Two problems of binomial sums involving harmonic numbers

Abstract

Two open problems recently proposed by Xi and Luo (Adv. Differ. Equ. 2021:38, 2021) are resolved by evaluating explicitly three binomial sums involving harmonic numbers, that are realized mainly by utilizing the generating function method and symmetric functions.

1 Introduction and outline

Denote by $\mathbb{N}$ the set of natural numbers with $\mathbb{N}_{0}=\mathbb{N}\cup \{0\}$. For an indeterminate x, define the rising and falling factorials by $(x)_{0}=\langle {x}\rangle _{0}\equiv 1$ and

$$\begin{aligned} &(x)_{n}=x(x+1)\cdots (x+n-1) \quad \text{for } n\in \mathbb{N}, \\ &\langle {x}\rangle _{n}=x(x-1)\cdots (x-n+1) \quad \text{for } n \in \mathbb{N}. \end{aligned}$$

The harmonic numbers of higher order are given by

$$ H_{0}^{(\lambda )}=1 \quad \text{and}\quad H_{n}^{(\lambda )}= \sum_{k=1}^{n} \frac{1}{k^{\lambda }} \quad \text{for } n,\lambda \in \mathbb{N}. $$

In order to reduce lengthy expressions, we shall employ the notations of elementary and complete symmetric functions. For a finite set S of real numbers, we define these functions by $\Phi _{0}(x|S)=\Psi _{0}(x|S)\equiv 1$ and

$$\begin{aligned} &\Phi _{n}(x|S)= \sum_{ \substack{ \sum _{\alpha \in S}k_{\alpha }=n\\{ 0\le k_{\alpha }\le 1}}} \prod_{\substack{\alpha \in S}}\frac{1}{(x+\alpha )^{k_{\alpha }}} \quad \text{for } n\in \mathbb{N}, \end{aligned}$$

(1)

$$\begin{aligned} &\Psi _{n}(x|S)= \sum_{ \substack{ \sum _{\alpha \in S}k_{\alpha }=n\\{ 0\le k_{\alpha }\le n}}} \prod_{\substack{\alpha \in S}}\frac{1}{(x+\alpha )^{k_{\alpha }}} \quad \text{for } n\in \mathbb{N}. \end{aligned}$$

(2)

We shall also need the signless Stirling numbers of the first kind (see [6]) which are determined by the connection coefficient of expanding the shifted factorials into monomials

$$ (y)_{n}= \sum_{k=0}^{n}{n \brack k}y^{k}.$$

(3)

There exist numerous summation formulae involving harmonic numbers (cf. [1–3, 7, 8]). In a recent paper [9], Xi and Luo proposed the following two open problems.

Problem I

Let x be an indeterminate. For $m,n\in \mathbb{N}_{0}$ with $m>n$, how to calculate the combinatorial sums

$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{m+k}{k} \quad \text{and} \quad \sum _{k=0}^{n}(-1)^{k}\binom{n}{k} \binom{m+k}{k} \frac{x}{x+k} ? $$

Problem II

Let x be an indeterminate. For $m,n,\lambda ,\rho \in \mathbb{N}_{0}$, what are the combinatorial sums

$$\begin{aligned} \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{m+k}{k} \biggl( \frac{x}{x+k} \biggr)^{\lambda }\quad \text{and} \quad \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{m+k}{k} \bigl\{ H_{\rho +k}^{(\lambda )}-H_{k}^{( \lambda )} \bigr\} ? \end{aligned}$$

The first binomial sum in Problem I can easily be evaluated by the Chu–Vandermonde convolution formula as follows:

$$\begin{aligned} \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{m+k}{k} &=\sum_{k=0}^{n} \binom{n}{n-k}\binom{-m-1}{k} \\ &=\binom{n-m-1}{n}=(-1)^{n}\binom{m}{n}. \end{aligned}$$

As the primary motivation, the aim of the present paper is to resolve these problems and evaluate the remaining three sums explicitly in the following theorems.

Theorem 1

Let x be an indeterminate. Then for $m,n\in \mathbb{N}_{0}$, the following algebraic identity holds:

$$\begin{aligned} \frac{n!}{(x)_{n+1}}\binom{m-x}{m} &=\sum_{k=0}^{n} \frac{(-1)^{k}}{x+k} \binom{n}{k}\binom{m+k}{k} \\ &\quad{}+\sum_{k=1}^{m-n}(-1)^{n+k} \binom{m}{n+k} \frac{(x+n+1)_{k-1}}{(n+1)_{k}}. \end{aligned}$$

We remark that when $m>n$, this theorem evaluates the second sum in Problem I by determining the polynomial part of the rational function explicitly as in the last line, which vanishes for $m\le n$, instead.

Theorem 2

Let x be an indeterminate. Then for $m,n,\lambda \in \mathbb{N}_{0}$, the following algebraic identity holds:

$$ \begin{aligned} &\sum_{k=0}^{n}\binom{n}{k} \binom{m+k}{k} \frac{(-1)^{k}}{(x+k)^{\lambda }} \\ &\quad =\frac{n!}{(x)_{n+1}} \binom{m-x}{m} \sum _{k=1}^{\lambda } \frac{\Phi _{k-1}(-x|[1,m])}{(k-1)!} \Psi _{\lambda -k}\bigl(x|[0,n]\bigr)\\ &\qquad {}+\sum_{k=1}^{m-n} \frac{(-1)^{n+k+\lambda }}{(\lambda -1)!} \binom{m}{n+k} \frac{(x+n+1)_{k-1}}{(n+1)_{k}} \Phi _{\lambda -1}\bigl(x+n|[1,k-1] \bigr). \end{aligned} $$

Theorem 3

Let x be an indeterminate. Then for $m,n,\lambda ,\rho \in \mathbb{N}_{0}$, the following algebraic identity holds:

$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{m+k}{k} \bigl\{ H_{\rho +k}^{( \lambda )}-H_{k}^{(\lambda )} \bigr\} \\ &\quad =\frac{n!}{m!} \sum_{k=1}^{\lambda } \sum_{j=1}^{m}\sum _{i=1}^{k} \frac{(-1)^{i+j}}{(j)_{n+1}} {j\brack i} {m-j+1\brack k-i} \frac{\Psi _{\lambda -k}(j|[0,n])}{(k-1)!} \\ &\qquad{}+\sum_{k=1}^{\lambda } \sum _{j=m+1}^{\rho }\frac{n!}{(j)_{n+1}} \frac{\Psi _{\lambda -k}(j|[0,n])}{(k-1)!} \binom{m-j}{m}\Phi _{k-1}\bigl(-j|[1,m]\bigr) \\ &\qquad{}+\sum_{k=1}^{m-n}\sum _{j=1}^{\rho } \frac{(-1)^{n+k+\lambda }}{(\lambda -1)!} \binom{m}{n+k} \frac{(j+n+1)_{k-1}}{(n+1)_{k}} \Phi _{\lambda -1}\bigl(j+n|[1,k-1]\bigr). \end{aligned}$$

The rest paper will be organized as follows. In the next section, we shall prove Theorem 1 by determining explicitly the polynomial part of a rational function when its numerator degree is greater than that of the denominator. Then Theorems 2 and 3 will be shown in Sect. 3 by establishing two analytical formulae of the derivatives of higher order for a polynomial function of the rising factorial and its reciprocal. The informed reader will notice that by employing symmetric functions Φ and Ψ, several involved expressions become simpler than those appearing in [9], where the Bell polynomials were employed.

2 Proof of Theorem 1

Observe that the rational function below can be decomposed into partial fractions

$$ \frac{n!}{(x)_{n+1}}\binom{m-x}{m} =P^{m}_{n}(x)+\sum _{k=0}^{n} \frac{A_{k}}{x+k}, $$

where $P^{m}_{n}(x)$ is a polynomial of degree $m-n-1$ in x which reduces to zero when $m\le n$, and the coefficients $A_{k}$ are determined by the limits

$$ A_{k}=\lim_{x\to -k}(x+k) \biggl\{ \frac{n!}{(x)_{n+1}} \binom{m-x}{m} \biggr\} =(-1)^{k}\binom{n}{k}\binom{m+k}{m}. $$

Therefore, we have found the equality

$$ \frac{n!}{(x)_{n+1}}\binom{m-x}{m} =P^{m}_{n}(x)+ \sum_{k=0}^{n} \frac{(-1)^{k}}{x+k} \binom{n}{k}\binom{m+k}{k}.$$

(4)

By scaling down m and then making use of

$$ \frac{m-x}{m}=\frac{m+k}{m}-\frac{k+x}{m}, $$

we can rewrite the last equality as

$$\begin{aligned} &\frac{n!}{(x)_{n+1}}\binom{m-x}{m} \\ &\quad =\frac{m-x}{m} \times \frac{n!}{(x)_{n+1}} \binom{m-1-x}{m-1} \\ &\quad =\frac{m-x}{m} \Biggl\{ P^{m-1}_{n}(x) +\sum _{k=0}^{n} \frac{(-1)^{k}}{x+k} \binom{n}{k}\binom{m-1+k}{k} \Biggr\} \\ &\quad =\frac{m-x}{m}P^{m-1}_{n}(x) +\sum _{k=0}^{n}\frac{(-1)^{k}}{x+k} \binom{n}{k} \binom{m+k}{k} -\sum_{k=0}^{n} \frac{(-1)^{k}}{m} \binom{n}{k}\binom{m-1+k}{k}. \end{aligned}$$

Evaluating the last sum by means of the Chu–Vandemonde formula and then comparing the resultant expression with (4), we get the following recurrence relation:

$$ P^{m}_{n}(x)= \frac{m-x}{m}P^{m-1}_{n}(x)- \frac{(-1)^{n}}{m}\binom{m-1}{n}.$$

(5)

In order to find an explicit expression for $P^{m}_{n}(x)$, let $Q_{m}:=P^{m+n}_{n}(x)$. Then the equality corresponding to (5) becomes

$$ Q_{m}= \frac{m+n-x}{m+n}Q_{m-1} - \frac{(-1)^{n}}{m+n}\binom{m+n-1}{n}.$$

(6)

It is routine to figure out the initial values $Q_{0}=0$ and $Q_{1}=\frac{(-1)^{n+1}}{n+1}$. Then we can manipulate the generating function

$$\begin{aligned} Q(y)&:=\sum_{m=1}^{\infty }Q_{m}y^{m+n}\\ &=\sum_{m=1}^{\infty } \biggl(1- \frac{x}{m+n} \biggr)Q_{m-1}y^{m+n} \\ &\quad{}-\sum_{m=1}^{\infty }(-1)^{n} \binom{m+n-1}{n}\frac{y^{m+n}}{m+n}. \end{aligned}$$

By differentiating the last equation with respect to y,

$$ Q'(y)=\frac{d}{dy} \bigl\{ yQ(y) \bigr\} -xQ(y) -\sum _{m=1}^{\infty }(-1)^{n} \binom{m+n-1}{n}y^{m+n-1}, $$

and then evaluating the binomial series on the right, we find, after some simplification, that $Q(y)$ satisfies the following differential equation:

$$ (1-y)Q'(y)-(1-x)Q(y)=\frac{y^{n}}{(y-1)^{n+1}}.$$

(7)

It is trivial to check that the corresponding homogeneous equation

$$ \frac{Q'(y)}{Q(y)}=\frac{1-x}{1-y} $$

has the binomial solution

$$ Q(y)= \Omega (1-y)^{x-1}$$

(8)

where Ω is an arbitrary constant. When $\Omega :=\Omega (y)$ is considered as a function of y, substituting the above solution into (7) gives rise to

$$ \Omega '(y)=(-1)^{n+1}y^{n}(1-y)^{-x-n-1}. $$

Therefore, we have the integral representation

$$ \Omega (y)=(-1)^{n+1} \int _{0}^{y}T^{n}(1-T)^{-x-n-1}\,dT. $$

Define for simplicity

$$ J_{n}:= \int _{0}^{y}T^{n}(1-T)^{-x-n-1}\,dT \quad \text{with } J_{0}= \frac{(1-y)^{-x}-1}{x}. $$

According to integration by parts, we can calculate $J_{n}$ as follows:

$$\begin{aligned} J_{n}&= \int _{0}^{y}T^{n}(1-T)^{-x-n-1}\,dT =\frac{y^{n}}{x+n}(1-y)^{-x-n} -\frac{n}{x+n}J_{n-1} \\ &=\frac{y^{n}}{x+n}(1-y)^{-x-n} - \frac{ny^{n-1}}{\langle {x+n}\rangle _{2}}(1-y)^{1-x-n} + \frac{\langle {n}\rangle _{2}}{\langle {x+n}\rangle _{2}}J_{n-2}. \end{aligned}$$

By means of the induction principle, we can show that

$$\begin{aligned} J_{n}&=\sum_{k=0}^{n-1} \frac{(-1)^{k}\langle {n}\rangle _{k}}{\langle {x+n}\rangle _{k+1}} y^{n-k}(1-y)^{k-x-n} +\frac{(-1)^{n}\langle {n}\rangle _{n}}{\langle {x+n}\rangle _{n}}J_{0} \\ &=\sum_{k=0}^{n} \frac{(-1)^{k}\langle {n}\rangle _{k}}{\langle {x+n}\rangle _{k+1}} y^{n-k}(1-y)^{k-x-n} +\frac{(-1)^{n+1}n!}{(x)_{n+1}}, \end{aligned}$$

which is equivalent to the expression

$$ \Omega (y)=\frac{n!}{(x)_{n+1}}-\sum_{k=0}^{n} \frac{(-1)^{n+k}\langle {n}\rangle _{k}}{\langle {x+n}\rangle _{k+1}} y^{n-k}(1-y)^{k-x-n}. $$

Substituting this into (8), we obtain the explicit generating function

$$\begin{aligned} Q(y)=\frac{n!}{(x)_{n+1}}(1-y)^{x-1} -\sum_{k=0}^{n} \frac{(-1)^{n+k}\langle {n}\rangle _{k}}{\langle {x+n}\rangle _{k+1}} y^{n-k}(1-y)^{k-1-n}. \end{aligned}$$

Extracting the coefficient of $y^{m+n}$ across the last equation yields

$$ Q_{m}=\bigl[y^{m+n}\bigr]Q(y) =\binom{m+n-x}{m+n} \frac{n!}{(x)_{n+1}} -\sum_{k=0}^{n} \frac{(-1)^{n+k}\langle {n}\rangle _{k}}{\langle {x+n}\rangle _{k+1}}\binom{m+n}{n-k}. $$

By reformulating the last sum with respect to k as

$$\begin{aligned} &\sum_{k=0}^{n}\binom{m+n}{n-k} \frac{(-1)^{n-k}\langle {n}\rangle _{k}}{\langle {x+n}\rangle _{k+1}} \\ &\quad =\frac{n!}{(x)_{n+1}} \sum_{k=0}^{n} \binom{m+n}{m+k} \binom{-x}{n-k} \\ &\quad =\frac{n!}{(x)_{n+1}} \Biggl\{ \sum_{k=-m}^{n} \binom{m+n}{m+k} \binom{-x}{n-k} -\sum_{k=-m}^{-1} \binom{m+n}{m+k}\binom{-x}{n-k} \Biggr\} \\ &\quad =\frac{n!}{(x)_{n+1}}\binom{m+n-x}{m+n} -\frac{n!}{(x)_{n+1}} \sum _{k=1}^{m} \binom{m+n}{m-k}\binom{-x}{n+k}, \end{aligned}$$

we find finally the binomial expression

$$ Q_{m}=\frac{n!}{(x)_{n+1}}\sum_{k=1}^{m} \binom{m+n}{m-k} \binom{-x}{n+k} =\sum_{k=1}^{m}(-1)^{n+k} \binom{m+n}{n+k} \frac{(x+n+1)_{k-1}}{(n+1)_{k}}. $$

This gives consequently the desired formula stated in Theorem 1:

$$ P^{m}_{n}(x)=Q_{m-n}(x) =\sum _{k=1}^{m-n}(-1)^{n+k}\binom{m}{n+k} \frac{(x+n+1)_{k-1}}{(n+1)_{k}}. $$

3 Proofs of Theorems 2 and 3

For the derivative operator $\mathcal{D}$ with respect to x, we have the following analytical formulae of higher order derivatives:

$$\begin{aligned} &\mathcal{D}^{n}\prod_{\substack{\alpha \in S}}(x+\alpha ) = \Phi _{n}(x|S) \prod_{\substack{\alpha \in S}}(x+ \alpha ), \end{aligned}$$

(9)

$$\begin{aligned} &\mathcal{D}^{n}\prod_{\substack{\alpha \in S}} \frac{1}{x+\alpha } = \Psi _{n}(x|S) \frac{n!(-1)^{n}}{\prod_{\substack{\alpha \in S}}(x+\alpha )}. \end{aligned}$$

(10)

The first one in (9) can be evaluated easily by induction on n. In order to prove the second one in (10), define

$$ R(x):= \prod_{\substack{\alpha \in S}}\frac{1}{x+\alpha } \quad \text{and} \quad \mathcal{D}^{n}R(x)=R(x)G_{n},$$

(11)

where the function $G_{n}$ remains to be determined with the initial values

$$ G_{0}=1\quad \text{and}\quad G_{1}=-\Psi _{1}(x|S). $$

Then by making use of the Leibniz rule, we have

$$\begin{aligned} \mathcal{D}^{\lambda +1}R(x)&=-\mathcal{D}^{\lambda } \bigl\{ R(x) \Psi _{1}(x|S) \bigr\} \\ &=-\sum_{k=0}^{\lambda }\binom{\lambda }{k} \mathcal{D}^{\lambda -k}R(x) \mathcal{D}^{k}\Psi _{1}(x|S) \\ &=-R(x)\sum_{k=0}^{\lambda }\binom{\lambda }{k}G_{\lambda -k} \mathcal{D}^{k}\Psi _{1}(x|S), \end{aligned}$$

which leads us to the binomial recursion

$$ G_{\lambda +1}=- \sum_{k=0}^{\lambda } \binom{\lambda }{k}G_{\lambda -k}\mathcal{D}^{k} \Psi _{1}(x|S).$$

(12)

In order to find an explicit expression for $G_{\lambda }$, we examine the exponential generating function defined by

$$ G(y):=\sum_{\lambda =0}^{\infty }\frac{y^{\lambda }}{\lambda !}G_{ \lambda }. $$

According to (12), its derivative with respect to y can be expressed as

$$\begin{aligned} G'(y)&=\sum_{\lambda =0}^{\infty } \frac{y^{\lambda }}{\lambda !}G_{ \lambda +1} =-\sum_{\lambda =0}^{\infty } \frac{y^{\lambda }}{\lambda !} \sum_{k=0}^{\lambda } \binom{\lambda }{k}G_{\lambda -k}\mathcal{D}^{k}\Psi _{1}(x|S) \\ &=-\sum_{k=0}^{\infty }\frac{y^{k}}{k!} \mathcal{D}^{k}\Psi _{1}(x|S) \sum _{\lambda =k}^{\infty }\frac{y^{\lambda -k}}{(\lambda -k)!}G_{ \lambda -k}. \end{aligned}$$

We therefore get the differential equation

$$ G'(y)=-G(y)\sum_{k=0}^{\infty } \frac{y^{k}}{k!}\mathcal{D}^{k}\Psi _{1}(x|S) $$

whose solution is given by the exponential function

$$ G(y)=\exp \Biggl\{ - \int _{0}^{y} \sum_{k=0}^{\infty } \frac{y^{k}}{k!} \mathcal{D}^{k}\Psi _{1}(x|S)\,dy \Biggr\} = \exp \Biggl\{ -\sum_{k=0}^{ \infty } \frac{y^{k+1}}{(k+1)!} \mathcal{D}^{k}\Psi _{1}(x|S) \Biggr\} . $$

Evaluating the last sum with respect to k explicitly as

$$\begin{aligned} \begin{aligned} \sum_{k=0}^{\infty }\frac{y^{k+1}}{(k+1)!} \mathcal{D}^{k}\Psi _{1}(x|S) &=\sum _{k=0}^{\infty }\frac{(-1)^{k}}{k+1} \sum _{\alpha \in S} \frac{y^{k+1}}{(x+\alpha )^{k+1}} \\ &=\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k} \sum _{\alpha \in S} \frac{y^{k}}{(x+\alpha )^{k}} \\ &=\sum_{\alpha \in S}\ln \biggl(1+\frac{y}{x+\alpha } \biggr), \end{aligned} \end{aligned}$$

we find the simplified generating function

$$ G(y)= \prod_{\alpha \in S} \biggl(1+ \frac{y}{x+\alpha } \biggr)^{-1}.$$

(13)

By extracting the coefficient of $y^{n}$, we confirm the formula (10) as follows:

$$ G_{n}=n!\bigl[y^{n}\bigr]G(y) =n!(-1)^{n}\Psi _{n}(x|S). $$

3.1 Proof of Theorem 2

This can be done by differentiating $\lambda -1$ times the equality displayed in Theorem 1. Firstly, it is trivial to have

$$ \mathcal{D}^{\lambda -1}\frac{1}{x+k} =(-1)^{\lambda -1} \frac{(\lambda -1)!}{(x+k)^{\lambda }}. $$

Then by making use of the Leibniz rule, we can compute

$$\begin{aligned} \mathcal{D}^{\lambda -1}\frac{(1-x)_{m}}{(x)_{n+1}} &=(-1)^{m} \mathcal{D}^{\lambda -1}\frac{\langle {x-1}\rangle _{m}}{(x)_{n+1}} \\ &=(-1)^{m}\sum_{k=1}^{\lambda } \binom{\lambda -1}{k-1} \mathcal{D}^{k-1} \langle {x-1}\rangle _{m} \mathcal{D}^{\lambda -k} \frac{1}{(x)_{n+1}} \\ &=\frac{(1-x)_{m}}{(x)_{n+1}} \sum_{k=1}^{\lambda }(-1)^{\lambda -k} \frac{(\lambda -1)!}{(k-1)!} \Phi _{k-1}\bigl(x|[-m,-1]\bigr)\Psi _{\lambda -k} \bigl(x|[0,n]\bigr) \\ &=(-1)^{\lambda -1}\frac{(1-x)_{m}}{(x)_{n+1}} \sum_{k=1}^{\lambda } \frac{(\lambda -1)!}{(k-1)!} \Phi _{k-1}\bigl(-x|[1,m]\bigr)\Psi _{\lambda -k} \bigl(x|[0,n]\bigr), \end{aligned}$$

where we have invoked two derivative formulae (9) and (10). Finally,

$$\begin{aligned} \mathcal{D}^{\lambda -1}(x+n+1)_{k-1} &=(x+n+1)_{k-1}\Phi _{\lambda -1}\bigl(x|[n+1,n+k-1]\bigr) \\ &=(x+n+1)_{k-1}\Phi _{\lambda -1}\bigl(x+n|[1,k-1]\bigr). \end{aligned}$$

Substituting the above three expressions into the equality of Theorem 1 and then making some simplifications, we find the algebraic identity in Theorem 2.

3.2 Proof of Theorem 3

Recalling (3), we can deduce, for the signless Stirling numbers, the symmetric function expression (see [4, Chap. V] and [5, §6.1])

$$\begin{aligned} {n+1\brack k+1}&=\bigl[y^{k}\bigr](1+y)_{n} =\sum _{ \substack{1\le i_{1}< i_{2}< \cdots < i_{n-k}\le n}}i_{1}i_{2}\cdots i_{n-k} \\ &=n!\sum_{\substack{1\le j_{1}< j_{2}< \cdots < j_{k}\le n}} \frac{1}{j_{1}j_{2}\cdots j_{k}}. \end{aligned}$$

This gives rise to the following identity:

$$ \Phi _{k}\bigl(0|[1,n]\bigr)= \frac{1}{n!}{n+1 \brack k+1}.$$

(14)

Let ρ be a natural number. When $x\to j$ with $1\le j\le \rho$, the limiting case of the equation displayed in Theorem 2 reads as

$$ \begin{aligned}[c] &\sum_{k=0}^{n} \binom{n}{k}\binom{m+k}{k} \frac{(-1)^{k}}{(j+k)^{\lambda }} \\ &\quad =\frac{n!}{(j)_{n+1}} \sum_{k=1}^{\lambda } \frac{\Psi _{\lambda -k}(j|[0,n])}{(k-1)!} \lim_{x\to j} \binom{m-x}{m} \Phi _{k-1}\bigl(-x|[1,m]\bigr) \\ &\qquad{}+\sum_{k=1}^{m-n} \frac{(-1)^{n+k+\lambda }}{(\lambda -1)!} \binom{m}{n+k} \frac{(j+n+1)_{k-1}}{(n+1)_{k}} \Phi _{\lambda -1}\bigl(j+n|[1,k-1]\bigr). \end{aligned} $$

(15)

When $j>m$, the limit in the middle line is given directly by letting $x=j$

$$ \lim_{x\to j}\binom{m-x}{m}\Phi _{k-1}\bigl(-x|[1,m] \bigr) =\binom{m-j}{m} \Phi _{k-1}\bigl(-j|[1,m]\bigr) $$

since the two factors on the right are well defined. Instead, for $1\le j\le m$, that limit can be determined as

$$\begin{aligned} &\lim_{x\to j}\binom{m-x}{m}\Phi _{k-1}\bigl(-x|[1,m] \bigr) \\ &\quad =\lim_{x\to j} \frac{(1-x)_{m}}{m!} \Phi _{k-1} \bigl(-x|[1,m]\bigr) \\ &\quad =\frac{(-1)^{j-1}}{j\binom{m}{j}} \sum_{i=1}^{k-1} \Phi _{i-1}\bigl(-j|[1,j-1]\bigr) \Phi _{k-i-1}\bigl(-j|[j+1,m] \bigr) \\ &\quad =\frac{(-1)^{j}}{j\binom{m}{j}} \sum_{i=1}^{k-1}(-1)^{i} \Phi _{i-1}\bigl(0|[1,j-1]\bigr) \Phi _{k-i-1}\bigl(0|[1,m-j] \bigr) \\ &\quad =\frac{(-1)^{j}}{m!} \sum_{i=1}^{k-1}(-1)^{i}{j \brack i} {m-j+1 \brack k-i}, \end{aligned}$$

where the last line is justified by (14). Finally summing equation (15) over j from 1 to ρ, we obtain the following equality involving harmonic numbers:

$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{m+k}{k} \bigl\{ H_{\rho +k}^{( \lambda )}-H_{k}^{(\lambda )} \bigr\} \\ &\quad =\frac{n!}{m!} \sum_{j=1}^{m} \frac{(-1)^{j}}{(j)_{n+1}} \sum_{k=1}^{ \lambda } \frac{\Psi _{\lambda -k}(j|[0,n])}{(k-1)!} \sum_{i=1}^{k-1}(-1)^{i}{j \brack i} {m-j+1\brack k-i} \\ &\qquad{}+\sum_{j=m+1}^{\rho } \frac{n!}{(j)_{n+1}} \sum_{k=1}^{\lambda } \frac{\Psi _{\lambda -k}(j|[0,n])}{(k-1)!} \binom{m-j}{m}\Phi _{k-1}\bigl(-j|[1,m]\bigr) \\ &\qquad{}+\sum_{k=1}^{m-n}\sum _{j=1}^{\rho } \frac{(-1)^{n+k+\lambda }}{(\lambda -1)!} \binom{m}{n+k} \frac{(j+n+1)_{k-1}}{(n+1)_{k}} \Phi _{\lambda -1}\bigl(j+n|[1,k-1]\bigr), \end{aligned}$$

which is equivalent to the formula stated in Theorem 3.