Parameterized Finite Binomial Sums

Abstract

We offer intriguing new insights into parameterized finite binomial sums, revealing elegant identities such as ${\displaystyle \sum _{k=0,k\ne n}^{m+n}}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right)={(-1)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{n}}\right)(H_m-H_n)$, where $n,m$ are non-negative integers and $H_n$ is the harmonic number. These formulas beautifully capture the intricate relationship between harmonic numbers and binomial coefficients, providing a fresh and captivating perspective on combinatorial sums.

1. Introduction and Preliminaries

By using a known identity and mathematical induction, Sun [1] proved the following remarkable identity: For $n\in \mathbb{N}$,

$$\sum _{k=1}^n{\displaystyle \frac{1}{k^2\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=3\sum _{k=1}^n{\displaystyle \frac{1}{k^2\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}}-\sum _{k=1}^n{\displaystyle \frac{1}{k^2}}.$$

(1)

Here and elsewhere, $\mathbb{N}$ denotes the set of positive integers. Sofo and Batır [2] provided an alternate representation to Equation (1): For $n\in \mathbb{N}$, the following identity holds:

$$\sum _{k=1}^n{\displaystyle \frac{1}{k^2\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=-H_n^2+\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k+1}}{k}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)H_{n+k}.$$

(2)

Here and throughout, $H_n$ denotes the classical harmonic number, defined as

$$H_n=\sum _{k=1}^n{\displaystyle \frac{1}{k}}(n\in \mathbb{N}),$$

(3)

and $H_0:=0$. In this article, an empty sum is conventionally interpreted as zero wherever it appears. Indeed, they went further by deriving a general formula for the sums

$$\sum _{k=1}^n{\displaystyle \frac{1}{k^m\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}\left(n\in \mathbb{N};m\in {\mathbb{Z}}_{⩾0}\right).$$

(4)

Here and throughout, ${\mathbb{Z}}_{⩾\eta }$ denotes the set of integers greater than or equal to some integer $\eta$. Additional specific instances of (4), including (2), when $m=0$ and $m=1$, yield the following identities:

$$\sum _{k=1}^n{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{1}{n-1}}\left(1-{\displaystyle \frac{n+1}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\right)\left(n\in {\mathbb{Z}}_{⩾2}\right),$$

(5)

and

$$\sum _{k=1}^n{\displaystyle \frac{1}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}(n\in \mathbb{N}).$$

(6)

1.1. Main Results

In this work, we present a new perspective on parameterized finite binomial sums, with the main results provided herein.

Theorem 1 extends identity (6) by incorporating a parameter t, which can be any complex number. This extension allows for the derivation of numerous other intriguing identities based on different choices of the parameter t.

Theorem 1. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$ and $t\in \mathbb{C}$. Then, the following identity holds:

$$\begin{array}{r}{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={(1+t)}^n\left\{H_m-H_n+{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{1}{k{\left(1+\frac{1}{t}\right)}^k}}\right\}\\ -{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){\displaystyle \frac{{(1+t)}^k}{n-k}}.\end{array}$$

(7)

The particular case when $m=n$ gives the following identity:

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=\sum _{k=1}^n{\displaystyle \frac{{(1+t)}^n}{k{\left(1+\frac{1}{t}\right)}^k}}-{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){(1+t)}^k.$$

Theorem 2 presents a novel finite summation identity involving the binomial coefficient, the harmonic number, and a parameter t, which can be any complex number. This identity enables the derivation of numerous other intriguing identities through various choices of the parameter t.

Theorem 2. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$ and $t\in \mathbb{C}$. Then,

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}& {\displaystyle \frac{{(-1)}^k{H}_{k-1}{t}^{k+n}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \sum _{k=0}^{m-1}}{\displaystyle \frac{{(-1)}^{k+1}{H}_k{t}^{k+1+n}}{k\left(\genfrac{}{}{0pt}{}{n+k+1}{k+1}\right)}}\\ =& {\displaystyle \frac{1}{2}}\left\{H_m^2-H_n^2-H_m^{\left(2\right)}-H_n^{\left(2\right)}\right\}{(1+t)}^n\\ & +{(1+t)}^n{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{H_n-H_{k-1}}{k{\left(1+\frac{1}{t}\right)}^k}}\\ & +{\displaystyle \frac{{(-1)}^{n+1}{H}_m}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){(1+t)}^k\\ & +{\displaystyle \frac{{(-1)}^{n+1}}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{{(n-k)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){(1+t)}^k.\end{array}$$

(8)

The particular case $m=n$ gives

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{(-1)}^k{H}_{k-1}{t}^{k+n}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=& -H_n^{\left(2\right)}{(1+t)}^n+{(1+t)}^n{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{H_n-H_{k-1}}{k{\left(1+\frac{1}{t}\right)}^k}}\\ & +{\displaystyle \frac{{(-1)}^{n+1}{H}_n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){(1+t)}^k\\ & +{\displaystyle \frac{{(-1)}^{n+1}}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}}{\displaystyle \frac{{(-1)}^k}{{(n-k)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){(1+t)}^k.\end{array}$$

In addition to providing alternative proofs for Equations (5) and (6), we also derive formulas for the following similar identities as particular cases of our main results:

$$\begin{array}{c}\hfill \sum _{k=1}^m{\displaystyle \frac{{ϵ}^k}{k^p\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}\mathrm{and}\sum _{k=1}^m{\displaystyle \frac{{ϵ}^k{H}_k}{k^p\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}},\end{array}$$

where $m,n\in {\mathbb{Z}}_{⩾0}$, $ϵ\in \{-1,1\}$, and $p=0,1$.

1.2. Required Functions, Notations, and Lemmas

To support our objectives, we revisit the definitions and properties of certain special functions and numbers. The well-known gamma function, $\Gamma \left(z\right)$, is given by

$$\Gamma \left(z\right)={\int }_0^{\infty }{u}^{z-1}{e}^{-u}du(\Re \left(z\right)>0),$$

where $\Re \left(z\right)$ denotes the real part of a complex number z. The gamma function $\Gamma \left(z\right)$ can be analytically continued to the entire complex plane, except at the non-positive integers, where it has simple poles (refer to [3] (Section 1.1)). The psi function, also known as the digamma function, is denoted by $\psi \left(z\right)$, and is defined as the logarithmic derivative of the gamma function:

$$\psi \left(z\right)={\displaystyle \frac{{\Gamma }^{\prime }\left(z\right)}{\Gamma \left(z\right)}}\left(z\in \mathbb{C}\setminus {\mathbb{Z}}_{⩽0}\right).$$

The polygamma function ${\psi }^{\left(q\right)}\left(z\right)$ is defined by

$$\begin{array}{ll}{\psi }^{\left(q\right)}\left(z\right)& :={\displaystyle \frac{d^q}{d{z}^q}}\left\{\psi \left(z\right)\right\}={\left(-1\right)}^{q+1}q!{\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{1}{{\left(r+z\right)}^{q+1}}}\\ & ={\left(-1\right)}^{q+1}q!\zeta (q+1,z)\left(q\in \mathbb{N},z\in \mathbb{C}\setminus {\mathbb{Z}}_{⩽0}\right),\end{array}$$

(9)

where ${\psi }^{\left(0\right)}\left(z\right)=\psi \left(z\right)$, and $\zeta (s,z)$ is the generalized (or Hurwitz) zeta function, defined by

$$\zeta (s,z):=\sum _{j=0}^{\infty }{\displaystyle \frac{1}{{(j+z)}^s}}\left(\Re \left(s\right)>1,z\in \mathbb{C}\setminus {\mathbb{Z}}_{⩽0}\right),$$

and $\zeta (s,1)=:\zeta \left(s\right)$ is the Riemann zeta function. Here and throughout, ${\mathbb{Z}}_{⩽\ell }$ denotes the set of integers less than or equal to some integer ℓ. The polygamma functions satisfy the following functional equation (refer to [4] (Equation (1.13))):

$$\begin{gathered} {\psi }^{\left(q\right)}(z+\ell )-{\psi }^{\left(q\right)}\left(z\right)={(-1)}^{q}q!{\displaystyle \sum _{j=1}^{\ell }}{\displaystyle \frac{1}{{(z+j-1)}^{q+1}}} \\ \hspace{1em}\hspace{1em}\left(\ell ,q\in {\mathbb{Z}}_{⩾0}\right). \end{gathered}$$

(10)

The harmonic numbers of order s, signified as $H_n^{\left(s\right)}$, are defined by

$$H_n^{\left(s\right)}=\sum _{k=1}^n{\displaystyle \frac{1}{k^s}}(n\in \mathbb{N};s\in \mathbb{C}).$$

Here, and in subsequent discussions, $\mathbb{C}$ represents the set of complex numbers. Also, $H_0^{\left(s\right)}:=0$. When $s=1$, we obtain the classical harmonic numbers $H_n^{\left(1\right)}:=H_n$ in (3). The following relations are recalled: for $n\in {\mathbb{Z}}_{⩾0}$,

$$\psi (n+1)=\psi \left(1\right)+H_n=-\gamma +H_n,$$

(11)

where $\gamma$ is Euler–Mascheroni constant (see, for instance, [3] (Section 1.2)), and

$$H_n^{(m+1)}=\zeta (m+1)+{\displaystyle \frac{{(-1)}^m}{m!}}{\psi }^{\left(m\right)}(n+1),$$

(12)

where $m\in \mathbb{N}$ and $n\in {\mathbb{Z}}_{⩾0}$ (see, for example, [5] (Equation (1.25))).

A generalized binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right)$ can be defined, in terms of the gamma function, by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{s}{t}}\right)={\displaystyle \frac{\Gamma (s+1)}{\Gamma (t+1)\Gamma (s-t+1)}}(s\in \mathbb{C}\setminus {\mathbb{Z}}_{⩽-1},t\in \mathbb{C}).$$

(13)

It is clear from (13) that, for $n,k\in {\mathbb{Z}}_{⩾0}$,

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)=\left\{\begin{array}{l}{\displaystyle \frac{n!}{k!(n-k)!}}\mathrm{if}n⩾k,\\ 0\mathrm{if}k>n.\end{array}\right.$$

(14)

We now present a series of lemmas, which will be used in the next section to prove the main results.

Lemma 1. 

Let $k\in {\mathbb{Z}}_{⩾0}$. Also, let $m,n\in {\mathbb{Z}}_{⩾0}$ with $m>n$. Then, the following identities hold:

$$\underset{x\to 0}{lim}x\Gamma (x-k)={\displaystyle \frac{{(-1)}^k}{k!}};$$

(15)

$$\begin{array}{c}\hfill \underset{x\to -k}{lim}{\displaystyle \frac{\psi \left(x\right)}{\Gamma \left(x\right)}}={(-1)}^{k-1}k!;\end{array}$$

(16)

$$\underset{x\to -k}{lim}{\displaystyle \frac{{\psi }^2\left(x\right)-{\psi }^{\prime }\left(x\right)}{\Gamma \left(x\right)}}=2{(-1)}^{k-1}k!\psi (k+1);$$

(17)

$$\begin{array}{ll}\underset{x\to n}{lim}{\displaystyle \frac{d}{dx}}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)& =\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)(\psi (x+1)-\psi (x-m+1))\\ & ={\displaystyle \frac{{(-1)}^{m-n-1}}{(m-n)\left(\genfrac{}{}{0pt}{}{m}{n}\right)}};\end{array}$$

(18)

$$\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\psi (x-m+1)={\displaystyle \frac{{(-1)}^{m-n}}{(m-n)\left(\genfrac{}{}{0pt}{}{m}{n}\right)}};$$

(19)

$$\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\displaystyle \frac{x-m}{x-n}}={\displaystyle \frac{{(-1)}^{m-n}}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}};$$

(20)

$$\begin{gathered} \underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\left(\psi (x-m+1){\displaystyle \frac{x-m}{x-n}}-{\displaystyle \frac{m-n}{{(x-n)}^2}}\right) \\ \hspace{1em}\hspace{1em}={\displaystyle \frac{{(-1)}^{m-n}\psi (m-n+1)}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}. \end{gathered}$$

(21)

Proof. 

The proofs of (15), (16), and (17) can be found in [4] (Lemma 1.2) (see also [6]).

To prove (18), let ${\mathcal{L}}_1$ represent the left-hand side of (18). Using (14), we have

$$\begin{array}{ll}{\mathcal{L}}_1& =\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)(\psi (x+1)-\psi (x-m+1))\\ & =-\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\psi (x-m+1)\\ & =-{\displaystyle \frac{n!}{m!}}\underset{x\to n}{lim}{\displaystyle \frac{\psi (x-m+1)}{\Gamma (x-m+1)}}\\ & =-{\displaystyle \frac{n!}{m!}}\underset{y\to -(m-n-1)}{lim}{\displaystyle \frac{\psi \left(y\right)}{\Gamma \left(y\right)}},\end{array}$$

which, upon employing (16), leads to

$${\mathcal{L}}_1={\displaystyle \frac{{(-1)}^{m-n-1}}{(m-n)\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}.$$

Formula (19) is proved in the course of the proof of (18).

To prove (20), let ${\mathcal{L}}_2$ represent the left-hand side of (20). Using (13), we find

$${\mathcal{L}}_2=-{\displaystyle \frac{n!(m-n)}{m!}}\underset{x\to n}{lim}{\displaystyle \frac{1}{(x-n)\Gamma (x-m+1)}}.$$

Setting $x-n=y$ in the last limit and using (15), we obtain

$$\begin{array}{ll}{\mathcal{L}}_2& =-{\displaystyle \frac{n!(m-n)}{m!}}\underset{y\to 0}{lim}{\displaystyle \frac{1}{y\Gamma (y-(m-n-1\left)\right)}}\\ & ={\displaystyle \frac{{(-1)}^{m-n}}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}.\end{array}$$

To prove (21), let ${\mathcal{L}}_3$ represent the left-hand side of (21). By using (13), we find

$$\begin{array}{ll}{\mathcal{L}}_3=& {\displaystyle \frac{n!}{m!}}\underset{I_1}{\underbrace{\underset{x\to n}{lim}{\displaystyle \frac{1}{(x-n)\Gamma (x-m+1)}}}}\\ & \times \underset{I_2}{\underbrace{\underset{x\to n}{lim}{\displaystyle \frac{\psi (x-m+1)(x-n)(x-m)-(m-n)}{x-n}}}}.\end{array}$$

(22)

Applying (15), we have

$$I_1={(-1)}^{m-n-1}(m-n-1)!.$$

(23)

Letting $x-n=u$, we obtain

$$\begin{array}{cc}\hfill & I_2=\underset{u\to 0}{lim}{\displaystyle \frac{\psi (u-\eta )u(u-\eta -1)-\eta -1}{u}},\hfill \end{array}$$

where $\eta :=m-n-1$. From (10), we derive the following specific identities: for $k\in {\mathbb{Z}}_{⩾0}$,

$$\psi (u-k)=\psi (u+1)-{\displaystyle \frac{1}{u}}-\sum _{j=1}^k{\displaystyle \frac{1}{u-j}},$$

(24)

and

$${\psi }^{\prime }(u-k)={\psi }^{\prime }(u+1)+{\displaystyle \frac{1}{u^2}}+\sum _{j=1}^k{\displaystyle \frac{1}{{(u-j)}^2}}.$$

(25)

Using (24), we obtain

$$\begin{array}{ll}{I}_2& =\underset{u\to 0}{lim}{\displaystyle \frac{1}{u}}\left\{u(u-\eta -1)\left[\psi (u+1)-{\displaystyle \frac{1}{u}}-{\displaystyle \sum _{j=1}^{\eta }}{\displaystyle \frac{1}{u-j}}\right]-{\displaystyle \frac{\eta +1}{1}}\right\}\\ & =\underset{u\to 0}{lim}(u-\eta -1)\psi (u+1)\\ & \hspace{1em}-\underset{u\to 0}{lim}\left[{\displaystyle \frac{u-\eta -1}{u}}+{\displaystyle \frac{\eta +1}{u}}+{\displaystyle \sum _{j=1}^{\eta }}{\displaystyle \frac{u-\eta -1}{u-j}}\right]\\ & =-(\eta +1)\psi \left(1\right)-(\eta +1)H_{\eta }-1.\end{array}$$

Using (11) and the following known (easily derivable) identity,

$$\psi (z+1)=\psi \left(z\right)+{\displaystyle \frac{1}{z}},$$

we obtain

$$\begin{array}{ll}{I}_2& =-(\eta +1)\psi \left(1\right)-(\eta +1)H_{\eta }-1=-(\eta +1)\psi (\eta +2)\\ & =-(m-n)\psi (m-n+1).\end{array}$$

(26)

Finally, substituting the results in (23) and (26) into (22), we prove the identity (21). □

Lemma 2. 

Let $x\in \mathbb{R}\setminus {\mathbb{Z}}_{⩽-1}$ and $t\in \mathbb{C}$. Also let $m,n\in {\mathbb{Z}}_{⩾0}$ with $m>n$. Then, the following identities hold:

$$\begin{array}{ll}\underset{x\to n}{lim}& {\displaystyle \sum _{k=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k\\ & ={(1+t)}^n{H}_n-{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)H_k{t}^{n-k}-{\displaystyle \sum _{k=1}^{m-n}}{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}},\end{array}$$

(27)

where

$$\begin{array}{c}\hfill {\varphi }_k\left(x\right):=\psi (x+1)-\psi (x-k+1);\end{array}$$

(28)

$$\begin{array}{ll}\underset{x\to n}{lim}& {\displaystyle \sum _{k=n+1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\left[{\psi }^2(x-k+1)-{\psi }^{\prime }(x-k+1)\right]t^k\\ & =2{\displaystyle \sum _{k=1}^{m-n}}{\displaystyle \frac{{(-1)}^k\psi \left(k\right)t^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}.\end{array}$$

(29)

Here and elsewhere, let $\mathbb{R}$ represent the set of real numbers.

Proof. 

Split the sum ${\displaystyle \sum _{k=0}^m}{\varphi }_k\left(x\right)t^k$ into two parts as follows:

$$\begin{array}{cc}\hfill & \sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\varphi }_k\left(x\right)t^k=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k+\sum _{k=n+1}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k.\hfill \end{array}$$

(30)

Taking the limit $x\to n$ on the first sum of the right-hand side of (30), with the aid of (11) and $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n-k}}\right)$, we have

$$\begin{array}{cc}\hfill \underset{x\to n}{lim}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k& =\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left[\psi (n+1)-\psi (n-k+1)\right]t^k\hfill \\ \hfill & ={(1+t)}^n{H}_n-\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)H_k{t}^{n-k}.\hfill \end{array}$$

(31)

For the second sum of the right-hand side of (30), by virtue of (14), we have

$$\begin{array}{c}\hfill \underset{x\to n}{lim}\sum _{k=n+1}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\psi (x+1)t^k=0.\end{array}$$

Therefore, it follows from (13) that

$$\begin{array}{cc}\hfill \underset{x\to n}{lim}\sum _{k=n+1}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k& =-\underset{x\to n}{lim}\sum _{k=n+1}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\psi (x-k+1)t^k\hfill \\ \hfill & =-\underset{x\to n}{lim}\sum _{k=n+1}^m{\displaystyle \frac{n!}{k!}}{\displaystyle \frac{\psi (x-k+1)}{\Gamma (x-k+1)}}{t}^k.\hfill \end{array}$$

Setting $k-n-1=k^{\prime }$ and then deleting the prime on k, we obtain

$$\begin{array}{cc}\hfill & \underset{x\to n}{lim}\sum _{k=n+1}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k=-\sum _{k=0}^{m-n-1}{\displaystyle \frac{n!}{(n+k+1)!}}\underset{x\to n}{lim}{\displaystyle \frac{\psi (x-n-k)}{\Gamma (x-n-k)}}{t}^{n+k+1}\hfill \\ \hfill & =-\sum _{k=0}^{m-n-1}{\displaystyle \frac{n!}{(n+k+1)!}}\underset{y\to -k}{lim}{\displaystyle \frac{\psi \left(y\right)}{\Gamma \left(y\right)}}{t}^{n+k+1}(y:=x-n-k).\hfill \end{array}$$

Utilizing (16) to simplify the last sum gives

$$\begin{array}{cc}\hfill \underset{x\to n}{lim}\sum _{k=n+1}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k& =\sum _{k=0}^{m-n-1}{\displaystyle \frac{{(-1)}^k}{(n+k+1)\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}{t}^{n+k+1}\hfill \\ \hfill & =-\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}{t}^{n+k}.\hfill \end{array}$$

(32)

Taking the limit on both sides of (30) as x tends to n, and using (31) and (32), we obtain the identity (27).

The detailed steps are omitted. □

Lemma 3. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$ with $n<m$. Then, the following identity holds:

$$\begin{array}{ll}\underset{x\to n}{lim}& \{{\psi }^2(x-m+1)-{\psi }^{\prime }(x-m+1)\\ & -{\displaystyle \frac{2(m-n)\psi (x-m+1)}{(x-n)(x-m)}}-{\displaystyle \frac{2(m-n)}{{(x-n)}^2(x-m)}}\}\\ =& {\psi }^2(m-n)-\zeta \left(2\right)-H_{m-n-1}^{\left(2\right)}+{\displaystyle \frac{2\psi (m-n)}{m-n}}.\end{array}$$

(33)

Proof. 

Let ${\mathcal{L}}_4$ be the left-hand side of (33). Setting $x-n=y$ and $L=m-n-1$, we obtain

$${\mathcal{L}}_4=\underset{y\to 0}{lim}\left\{{\psi }^2(y-L)-{\psi }^{\prime }(y-L)-{\displaystyle \frac{2(L+1)\psi (y-L)}{y(y-L-1)}}-{\displaystyle \frac{2(L+1)}{y^2(y-L-1)}}\right\}.$$

Using (24) and (25), we obtain

$$\begin{array}{ll}{\mathcal{L}}_4& =\underset{y\to 0}{lim}[{\psi }^2(y+1)-{\psi }^{\prime }(y+1)+{\left({\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{y-j}}\right)}^2-2\psi (y+1){\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{y-j}}\\ & \hspace{1em}\hspace{1em}\hspace{1em}-{\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{{(y-j)}^2}}+{\displaystyle \frac{2}{y-L-1}}\left({\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{y-j}}-\psi (y+1)\right)]\\ & =\underset{y\to 0}{lim}[{\left({\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{y-j}}-\psi (y+1)\right)}^2-{\psi }^{\prime }(y+1)\\ & \hspace{1em}\hspace{1em}\hspace{1em}-{\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{{(y-j)}^2}}+{\displaystyle \frac{2}{y-L-1}}\left({\displaystyle \sum _{j=1}^L}{\displaystyle \frac{1}{y-j}}-\psi (y+1)\right)].\end{array}$$

Using (11), we have

$${\mathcal{L}}_4={\psi }^2(L+1)-{\psi }^{\prime }\left(1\right)-H_L^{\left(2\right)}+{\displaystyle \frac{2}{L+1}}\psi (L+1),$$

which, upon setting $L=m-n-1$ with the aid of (9), leads to the right-hand side of (33). □

2. Proofs of Main Results

This section provides proofs of our main results, Theorems 1 and 2.

Proof of Theorem 1. 

We begin by recalling the following known identity (see [7] (Equation (1), p. 186)): for $m\in {\mathbb{Z}}_{⩾0}$, $x\in \mathbb{C}$, and $t\in \mathbb{C}$,

$$\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)t^k={(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\sum _{k=0}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{x-m}{x-k}}{(1+t)}^k.$$

Differentiating both sides of this equality with respect to the variable x gives

$$\begin{array}{l}{\displaystyle \sum _{k=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k\\ \hspace{1em}={(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\varphi }_m\left(x\right){\displaystyle \sum _{k=0}^m}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{x-m}{x-k}}{(1+t)}^k\\ \hspace{1em}\hspace{1em}\hspace{1em}+{(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\displaystyle \sum _{k=0}^m}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(x-k)}^2}}{(1+t)}^k,\end{array}$$

(34)

where ${\varphi }_k\left(x\right)$ is given as in (28). Let $n\in {\mathbb{Z}}_{⩾0}$ with $n<m$. Let us separate the nth term from the two sums on the right-hand side of (34), rearrange the extracted terms and the remaining sums, and then take the limit of both sides of the resulting expressions as $x\to n$. Then, we obtain the following identity:

$$\begin{array}{ll}\underset{x\to n}{lim}& {\displaystyle \sum _{k=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right){\varphi }_k\left(x\right)t^k={(-1)}^m\underset{x\to n}{lim}{f}_4\left(x\right)\\ & +{(-1)}^{m+n}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n}}\right){(1+t)}^n\underset{x\to n}{lim}\left\{f_1\left(x\right)-f_2\left(x\right)\right\}\\ & +{(-1)}^m\underset{x\to n}{lim}{f}_3\left(x\right){\displaystyle \sum _{k=0}^{n-1}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{n-m}{n-k}}{(1+t)}^k\\ & +{(-1)}^m\underset{x\to n}{lim}{f}_3\left(x\right){\displaystyle \sum _{k=n+1}^m}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{n-m}{n-k}}{(1+t)}^k,\end{array}$$

(35)

where

$$f_1\left(x\right)=\psi (x+1)\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\displaystyle \frac{x-m}{x-n}},$$

$$f_2\left(x\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\left\{\psi (x-m+1){\displaystyle \frac{x-m}{x-n}}-{\displaystyle \frac{m-n}{{(x-n)}^2}}\right\},$$

$$f_3\left(x\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\varphi }_m\left(x\right),$$

and

$$\begin{array}{cc}\hfill f_4\left(x\right)=& \left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\{\sum _{k=0}^{n-1}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(x-k)}^2}}{(1+t)}^k\hfill \\ \hfill & +\sum _{k=n+1}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(x-k)}^2}}{(1+t)}^k\}.\hfill \end{array}$$

By making respective use of (14), (18), (20), and (21), we obtain the following four identities:

$$\underset{x\to n}{lim}{f}_1\left(x\right)=\psi (n+1){\displaystyle \frac{{(-1)}^{m-n}}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}};$$

(36)

$$\underset{x\to n}{lim}{f}_2\left(x\right)={\displaystyle \frac{{(-1)}^{m-n}\psi (m-n+1)}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}};$$

(37)

$$\underset{x\to n}{lim}{f}_3\left(x\right)={\displaystyle \frac{{(-1)}^{m-n-1}}{(m-n)\left(\genfrac{}{}{0pt}{}{m}{n}\right)}};$$

(38)

$$\underset{x\to n}{lim}{f}_4\left(x\right)=0.$$

(39)

Substituting the results in (36), (37), (38), and (39) into (35), with the aid of (11) and (27), we obtain

$$\begin{array}{l} {\displaystyle \sum _{k=1}^{m-n}}{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={(1+t)}^n{H}_{m-n}-{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)H_k{t}^{n-k}\\ -{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\left[{\displaystyle \sum _{k=0}^{n-1}}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(1+t)}^k+{\displaystyle \sum _{k=n+1}^m}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(1+t)}^k\right].\end{array}$$

(40)

Replacing m with $m+n$ in Equation (40) and using the known formula (see [8] (Equation (1)); see also [9]):

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)H_k{t}^k={(1+t)}^n\left\{H_n-\sum _{k=1}^n{\displaystyle \frac{1}{k{(1+t)}^k}}\right\},$$

(41)

we obtain the desired identity (7). □

Proof of Theorem 2. 

Differentiating both sides of (34) with respect to x, we find

$$\begin{array}{cc}\hfill & \sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\left\{{\left({\varphi }_k\left(x\right)\right)}^2+{\varphi }_k^{\prime }\left(x\right)\right\}{t}^k\hfill \\ \hfill & ={(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\left[{\left({\varphi }_m\left(x\right)\right)}^2+{\varphi }_m^{\prime }\left(x\right)\right]\sum _{k=0}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{x-m}{x-k}}{(1+t)}^k\hfill \\ \hfill & +2{(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\varphi }_m\left(x\right)\sum _{k=0}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(x-k)}^2}}{(1+t)}^k\hfill \\ \hfill & -2{(-1)}^m\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\sum _{k=0}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(x-k)}^3}}{(1+t)}^k,\hfill \end{array}$$

(42)

where ${\varphi }_k\left(x\right)$ is given as in (28).

Let $m,n\in {\mathbb{Z}}_{⩾0}$ with $n<m$. Breaking up the sum on the left-hand side of (42) into two summations, extracting the nth terms of the three sums on the right-hand side of the equation separately, and finally taking the limit of both sides of the resultant equality as x tends to n, we obtain

$$\begin{array}{cc}\hfill & \underset{T_1}{\underbrace{\underset{x\to n}{lim}{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\left\{{\left({\varphi }_k\left(x\right)\right)}^2+{\varphi }_k^{\prime }\left(x\right)\right\}{t}^k}}+\underset{T_2}{\underbrace{\underset{x\to n}{lim}{\displaystyle \sum _{k=n+1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\left\{{\left({\varphi }_k\left(x\right)\right)}^2+{\varphi }_k^{\prime }\left(x\right)\right\}}}{t}^k\hfill \\ \hfill & ={(-1)}^{m+n}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n}}\right){(1+t)}^n\underset{Q_1}{\underbrace{\underset{x\to n}{lim}\left\{g_1\left(x\right)-2g_2\left(x\right)+g_3\left(x\right)\right\}}}\hfill \\ \hfill & +{(-1)}^m\underset{Q_2}{\underbrace{\underset{x\to n}{lim}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\left\{{\left({\varphi }_m\left(x\right)\right)}^2+{\varphi }_m^{\prime }\left(x\right)\right\}\right]}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{n-m}{n-k}}{(1+t)}^k\hfill \\ \hfill & +2{(-1)}^m\underset{Q_3}{\underbrace{\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\varphi }_m\left(x\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(n-k)}^2}}{(1+t)}^k\hfill \\ \hfill & -2{(-1)}^m\underset{0}{\underbrace{\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{m-k}{{(n-k)}^3}}{(1+t)}^k,\hfill \end{array}$$

(43)

where

$$g_1\left(x\right):=\left[{\psi }^2(x+1)+{\psi }^{\prime }(x+1)\right]\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\displaystyle \frac{x-m}{x-n}},$$

$$\begin{array}{c}\hfill g_2\left(x\right):=\psi (x+1)\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right)\left[\psi (x-m+1){\displaystyle \frac{x-m}{x-n}}-{\displaystyle \frac{m-n}{{(x-n)}^2}}\right],\end{array}$$

$$\begin{array}{cc}\hfill g_3\left(x\right):& =\left({\displaystyle \genfrac{}{}{0pt}{}{x}{m}}\right){\displaystyle \frac{x-m}{x-n}}[{\psi }^2(x-m+1)-{\psi }^{\prime }(x-m+1)\hfill \\ \hfill & -{\displaystyle \frac{2(m-n)\psi (x-m+1)}{(x-n)(x-m)}}-{\displaystyle \frac{2(m-n)}{{(x-n)}^2(x-m)}}].\hfill \end{array}$$

Using (11) and (12) in the limiting result in $T_1$, with the aid of [10] (Equation (40)), we obtain

$$\begin{array}{cc}\hfill T_1& =\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left\{{\left(H_n-H_{n-k}\right)}^2+H_{n-k}^{\left(2\right)}-H_n^{\left(2\right)}\right\}{t}^k\hfill \\ \hfill & =2{(1+t)}^n\sum _{k=1}^n{\displaystyle \frac{H_{k-1}}{k{(1+\frac{1}{t})}^k}}.\hfill \end{array}$$

(44)

Employing the definition of ${\varphi }_k\left(x\right)$ in (28), we have

$$\begin{array}{cc}\hfill & T_2=\left[{\psi }^2(n+1)-{\psi }^{\prime }(n+1)\right]\underset{0}{\underbrace{{\displaystyle \sum _{k=n+1}^m}\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)}}{t}^k\hfill \\ \hfill & -2\psi (n+1)\underset{I_1}{\underbrace{{\displaystyle \sum _{k=n+1}^m}\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\psi (x-k+1)t^k}}\hfill \\ \hfill & +\underset{I_2}{\underbrace{{\displaystyle \sum _{k=n+1}^m}\underset{x\to n}{lim}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)\left\{{\psi }^2(x-k+1)-{\psi }^{\prime }(x-k+1)\right\}{t}^k}}.\hfill \end{array}$$

(45)

Letting $k-n-1=k^{\prime }$ in the sum $I_1$, and then deleting the prime on k, with the aid of (16), we obtain

$$\begin{array}{cc}\hfill I_1& =\sum _{k=0}^{m-n-1}{\displaystyle \frac{n!}{(n+k+1)!}}\underset{x\to n}{lim}{\displaystyle \frac{\psi (x-n-k)}{\Gamma (x-n-k)}}{t}^{n+k+1}\hfill \\ \hfill & =\sum _{k=0}^{m-n-1}{\displaystyle \frac{n!}{(n+k+1)!}}·\underset{u\to -k}{lim}{\displaystyle \frac{\psi \left(u\right)}{\Gamma \left(u\right)}}{t}^{n+k+1}(u=x-n-k)\hfill \\ \hfill & =\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}.\hfill \end{array}$$

(46)

Similarly, we derive

$$\begin{array}{cc}\hfill I_2=& \sum _{k=n+1}^m{\displaystyle \frac{n!}{k!}}\underset{x\to n}{lim}{\displaystyle \frac{{\psi }^2(x-k+1)-{\psi }^{\prime }(x-k+1)}{\Gamma (x-k+1)}}{t}^k\hfill \\ \hfill =& \sum _{k=0}^{m-n-1}{\displaystyle \frac{n!}{(n+k+1)!}}\underset{y\to -k}{lim}{\displaystyle \frac{{\psi }^2\left(y\right)-{\psi }^{\prime }\left(y\right)}{\Gamma \left(y\right)}}{t}^{n+k+1}.\hfill \end{array}$$

Using (17) in the limit of the last equality gives

$$\begin{array}{c}\hfill I_2=2\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k\psi \left(k\right)}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}{t}^{n+k}.\end{array}$$

(47)

Substituting the values of $I_1$ and $I_2$ in (46) and (47) into (45), we obtain

$$\begin{array}{c}\hfill T_2=2\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k{H}_{k-1}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}{t}^{n+k}-2H_n\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}.\end{array}$$

Applying (7) to the second sum in the above $T_2$, we have

$$\begin{array}{cc}\hfill T_2=& 2\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k{H}_{k-1}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}{t}^{n+k}+{\displaystyle \frac{2H_n{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^m{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(1+t)}^k\hfill \\ \hfill & -2H_n{(1+t)}^n\left[H_{m-n}-H_n+\sum _{k=1}^n{\displaystyle \frac{1}{k{(1+\frac{1}{t})}^k}}\right].\hfill \end{array}$$

(48)

By using (15), we have

$$\underset{x\to n}{lim}{g}_1\left(x\right)=\left[{\psi }^2(n+1)+{\psi }^{\prime }(n+1)\right]{\displaystyle \frac{{(-1)}^{m-n}}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}.$$

(49)

By using (21), we have

$$\underset{x\to n}{lim}{g}_2\left(x\right)=\psi (n+1){\displaystyle \frac{{(-1)}^{m-n}\psi (m-n+1)}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}.$$

(50)

By using (20) and (33), we have

$$\begin{array}{cc}\hfill \underset{x\to n}{lim}{g}_3\left(x\right)& ={\displaystyle \frac{{(-1)}^{m-n}}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\left[{\psi }^2(m-n)-\zeta \left(2\right)-H_{m-n-1}^{\left(2\right)}+{\displaystyle \frac{2\psi (m-n)}{m-n}}\right].\hfill \end{array}$$

(51)

Using (49), (50) and (51), we obtain

$$\begin{array}{cc}\hfill & Q_1={\displaystyle \frac{{(-1)}^{m-n}}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\left[{\left(H_n-H_{m-n}\right)}^2-H_n^{\left(2\right)}-H_{m-n}^{\left(2\right)}\right].\hfill \end{array}$$

(52)

We find that

$$\begin{array}{ll}{Q}_2={\displaystyle \frac{n!}{m!}}& \{\underset{0}{\underbrace{\underset{x\to n}{lim}{\displaystyle \frac{{\psi }^2(x+1)+{\psi }^{\prime }(x+1)}{\Gamma (x-m+1)}}}}-2\psi (n+1)\underset{x\to n}{lim}{\displaystyle \frac{\psi (x-m+1)}{\Gamma (x-m+1)}}\\ & +\underset{x\to n}{lim}{\displaystyle \frac{{\psi }^2(x-m+1)-{\psi }^{\prime }(x-m+1)}{\Gamma (x-m+1)}}\}.\end{array}$$

Setting $K=m-n-1$, we have

$$Q_2={\displaystyle \frac{n!}{m!}}\left\{-2\psi (n+1)\underset{y\to -K}{lim}{\displaystyle \frac{\psi \left(y\right)}{\Gamma \left(y\right)}}+\underset{y\to -K}{lim}{\displaystyle \frac{{\psi }^2\left(y\right)-{\psi }^{\prime }\left(y\right)}{\Gamma \left(y\right)}}\right\},$$

which, upon employing (16) and (17), yields the following value:

$$Q_2={\displaystyle \frac{2{(-1)}^{m-n-1}}{(m-n)\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\left[\psi (n+1)-\psi (m-n)\right].$$

(53)

By using (16), we find that

$$\begin{array}{cc}\hfill Q_3& ={\displaystyle \frac{n!}{m!}}\underset{0}{\underbrace{\underset{x\to n}{lim}{\displaystyle \frac{\psi (x+1)}{\Gamma (x-m+1)}}}}-{\displaystyle \frac{n!}{m!}}\underset{x\to n}{lim}{\displaystyle \frac{\psi (x-m+1)}{\Gamma (x-m+1)}}\hfill \\ \hfill & =-{\displaystyle \frac{n!}{m!}}\underset{y\to -(m-n-1)}{lim}{\displaystyle \frac{\psi \left(y\right)}{\Gamma \left(y\right)}}(y=x-m+1)\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{m-n-1}}{(m-n)\left(\genfrac{}{}{0pt}{}{m}{n}\right)}},\hfill \end{array}$$

(54)

where we have used (14) in the first line.

Substituting the values of $T_1,T_2,Q_1,Q_2$, and $Q_3$ obtained in (44), (48), (52), (53), and (54) into (43), respectively, we obtain

$$\begin{array}{cc}\hfill & 2{(1+t)}^n\sum _{k=1}^n{\displaystyle \frac{H_{k-1}}{k{\left(1+\frac{1}{t}\right)}^k}}+2\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k{H}_{k-1}{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}-2H_n\sum _{k=1}^{m-n}{\displaystyle \frac{{(-1)}^k{t}^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}\hfill \\ \hfill & =\left[{\left(H_n-H_{m-n}\right)}^2-H_n^{\left(2\right)}-H_{m-n}^{\left(2\right)}\right]{(1+t)}^n\hfill \\ \hfill & +{\displaystyle \frac{2{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\left[H_n-H_{m-n}\right]\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^m{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(1+t)}^k\hfill \\ \hfill & -{\displaystyle \frac{2{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^m{\displaystyle \frac{{(-1)}^k}{{(n-k)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(1+t)}^k.\hfill \end{array}$$

Finally, employing Theorem 1, and replacing m by $m+n$ in the resultant identity, we obtain the desired result (8). □

3. Particular Cases and Remarks

This section offers specific instances of the main results along with pertinent remarks.

Setting $t=-1$ in Equation (7) yields the identity presented in the following corollary.

Corollary 1. 

Let $m,n\in \mathbb{N}$. Then,

$$\sum _{k=1}^m{\displaystyle \frac{1}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{1}{n}}\left\{1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}\right\}.$$

The special case $m=n$ leads to the following result:

$$\sum _{k=1}^n{\displaystyle \frac{1}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{1}{n}}\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\right),$$

which recovers the result (6).

Setting $t=1$ in Equation (7) yields the identity presented in the following corollary.

Corollary 2. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$. Then,

$$\sum _{k=1}^m{\displaystyle \frac{{(-1)}^k}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=2^n\left(H_m-H_n+\sum _{k=1}^n{\displaystyle \frac{1}{k{2}^k}}\right)-{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){\displaystyle \frac{{(-2)}^k}{n-k}}.$$

In the particular case when $m=n$, this gives

$$\begin{array}{c}\hfill \sum _{k=1}^n{\displaystyle \frac{{(-1)}^k}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=\sum _{k=1}^n{\displaystyle \frac{2^{n-k}}{k}}-{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){\displaystyle \frac{{(-2)}^k}{n-k}}.\end{array}$$

Corollary 3. 

Let $m\in \mathbb{N}$ and $n\in {\mathbb{Z}}_{⩾2}$. Then,

$$\sum _{k=0}^m{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{n}{n-1}}\left\{1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{m+n}{n-1}\right)}}\right\}.$$

(55)

The particular case when $m=n$ gives the following identity:

$$\sum _{k=0}^n{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{n}{n-1}}\left\{1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{2n}{n-1}\right)}}\right\},$$

which recovers the result (5).

Proof. 

Differentiating both sides of (7) with respect to t, and setting $t=-1$ in the resulting identity, we obtain

$$\sum _{k=0}^{m-1}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{n}{n-1}}\left\{1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{m+n-1}{n-1}\right)}}\right\},$$

which, upon replacing m with $m+1$, leads to the desired identity (55). □

Corollary 4. 

Let $m\in {\mathbb{Z}}_{⩾0}$ and $n\in {\mathbb{Z}}_{⩾2}$. Then,

$$\begin{array}{ll}{\displaystyle \sum _{k=0}^m}{\displaystyle \frac{{(-1)}^k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=& -n{2}^{n-1}\left(H_{m+1}-H_{n-1}+{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{1}{k{2}^k}}\right)\\ & -{\displaystyle \frac{n{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m+n}{n-1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n-1\end{array}}^{m+n}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){\displaystyle \frac{{(-2)}^k}{n-1-k}}.\end{array}$$

(56)

The particular case $m=n$ yields

$$\begin{array}{ll}{\displaystyle \sum _{k=0}^n}{\displaystyle \frac{{(-1)}^k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=& -n{2}^{n-1}\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n+1}}+{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{1}{k{2}^k}}\right)\\ & -{\displaystyle \frac{n{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n-1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n-1\end{array}}^{2n}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){\displaystyle \frac{{(-2)}^k}{n-1-k}}.\end{array}$$

Proof. 

Differentiating both sides of (7) with respect to t, using the following identities:

$${\displaystyle \frac{n+1}{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right)\mathrm{and}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n-k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right),$$

and putting $t=1$ in the resulting identity, we have

$$\begin{array}{cc}\hfill \sum _{k=1}^m{\displaystyle \frac{{(-1)}^k}{\left(\genfrac{}{}{0pt}{}{n+k-1}{k-1}\right)}}=& n{2}^{n-1}{H}_m-n\sum _{k=1}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k}}\right)H_k\hfill \\ \hfill & -{\displaystyle \frac{n{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m+n-1}{n-1}\right)}}\sum _{\begin{array}{c}k=1\\ k\ne n\end{array}}^{m+n}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+n-1}{k-1}}\right){\displaystyle \frac{2^{k-1}}{n-k}}.\hfill \end{array}$$

Set $k-1=k^{\prime }$ in the sum on the left hand side and in the third sum on the right hand side, and then delete the prime on k, and use (41) with $u=1$. Finally, replacing m with $m+1$ in both sides of the last resulting identity, we obtain the desired identity (56). □

Setting $t=0$ in (7) gives the identity stated in the following corollary.

Corollary 5. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$. Then,

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right)={(-1)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{n}}\right)(H_m-H_n).\end{array}$$

(57)

The special case $m=n$ gives

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}{\displaystyle \frac{{(-1)}^k}{n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)=0.\end{array}$$

Remark 1. 

Putting $n=0$ in Equation (57) yields the classical identity due to Euler: For $m\in {\mathbb{Z}}_{⩾0}$,

$$\begin{array}{c}\hfill \sum _{k=1}^m{\displaystyle \frac{{(-1)}^{k-1}}{k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=H_m.\end{array}$$

(58)

In this context, Equation (57) offers an intriguing and novel generalization of Equation (58).

By substituting $t=-1$ into (8), we obtain the identity presented in the following corollary.

Corollary 6. 

Let $m\in {\mathbb{Z}}_{⩾0}$ and $n\in \mathbb{N}$. Then,

$$\begin{array}{c}\hfill \sum _{k=1}^m{\displaystyle \frac{H_{k-1}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{1}{n^2}}-{\displaystyle \frac{n{H}_m+1}{n^2\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}.\end{array}$$

The particular case $m=n$ gives

$$\begin{array}{c}\hfill {\displaystyle \sum _{k=1}^n}{\displaystyle \frac{H_{k-1}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{1}{n^2}}-{\displaystyle \frac{n{H}_n+1}{n^2\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}.\end{array}$$

(59)

Remark 2. 

Applying $H_{k-1}=H_k-{\displaystyle \frac{1}{k}}$ in (59) and using (1), we derive

$$\begin{array}{c}\hfill \sum _{k=1}^n{\displaystyle \frac{H_k}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=-{\displaystyle \frac{n{H}_n+1}{n^2\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}-H_{n-1}^{\left(2\right)}+3\sum _{k=1}^n{\displaystyle \frac{1}{k^2\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}}.\end{array}$$

By setting $t=1$ into (8), we derive the identity shown in the following corollary.

Corollary 7. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$. Then,

$$\begin{array}{cc}\hfill & \sum _{k=1}^m{\displaystyle \frac{{(-1)}^k{H}_{k-1}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=\left\{H_m^2-H_n^2-H_m^{\left(2\right)}-H_n^{\left(2\right)}\right\}{2}^{n-1}\hfill \\ \hfill & +2^n\sum _{k=1}^m{\displaystyle \frac{H_n-H_{k-1}}{k{2}^k}}-{\displaystyle \frac{{(-1)}^n{H}_m}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){\displaystyle \frac{{(-2)}^k}{n-k}}\hfill \\ \hfill & -{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){\displaystyle \frac{{(-2)}^k}{{(n-k)}^2}}.\hfill \end{array}$$

The particular case when $m=n$ gives

$$\begin{array}{cc}\hfill \sum _{k=1}^n{\displaystyle \frac{{(-1)}^k{H}_{k-1}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=& -H_n^{\left(2\right)}{2}^{n-1}+2^n\sum _{k=1}^m{\displaystyle \frac{H_n-H_{k-1}}{k{2}^k}}\hfill \\ \hfill & -{\displaystyle \frac{{(-1)}^n{H}_n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){\displaystyle \frac{{(-2)}^k}{n-k}}\hfill \\ \hfill & -{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right){\displaystyle \frac{{(-2)}^k}{{(n-k)}^2}}.\hfill \end{array}$$

Corollary 8. 

Let $n\in {\mathbb{Z}}_{⩾0}$ and $t\in \mathbb{C}$. Then,

$$\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{t^k}{k^2}}=\sum _{k=1}^n{\displaystyle \frac{(H_n-H_{k-1}){(1+t)}^k}{k}}-{\displaystyle \frac{H_n^2+H_n^{\left(2\right)}}{2}}.$$

(60)

Proof. 

Setting $m=0$ in (8), we obtain

$$\sum _{k=1}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(1+t)}^{-k}}{k^2}}=\sum _{k=1}^n{\displaystyle \frac{H_n-H_{k-1}}{k{\left(1+\frac{1}{t}\right)}^k}}-{\displaystyle \frac{{\left(H_n\right)}^2+H_n^{\left(2\right)}}{2}}.$$

Then, replacing t with $-{\displaystyle \frac{1+t}{t}}$ in the both sides of the above identity, we obtain the desired identity (60). □

Corollary 9. 

Let $n\in \mathbb{N}$ and $t\in \mathbb{C}$. Then,

$$\begin{array}{c}\hfill \sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{t^k}{k^3}}=\sum _{k=1}^n{\displaystyle \frac{H_n-H_{k-1}}{k}}\sum _{j=1}^k{\displaystyle \frac{{(1+t)}^j-1}{j}}.\end{array}$$

Proof. 

Integrating both sides of Equation (60) over $[0,t]$, after dividing both sides by t, the proof follows. We omit the details. □

Setting $t=0$ in (8) and using (57), we obtain the identity offered in the following corollary.

Corollary 10. 

Let $m,n\in {\mathbb{Z}}_{⩾0}$. Then,

$$\begin{array}{ll}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}& {\displaystyle \frac{{(-1)}^k}{{(n-k)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right)\\ & ={\displaystyle \frac{{(-1)}^{n+1}\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}{2}}\left\{{\left(H_m-H_n\right)}^2+H_n^{\left(2\right)}+H_m^{\left(2\right)}\right\}.\end{array}$$

The particular case when $m=n$ provides

$$\begin{array}{c}\hfill \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{2n}{\displaystyle \frac{{(-1)}^k}{{(n-k)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)={(-1)}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)H_n^{\left(2\right)}.\end{array}$$

Corollary 11. 

Let $m\in \mathbb{N}$ and $n\in {\mathbb{Z}}_{⩾2}$. Then,

$$\begin{array}{c}\hfill \sum _{k=1}^m{\displaystyle \frac{H_k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{n}{n-1}}\left\{{\displaystyle \frac{1}{n-1}}-{\displaystyle \frac{H_{m+1}}{\left(\genfrac{}{}{0pt}{}{n+m}{n-1}\right)}}-{\displaystyle \frac{1}{(n-1)\left(\genfrac{}{}{0pt}{}{n+m}{n-1}\right)}}\right\}.\end{array}$$

(61)

The special case when $m=n$ gives

$$\sum _{k=1}^n{\displaystyle \frac{H_k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{n}{n-1}}\left({\displaystyle \frac{1}{n-1}}-{\displaystyle \frac{H_{n+1}}{\left(\genfrac{}{}{0pt}{}{2n}{n-1}\right)}}-{\displaystyle \frac{1}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n-1}\right)}}\right).$$

Proof. 

Differentiating both sides of (8) with respect to t and setting $t=-1$ in the resulting identity, we obtain

$$\sum _{k=1}^{m-1}{\displaystyle \frac{H_k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={\displaystyle \frac{n}{{(n-1)}^2}}-{\displaystyle \frac{n{H}_m}{(n-1)\left(\genfrac{}{}{0pt}{}{m+n-1}{n-1}\right)}}-{\displaystyle \frac{n}{{(n-1)}^2\left(\genfrac{}{}{0pt}{}{m+n-1}{n-1}\right)}}.$$

Replacing m with $m+1$ in the just preceding identity, we obtain the desired identity (61). □

Corollary 12. 

Let $m,n\in \mathbb{N}$. Then,

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}& {\displaystyle \frac{{(-1)}^{k-1}{H}_k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=n{2}^{n-2}\left\{H_{m+1}^2-H_n^2-H_{m+1}^{\left(2\right)}-H_n^{\left(2\right)}\right\}\\ & +2^{n-1}{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{(n+k)(H_n-H_{k-1})}{k{2}^k}}\\ & +{\displaystyle \frac{{(-1)}^{n}n{H}_{m+1}}{\left(\genfrac{}{}{0pt}{}{m+n}{n-1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k+1\ne n\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{n-k-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right)2^k\\ & +{\displaystyle \frac{{(-1)}^{n}n}{\left(\genfrac{}{}{0pt}{}{m+n}{n-1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k+1\ne n\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{{(n-k-1)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right)2^k.\end{array}$$

(62)

The special case when $m=n$ offers

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^n}& {\displaystyle \frac{{(-1)}^{k-1}{H}_k}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=n{2}^{n-1}\left({\displaystyle \frac{H_n}{n+1}}-H_n^{\left(2\right)}\right)\\ & +2^{n-1}{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{(n+k)(H_n-H_{k-1})}{k{2}^k}}\\ & +{\displaystyle \frac{{(-1)}^{n}n{H}_{n+1}}{\left(\genfrac{}{}{0pt}{}{2n}{n-1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k+1\ne n\end{array}}^{2n}}{\displaystyle \frac{{(-1)}^k}{n-k-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)2^k\\ & +{\displaystyle \frac{{(-1)}^{n}n}{\left(\genfrac{}{}{0pt}{}{2n}{n-1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k+1\ne n\end{array}}^{2n}}{\displaystyle \frac{{(-1)}^k}{{(n-k-1)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)2^k.\end{array}$$

Proof. 

By differentiating both sides of Equation (8) with respect to t, then setting $t=1$ in the resulting expression, and finally replacing m with $m+1$ in the last resulting identity, we derive the desired identity (62). □

Corollary 13. 

Let $n\in {\mathbb{Z}}_{⩾0}$ and $m\in {\mathbb{Z}}_{⩾1}$. Then,

$$\begin{array}{r}{\displaystyle \sum _{k=2}^m}{\displaystyle \frac{1}{k(k-1)\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}={(-1)}^n{\displaystyle \frac{H_{m-1}-H_n}{n+1}}+{\displaystyle \frac{{(-1)}^n}{n+1}}{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{(-1)}^k}{k\left(\genfrac{}{}{0pt}{}{n}{k}\right)}}\\ +{\displaystyle \frac{1}{(n+1)\left(\genfrac{}{}{0pt}{}{m+n}{1+n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=1\\ k\ne n+1\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{n-k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right).\end{array}$$

(63)

The particular case when $m=1$ offers

$$\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{n-k}}{n-k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k}}\right)=H_n-\sum _{k=1}^n{\displaystyle \frac{{(-1)}^k}{k\left(\genfrac{}{}{0pt}{}{n}{k}\right)}}.$$

(64)

Proof. 

Replacing t with $-t$ in (7), we have

$$\begin{array}{ll}{(-1)}^n{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{t^{n+k}}{k\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=& \left(H_m-H_n\right){(1-t)}^n+{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{(-1)}^k{t}^k{(1-t)}^{n-k}}{k}}\\ & -{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right){\displaystyle \frac{{(1-t)}^k}{n-k}}.\end{array}$$

Integrating both sides of this equality from $t=0$ to $t=1$ and using the well-known gamma-beta functional equation, we obtain

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{1}{k(n+k+1)\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}& ={(-1)}^n{\displaystyle \frac{H_m-H_n}{n+1}}+{\displaystyle \frac{{(-1)}^n}{n+1}}{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{(-1)}^k}{k\left(\genfrac{}{}{0pt}{}{n}{k}\right)}}\\ & -{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{(k+1)n-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right).\end{array}$$

(65)

Using ${\displaystyle \frac{n+1}{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right)$, we obtain

$$\sum _{k=1}^m{\displaystyle \frac{1}{k(n+k+1)\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}=\sum _{k=1}^m{\displaystyle \frac{1}{k(k+1)\left(\genfrac{}{}{0pt}{}{n+k+1}{k+1}\right)}}=\sum _{k=2}^{m+1}{\displaystyle \frac{1}{k(k-1)\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}$$

(66)

and

$$\begin{array}{ll}{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{m+n}{n}\right)}}& {\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}{\displaystyle \frac{{(-1)}^k}{(k+1)(n-k)}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{k}}\right)\\ & ={\displaystyle \frac{1}{(n+1)\left(\genfrac{}{}{0pt}{}{m+n+1}{n+1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n\end{array}}^{m+n}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n+1}{k+1}}\right){\displaystyle \frac{{(-1)}^k}{n-k}}\\ & ={\displaystyle \frac{1}{(n+1)\left(\genfrac{}{}{0pt}{}{m+n+1}{n+1}\right)}}{\displaystyle \sum _{\begin{array}{c}k=1\\ k\ne n+1\end{array}}^{m+n}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+n+1}{k}}\right){\displaystyle \frac{{(-1)}^{k-1}}{n-k+1}}.\end{array}$$

(67)

Using (66) and (67) in (65) and replacing m with $m-1$, the desired identity (63) follows. □

Remark 3. 

Setting $t=-1$ in (60), we obtain

$$\sum _{k=1}^m{\displaystyle \frac{{(-1)}^{k-1}}{k^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)={\displaystyle \frac{1}{2}}\left(H_m^2+H_m^{\left(2\right)}\right)(m\in {\mathbb{Z}}_{⩾0}).$$

(68)

By using (64) and (68), along with the partial fractions

$${\displaystyle \frac{1}{k^2(k+1)}}={\displaystyle \frac{1}{k^2}}-{\displaystyle \frac{1}{k}}+{\displaystyle \frac{1}{k+1}},$$

and

$${\displaystyle \frac{1}{k{(k+1)}^2}}={\displaystyle \frac{1}{k}}-{\displaystyle \frac{1}{{(k+1)}^2}}-{\displaystyle \frac{1}{k+1}},$$

we obtain the following identities: For $m\in {\mathbb{Z}}_{⩾0}$,

$$\sum _{k=1}^m{\displaystyle \frac{{(-1)}^{k-1}}{k^2(k+1)}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=1-{\displaystyle \frac{1}{m+1}}-H_m+{\displaystyle \frac{1}{2}}\left(H_m^2+H_m^{\left(2\right)}\right);$$

(69)

$$\sum _{k=1}^m{\displaystyle \frac{{(-1)}^{k-1}}{k{(k+1)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=-2+{\displaystyle \frac{1}{m+1}}+{\displaystyle \frac{1}{{(m+1)}^2}}+{\displaystyle \frac{m+2}{m+1}}{H}_m.$$

(70)

To derive Equation (70), we utilize the subsequent identity

$$\sum _{k=1}^m{\displaystyle \frac{{(-1)}^{k-1}}{{(k+1)}^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=1-{\displaystyle \frac{1}{{(m+1)}^2}}-{\displaystyle \frac{H_m}{m+1}}(m\in {\mathbb{Z}}_{⩾0}),$$

for which we also use Equation (68) and the well-known binomial identity

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k-1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{k}}\right).$$

The following identities are similar to those found above:

$$\sum _{k=1}^m{\displaystyle \frac{{(-1)}^{k-1}}{k^3}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=\frac{1}{6}{H}_m^3+\frac{1}{2}{H}_m{H}_m^{\left(2\right)}+\frac{1}{3}{H}_m^{\left(3\right)};$$

(71)

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{{(-1)}^{k-1}}{k^4}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=& \frac{1}{24}{H}_m^4+\frac{1}{4}{H}_m^2{H}_m^{\left(2\right)}+\frac{1}{8}{\left\{H_m^{\left(2\right)}\right\}}^2\\ & +\frac{1}{3}{H}_m{H}_m^{\left(3\right)}+\frac{1}{4}{H}_m^{\left(4\right)};\end{array}$$

(72)

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{{(-1)}^{k-1}}{k^5}}& \left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=\frac{1}{120}{H}_m^5+\frac{1}{12}{H}_m^3{H}_m^{\left(2\right)}+\frac{1}{8}{H}_m{\left\{H_m^{\left(2\right)}\right\}}^2\\ & +\frac{1}{6}{H}_m^2{H}_m^{\left(3\right)}+\frac{1}{6}{H}_m^{\left(2\right)}{H}_m^{\left(3\right)}+\frac{1}{4}{H}_m{H}_m^{\left(4\right)}+\frac{1}{5}{H}_m^{\left(5\right)};\end{array}$$

(73)

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}& {\displaystyle \frac{{(-1)}^{k-1}}{k^6}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=\frac{1}{720}{H}_m^6+\frac{1}{48}{H}_m^4{H}_m^{\left(2\right)}+\frac{1}{16}{H}_m^2{\left\{H_m^{\left(2\right)}\right\}}^2\\ & +\frac{1}{48}{\left\{H_m^{\left(2\right)}\right\}}^3+\frac{1}{18}{H}_m^3{H}_m^{\left(3\right)}+\frac{1}{6}{H}_m{H}_m^{\left(2\right)}{H}_m^{\left(3\right)}+\frac{1}{18}{\left\{H_m^{\left(3\right)}\right\}}^2\\ & +\frac{1}{8}{H}_m^2{H}_m^{\left(4\right)}+\frac{1}{8}{H}_m^{\left(2\right)}{H}_m^{\left(4\right)}+\frac{1}{5}{H}_m{H}_m^{\left(5\right)}+\frac{1}{6}{H}_m^{\left(6\right)};\end{array}$$

(74)

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}{\displaystyle \frac{{(-1)}^{k-1}}{k^3(k+1)}}& \left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=\frac{1}{6}{H}_m^3+\frac{1}{2}{H}_m{H}_m^{\left(2\right)}+\frac{1}{3}{H}_m^{\left(3\right)}\\ & -\frac{1}{2}{H}_m^2-\frac{1}{2}{H}_m^{\left(2\right)}-1+{\displaystyle \frac{1}{m+1}}+H_m;\end{array}$$

(75)

$$\begin{array}{ll}{\displaystyle \sum _{k=1}^m}& {\displaystyle \frac{{(-1)}^{k-1}}{k^4(k+1)}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)=\frac{1}{24}{H}_m^4+\frac{1}{4}{H}_m^2{H}_m^{\left(2\right)}+\frac{1}{8}{\left\{H_m^{\left(2\right)}\right\}}^2\\ & \hspace{1em}+\frac{1}{3}{H}_m{H}_m^{\left(3\right)}+\frac{1}{4}{H}_m^{\left(4\right)}-\frac{1}{6}{H}_m^3-\frac{1}{2}{H}_m{H}_m^{\left(2\right)}-\frac{1}{3}{H}_m^{\left(3\right)}\\ & \hspace{1em}+\frac{1}{2}{H}_m^2+\frac{1}{2}{H}_m^{\left(2\right)}+1-{\displaystyle \frac{1}{m+1}}-H_m,\end{array}$$

(76)

where $m\in {\mathbb{Z}}_{⩾0}$.

Identities (71), (72), (73), and (74) can be derived by applying Equations (11) and (12) to the corresponding formulas obtained from MATHEMATICA 13.0. Indeed, a generalized formula that encompasses the identities (68), (71), (72), and (73) as special cases is presented in [11] (Corollary 3.2).

Formula (75) can be obtained by using the same method in (69), along with Equations (64), (68), and (71).

Formula (76) can be derived by using the same method in (69), along with Equations (64), (68), (71), and (72).

4. Concluding Remarks

We have conducted a thorough analysis of the following finite parameterized binomial sums:

$$\begin{array}{c}\hfill \sum _{k=1}^m{\displaystyle \frac{{ϵ}^k}{k^p\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}\mathrm{and}\sum _{k=1}^m{\displaystyle \frac{{ϵ}^k{H}_k}{k^p\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}},\end{array}$$

where $m,n\in {\mathbb{Z}}_{⩾0}$, $ϵ\in \{-1,1\}$, and $p=0,1$. Our study includes numerous corollaries that underscore the significance of our findings.

Our results appear to be novel in the existing literature, and we believe that our methodology is at least as important as the results themselves. By applying our method to the third, fourth, and higher derivatives of Equation (2) with respect to the variable x, many additional parameterized finite binomial sums involving harmonic numbers can be derived.

Infinite versions of the sums we explored can be found in [12]. For further parameterized finite binomial sums, we direct readers to [13].