INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A33

A COMMON GENERALIZATION OF SOME IDENTITIES^{1}

Zhizheng Zhang

Department of Mathematics, Luoyang Teachers’ College, Luoyang 471022, P. R. China College of Mathematics and Information Science, Henan University, Kaifeng 475001, P. R. China

                           zhzhzhang-yang@163.com

Jun Wang

Department of Applied Mathematics, Dalian University of Technology, Dalian 116024, P. R. China

                            junwang@dlut.edu.cn

Received: 8/12/05, Revised: 11/18/05, Accepted: 12/10/05, Published: 12/21/05

Abstract

By means of partial fraction decomposition, this paper provides an algebraic identity from which a lot of interesting identities follow.

1. Notation and Introduction

In [2], by the WZ method, the following identity was confirmed successfully.

∑_{k=1}^{n} \binom{n}{k}^{2}\binom{n + k}{k}^{2} \{1 + 2kH_{n+k} + 2kH_{n−k} − 4kH_{k}\} = 0,  (1)

where H_{0} = 0 and H_{n} = ∑_{k=1}^{n} \frac{1}{k}. Ahlgren and Ono [1] have shown that Beukers’ conjecture is implied by this beautiful binomial identity. See [3] or [1, Theorem 7] for Beukers’ conjecture. In [4] and [6], by means of partial fraction decomposition, W.-C. Chu gave beautiful proofs of (1) and a number of interesting combinatorial identities. At the same time, Identity (1) was also extended.

The purpose of this paper is to obtain an algebraic identity by means of partial fraction decomposition. As applications, we show a number of interesting identities.

Throughout the paper, f(x) is known for an arbitrary polynomial of degree < λn and the function A(T_{1}, T_{2}, …, T_{n}) is defined by

A(T_{1}, T_{2}, …, T_{n}) = \{ \begin{matrix}1, & if n = 0, \\ ∑ \frac{n!}{k_{1}!k_{2}!...k_{n}!} (\frac{T_{1}}{1} )^{k_{1}} (\frac{T_{2}}{2} )^{k_{2}} … (\frac{T_{n}}{n} )^{k_{n}} , & if n > 0, \end{matrix}

1Supported by the National Natural Science Foundation of China (Grant No. 10471016), the Natural Science Foundation of Henan Province (Grant No. 0511010300) and the Natural Science Foundation of the Education Department of Henan Province (Grant No. 200510482001).

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where the sum is taken over all nonnegative integers k_{i}’s such that k_{1} + 2k_{2} + … + nk_{n} = n. For example, A(T_{1}, T_{2}, …, T_{n}) for n ≤ 4 are:

\begin{matrix}A(T_{1}) = & T_{1} \\ A(T_{1}, T_{2}) = & T_{1}^{2} + T_{2} \\ A(T_{1}, T_{2}, T_{3}) = & T_{1}^{3} + 3T_{1}T_{2} + 2T_{3} \\ A(T_{1}, T_{2}, T_{3}, T_{4}) = & T_{1}^{4} + 8T_{1}T_{3} + 6T_{1}^{2}T_{2} + 3T_{2}^{2} + 6T_{4}. \end{matrix}

2. Main Result Theorem 2.1 Let a_{1}, a_{2}, …, a_{n} be a real sequence with a_{i} ≠ a_{j} (i ≠ j) and f(a_{i}) ≠ 0, (i = 1, …, n). Suppose that λ and r are two integers with λ ≥ 1 and r ≥ 0. Then

∑_{k=1}^{n} \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}f^{(ℓ−s)}(−a_{k})

×A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x)) \frac{\binom{λ−ℓ+r−1}{r}}{(x + a_{k})^{λ−ℓ+r}}

= \frac{(−1)^{r}}{r!(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}f^{(r−j)}(x)
×A (λS_{1}(x), λS_{2}(x), …, λS_{j}(x)) ,  (2)

where

T_{m}(x) = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{(x + a_{i})^{m}} , (1 ≤ m ≤ λ),

and

S_{m}(x) = ∑_{i=1}^{n} \frac{1}{(x + a_{i})^{m}} , (1 ≤ m ≤ r).

Proof. Applying the partial fraction decomposition, let

\frac{f(x)}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} = ∑_{k=1}^{n} ∑_{ℓ=0}^{λ−1} \frac{G(k, ℓ)}{(x + a_{k})^{λ−ℓ}} .

Multiply both sides in the preceding equation by (x + a_{k})^{λ−ℓ} and take x → a_{k}. Then

G(k, ℓ) = lim_{x→−a_{k}}(x + a_{k})^{λ−ℓ} \{\frac{f(x)}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} − ∑_{i=0}^{ℓ−1} \frac{G(k, i)}{(x + a_{k})^{λ−i}} \}

= lim_{x→−a_{k}} \{\frac{f(x)}{(x + a_{k})^{ℓ} ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (x + a_{j})^{λ}} − ∑_{i=0}^{ℓ−1} \frac{G(k, i)(x + a_{k})^{i}}{(x + a_{k})^{ℓ}} \}

= lim_{x→−a_{k}} \frac{_{\frac{f(x)}{n ; ∏ ; \begin{matrix}j = 1 \\ j ≠ k \end{matrix} (x+a_{j})^{λ}}} − ∑_{i=0}^{ℓ−1} G(k, i)(x + a_{k})^{i}}{(x + a_{k})^{ℓ}} .

Repeatedly applying L’Hôspital’s rule we obtain the coefficient as follows:

G(k, ℓ) = lim_{x→−a_{k}} \frac{1}{ℓ!} \frac{d^{ℓ}}{dx^{ℓ}} \frac{f(x)}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (x + a_{j})^{λ}}

= \frac{1}{ℓ!} lim_{x→−a_{k}} ∑_{r=0}^{ℓ} \binom{ℓ}{r}f^{(ℓ−r)}(x) ( \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (x + a_{j})^{λ}} )^{(r)} .

By using differentiation of composite functions (see [12]), we have

( \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (x + a_{j})^{λ}} )^{(r)} = \frac{(−1)^{r}}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (x + a_{j})^{λ}} A (λT_{1}(x), λT_{2}(x), …, λT_{r}(x)) ,

where

T_{m}(x) = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{(x + a_{i})^{m}} , (1 ≤ m ≤ ℓ).

Hence,

G(k, ℓ) = \frac{1}{ℓ! ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{r=0}^{ℓ}(−1)^{r}\binom{ℓ}{r}f^{(ℓ−r)}(−a_{k})A (λT_{1}(x), λT_{2}(x), …, λT_{r}(x)) ,

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A33 4

where

T_{m}(x) = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{(a_{i} − a_{k})^{m}} , (1 ≤ m ≤ ℓ).

So we get

\frac{f(x)}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} = ∑_{k=1}^{n} \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}

×f^{(ℓ−s)}(−a_{k})A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x)) \frac{1}{(x + a_{k})^{λ−ℓ}} .

By differentiating the above equation r times with respect to x, we have

∑_{k=1}^{n} \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}f^{(ℓ−s)}(−a_{k})A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x))

×\frac{(−1)^{r}(λ − ℓ + r − 1)_{r}}{(x + a_{k})^{λ−ℓ+r}}

= \frac{d^{r}}{dx^{r}} (\frac{f(x)}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} )
= ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(x) (\frac{1}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} )^{(j)}
= \frac{1}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}f^{(r−j)}(x)A (λS_{1}(x), λS_{2}(x), …, λS_{j}(x)) ,

where

S_{m}(x) = ∑_{i=1}^{n} \frac{1}{(x + a_{i})^{m}} , (1 ≤ m ≤ r).

This completes the proof. □

3. Some Corollaries Corollary 3.2

∑_{k=1}^{n} \frac{f(−a_{k})}{(x + a_{k})^{r+1} ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})}

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A33 5

= \frac{(−1)^{r}}{r!(x + a_{1})(x + a_{2}) … (x + a_{n})} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}f^{(r−j)}(x)A (S_{1}(x), S_{2}(x), …, S_{j}(x)) ,

(3)

∑_{k=1}^{n} \frac{f(−a_{k})}{(x + a_{k})^{r+1} ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{2}} \{\frac{r + 1}{x + a_{k}} + \frac{f′(−a_{k})}{f(−a_{k})} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x + a_{i}} \}
= \frac{(−1)^{r}}{r!(x + a_{1})^{2}(x + a_{2})^{2} … (x + a_{n})^{2}} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}f^{(r−j)}(x)A (2S_{1}(x), 2S_{2}(x), …, 2S_{j}(x))

(4)

and

∑_{k=1}^{n} \frac{f(−a_{k})}{(x + a_{k})^{r+1} ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{3}} \{\frac{(r + 2)(r + 1)}{(x + a_{k})^{2}} + 2(r + 1) \frac{\frac{f′(−a_{k})}{f(−a_{k})} − 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x+a_{i}}}{x + a_{k}}

+ \frac{f′′(−a_{k})}{f(−a_{k})} − 6\frac{f′(−a_{k})}{f(−a_{k})} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x + a_{i}} + 9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x + a_{i}} )^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{(x + a_{i})^{2}} \}
= \frac{2(−1)^{r}}{r!(x + a_{1})^{3}(x + a_{2})^{3} … (x + a_{n})^{3}} ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}f^{(r−j)}(x)A (3S_{1}(x), 3S_{2}(x), …, 3S_{j}(x)) .

(5)

Proof. In Theorem 2.1, take λ = 1, 2, 3. □ Note. The first formula (3) was given in [14] and an extension of (3) was obtained in [15]. Corollary 3.3

∑_{k=1}^{n} \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}f^{(ℓ−s)}(−a_{k})

×A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x)) \frac{1}{(x + a_{k})^{λ−ℓ}}

= \frac{f(x)}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} ,  (6)

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∑_{k=1}^{n} \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}f^{(ℓ−s)}(−a_{k})

×A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x)) \frac{λ − ℓ}{(x + a_{k})^{λ−ℓ+1}}

= −\frac{1}{(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} \{f′(x) − λf(x) ∑_{i=1}^{n} \frac{1}{x + a_{i}} \}  (7)

and

∑_{k=1}^{n} \frac{1}{∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{λ}} ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}f^{(ℓ−s)}(−a_{k})

×A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x)) \frac{(λ − ℓ + 1)(λ − ℓ)}{2(x + a_{k})^{λ−ℓ+2}}

= \frac{1}{2(x + a_{1})^{λ}(x + a_{2})^{λ} … (x + a_{n})^{λ}} \{f′′(x) − 2λf′(x) ∑_{i=1}^{n} \frac{1}{x + a_{i}}
+λ^{2}f(x) (∑_{i=1}^{n} \frac{1}{x + a_{i}} )^{2} + λf(x) ∑_{i=1}^{n} \frac{1}{(x + a_{i})^{2}} \} ,  (8)

where

T_{m}(x) = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{(x + a_{i})^{m}} , (1 ≤ m ≤ λ).

Proof. In Theorem 2.1, take r = 0, 1, 2. □ Corollary 3.4

∑_{k=1}^{n} \frac{f(−a_{k})}{(x + a_{k}) ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})} = \frac{f(x)}{(x + a_{1})(x + a_{2}) … (x + a_{n})},  (9)

∑_{k=1}^{n} \frac{f(−a_{k})}{(x + a_{k})^{2} ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{2}} \{\frac{r + 1}{x + a_{k}} + \frac{f′(−a_{k})}{f(−a_{k})} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x + a_{i}} \}
= −\frac{1}{(x + a_{1})^{2}(x + a_{2})^{2} … (x + a_{n})^{2}} \{f′(x) − 2f(x) ∑_{i=1}^{n} \frac{1}{x + a_{i}} \}  (10)

and

∑_{k=1}^{n} \frac{f(−a_{k})}{(x + a_{k})^{3} ∏_{\begin{matrix}j = 1 \\ j ≠ k \end{matrix}}^{n} (a_{j} − a_{k})^{3}} \{\frac{(r + 2)(r + 1)}{(x + a_{k})^{2}} + 2(r + 1) \frac{\frac{f′(−a_{k})}{f(−a_{k})} − 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x+a_{i}}}{x + a_{k}}

+ \frac{f′′(−a_{k})}{f(−a_{k})} − 6\frac{f′(−a_{k})}{f(−a_{k})} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x + a_{i}} + 9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{x + a_{i}} )^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{(x + a_{i})^{2}} \}

= \frac{1}{2(x + a_{1})^{3}(x + a_{2})^{3} … (x + a_{n})^{3}} \{f′′(x) − 6f′(x) ∑_{i=1}^{n} \frac{1}{x + a_{i}}

+9f(x) (∑_{i=1}^{n} \frac{1}{x + a_{i}} )^{2} + 3f(x) ∑_{i=1}^{n} \frac{1}{(x + a_{i})^{2}} \} .  (11)

Proof. Take r = 0, 1, 2 in Corollary 3.2 or λ = 1, 2, 3 in Corollary 3.3. □

4. Some Applications

It is seen that the identities in Corollaries 3.2, 3.3 and 3.4 are independent of the choice of a_{k} and f(x). So we can obtain some interesting identities by taking special values for a_{k} and f(x).

4.1 The case: a_{k} = −k

In [5, 11], some interesting identities involving the harmonic numbers were studied. In this section, by taking a_{k} = −k, we show a number of identities involving the harmonic numbers.

Theorem 4.5

∑_{k=1}^{n}(−1)^{(k+1)λ}\binom{n}{k}^{λ} ∑_{ℓ=0}^{λ−1} \frac{(−1)^{ℓ}}{ℓ!} ∑_{s=0}^{ℓ} \binom{ℓ}{s}f^{(ℓ−s)}(k)A (λT_{1}, λT_{2}, …, λT_{s}) \frac{\binom{λ−ℓ+r−1}{r}}{k^{λ−ℓ+r−1}}

= \frac{1}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (λS_{1}, λS_{2}, …, λS_{j}) ,  (12)

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A33 8

where

T_{m} = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{m}} , (1 ≤ m ≤ λ),

and

S_{m} = ∑_{i=1}^{n} \frac{1}{i^{m}} , (1 ≤ m ≤ r).

Proof. Take a_{k} = −k and x = 0 in Theorem 2.1. □

When λ = 1, then we can obtain the main results of paper [9] by Mercier:

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}f(k)}{k^{r}} = \frac{1}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (S_{1}, S_{2}, …, S_{j}) .  (13)

In [9], from this identity, Mercier obtained a lot of combinatorial identities. For example,

∑_{k=1}^{n}(−1)^{k+1}\binom{n}{k}f(k) = f(0),  (14)
∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}}{k}f(k) = f′(0) + f(0) ∑_{i=1}^{n} \frac{1}{i},  (15)
∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}}{k^{2}}f(k) = \frac{1}{2} \{f′′(0) + 2f′(0) ∑_{i=1}^{n} \frac{1}{i} + f(0) ((∑_{i=1}^{n} \frac{1}{i})^{2} + ∑_{i=1}^{n} \frac{1}{i^{2}} )\} . (16)

For λ = 2, 3, we have

∑_{k=1}^{n} \frac{\binom{n}{k}^{2}f(k)}{k^{r}} \{\frac{r + 1}{k} − \frac{f′(k)}{f(k)} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\}
= \frac{1}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (2S_{1}, 2S_{2}, …, 2S_{j})  (17)

and

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}^{3}f(k)}{k^{r}} \{\frac{(r + 2)(r + 1)}{k^{2}} − 2\frac{r + 1}{k} (\frac{f′(k)}{f(k)} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}) + \frac{f′′(k)}{f(k)}

+6\frac{f′(k)}{f(k)} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i} + 9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \}

= \frac{2}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (3S_{1}, 3S_{2}, …, 3S_{j}) .  (18)

In particular,

∑_{k=1}^{n} \binom{n}{k}^{2}f(k) \{\frac{1}{k} − \frac{f′(k)}{f(k)} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\} = f(0),  (19)

∑_{k=1}^{n} \frac{\binom{n}{k}2}{k} f(k) \{\frac{2}{k} − \frac{f′(k)}{f(k)} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\} = f′(0) + 2f(0) ∑_{i=1}^{n} \frac{1}{i},  (20)

∑_{k=1}^{n} \frac{\binom{n}{k}2}{k^{2}} f(k) \{\frac{3}{k} − \frac{f′(k)}{f(k)} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\}

= \frac{1}{2} \{f′′(0) + 4f′(0) ∑_{i=1}^{n} \frac{1}{i} + f(0) ((4 ∑_{i=1}^{n} \frac{1}{i})^{2} + 2 ∑_{i=1}^{n} \frac{1}{i^{2}} )\} ,  (21)

∑_{k=1}^{n}(−1)^{k+1}\binom{n}{k}^{3}f(k) \{\frac{2}{k^{2}} − \frac{2}{k} (\frac{f′(k)}{f(k)} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}) + \frac{f′′(k)}{f(k)} + 6\frac{f′(k)}{f(k)} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}

+9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \}

= 2f(0),  (22)

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}^{3}}{k} f(k) \{\frac{6}{k^{2}} − \frac{4}{k} (\frac{f′(k)}{f(k)} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}) + \frac{f′′(k)}{f(k)} + 6\frac{f′(k)}{f(k)} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}

+9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \}

= 2f′(0) + 6f(0) ∑_{i=1}^{n} \frac{1}{i}.  (23)

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}^{3}}{k^{2}} f(k) \{\frac{12}{k^{2}} − \frac{6}{k} (\frac{f′(k)}{f(k)} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}) + \frac{f′′(k)}{f(k)} + 6\frac{f′(k)}{f(k)} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}

+9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \}

= f′′(0) + 6f′(0) ∑_{i=1}^{n} \frac{1}{i} + f(0) (9 (∑_{i=1}^{n} \frac{1}{i})^{2} + 3 ∑_{i=1}^{n} \frac{1}{i^{2}} ) .  (24)

In identities (15)-(16) and (19)-(24), according to different choices of f(x), we can get a lot of identities. For example, if f(x) = 1, then we have

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}}{k} = ∑_{i=1}^{n} \frac{1}{i},  (25)

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}}{k^{2}} = \frac{1}{2} \{((∑_{i=1}^{n} \frac{1}{i})^{2} + ∑_{i=1}^{n} \frac{1}{i^{2}} )\} ,  (26)

∑_{k=1}^{n} \binom{n}{k}^{2} \{\frac{1}{k} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\} = 1,  (27)

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∑_{k=1}^{n} \frac{\binom{n}{k}2}{k} \{\frac{2}{k} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\} = 2 ∑_{i=1}^{n} \frac{1}{i},  (28)
∑_{k=1}^{n} \frac{\binom{n}{k}2}{k^{2}} \{\frac{3}{k} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i}\} = 2 (∑_{i=1}^{n} \frac{1}{i})^{2} + ∑_{i=1}^{n} \frac{1}{i^{2}} ,  (29)
∑_{k=1}^{n}(−1)^{k+1}\binom{n}{k}^{3} \{\frac{2}{k^{2}} − \frac{6}{k} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i} + 9 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + 3 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \} = 2, (30)
∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}^{3}}{k} \{\frac{2}{k^{2}} − \frac{4}{k} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i} + 3 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \} = 2 ∑_{i=1}^{n} \frac{1}{i}. (31)

∑_{k=1}^{n} \frac{(−1)^{k+1}\binom{n}{k}^{3}}{k^{2}} \{\frac{4}{k^{2}} − \frac{6}{k} ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i} + 3 ( ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i})^{2} + ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{i^{2}} \} = 3 (∑_{i=1}^{n} \frac{1}{i})^{2} + ∑_{i=1}^{n} \frac{1}{i^{2}} .

(32)

4.2 The case: a_{k} = −\frac{1−q^{k}}{1−q} In this section, taking a_{k} = −\frac{1−q^{k}}{1−q} we obtain some identities involving the Gaussian binomial coefficient. Theorem 4.6

∑_{k=1}^{n} \frac{(−1)^{(k+1)λ} [ \begin{matrix}n \\ k \end{matrix} ]^{λ}}{q^{λ(nk− \frac{1}{2} k(k+1))} (\frac{1−q^{k}}{1−q} )^{λ+r−1}} ∑_{ℓ=0}^{λ−1} \frac{(−1)^{ℓ}}{ℓ!} ∑_{s=0}^{ℓ} \binom{ℓ}{s}f^{(ℓ−s)} (\frac{1 − q^{k}}{1 − q} )

×A (λT_{1}, λT_{2}, …, λT_{s}) \binom{λ − ℓ + r − 1}{r} (\frac{1 − q^{k}}{1 − q} )^{ℓ}

= \frac{1}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (λS_{1}, λS_{2}, …, λS_{j})  (33)

where

T_{m} = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} (\frac{1 − q}{1 − q^{i}} )^{m} , (1 ≤ m ≤ λ),

and

S_{m} = ∑_{i=1}^{n} (\frac{1 − q}{1 − q^{i}} )^{m} , (1 ≤ m ≤ r).

Proof. Take a_{k} = −\frac{1−q^{k}}{1−q} and x = 0 in Theorem 2.1. □ When λ = 1, this reduces to an identity due to Mercier [10]:

∑_{k=1}^{n} \frac{(−1)^{k+1} [ \begin{matrix}n \\ k \end{matrix} ] f (\frac{1−q^{k}}{1−q} )}{q^{nk− \frac{1}{2} k(k+1)} (\frac{1−q^{k}}{1−q} )^{r}} = \frac{1}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (S_{1}, S_{2}, …, S_{j}) .  (34)

In particular, for r = 0, 1, we have

∑_{k=1}^{n} \frac{(−1)^{k+1} [ \begin{matrix}n \\ k \end{matrix} ] f (\frac{1−q^{k}}{1−q} )}{q^{nk− \frac{1}{2} k(k+1)}} = f(0),  (35)
∑_{k=1}^{n} \frac{(−1)^{k+1} [ \begin{matrix}n \\ k \end{matrix} ] f (\frac{1−q^{k}}{1−q} )}{q^{nk− \frac{1}{2} k(k+1)} (\frac{1−q^{k}}{1−q} )} = f′(0) + f(0) ∑_{i=1}^{n} \frac{1 − q}{1 − q^{i}} .  (36)

Take f(x) = 1 and let q → 1/q. Then (36) becomes the result of Van Hamme [13]:

∑_{k=1}^{n} [ \begin{matrix}n \\ k \end{matrix} ] \frac{(−1)^{k−1}q^{\binom{k+1}{2}}}{1 − q^{k}} = ∑_{i=1}^{n} \frac{q^{i}}{1 − q^{i}} .  (37)

When λ = 2, we obtain

∑_{k=1}^{n} \frac{[ \begin{matrix}n \\ k \end{matrix} ]^{2} f (\frac{1−q^{k}}{1−q} )}{q^{2nk−k(k+1)} (\frac{1−q^{k}}{1−q} )^{r}} \{(r + 1)\frac{1 − q}{1 − q^{k}} − \frac{f′ (\frac{1−q^{k}}{1−q} )}{f (\frac{1−q^{k}}{1−q} )} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1 − q}{1 − q^{i}} \}
= \frac{1}{r!} ∑_{j=0}^{r} \binom{r}{j}f^{(r−j)}(0)A (2S_{1}, 2S_{2}, …, 2S_{j}) .  (38)

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In particular, for r = 0, 1 in (38), the following identities hold.

∑_{k=1}^{n} [ \begin{matrix}n \\ k \end{matrix} ]^{2} q^{k(k+1)−2nk}f (\frac{1 − q^{k}}{1 − q} ) \{\frac{1 − q}{1 − q^{k}} − \frac{f′ (\frac{1−q^{k}}{1−q} )}{f (\frac{1−q^{k}}{1−q} )} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1 − q}{1 − q^{i}} \} = f(0), (39)

and

∑_{k=1}^{n} \frac{[ \begin{matrix}n \\ k \end{matrix} ]^{2} f (\frac{1−q^{k}}{1−q} )}{q^{2nk−k(k+1)} (\frac{1−q^{k}}{1−q} )} \{2\frac{1 − q}{1 − q^{k}} − \frac{f′ (\frac{1−q^{k}}{1−q} )}{f (\frac{1−q^{k}}{1−q} )} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1 − q}{1 − q^{i}} \}
= f′(0) + 2f(0) ∑_{i=1}^{n} \frac{1 − q}{1 − q^{i}} .  (40)

Specifically, for f(x) = 1, we have

∑_{k=1}^{n} [ \begin{matrix}n \\ k \end{matrix} ]^{2} q^{k(k+1)−2nk} \{\frac{1}{1 − q^{k}} − 2 ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{1 − q^{i}} \} = \frac{1}{1 − q} ,  (41)

∑_{k=1}^{n} [ \begin{matrix}n \\ k \end{matrix} ]^{2} \frac{q^{k(k+1)−2nk}}{1 − q^{k}} \{\frac{1}{1 − q^{k}} − ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{1}{1 − q^{i}} \} = \frac{1}{1 − q} ∑_{i=1}^{n} \frac{1}{1 − q^{i}} .  (42)

4.3 An identity of Fu and Lascoux Theorem 4.7

\frac{a^{nλ} (\frac{bq}{a}; q)_{n}^{λ}}{(az − bc)^{(n−1)λ}(q; q)_{n}} ∑_{k=1}^{n}(−1)^{(k−1)λ} [ \begin{matrix}n \\ k \end{matrix} ]^{λ} q^{λ\binom{k+1}{2}−λnk}(1 − q^{k})(a − bq^{k})^{λ(n−2)}

× ∑_{ℓ=0}^{λ−1} \frac{1}{ℓ!} ∑_{s=0}^{ℓ}(−1)^{s}\binom{ℓ}{s}f^{(ℓ−s)} (−\frac{c − zq^{k}}{a − bq^{k}} ) A (λT_{1}(x), λT_{2}(x), …, λT_{s}(x)) \begin{matrix}( & {λ−ℓ+r−1 \atop r} & ) \\ (x + \frac{c−zq^{k}}{a−bq^{k}} ) & λ−ℓ+r \end{matrix}

= \frac{(−1)^{r}}{r! (x + \frac{c−zq}{a−bq})^{λ} (x + \frac{c−zq^{2}}{a−bq^{2}} )^{λ} … (x + \frac{c−zq^{n}}{a−bq^{n}} )^{λ}}
× ∑_{j=0}^{r}(−1)^{j}\binom{r}{j}f^{(r−j)}(x)A (λS_{1}(x), λS_{2}(x), …, λS_{j}(x)) ,  (43)

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where

T_{m}(x) = ∑_{\begin{matrix}i = 1 \\ i ≠ k \end{matrix}}^{n} \frac{(a − bq^{i})^{m}}{(ax + c − q^{i}(bx + z))^{m}} , (1 ≤ m ≤ λ),

and

S_{m}(x) = ∑_{i=1}^{n} \frac{(a − bq^{i})^{m}}{(ax + c − q^{i}(bx + z))^{m}} , (1 ≤ m ≤ r).

Proof. In Theorem 2.1, take a_{k} = \frac{c−zq^{k}}{a−bq^{k}}. □

When λ = 1, r = 0, x → −x and f(x) = 1, this reduces to

\frac{c^{n} (\frac{zq}{c}; q)_{n}}{(az − bc)^{n−1}(q; q)_{n}} ∑_{k=1}^{n}(−1)^{k−1} [ \begin{matrix}n \\ k \end{matrix} ] q^{\binom{k+1}{2}−nk}(1 − q^{k})\frac{(a − bq^{k})^{n−1}}{c − zq^{k}} · \frac{1}{1 − \frac{a−bq^{k}}{c−zq^{k}} x}

= \frac{1}{(1 − \frac{a−bq}{c−zq}x) (1 − \frac{a−bq^{2}}{c−zq^{2}} x) … (1 − \frac{a−bq^{n}}{c−zq^{n}} x)}.  (44)

Comparing the coefficients of x^{τ} on both sides of the above equation, we have an identity of Fu and Lascoux [8]:

\frac{c^{n} (\frac{zq}{c}; q)_{n}}{(az − bc)^{n−1}(q; q)_{n}} ∑_{k=1}^{n}(−1)^{k−1} [ \begin{matrix}n \\ k \end{matrix} ] q^{\binom{k+1}{2}−nk}(1 − q^{k})\frac{(a − bq^{k})^{n−1+τ}}{(c − zq^{k})^{τ+1}}

= h_{τ} (\frac{a − bq}{c − zq} , \frac{a − bq^{2}}{c − zq^{2}} , …, \frac{a − bq^{n}}{c − zq^{n}} ) ,  (45)

where h_{τ} (a_{1}, a_{2}, …, a_{n}) is the τ-th complete symmetric function defined by

h_{τ} (a_{1}, a_{2}, …, a_{n}) = ∑_{1≤i_{1}≤i_{2}≤...≤i_{τ} ≤n} a_{i_{1}}a_{i_{2}} … a_{i_{τ}} .

Taking a = 0, c = 1 and b = −1 in (45) we have

∑_{k=1}^{n}(−1)^{k−1} [ \begin{matrix}n \\ k \end{matrix} ] q^{\binom{k}{2}+τk} \frac{1 − q^{k}}{(1 − zq^{k})^{τ+1}} = \frac{(q; q)_{n}}{(zq; q)_{n}} h_{τ} (\frac{q}{1 − zq} , \frac{q^{2}}{1 − zq^{2}} , …, \frac{q^{n}}{1 − zq^{n}} ) . (46)

When z = 1, this identity reduces to Dilcher’s identity [7]:

∑_{k=1}^{n}(−1)^{k−1} [ \begin{matrix}n \\ k \end{matrix} ] q^{\binom{k}{2}+τk} \frac{1}{(1 − q^{k})^{τ}} = h_{τ} (\frac{q}{1 − q} , \frac{q^{2}}{1 − q^{2}} , …, \frac{q^{n}}{1 − q^{n}} ) .  (47)

Acknowledgements

We would like to thank the anonymous referee for his valuable suggestions.

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