n)^{3}} − \frac{n}{(s + n)^{2}} ∑_{k=1}^{n} \frac{(−1)^{k}\binom{s+k−1}{k−1}}{k^{2}\binom{n}{k}}

+ \frac{s}{s + n} ∑_{k=1}^{n} \frac{(−1)^{k}\binom{s+k−1}{k−1}}{k^{2}\binom{n}{k}} (ψ(s + k) − ψ(s + 1)) .  (25)

Theorem 9. Let s ∈ ℂ\ℤ^{−} and n ∈ ℕ. Then we have

∑_{k=1}^{n} \binom{n + s}{k}H_{k}^{−} = 2^{n} ∑_{k=1}^{n} \binom{s + k − 1}{k}\frac{H_{k}^{−}}{2^{k}} + 2^{n} ∑_{k=1}^{n} \frac{2^{−k}}{s + k}
+ 2^{n} ∑_{k=1}^{n} \frac{(−1)^{k+1}}{s + k}\binom{s + k − 1}{k}\frac{1}{2^{k}} .  (26)

Proof. Setting a_{k} = \frac{(−1)^{k+1}}{k} in (14), we obtain

∑_{k=0}^{n} \binom{n + s}{k}H_{k}^{−} = 2^{n} ∑_{k=0}^{n−1} \binom{s + k}{1 + k}\frac{H_{k+1}^{−}}{2^{k+1}}
+ 2^{n} ∑_{k=0}^{n−1} ( ∑_{j=0}^{k} \binom{s + k}{j}\frac{(−1)^{j}}{j + 1} )\frac{1}{2^{k+1}}.  (27)

According to (10) and using (16), we deduce

∑_{j=0}^{k} \binom{s + k}{j}\frac{(−1)^{j}}{j + 1} = \frac{1}{s + k + 1} ∑_{j=1}^{k+1}(−1)^{j+1}\binom{s + k + 1}{j}

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

Substituting this in (27), and letting k+1 = k′ (and then k′ = k), we complete the proof.

Theorem 10. Let n ∈ ℕ and s ∈ ℂ\ℤ^{−}. Then we have

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (H_{k}^{−})^{2} = (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{−})^{2}

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

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

Proof. Putting a_{k} = (H_{k}^{−})^{2} − (H_{k−1}^{−})^{2} in (14), we obtain

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (H_{k}^{−})^{2} = ∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} ∑_{j=1}^{k} ((H_{j}^{−})^{2} − (H_{j−1}^{−})^{2})

= (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{−})^{2}

− ∑_{k=0}^{n−1}(−1)^{k}\binom{s + n − 1}{k} ((H_{k+1}^{−})^{2} − (H_{k}^{−})^{2}) . (28)

But since

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

we deduce from (28)

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (H_{k}^{−})^{2} = (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{−})^{2}

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

Making use of the second identity in (10), and simplifying the result, It follows from this identity that

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (H_{k}^{−})^{2} = (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{−})^{2}

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

Letting k + 1 = k′ (and then k′ = k), this can be simplified to

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (H_{k}^{−})^{2} = (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{−})^{2}

− \frac{2}{s + n} ∑_{k=1}^{n} \binom{s + n}{k}H_{k}^{−} − \frac{1}{s + n} ∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k}\frac{1}{k}.  (29)

We therefore conclude from Theorem 9

∑_{k=1}^{n}(−1)^{k}\binom{s + n}{k} (H_{k}^{−})^{2} = (−1)^{n}\binom{s + n − 1}{n} (H_{n}^{−})^{2}

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

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

which is the desired conclusion.

3 Applications

In this section we present some applications of our results. Our first identity is well-known and due to Euler [13]. Identity 11. Let n ∈ ℕ. If we set s = 0 in Corollary 4, we get the well-known identity

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

Identity 12. Let n ∈ ℕ. Then setting a_{k} = \frac{1}{k}, x = 1 and s = 0 in Theorem 1, we obtain

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

Remark 13. This identity is also well-known; see [1, 3, 5, 17, 29]. Identity 14. For n ∈ ℕ the following identity holds.

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

Proof. Setting s = 0 in Corollary 5, we get

∑_{k=0}^{n}(−1)^{k}\binom{n}{k}\{ψ(n + 1) − ψ(n − k + 1)\}H_{k}

= (−1)^{n}H_{n}\binom{n − 1}{n}ψ(n) − (−1)^{n}H_{n}\binom{s + n − 1}{n}ψ(s)|_{s=0} + \frac{1}{n^{2}}

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

− \frac{(−1)^{n}}{n}\binom{s + n − 1}{n}ψ(s)|_{s=0}.  (30)

Since \binom{n−1}{n} = 0, ψ(n + 1) − ψ(n − k + 1) = H_{n} − H_{n−k} by (6), and

\binom{s + n − 1}{n}ψ(s)|_{s=0} = \frac{Γ(s + n)}{n!} \frac{ψ(s)}{Γ(s)}|_{s=0}
= \frac{1}{n} lim_{s→0} \frac{sψ(s + 1) − 1}{Γ(s + 1)} = −\frac{1}{n},  (31)

we conclude from (30) and (31)

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

Now the proof follows from Identity 11. Identity 15. Let n be a positive integer. Then we have

∑_{k=1}^{n} \frac{H_{k}C_{k}}{4^{k}} = 2 + \frac{(H_{n} − 2)(2n + 1)C_{n}}{4^{n}}.

Proof. Setting s = −\frac{1}{2} − n in Corollary 3 we find

∑_{k=1}^{n}(−1)^{k}\binom{−\frac{1}{2}}{k}H_{k} = (−1)^{n}\binom{−\frac{3}{2}}{n}H_{n} − 2(−1)^{n}\binom{−\frac{3}{2}}{n} + 2.

The proof immediately follows from (5). Remark 16. Identity 15 is not new and it has been proved by Chen et al. [7] by a method combining the Abel’s summation formula and Gosper’s algorithm elegantly. Identity 17. Let n be a positive integer. Then we have

∑_{k=1}^{n} \frac{H_{2k}C_{k}}{4^{k}} = 1 + \frac{(2n + 1)(H_{2n+1} − 2)C_{n}}{4^{n}}.

Proof. In [25, Identity 3] it was proved that

∑_{k=1}^{n} \frac{(2H_{2k} − H_{k})C_{k}}{4^{k}} = \frac{(2n + 1)(2H_{2n+1} − H_{n} − 2)C_{n}}{4^{n}}.  (32)

Now we complete the proof by combining Identity 15 and (32).

Setting s = 0 in (23), we get the following known result; see [1, 3, 20, 32]. Identity 18. Let n be a positive integer. Then we have

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

Identity 19. Let n be a positive integer. Then we have

∑_{k=1}^{n} \frac{H_{k}^{2}C_{k}}{4^{k}} = −2H_{n} − 8

+ \frac{(2n + 1)C_{n}}{4^{n}}(H_{n}^{2} − 4H_{n} + 8 + 2 ∑_{k=0}^{n−1} \frac{4^{k}}{(n − k)(2k + 1)C_{k}} ).

Proof. Setting s = \frac{1}{2} − n in (23) we get

∑_{k=1}^{n}(−1)^{k}\binom{−\frac{1}{2}}{k}H_{k}^{2} = (−1)^{n}\binom{−\frac{3}{2}}{n} (H_{n}^{2} − 4H_{n} + 8) − 2H_{n} − 8

− \frac{2n + 1}{2} ∑_{k=1}^{n} \frac{(−1)^{k}}{k^{2}\binom{n}{k}} \binom{−\frac{3}{2} − n + k}{k − 1}.  (33)

Using (9) we get, after simplifying

\binom{−\frac{3}{2} − n + k}{k − 1} = \frac{Γ (−\frac{1}{2} − n + k)}{(k − 1)!Γ (\frac{1}{2} − n)} = \frac{−2}{2n − 2k + 1} \frac{Γ (\frac{1}{2} − n + k)}{(k − 1)!Γ (\frac{1}{2} − n)}

= \frac{2(−1)^{k+1}}{2n − 2k + 1} \frac{(n − k)!}{(k − 1)!(2n − 2k)!} \frac{(2n)!}{n!4^{k}}.

Thus,

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

= −\frac{2}{2n − 2k + 1} · \frac{(n − k)!}{(k − 1)!(2n − 2k)!} · \frac{(2n)!}{n!4^{k}} · \frac{k!(n − k)!}{k^{2}n!}

Upon simplifying this becomes

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

Substituting this in (33), and changing the index, we arrive at the conclusion by taking into account (10).

Identity 20. For n ∈ ℕ the following identity holds.

∑_{k=1}^{n}(−1)^{k}\binom{n}{k}H_{n−k}H_{k}^{2}

= \frac{(1 − (−1)^{n}) H_{n}^{2}}{n} − \frac{2 (1 + (−1)^{n}) H_{n}}{n^{2}} − \frac{4 + (−1)^{n}}{n^{3}} − \frac{1}{n} ∑_{k=1}^{n} \frac{(−1)^{k}}{k^{2}\binom{n}{k}} .

Proof. The proof is done by the same way with the proof of Identity 18 if we take s = 0 in (25).

Identity 21. For n ∈ ℕ we have

∑_{k=1}^{n} \frac{C_{k}H_{k}^{(2)}}{4^{k}} = 2H_{n} + \frac{(2n + 1)C_{n}}{4^{n}}(H_{n}^{(2)} − 2 ∑_{k=0}^{n−1} \frac{4^{k}}{(n − k)(2k + 1)C_{k}} ).

Proof. The proof can be done by the same method with the proof of Identity 18 by setting s = −\frac{1}{2} − n in (19).

Identity 22. Let n ∈ ℕ. Then

∑_{k=1}^{n} \frac{(H_{k}^{2} + H_{k}^{(2)}) C_{k}}{4^{k}} = \frac{(2n + 1)C_{n}}{4^{n}} (H_{n}^{2} + H_{n}^{(2)} − 4H_{n} + 8) − 8.

Proof. If we sum the equations given in Identity 19 and Identity 21 side by side, the proof isimmediately follows.

Putting s = −\frac{1}{2} − n in (18), we get

Identity 23. For n ∈ ℕ we have

∑_{k=1}^{n} \frac{(H_{k} − 2H_{2k}) H_{k}C_{k}}{4^{k}}

= \frac{(2n + 1)C_{n}}{4^{n}}(H_{n} − 2H_{2n} + \frac{4n}{2n + 1})(H_{n} − 2) − \frac{(2n + 1)C_{n}}{4^{n−1}} + 4.

Combining Identity 19 and Identity 23, we get the following conclusion.

Identity 24. For n ∈ ℕ we have

∑_{k=1}^{n} \frac{H_{k}H_{2k}C_{k}}{4^{k}} = \frac{(2n + 1)C_{n}}{4^{n}}(H_{n}H_{2n} − H_{n} − 2H_{2n} + 6

+ \frac{2n(2 − H_{n})}{2n + 1} + ∑_{k=0}^{n−1} \frac{4^{k}}{(n − k)(2k + 1)C_{k}} ) − H_{n} − 6.

Identity 25. Let n ∈ ℕ. Then

∑_{k=1}^{n} \frac{(−1)^{k}H_{k}^{−}C_{k}}{4^{k}} = \frac{(−1)^{n}(2n + 1)C_{n}}{4^{n}} ∑_{k=0}^{n−1} \frac{(−8)^{k}\binom{n}{k}H_{n−k}^{−}}{(2k + 1)C_{k}}

+ \frac{2(2n + 1)C_{n}}{4^{n}} ∑_{k=0}^{n−1} \frac{\binom{n}{k}8^{k}}{(2k + 1)^{2}C_{k}} − 2 ∑_{k=0}^{n−1} \frac{2^{k}}{2k + 1}.

Proof. The proof follows from (26) by putting s = −\frac{1}{2} − n. We omit the details.

Identity 26. Let n ∈ ℕ. If we set s = 0 in (26), we find

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

Remark 27. This identity is not new and an inductive proof of it can be found in [6, Eq. (14)] and [4, Eq. (9.20)]. See also [1, Remark 21].

Identity 28. For n ∈ ℕ we have

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

Proof. Differentiating both sides of (26) with respect to s and then setting s = 0, we get by (6)

∑_{k=0}^{n} \binom{n}{k}(ψ(n + 1) − ψ(n − k + 1))H_{k}^{−}

H_{n} ∑_{k=0}^{n} \binom{n}{k}H_{k}^{−} − ∑_{k=0}^{n} \binom{n}{k}H_{n−k}H_{k}^{−}

= −2^{n} ∑_{k=1}^{n} \binom{s + k − 1}{k}ψ(s)|_{s=0} \frac{H_{k}^{−}}{2^{k}} − 2^{n} ∑_{k=1}^{n} \frac{1}{2^{k}k^{2}}

+ 2^{n} ∑_{k=1}^{n} \frac{(−1)^{k}}{k} \binom{s + k − 1}{k}ψ(s)|_{s=0} \frac{1}{2^{k}} ,

where we have used \binom{k−1}{k} = 0. Using (31) with n = k it follows that

H_{n} ∑_{k=1}^{n} \binom{n}{k}H_{k}^{−} − ∑_{k=0}^{n} \binom{n}{k}H_{n−k}H_{k}^{−} = 2^{n} ∑_{k=1}^{n} \frac{H_{k}^{−}}{k2^{k}} − 2^{n} ∑_{k=1}^{n} \frac{1}{2^{k}k^{2}}

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

Employing Identity 26 the conclusion follows. Identity 29. Let n ∈ ℕ. If we set s = 0 in (28) we deduce by Identity 11

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

Remark 30. Identity 29 can be found in [1, Identity 25].

Setting s = \frac{1}{2} in Corollary 3 and using the formula (5), after a short computation, we get Identity 31. Let n ∈ ℕ. Then

∑_{k=0}^{n} \frac{(−1)^{k}4^{k}\binom{n}{k}H_{n−k}}{(2k + 1)C_{k}} = \frac{H_{n}}{2n + 1} + \frac{2}{(2n + 1)^{2}} − \frac{2(−1)^{n}4^{n}}{(2n + 1)^{2}C_{n}}.

If we set s = \frac{1}{2} − n in Corollary 3, and using the first identity in (10) we obtain

Identity 32. For n ∈ ℕ we have

∑_{k=0}^{n−1} \frac{C_{k}H_{k+1}}{(k + 1)4^{k}} = 4 − \frac{2(H_{n} + 2)C_{n}}{4^{n}}.

Identity 33. Setting s = −\frac{1}{2} in Corollary 3, we get for n ∈ ℕ

∑_{k=0}^{n} \frac{(−1)^{k+1}4^{k}\binom{n}{k}H_{n−k}}{C_{k}} = \frac{H_{n}}{2n − 1} + \frac{2}{(2n − 1)^{2}} + \frac{2 · 4^{n}(−1)^{n}}{(2n − 1)C_{n}}.

Identity 34. Setting s = −\frac{1}{2} in Proposition 6, we get

∑_{k=0}^{n} \frac{(−1)^{k}4^{k}\binom{n}{k}H_{n−k}^{(2)}}{C_{k}} = \frac{2(−1)^{n+1}4^{n}}{(2n − 1)C_{n}} (H_{n} + ∑_{k=1}^{n} \frac{(−1)^{k}C_{k}}{k(2k − 1)4^{k}\binom{n}{k}}) − \frac{H_{n}^{(2)}}{2n − 1}.

Identity 35. Let n be a non-negative integer. Then

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

Proof. If we differentiate both sides of (17) with respect to s, and put s = 0 and a_{k} = \frac{1}{k}, we can easily obtain the desired result after some easy calculations.

Remark 36. Applying our results given in this paper, we may obtain many other finite binomial and central binomial sum identities involving harmonic numbers, but for briefness we are satisfied with these examples.

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2010 Mathematics Subject Classification: Primary 05A10; Secondary 05A19. Keywords: Binomial coefficient, finite sum, combinatorial identity, harmonic number, central binomial coefficient, harmonic number sum.