On the Application of a Hypergeometric Identity to Generate Generalized Hypergeometric Reduction Formulas

Abstract

We systematically exploit a new generalized hypergeometric identity to obtain new hypergeometric summation formulas. As a consistency test, alternative proofs for some special cases are also provided. As a byproduct, new summation formulas with finite sums involving the psi function and a recursive formula for Bateman’s G function are derived. Finally, all the results have been numerically checked with MATHEMATICA.

1. Introduction

Mathematical applications of generalized hypergeometric functions are abundant in the existing literature. For instance, a variety of problems in classical mechanics and mathematical physics led to Picard–Fuchs equations. These equations are frequently solvable in terms of generalized hypergeometric functions [1]. Also, many combinatorial identities, especially ones involving binomial and related coefficients, are special cases of hypergeometric identities [2] (Sect. 2.7). Another example is found in the calculation of the moments of certain probability distributions, which are given in terms of generalized hypergeometric functions [3]. Therefore, the calculation of generalized hypergeometric functions for particular values of the argument and parameters in terms of more simple functions is of great relevance. The classical compilation of these summation formulas is found in [4] (Chap. 7). A revision, as well as an extension of these tables, was carried out in [5]. More recently, we found a new compilation of representations of generalized hypergeometric functions in [6] (Chap. 8).

In the existing literature, we found several papers devoted to the calculation of hypergeometric summation formulas for a given number of parameters and particular values of the argument (see, e.g., [7,8,9,10,11]). However, the number of papers devoted to the calculation of hypergeometric summation formulas for an arbitrary number of parameters as well as an arbitrary argument is relatively scarce (see, e.g., [12,13]). The aim of this paper is to contribute to articles of this last type.

This paper is organized as follows. Section 2 presents the special functions that will be used throughout the paper, as well as some of their basic properties. Section 3 proves the main hypergeometric identity on which most of the results obtained in the article are based. In Section 4 we apply this hypergeometric identity to obtain reduction formulas for particular arguments, while in Section 5 and Section 6, we apply it to arbitrary arguments. Finally, we provide our conclusions in Section 7.

All the results presented in this paper have been tested with MATHEMATICA. The corresponding MATHEMATICA notebook is available at https://shorturl.at/tGOwb (accessed on 11 October 2025).

2. Preliminaries

Let $\mathbb{C}$, $\mathbb{Z}$, $\mathbb{N}$, ${\mathbb{Z}}_{\ge 0}$, and ${\mathbb{Z}}_{\le 0}$ denote the sets of complex numbers, integers, positive integers, non-negative integers, and non-positive integers, respectively.

The gamma function is commonly defined as [14] (Eqn. 1.1.1)

$$\Gamma \left(z\right):={\int }_0^{\infty }{e}^{-t}{t}^{z-1}dt.\Re \left(z\right)>0,$$

(1)

and satisfies the reflection Formula [14] (Eqn. 1.2.2)

$$\Gamma \left(z\right)\Gamma \left(1-z\right)={\displaystyle \frac{\pi }{\sin \pi z}},z\in \mathbb{C}\setminus \mathbb{Z}.$$

(2)

The incomplete gamma function is defined as [15] (Eqn. 8.2.1)

$$\gamma \left(\nu ,z\right):={\int }_0^z{e}^{-t}{t}^{\nu -1}dt,\left(z\in \mathbb{C},\Re \left(\nu \right)>0\right).$$

(3)

Also, the logarithmic derivative of $\Gamma \left(z\right)$ is defined as [14] (Eqn. 1.3.1)

$$\psi \left(z\right):={\displaystyle \frac{{\Gamma }^{\prime }\left(z\right)}{\Gamma \left(z\right)}},z\in \mathbb{C}\setminus \mathbb{Z}$$

(4)

and Bateman’s G function is defined in terms of the $\psi \left(z\right)$ function as [16] (Eqn. 44:13:1)

$$\beta \left(z\right):={\displaystyle \frac{1}{2}}\left[\psi \left({\displaystyle \frac{z+1}{2}}\right)-\psi \left({\displaystyle \frac{z}{2}}\right)\right].$$

(5)

The beta function is defined as [16] (Eqn. 43:13:1-2)

$$\mathrm{B}\left(\alpha ,\beta \right):=\left\{\begin{array}{cc}{\displaystyle {\int }_0^1{t}^{\alpha -1}{\left(1-t\right)}^{\beta -1}dt,}\hfill & \Re \left(\alpha \right)>0,\Re \left(\beta \right)>0;\hfill \\ {\displaystyle \frac{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}{\Gamma \left(\alpha +\beta \right)},}\hfill & \alpha ,\beta \in \mathbb{C}\setminus {\mathbb{Z}}_{\le 0}.\hfill \end{array}\right.$$

(6)

and the incomplete beta function is defined as [15] (Eqn. 8.17.1)

$${\mathrm{B}}_z\left(\alpha ,\beta \right):={\int }_0^z{t}^{\alpha -1}{\left(1-t\right)}^{\beta -1}dt,\Re \left(\alpha \right)>0,\Re \left(\beta \right)>0.$$

(7)

The Pochhammer symbol ${\left(\alpha \right)}_k,\left(\alpha \in \mathbb{C}\right)$, can be expressed in terms of gamma functions as

$${\left(\alpha \right)}_k:=\prod _{j=0}^{k-1}\left(\alpha +j\right)={\displaystyle \frac{\Gamma \left(\alpha +k\right)}{\Gamma \left(\alpha \right)}},k\in \mathbb{N}$$

(8)

with ${\left(\alpha \right)}_0:=1$. Note that, according to [16] (Eqn. 18:5:1),

$${\left(-\alpha \right)}_k={\left(-1\right)}^k{\left(\alpha -k+1\right)}_k,$$

(9)

and thus, for $m,k\in {\mathbb{Z}}_{\ge 0}$, we have

$${\left(-m\right)}_k=\left\{\begin{array}{cc}{\displaystyle \frac{{\left(-1\right)}^{k}m!}{\left(m-k\right)!},}\hfill & 0\le k\le m;\hfill \\ 0,\hfill & k>m.\hfill \end{array}\right.$$

(10)

Also, applying the reflection Formula (2), we have [4] (Appendix II.2)

$${\left(\alpha \right)}_{-n}={\displaystyle \frac{{\left(-1\right)}^n}{{\left(1-\alpha \right)}_n}},\left(n\in \mathbb{N},\alpha \in \mathbb{C}\setminus \mathbb{Z}\right).$$

(11)

The generalized hypergeometric function is defined by the series [15] (Eqn. 16.2.1)

$$\begin{array}{ccc}\hfill {}_{p}F_q\left(\begin{array}{c}{a}_1,\dots ,a_p\\ b_1,\dots ,b_q\end{array};z\right)& =& \sum _{k=0}^{\infty }{\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k{z}^k}{k!{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k}}\hfill \\ & =& {}_{p}F_q\left(a_1,\dots ,a_p;b_1,\dots ,b_q;z\right),\hfill \end{array}$$

(12)

wherever this series converges, and by analytic continuation elsewhere. It is assumed that the variable z, the numerator parameters $a_1,\dots ,a_p$, and the denominator parameters $b_1,\dots ,b_q$ take on complex values, provided that

$$\left(b_j\in \mathbb{C}\setminus {\mathbb{Z}}_{\le 0};j=1,\dots ,q\right).$$

Many special functions can be expressed in terms of generalized hypergeometric series, such as the Bessel function of the first kind [15] (Eqn. 10.16.9)

$$J_{\nu }\left(z\right)={\displaystyle \frac{{(z/2)}^{\nu }}{\Gamma (\nu +1)}}{}_{0}F_1\left(;\nu +1;-{\displaystyle \frac{z^2}{4}}\right),$$

and the modified Bessel function [15] (Eqn. 10.39.9)

$$I_{\nu }\left(z\right)={\displaystyle \frac{{(z/2)}^{\nu }}{\Gamma (\nu +1)}}{}_{0}F_1\left(;\nu +1;{\displaystyle \frac{z^2}{4}}\right).$$

If a numerator parameter is a negative integer or zero, the ${}_{p}F_q$ series terminates. For instance, the generalized Laguerre polynomials can be defined as [15] (Eqn. 18.5.12)

$$L_n^{\alpha }\left(z\right):={\displaystyle \frac{{\left(\alpha +1\right)}_n}{n!}}{}_{1}F_1\left(\begin{array}{c}-n\\ \alpha +1\end{array};z\right).$$

(13)

According to [15] (Eqn. 16.3.2), the following differentiation formula is satisfied:

$$\begin{array}{ccc} \hfill & \hfill & {\displaystyle \frac{d^n}{d{z}^n}}\left[z^{\gamma }{}_{p}F_q\left(\begin{array}{c}{a}_1,\dots ,a_p\\ b_1,\dots ,b_q\end{array};z\right)\right]\hfill \\ \hfill & =& {\left(\gamma -n+1\right)}_n{z}^{\gamma -n}{}_{p+1}F_{q+1}\left(\begin{array}{c}\gamma +1,a_1,\dots ,a_p\\ \gamma +1-n,b_1,\dots ,b_q\end{array};z\right),\hfill \end{array}$$

(14)

and thus,

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+1}F_{q+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,a+n\\ b_1,\dots ,b_q,a\end{array};z\right)\hfill \\ & =& {\displaystyle \frac{z^{1-a}}{{\left(a\right)}_n}}{\displaystyle \frac{d^n}{d{z}^n}}\left[z^{n+a-1}{}_{p}F_q\left(\begin{array}{c}{a}_1,\dots ,a_p\\ b_1,\dots ,b_q\end{array};z\right)\right].\hfill \end{array}$$

(15)

Also, the Chu–Vandermonde summation formula is [15] (Eqn. 15.4.24)

$${}_{2}F_1\left(\begin{array}{c}-n,a\\ c\end{array};1\right)={\displaystyle \frac{{\left(c-a\right)}_n}{{\left(c\right)}_n}},\left(n\in {\mathbb{Z}}_{\ge 0},c\in \mathbb{C}\setminus {\mathbb{Z}}_{\le 0}\right).$$

(16)

Further, according to [15] (Eqn. 15.8.1), we have the linear transformation formula

$${}_{2}F_1\left(\begin{array}{c}a,b\\ c\end{array};z\right)={\left(1-z\right)}^{-a}{}_{2}F_1\left(\begin{array}{c}a,c-b\\ c\end{array};{\displaystyle \frac{z}{z-1}}\right).$$

(17)

Finally, Leibniz’s differentiation Formula [15] (Eqn. 1.4.12) is given by

$${\displaystyle \frac{d^n}{d{z}^n}}\left[f\left(z\right)g\left(z\right)\right]=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{d^k}{d{z}^k}}\left[f\left(z\right)\right]{\displaystyle \frac{d^{n-k}}{d{z}^{n-k}}}\left[g\left(z\right)\right].$$

(18)

3. Main Result

Lemma 1.

The following representation of the generalized hypergeometric function ${}_{p}F_q$ holds true for $n\in {\mathbb{Z}}_{\ge 0}$:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p}F_q\left(\begin{array}{c}{a}_1,\dots ,a_p\\ b_1,\dots ,b_q\end{array};z\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k{z}^k}{{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k{\left(c+n\right)}_k}}{}_{p+1}F_{q+1}\left(\begin{array}{c}{a}_1+k,\dots ,a_p+k,c+k\\ b_1+k,\dots ,b_q+k,c+n+k\end{array};z\right).\hfill \end{array}$$

(19)

Proof. 

According to the Chu–Vandermonde summation Formula (16), the definition of the hypergeometric sum (12), and property (10), we have

$$\begin{array}{ccc}\hfill 1& =& {\displaystyle \frac{{\left(c\right)}_j}{{\left(c+n\right)}_j}}{}_{2}F_1\left(\begin{array}{c}-n,-j\\ c\end{array};1\right)={\displaystyle \frac{{\left(c\right)}_j}{{\left(c+n\right)}_j}}\sum _{k=0}^n{\displaystyle \frac{{\left(-n\right)}_k{\left(-j\right)}_k}{k!{\left(c\right)}_k}}\hfill \\ & =& {\displaystyle \frac{{\left(c\right)}_j}{{\left(c+n\right)}_j}}\sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(-1\right)}^k{\left(-j\right)}_k}{{\left(c\right)}_k}}.\hfill \end{array}$$

(20)

Insert (20) into (12) and exchange the summation order to arrive at

$$\begin{array}{ccc} \hfill & \hfill & {}_{p}F_q\left(\begin{array}{c}{a}_1,\dots ,a_p\\ b_1,\dots ,b_q\end{array};z\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(-1\right)}^k}{{\left(c\right)}_k}}\underset{S}{\underbrace{\sum _{j=k}^{\infty }{\displaystyle \frac{{\left(-j\right)}_k{\left(a_1\right)}_j\cdots {\left(a_p\right)}_j}{j!{\left(b_1\right)}_j\cdots {\left(b_q\right)}_j}}{\displaystyle \frac{{\left(c\right)}_j}{{\left(c+n\right)}_j}}{z}^j}},\hfill \end{array}$$

(21)

where the summation indices are $k=0,1,2,\dots ,n$ and $j=k,k+1,\dots$. Now, perform the change $\ell =j-k$; thus,

$$S=\sum _{\ell =0}^{\infty }{\displaystyle \frac{{\left(-k-\ell \right)}_k{\left(a_1\right)}_{k+\ell }\cdots {\left(a_p\right)}_{k+\ell }}{\left(k+\ell \right)!{\left(b_1\right)}_{k+\ell }\cdots {\left(b_q\right)}_{k+\ell }}}{\displaystyle \frac{{\left(c\right)}_{k+\ell }}{{\left(c+n\right)}_{k+\ell }}}{z}^{k+\ell }.$$

From (9), we have

$${\left(-k-\ell \right)}_k={\left(-1\right)}^k{\left(\ell +1\right)}_k,$$

(22)

and from (8),

$${\left(a\right)}_{k+\ell }={\left(a\right)}_k{\left(a+k\right)}_{\ell },$$

(23)

and

$$\left(k+\ell \right)!=\ell !{\left(1+\ell \right)}_k.$$

(24)

Take into account (22)–(24), as well as definition (12), to obtain

$$\begin{array}{ccc} & & S\hfill \\ & =& {\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k}{{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k}}{\displaystyle \frac{{\left(c\right)}_k{\left(-z\right)}^k}{{\left(c+n\right)}_k}}\sum _{\ell =0}^{\infty }{\displaystyle \frac{{\left(a_1+k\right)}_{\ell }\cdots {\left(a_p+k\right)}_{\ell }}{\ell !{\left(b_1+k\right)}_{\ell }\cdots {\left(b_q\right)}_{k+\ell }}}{\displaystyle \frac{{\left(c+k\right)}_{\ell }}{{\left(c+n+k\right)}_{\ell }}}{z}^{\ell }\hfill \\ & =& {\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k}{{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k}}{\displaystyle \frac{{\left(c\right)}_k{\left(-z\right)}^k}{{\left(c+n\right)}_k}}{}_{p+1}F_{q+1}\left(\begin{array}{c}{a}_1+k,\dots ,a_p+k,c+k\\ b_1+k,\dots ,b_q+k,c+n+k\end{array};z\right).\hfill \end{array}$$

(25)

Finally, substitute (25) into (21) to complete the proof. □

Theorem 1.

For $n\in {\mathbb{Z}}_{\ge 0}$ the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+1}F_{q+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,c+n\\ b_1,\dots ,b_q,c\end{array};z\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k{z}^k}{{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k{\left(c\right)}_k}}{}_{p}F_q\left(\begin{array}{c}{a}_1+k,\dots ,a_p+k\\ b_1+k,\dots ,b_q+k\end{array};z\right).\hfill \end{array}$$

(26)

Proof. 

Rewrite (19) for ${}_{p+1}F_{q+1}\left(a_1,\dots ,a_p,a_{p+1};b_1,\dots ,b_q,b_{q+1};z\right)$ and consider $a_{p+1}=c+n$, $b_{q+1}=c$ to obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+1}F_{q+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,c+n\\ b_1,\dots ,b_q,c\end{array};z\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k{z}^k}{{\left(b_1\right)}_k\cdots {\left(b_q\right)}_k{\left(c\right)}_k}}{}_{p+2}F_{q+2}\left(\begin{array}{c}{a}_1+k,\dots ,a_p+k,c+n+k,c+k\\ b_1+k,\dots ,b_q+k,c+k,c+n+k\end{array};z\right).\hfill \end{array}$$

According to the definition of the hypergeometric series (12), simplify the result to complete the proof. □

4. Application to Reduction Formulas with Arguments $z={\displaystyle \frac{1}{2}},1$

Theorem 2.

For $a,c\in \mathbb{C}$, and $n\in {\mathbb{Z}}_{\ge 0}$, the following reduction formula holds true:

$${}_{3}F_2\left(\begin{array}{c}a,a,c+n\\ a+1,c\end{array};{\displaystyle \frac{1}{2}}\right)=a{2}^a\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a\right)}_k}{{\left(c\right)}_k}}\beta \left(a+k\right).$$

(27)

where $\beta \left(z\right)$ denotes Bateman’s G-function defined in (5).

Proof. 

According to (26), we have

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}a,a,c+n\\ a+1,c\end{array};{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left[{\left(a\right)}_k\right]}^2{2}^{-k}}{{\left(a+1\right)}_k{\left(c\right)}_k}}{}_{2}F_1\left(\begin{array}{c}a+k,a+k\\ a+1+k\end{array};{\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$

Apply the identity

$${\displaystyle \frac{{\left(a\right)}_k}{{\left(a+1\right)}_k}}={\displaystyle \frac{a}{a+k}},$$

(28)

and the reduction Formula [4] (Eqn. 7.3.7(16))

$${}_{2}F_1\left(\begin{array}{c}a,a\\ a+1\end{array};{\displaystyle \frac{1}{2}}\right)=2^{a}a\beta \left(a\right),$$

to complete the proof. □

Remark 1.

For $c=a$, (27) reduces to

$${}_{2}F_1\left(\begin{array}{c}a,a+n\\ a+1\end{array};{\displaystyle \frac{1}{2}}\right)=a{2}^a\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\beta \left(a+k\right).$$

Applying the linear transformation Formula (17) and the change $n\to n+1$, we obtain

$${}_{2}F_1\left(\begin{array}{c}-n,a\\ a+1\end{array};-1\right)=a\sum _{k=0}^{n+1}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k}}\right)\beta \left(a+k\right).$$

(29)

However, applying (10) and (28) to the definition of the hypergeometric function (12), we arrive at

$${}_{2}F_1\left(\begin{array}{c}-n,a\\ a+1\end{array};-1\right)=a\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{1}{a+k}}.$$

(30)

From (29) and (30), we obtain the following recursive equation for Bateman’s G function:

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

(31)

Theorem 3.

For $a,b,c\in \mathbb{C}$, and $n,m\in {\mathbb{Z}}_{\ge 0}$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}a,b+m,c+n\\ {\displaystyle \frac{a+b+1}{2}},c\end{array};{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & =& {\displaystyle \frac{2^{a-1}\Gamma \left(\frac{a+b+1}{2}\right)}{\Gamma \left(a\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b+m\right)}_k}{{\left(c\right)}_k}}\sum _{s=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{s}}\right){\displaystyle \frac{\Gamma \left(\frac{a+k+s}{2}\right)}{\Gamma \left(\frac{1+b+k+s}{2}\right)}}.\hfill \end{array}$$

(32)

Proof. 

According to (26), we have

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}a,b+m,c+n\\ {\displaystyle \frac{a+b+1}{2}},c\end{array};{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a\right)}_k{\left(b+m\right)}_k}{{\left(\frac{a+b+1}{2}\right)}_k{\left(c\right)}_k}}{2}^{-k}{}_{2}F_1\left(\begin{array}{c}a+k,b+m+k\\ {\displaystyle \frac{a+b+1}{2}}+k\end{array};{\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$

Apply the reduction Formula [4] (7.3.7(2))

$${}_{2}F_1\left(\begin{array}{c}a,b+m\\ {\displaystyle \frac{a+b+1}{2}}\end{array};{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{2^{a-1}\Gamma \left(\frac{a+b+1}{2}\right)}{\Gamma \left(a\right)}}\sum _{s=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{s}}\right){\displaystyle \frac{\Gamma \left(\frac{a+s}{2}\right)}{\Gamma \left(\frac{1+b+s}{2}\right)}},$$

and simplify the result to complete the proof. □

Theorem 4.

For $a,b,c\in \mathbb{C}$ and $n,m\in {\mathbb{Z}}_{\ge 0}$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}a,b-m,c+n\\ {\displaystyle \frac{a+b+1}{2}},c\end{array};{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & =& {\displaystyle \frac{2^{a-1}\Gamma \left(\frac{a+b+1}{2}\right)\Gamma \left(\frac{b-a+1}{2}-m\right)}{\Gamma \left(a\right)\Gamma \left(\frac{b-a+1}{2}\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b-m\right)}_k}{{\left(c\right)}_k}}\sum _{s=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{s}}\right){\displaystyle \frac{\Gamma \left(\frac{a+k+s}{2}\right)}{\Gamma \left(\frac{1+b+k+s}{2}-m\right)}}.\hfill \end{array}$$

(33)

Proof. 

According to (26), we have

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}a,b-m,c+n\\ {\displaystyle \frac{a+b+1}{2}},c\end{array};{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a\right)}_k{\left(b-m\right)}_k}{{\left(\frac{a+b+1}{2}\right)}_k{\left(c\right)}_k}}{2}^{-k}{}_{2}F_1\left(\begin{array}{c}a+k,b-m+k\\ {\displaystyle \frac{a+b+1}{2}}+k\end{array};{\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$

Apply the reduction Formula [7]

$${}_{2}F_1\left(\begin{array}{c}a,b-m\\ {\displaystyle \frac{a+b+1}{2}}\end{array};{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{2^{a-1}\Gamma \left(\frac{a+b+1}{2}\right)\Gamma \left(\frac{b-a+1}{2}-m\right)}{\Gamma \left(a\right)\Gamma \left(\frac{b-a+1}{2}-m\right)}}\sum _{s=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{s}}\right){\displaystyle \frac{{\left(-1\right)}^s\Gamma \left(\frac{a+s}{2}\right)}{\Gamma \left(\frac{1+b+s}{2}-m\right)}},$$

and simplify the result to complete the proof. □

For the next result, we first need to prove the following Lemma.

Lemma 2.

For $b\ne c$, the following finite sum holds true:

$$\begin{array}{ccc} \hfill & \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\psi \left(b+k\right)\hfill \\ \hfill & =& {\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}}\left[\psi \left(1+b-c\right)+\psi \left(b\right)-\psi \left(1+b-c-n\right)\right].\hfill \end{array}$$

(34)

Proof. 

Apply (26) to obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b,c+n\\ b+1,b+1,c\end{array};1\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a\right)}_k}{{\left(c\right)}_k}}{\left({\displaystyle \frac{{\left(b\right)}_k}{{\left(b+1\right)}_k}}\right)}^2{}_{3}F_2\left(\begin{array}{c}a+k,b+k,b+k\\ b+k+1,b+k+1\end{array};1\right).\hfill \end{array}$$

Taking into account (28) and the reduction Formula [17],

$${}_{3}F_2\left(\begin{array}{c}a,b,b\\ b+1,b+1\end{array};1\right)=b^2\mathrm{B}\left(1-a,b\right)\left[\psi \left(1+b-a\right)-\psi \left(b\right)\right],$$

(35)

after simplification, we arrive at

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b,c+n\\ b+1,b+1,c\end{array};1\right)\hfill \\ \hfill & =& b^2\mathrm{B}\left(1-a,b\right)\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a\right)}_k{\left(b\right)}_k{\left(1-a\right)}_{-k}}{{\left(c\right)}_k}}\left[\psi \left(1+b-a\right)-\psi \left(b+k\right)\right].\hfill \end{array}$$

Now, applying (11), we obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b,c+n\\ b+1,b+1,c\end{array};1\right)=b^2\mathrm{B}\left(1-a,b\right)\hfill \\ \hfill & \hfill & \left[\psi \left(1+b-a\right)\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}-\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\psi \left(b+k\right)\right]\hfill \end{array}$$

(36)

However, according to (10) and the Chu–Vandermonde summation Formula (16), we have

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}={}_{2}F_1\left(\begin{array}{c}-n,b\\ c\end{array};1\right)={\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}},$$

(37)

and hence, inserting (37) into (36), we obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b,c+n\\ b+1,b+1,c\end{array};1\right)\hfill \\ \hfill & =& b^2\mathrm{B}\left(1-a,b\right)\left[\psi \left(1+b-a\right){\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}}-\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\psi \left(b+k\right)\right].\hfill \end{array}$$

(38)

Notice that for $a=c$, (38) reduces to

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}b,b,c+n\\ b+1,b+1\end{array};1\right)\hfill \\ \hfill & =& b^2\mathrm{B}\left(1-c,b\right)\left[\psi \left(1+b-c\right){\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}}-\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\psi \left(b+k\right)\right],\hfill \end{array}$$

(39)

but, according to (35), we have

$${}_{3}F_2\left(\begin{array}{c}b,b,c+n\\ b+1,b+1\end{array};1\right)=b^2\mathrm{B}\left(1-c-n,b\right)\left[\psi \left(1+b-c-n\right)-\psi \left(b\right)\right].$$

(40)

Compare (39) with (40) to obtain

$$\begin{array}{ccc} \hfill & \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\psi \left(b+k\right)\hfill \\ \hfill & =& {\displaystyle \frac{\mathrm{B}\left(1-c-n,b\right)}{\mathrm{B}\left(1-c,b\right)}}\left[\psi \left(b\right)-\psi \left(1+b-c-n\right)\right]+\psi \left(1+b-c\right){\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}}.\hfill \end{array}$$

Finally, complete the proof by applying the identity

$${\displaystyle \frac{\mathrm{B}\left(1-c-n,b\right)}{\mathrm{B}\left(1-c,b\right)}}={\displaystyle \frac{{\left(1-c\right)}_{-n}}{{\left(1-c+b\right)}_{-n}}}={\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}}.$$

□

Theorem 5.

For $a,b,c\in \mathbb{C}$, with $b\ne c$ and $n\in {\mathbb{Z}}_{\ge 0}$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b,c+n\\ b+1,b+1,c\end{array};1\right)=b^2\mathrm{B}\left(1-a,b\right)\hfill \\ \hfill & \hfill & {\displaystyle \frac{{\left(c-b\right)}_n}{{\left(c\right)}_n}}\left[\psi \left(1+b-a\right)-\psi \left(1+b-c\right)-\psi \left(b\right)+\psi \left(1+b-c-n\right)\right].\hfill \end{array}$$

(41)

Proof. 

Insert (34) into (38) to arrive at the desired result. □

Note that the case $b=c$ is not included in (34). Next, we derive this case.

Lemma 3.

For $n\in \mathbb{N}$ and $b\in \mathbb{C}\setminus {\mathbb{Z}}_{\le 0}$, the following finite sum holds true:

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

(42)

Proof. 

Take $c=b$ in (41) to obtain

$${}_{3}F_2\left(\begin{array}{c}a,b,b+n\\ b+1,b+1\end{array};1\right)=-b^2\mathrm{B}\left(1-a,b\right)\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k\psi \left(b+k\right).$$

(43)

Further, take $a=b$; thus,

$${}_{3}F_2\left(\begin{array}{c}b,b,b+n\\ b+1,b+1\end{array};1\right)=-b^2\mathrm{B}\left(1-b,b\right)\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k\psi \left(b+k\right).$$

(44)

However, from (35), we have

$${}_{3}F_2\left(\begin{array}{c}b,b,b+n\\ b+1,b+1\end{array};1\right)=b^2\mathrm{B}\left(1-b-n,b\right)\left[\psi \left(1-n\right)-\psi \left(b\right)\right].$$

(45)

Compare (44) with (45) and take into account (6) and (11) to arrive at

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k\psi \left(b+k\right)=-{\displaystyle \frac{{\left(-1\right)}^n}{{\left(b\right)}_n}}\left[{\displaystyle \frac{\psi \left(1-n\right)}{\Gamma \left(1-n\right)}}-{\displaystyle \frac{\psi \left(b\right)}{\Gamma \left(1-n\right)}}\right].$$

Finally, for $m\in {\mathbb{Z}}_{\ge 0}$, apply [14] (Eqns. 1.1.5 and 1.3.2)

$$\begin{array}{ccc}\hfill \underset{z\to -m}{lim}\Gamma \left(z\right)& =& {\displaystyle \frac{{\left(-1\right)}^m}{m!}}\underset{z\to -m}{lim}{\displaystyle \frac{1}{z+m}},\hfill \\ \hfill \underset{z\to -m}{lim}\psi \left(z\right)& =& -\underset{z\to -m}{lim}{\displaystyle \frac{1}{z+m}},\hfill \end{array}$$

to complete the proof. □

Corollary 1.

Insert (42) into (43) to obtain, for $n\in \mathbb{N}$,

$${}_{3}F_2\left(\begin{array}{c}a,b,b+n\\ b+1,b+1\end{array};1\right)=\left(n-1\right)!{\displaystyle \frac{b^2\mathrm{B}\left(1-a,b\right)}{{\left(b\right)}_n}}.$$

(46)

Theorem 6.

For $n\in \mathbb{N}$ and $m\in {\mathbb{Z}}_{\ge 0}$, the following reduction formula holds true:

$${}_{4}F_3\left(\begin{array}{c}a,b,b+n,c+m\\ b+1,b+1,c\end{array};1\right)=\left(n-1\right)!{\displaystyle \frac{b^2\mathrm{B}\left(1-a,b\right)}{{\left(b\right)}_n}}{\displaystyle \frac{{\left(c-b\right)}_m}{{\left(c\right)}_m}}.$$

(47)

Proof. 

Application of (26) leads to

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b+n,c+m\\ b+1,b+1,c\end{array};1\right)\hfill \\ \hfill & =& \sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{{\left(a\right)}_k{\left(b\right)}_k{\left(b+n\right)}_k}{{\left[{\left(b+1\right)}_k\right]}^2{\left(c\right)}_k}}{}_{3}F_2\left(\begin{array}{c}a+k,b+k,b+n+k\\ b+1+k,b+1+k\end{array};1\right).\hfill \end{array}$$

Take into account (46) and simplify the result to arrive at

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b+n,c+m\\ b+1,b+1,c\end{array};1\right)\hfill \\ \hfill & =& \left(n-1\right)!{\displaystyle \frac{b\Gamma \left(b+1\right)}{\Gamma \left(1-a+b\right)}}\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\displaystyle \frac{{\left(a\right)}_k{\left(b+n\right)}_k}{{\left(c\right)}_k{\left(b+k\right)}_n}}\Gamma \left(1-a-k\right).\hfill \end{array}$$

(48)

Note that, according to (8),

$${\displaystyle \frac{{\left(b+n\right)}_k}{{\left(b+k\right)}_n}}={\displaystyle \frac{{\left(b\right)}_k}{{\left(b\right)}_n}},$$

and according to (2),

$${\displaystyle \frac{\Gamma \left(1-a-k\right)\Gamma \left(a+k\right)}{\Gamma \left(a\right)\Gamma \left(1-a\right)}}={\left(-1\right)}^k,$$

and thus, (48) reduces to

$$\begin{array}{ccc} \hfill & \hfill & {}_{4}F_3\left(\begin{array}{c}a,b,b+n,c+m\\ b+1,b+1,c\end{array};1\right)\hfill \\ \hfill & =& \left(n-1\right)!{\displaystyle \frac{b^2\mathrm{B}\left(1-a,b\right)}{{\left(b\right)}_n}}\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{{\left(a\right)}_k}{{\left(c\right)}_k}}.\hfill \end{array}$$

Finally, apply (37) to complete the proof. □

5. Application to ${}_{p}F_{p+1}$ and ${}_{p}F_p$ Reduction Formulas with $p=1,2,3$ and Arbitrary z

Theorem 7.

For $n\in {\mathbb{Z}}_{\ge 0}$, $b,c\in \mathbb{C}\setminus {\mathbb{Z}}_{\ge 0}$, and $z\in \mathbb{C}\setminus \left(-\infty ,0\right]$, the following reduction formula holds true:

$${}_{1}F_2\left(\begin{array}{c}c+n\\ b,c\end{array};z\right)=z^{\left(1-b\right)/2}\Gamma \left(b\right)\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{z^{k/2}}{{\left(c\right)}_k}}{I}_{b+k-1}\left(2\sqrt{z}\right).$$

(49)

Proof. 

According to (26), we have

$${}_{1}F_2\left(\begin{array}{c}c+n\\ b,c\end{array};z\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{z^k}{{\left(b\right)}_k{\left(c\right)}_k}}{}_{0}F_1\left(\begin{array}{c}-\\ b+k\end{array};z\right).$$

Apply the reduction Formula [4] (Eqn. 7.13.1(1))

$${}_{0}F_1\left(\begin{array}{c}-\\ b\end{array};z\right)=z^{\left(1-b\right)/2}\Gamma \left(b\right)I_{b-1}\left(2\sqrt{z}\right),$$

(50)

and simplify the result to complete the proof. □

We obtain an alternative proof of (49) as follows.

Proof. 

From (50), we have

$${\displaystyle \frac{d^n}{d{z}^n}}\left[z^{c-1}{}_{0}F_1\left(\begin{array}{c}-\\ b\end{array};z\right)\right]=\Gamma \left(b\right){\displaystyle \frac{d^n}{d{z}^n}}\left[z^{c-1}{z}^{\left(1-b\right)/2}{I}_{b-1}\left(2\sqrt{z}\right)\right].$$

Apply Leibniz’s differentiation Formula (18) and the hypergeometric differentiation Formula (14), as well as [6] (Eqn. 1.13.1(5))

$${\displaystyle \frac{d^n}{d{z}^n}}\left[z^{\pm \nu /2}{I}_{\nu }\left(a\sqrt{z}\right)\right]={\left({\displaystyle \frac{a}{2}}\right)}^n{z}^{\left(\pm \nu -n\right)/2}{I}_{\nu \mp n}\left(a\sqrt{z}\right),$$

to arrive at

$$\begin{array}{ccc} \hfill & \hfill & {}_{1}F_2\left(\begin{array}{c}c\\ b,c-n\end{array};z\right)\hfill \\ \hfill & =& {\left(-1\right)}^n{\displaystyle \frac{\Gamma \left(b\right)z^{\left(1-b\right)/2}}{{\left(c-n\right)}_n}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\left(1-c\right)}_{n-k}{z}^{k/2}{I}_{b+k-1}\left(2\sqrt{z}\right).\hfill \end{array}$$

Perform the change of variables $c\to c+n$ and take into account the identity

$${\left(-1\right)}^{n+k}{\displaystyle \frac{{\left(1-c-n\right)}_{n-k}}{{\left(c\right)}_n}}={\displaystyle \frac{1}{{\left(c\right)}_k}},$$

to obtain (49), as we wanted to prove. □

Theorem 8.

For $n,m\in {\mathbb{Z}}_{\ge 0}$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{2}F_3\left(\begin{array}{c}c+n,d+m\\ b,c,d\end{array};z\right)\hfill \\ \hfill & =& z^{\left(1-b\right)/2}\Gamma \left(b\right)\Gamma \left(d\right)\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(d+m\right)}_k}{{\left(c\right)}_k}}\sum _{\ell =0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{\ell }}\right){\displaystyle \frac{z^{\left(k+\ell \right)/2}}{\Gamma \left(d+k+\ell \right)}}{I}_{b+k+\ell -1}\left(2\sqrt{z}\right).\hfill \end{array}$$

(51)

Proof. 

According to (26), we have

$$\begin{array}{ccc} \hfill & \hfill & {}_{2}F_3\left(\begin{array}{c}c+n,d+m\\ b,c,d\end{array};z\right)\hfill \\ \hfill & =& \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(d+m\right)}_k{z}^k}{{\left(b\right)}_k{\left(c\right)}_k{\left(d\right)}_k}}{}_{1}F_2\left(\begin{array}{c}d+m+k\\ b+k,d+k\end{array};z\right).\hfill \end{array}$$

Apply (49) and simplify the result to complete the proof. □

Remark 2.

The reduction Formulas (49) and (51) can be rewritten in terms of the Bessel function of the first kind, $J_{\nu }\left(z\right)$, taking into account the reduction Formula [4] (Eqn. 7.13.1(1)):

$${}_{0}F_1\left(\begin{array}{c}-\\ b\end{array};-z\right)=z^{\left(1-b\right)/2}\Gamma \left(b\right)J_{b-1}\left(2\sqrt{z}\right).$$

Theorem 9.

For $n\in {\mathbb{Z}}_{\ge 0}$ the following reduction formula holds true:

$${}_{2}F_2\left(\begin{array}{c}a,c+n\\ a+1,c\end{array};z\right)=a{\left(-z\right)}^a\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(-1\right)}^k{\displaystyle \frac{\gamma \left(a+k,-z\right)}{{\left(c\right)}_k}}.$$

Proof. 

According to (26), we have

$${}_{2}F_2\left(\begin{array}{c}a,c+n\\ a+1,c\end{array};z\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a\right)}_k{z}^k}{{\left(a+1\right)}_k{\left(c\right)}_k}}{}_{1}F_1\left(\begin{array}{c}a+k,\\ a+1+k,b+k\end{array};z\right).$$

Apply (28) and the reduction Formula [4] (Eqn. 7.11.3(1))

$${}_{1}F_1\left(\begin{array}{c}a,\\ a+1\end{array};-z\right)=a{z}^{-a}\gamma \left(a,z\right),$$

to complete the proof. □

Theorem 10.

For $n,m\in {\mathbb{Z}}_{\ge 0}$ the following reduction formula holds true:

$${}_{2}F_2\left(\begin{array}{c}a+n,b+m\\ a,b\end{array};z\right)={\displaystyle \frac{m!e^z}{{\left(b\right)}_m}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{z^k{L}_m^{b+k-1}\left(-z\right)}{{\left(a\right)}_k}}.$$

(52)

Proof. 

Performing the change $a\to a+n$ and taking into account (9), transform the reduction Formula [4] (Eqn. 7.11.1(8))

$${}_{1}F_1\left(\begin{array}{c}a\\ a-n\end{array};z\right)={\displaystyle \frac{{\left(-1\right)}^{n}n!}{{\left(1-a\right)}_n}}{e}^z{L}_n^{a-n-1}\left(-z\right),$$

into

$${}_{1}F_1\left(\begin{array}{c}a+n\\ a\end{array};z\right)={\displaystyle \frac{n!}{{\left(a\right)}_n}}{e}^z{L}_n^{a-1}\left(-z\right).$$

(53)

According to (26), we have

$${}_{2}F_2\left(\begin{array}{c}a+n,b+m\\ a,b\end{array};z\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a+n\right)}_k{z}^k}{{\left(a\right)}_k{\left(b\right)}_k}}{}_{1}F_1\left(\begin{array}{c}a+n+k\\ a+k\end{array};z\right).$$

Apply (53) and simplify the result to complete the proof. □

Remark 3.

Note that (52) provides an alternative expression to the one found in the literature [18]:

$${}_{2}F_2\left(\begin{array}{c}a+n,b+m\\ a,b\end{array};z\right)=e^z\sum _{k=0}^{n+m}{\displaystyle \frac{{\left(-n-m\right)}_k{\left(-z\right)}^k}{k!{\left(a\right)}_k}}{}_{3}F_2\left(\begin{array}{c}-k,-m,b-a-n\\ b,-n-m\end{array};1\right).$$

Theorem 11.

For $n,m,k\in {\mathbb{Z}}_{\ge 0}$ the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_3\left(\begin{array}{c}a+n,b+m,c+k\\ a,b,c\end{array};z\right)\hfill \\ \hfill & =& {\displaystyle \frac{m!e^z}{{\left(b\right)}_m{\left(a\right)}_n}}\sum _{\ell =0}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\ell }}\right){\displaystyle \frac{\Gamma \left(a+n+\ell \right)}{{\left(c\right)}_{\ell }}}\sum _{s=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{s}}\right)z^{\ell +s}{\displaystyle \frac{L_m^{b+\ell +s-1}\left(-z\right)}{\Gamma \left(a+\ell +s\right)}}.\hfill \end{array}$$

Proof. 

According to (26), we have

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_3\left(\begin{array}{c}a+n,b+m,c+k\\ a,b,c\end{array};z\right)\hfill \\ \hfill & =& \sum _{\ell =0}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\ell }}\right){\displaystyle \frac{{\left(a+n\right)}_{\ell }{\left(b+m\right)}_{\ell }{z}^{\ell }}{{\left(a\right)}_{\ell }{\left(b\right)}_{\ell }{\left(c\right)}_{\ell }}}{}_{2}F_2\left(\begin{array}{c}a+n+\ell ,b+m+\ell \\ a+\ell ,b+\ell \end{array};z\right).\hfill \end{array}$$

Apply (52) and simplify the result to complete the proof. □

Remark 4.

In [18], a conjecture is presented as the following reduction formula:

$${}_{p}F_p\left(\begin{array}{c}{a}_1+n_1,\dots ,a_p+n_p\\ a_1,\dots ,a_p\end{array};z\right)=e^z{\mathcal{P}}_p\left(n_1,\dots ,n_p;a_1,\dots ,a_p\right),$$

(54)

where ${\mathcal{P}}_p\left(n_1,\dots ,n_p;a_1,\dots ,a_p\right)$ is a polynomial in z of degree $n=n_1+\dots +n_p$, with $n_j\in {\mathbb{Z}}_{\ge 0}$$\left(j=1,\dots ,p\right)$. From Theorems 10 and 11, it is clear that the repeated application of (26) yields the form given in (54).

6. Application to ${}_{p+1}F_p$ Reduction Formulas with Arbitrary p and z

Recently, in [13] we found the following reduction formula for $p\in \mathbb{N}$ and $a_i\ne a_j$$\left(i\ne j\right)$:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+2}F_{p+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n\\ a_1+1,\dots ,a_p+1,c\end{array};z\right)\hfill \\ & =& {\displaystyle \frac{z^{1-c}}{{\left(c\right)}_n}}\prod _{s=1}^p{a}_s\sum _{k=0}^n{\displaystyle \frac{{\mathcal{H}}^n\left(a_k,1-b,n+c-a_k-1;z\right)}{{\prod }_{\ell \ne k}^p\left(a_{\ell }-a_k\right)}},\hfill \end{array}$$

(55)

where

$$\begin{array}{ccc} \hfill & \hfill & {\mathcal{H}}^n\left(\alpha ,\beta ,\gamma ;z\right):={\displaystyle \frac{d^n}{d{z}^n}}\left[z^{\gamma }{\mathrm{B}}_z\left(\alpha ,\beta \right)\right]\hfill \\ & =& {\left(-1\right)}^n{z}^{\gamma -n}\hfill \\ \hfill & \hfill & \left\{{\left(-\gamma \right)}_n{\mathrm{B}}_z\left(\alpha ,\beta \right)-z^{\alpha }\sum _{k=0}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k+1}}\right){\left(-\gamma \right)}_{n-k-1}{\left(1-\alpha \right)}_k{}_{2}F_1\left(\begin{array}{c}\alpha ,1-\beta \\ \alpha -k\end{array};z\right)\right\},\hfill \end{array}$$

(56)

which satisfies the following property for $\Re \left(\beta \right)>n$:

$${\mathcal{H}}^n\left(\alpha ,\beta ,\gamma ;1\right)={\left(-1\right)}^n{\left(-\gamma \right)}_n\mathrm{B}\left(\alpha ,\beta \right).$$

(57)

Next, with the aid of Theorem 1, we derive a much simpler reduction formula for the same case as the one given in (55).

Theorem 12.

For $n\in {\mathbb{Z}}_{\ge 0}$, $p\in \mathbb{N}$, and $a_i\ne a_j$$\left(i\ne j\right)$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+2}F_{p+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n\\ a_1+1,\dots ,a_p+1,c\end{array};z\right)\hfill \\ \hfill & =& \prod _{s=1}^p{a}_s\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\sum _{\ell =1}^p{\displaystyle \frac{z^{-a_{\ell }}{\mathrm{B}}_z\left(a_{\ell }+k,1-b-k\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}.\hfill \end{array}$$

(58)

Proof. 

According to (26), we have

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+2}F_{p+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n\\ a_1+1,\dots ,a_p+1,c\end{array};z\right)\hfill \\ \hfill & =& \sum _{k=0}^n{\displaystyle \frac{{\left(a_1\right)}_k\cdots {\left(a_p\right)}_k{\left(b\right)}_k{z}^k}{{\left(a_1+1\right)}_k\cdots {\left(a_p+1\right)}_k{\left(c\right)}_k}}{}_{p+1}F_p\left(\begin{array}{c}{a}_1+k,\dots ,a_p+k,b+k\\ a_1+1+k,\dots ,a_p+1+k\end{array};z\right).\hfill \end{array}$$

Take into account (28) and the results given in [4] (Eqns. 7.10.1(1) and 7.3.1(28)), i.e.,

$${}_{m+1}F_m\left(\begin{array}{c}a,b_1,\dots ,b_m\\ b_1+1,\dots ,b_m+1\end{array};z\right)=\prod _{s=1}^m{b}_s\sum _{j=1}^m{\displaystyle \frac{z^{-b_j}{\mathrm{B}}_z\left(b_j,1-a\right)}{{\prod }_{\ell \ne j}^m\left(b_{\ell }-b_j\right)}},$$

to complete the proof. □

Remark 5.

It is worth noting that the particular case $p=0$ is not included in (58), but it is given in the literature as [4] (Eqn. 7.3.1(21,140))

$${}_{2}F_1\left(\begin{array}{c}b,c+n\\ c\end{array};z\right)={\left(1-z\right)}^{-b}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}{\left({\displaystyle \frac{z}{1-z}}\right)}^k.$$

(59)

Despite the fact that (58) is quite different from the expression reported in the literature, i.e., (55), we can derive the same formula reported in [13] for $z=1$ from (58). Indeed, substitute $z=1$ into (58), exchange the sum order, and expand the corresponding beta function according to (6) to obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+2}F_{p+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n\\ a_1+1,\dots ,a_p+1,c\end{array};1\right)\hfill \\ & =& \prod _{s=1}^p{a}_s\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\sum _{\ell =1}^p{\displaystyle \frac{\mathrm{B}\left(a_{\ell }+k,1-b-k\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}\hfill \\ & =& \prod _{s=1}^p{a}_s\sum _{\ell =1}^p{\displaystyle \frac{1}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)\Gamma \left(a_{\ell }+1-b\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\Gamma \left(a_{\ell }+k\right)\Gamma \left(1-b-k\right).\hfill \end{array}$$

Now, take into account the reflection formula of the gamma function (2), the definition of the beta function (6), and the Pochhammer symbol (8); thus,

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+2}F_{p+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n\\ a_1+1,\dots ,a_p+1,c\end{array};1\right)\hfill \\ & =& {\displaystyle \frac{\pi }{\Gamma \left(1-b\right)\sin \pi b}}\prod _{s=1}^p{a}_s\sum _{\ell =1}^p{\displaystyle \frac{\Gamma \left(a_{\ell }\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)\Gamma \left(a_{\ell }+1-b\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a_{\ell }\right)}_k}{{\left(c\right)}_k}}{\left(-1\right)}^k\hfill \\ & =& \prod _{s=1}^p{a}_s\sum _{\ell =1}^p{\displaystyle \frac{\mathrm{B}\left(a_{\ell },1-b\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a_{\ell }\right)}_k}{{\left(c\right)}_k}}{\left(-1\right)}^k.\hfill \end{array}$$

Finally, we apply the result given in (37) to arrive at the desired result given in [13], i.e.,

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+2}F_{p+1}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n\\ a_1+1,\dots ,a_p+1,c\end{array};1\right)\hfill \\ & =& {\displaystyle \frac{1}{{\left(c\right)}_n}}\prod _{s=1}^p{a}_s\sum _{\ell =1}^p{\displaystyle \frac{{\left(c-a_{\ell }\right)}_n\mathrm{B}\left(a_{\ell },1-b\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}.\hfill \end{array}$$

(60)

Remark 6.

As a consistency test, note that for $p=1$, (60) reduces to

$${}_{3}F_2\left(\begin{array}{c}a,b,c+m\\ b+1,c\end{array};1\right)=b\mathrm{B}\left(1-a,b\right){\displaystyle \frac{{\left(c-b\right)}_m}{{\left(c\right)}_m}},$$

which coincides with (47) for $n=1$.

Theorem 13.

For $n,m\in {\mathbb{Z}}_{\ge 0}$, $p\in \mathbb{N}$, and $a_i\ne a_j$$\left(i\ne j\right)$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};z\right)\hfill \\ \hfill & =& {\displaystyle \frac{z^{1-d}{\prod }_{s=1}^p{a}_s}{{\left(d\right)}_m}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\sum _{\ell =1}^p{\displaystyle \frac{{\mathcal{H}}^m\left(a_{\ell }+k,1-b-k,m+d-a_{\ell }-1;z\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}.\hfill \end{array}$$

(61)

Proof. 

Apply the n-th derivative to both sides of (58), taking into account (15) and (56) to complete the proof. □

Alternatively, we can derive a much simpler expression for the reduction formula given in (61).

Theorem 14.

For $n,m\in {\mathbb{Z}}_{\ge 0}$, $p\in \mathbb{N}$, and $a_i\ne a_j$$\left(i\ne j\right)$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};z\right)\hfill \\ \hfill & =& {\displaystyle \frac{{\prod }_{s=1}^p{a}_s}{{\left(d\right)}_m}}\sum _{k=0}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}{{\left(c\right)}_k}}\sum _{s=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{s}}\right){\left(b+k\right)}_s{\left(d+k+s\right)}_{m-s}\hfill \\ \hfill & \hfill & \sum _{\ell =1}^p{\displaystyle \frac{{\mathrm{B}}_z\left(a_{\ell }+k+s,1-b-k-s\right)}{z^{a_{\ell }}{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}.\hfill \end{array}$$

(62)

Proof. 

First, note that for $p=1$, (58) reduces to

$${}_{3}F_2\left(\begin{array}{c}a,b,c+n\\ a+1,c\end{array};z\right)=a{z}^{-a}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}{\mathrm{B}}_z\left(a_{\ell }+k,1-b-k\right).$$

(63)

Next, apply the differentiation Formula (15) to (58) in order to obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};z\right)\hfill \\ \hfill & =& {\displaystyle \frac{z^{1-d}{\prod }_{s=1}^p{a}_s}{{\left(d\right)}_m}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\sum _{\ell =1}^p{\displaystyle \frac{1}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}\hfill \\ \hfill & \hfill & {\displaystyle \frac{d^m}{d{z}^m}}\left[z^{m+d-1-a_{\ell }}{\mathrm{B}}_z\left(a_{\ell }+k,1-b-k\right)\right]\hfill \end{array}$$

(64)

Now, according to [4] (Eqn. 7.3.1(28)), we have the reduction formula

$${}_{2}F_1\left(\begin{array}{c}a,b\\ a+1\end{array};z\right)=a{z}^{-a}{\mathrm{B}}_z\left(a,1-b\right),$$

and thus, applying the differentiation Formula (14) and the result given in (63), we arrive at

$$\begin{array}{ccc} \hfill & \hfill & {\displaystyle \frac{d^m}{d{z}^m}}\left[z^{m+d-1-a_{\ell }}{\mathrm{B}}_z\left(a_{\ell }+k,1-b-k\right)\right]\hfill \\ \hfill & =& {\displaystyle \frac{1}{a_{\ell }+k}}{\displaystyle \frac{d^m}{d{z}^m}}\left[z^{m+d-1+k}{}_{2}F_1\left(\begin{array}{c}a,b\\ a+1\end{array};z\right)\right]\hfill \\ \hfill & =& {\displaystyle \frac{{\left(d+k\right)}_m}{a_{\ell }+k}}{z}^{d-1+k}{}_{3}F_2\left(\begin{array}{c}m+d-1+k,a_{\ell }+k,b+k\\ d+k,a_{\ell }+k+1\end{array};z\right)\hfill \\ \hfill & =& {\left(d+k\right)}_m{z}^{d-1}\sum _{s=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{s}}\right){\displaystyle \frac{{\left(b+k\right)}_s}{{\left(d+k\right)}_s}}{z}^{-a_{\ell }}{\mathrm{B}}_z\left(a_{\ell }+k+s,1-b-k-s\right).\hfill \end{array}$$

(65)

Insert (65) into (64) and simplify the result, taking into account the property

$${\displaystyle \frac{\Gamma \left(\alpha +m\right)}{\Gamma \left(\alpha +s\right)}}={\left(\alpha \right)}_{m-s},$$

to complete the proof. □

Theorem 15.

For $p\in \mathbb{N}$, $a_i\ne a_j$$\left(i\ne j\right)$, and $\Re \left(b\right)<1-max\left(n,m\right)$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};1\right)\hfill \\ \hfill & =& {\displaystyle \frac{{\prod }_{s=1}^p{a}_s}{{\left(d\right)}_m{\left(c\right)}_n}}\sum _{\ell =1}^p{\displaystyle \frac{{\left(d-a_{\ell }\right)}_m{\left(c-a_{\ell }\right)}_n}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}\mathrm{B}\left(a_{\ell },1-b\right).\hfill \end{array}$$

(66)

Proof. 

Taking $z=1$ in (61) and applying (57), we obtain

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};1\right)\hfill \\ \hfill & =& {\displaystyle \frac{1}{{\left(d\right)}_m}}\prod _{s=1}^p{a}_s\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\sum _{\ell =1}^p{\displaystyle \frac{{\left(d-a_{\ell }\right)}_m\mathrm{B}\left(a_{\ell }+k,1-b-k\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)}}.\hfill \end{array}$$

Now, exchange the sum order and expand the beta function according to (6):

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};1\right)\hfill \\ \hfill & =& {\displaystyle \frac{{\prod }_{s=1}^p{a}_s}{{\left(d\right)}_m}}\sum _{\ell =1}^p{\displaystyle \frac{{\left(d-a_{\ell }\right)}_m}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)\Gamma \left(a_{\ell }+1-b\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}\Gamma \left(a_{\ell }+k\right)\Gamma \left(1-b-k\right).\hfill \end{array}$$

Apply the reflection Formula (2) and the definition of the Pochhammer symbol (8) to arrive at

$$\begin{array}{ccc} \hfill & \hfill & {}_{p+3}F_{p+2}\left(\begin{array}{c}{a}_1,\dots ,a_p,b,c+n,d+m\\ a_1+1,\dots ,a_p+1,c,d\end{array};1\right)\hfill \\ \hfill & =& {\displaystyle \frac{\pi {\prod }_{s=1}^p{a}_s}{\sin \pi b\Gamma \left(b\right){\left(d\right)}_m}}\sum _{\ell =1}^p{\displaystyle \frac{{\left(d-a_{\ell }\right)}_m\Gamma \left(a_{\ell }\right)}{{\prod }_{j\ne \ell }^p\left(a_j-a_{\ell }\right)\Gamma \left(a_{\ell }+1-b\right)}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(a_{\ell }\right)}_k}{{\left(c\right)}_k}}{\left(-1\right)}^k.\hfill \end{array}$$

(67)

Now, insert (37) into (67) and take into account the reflection Formula (2) and the definition of the beta function (6) to complete the proof. □

Note that the particular case $p=0$ is not included in (61). In order to derive the corresponding formula for $p=0$, we first prove the following lemma.

Lemma 4.

For $m\in {\mathbb{Z}}_{\ge 0}$ and $\alpha ,\beta \in \mathbb{R}$, the following derivative formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {\displaystyle \frac{d^m}{d{z}^m}}\left({\displaystyle \frac{z^{\alpha }}{{\left(1-z\right)}^{\beta }}}\right)\hfill \\ \hfill & =& {\displaystyle \frac{z^{\alpha -m}}{{\left(1-z\right)}^{\beta +m}}}{\left(\alpha -m+1\right)}_m{}_{2}F_1\left(\begin{array}{c}-m,1-m+\alpha -\beta \\ 1-m+\alpha \end{array};z\right).\hfill \end{array}$$

(68)

Proof. 

Apply Leibniz’s differentiation Formula (18) and the n-th derivative formula [6] (Eqn. 1.1.2(1))

$${\displaystyle \frac{d^n}{d{z}^n}}\left[z^{\alpha }\right]={\left(-1\right)}^n{\left(-\alpha \right)}^n{z}^{\alpha -n},$$

and hence,

$${\displaystyle \frac{d^n}{d{z}^n}}\left[{\left(1-z\right)}^{\alpha }\right]={\left(-\alpha \right)}^n{\left(1-z\right)}^{\alpha -n},$$

to obtain

$$\begin{array}{ccc} \hfill & \hfill & {\displaystyle \frac{d^m}{d{z}^m}}\left({\displaystyle \frac{z^{\alpha }}{{\left(1-z\right)}^{\beta }}}\right)\hfill \\ & =& \sum _{\ell =0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{\ell }}\right){\displaystyle \frac{d^{\ell }}{d{z}^{\ell }}}\left[{\left(1-z\right)}^{-\beta }\right]{\displaystyle \frac{d^{m-\ell }}{d{z}^{m-\ell }}}\left[z^{\alpha }\right]\hfill \\ & =& {\displaystyle \frac{{\left(-1\right)}^m{z}^{\alpha -m}}{{\left(1-z\right)}^{\beta }}}\sum _{\ell =0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{\ell }}\right){\left(\beta \right)}_{\ell }{\left(-\alpha \right)}_{m-\ell }{\left({\displaystyle \frac{z}{z-1}}\right)}^{\ell }.\hfill \end{array}$$

(69)

Now, according to (12) and (10), we have

$$\begin{array}{ccc}\hfill {}_{2}F_1\left(\begin{array}{c}-m,\beta \\ 1-m+\alpha \end{array};z\right)& =& \sum _{\ell =0}^{\infty }{\displaystyle \frac{{\left(-m\right)}_{\ell }{\left(\beta \right)}_{\ell }}{\ell !{\left(1-m+\alpha \right)}_{\ell }}}{z}^{\ell }\hfill \\ & =& \sum _{\ell =0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{\ell }}\right){\displaystyle \frac{{\left(\beta \right)}_{\ell }{\left(-z\right)}^{\ell }}{{\left(1-m+\alpha \right)}_{\ell }}}.\hfill \end{array}$$

(70)

However, from the definition of the Pochhammer symbol (8) and property (9), it is easy to prove the identity

$${\left(-\alpha \right)}_{m-\ell }={\displaystyle \frac{{\left(-1\right)}^{m+\ell }{\left(\alpha -m+1\right)}_m}{{\left(1-m+\alpha \right)}_{\ell }}},$$

(71)

and thus, substituting (71) into (69) and taking into account (70), we arrive at

$${\displaystyle \frac{d^m}{d{z}^m}}\left({\displaystyle \frac{z^{\alpha }}{{\left(1-z\right)}^{\beta }}}\right)={\displaystyle \frac{z^{\alpha -m}{\left(\alpha -m+1\right)}_m}{{\left(1-z\right)}^{\beta }}}{}_{2}F_1\left(\begin{array}{c}-m,\beta \\ 1-m+\alpha \end{array};{\displaystyle \frac{z}{z-1}}\right).$$

Finally, apply the linear transformation Formula (17) to complete the proof. □

Theorem 16.

For $b,c,d\in \mathbb{C}$, $n,m\in {\mathbb{Z}}_{\ge 0}$, and $z\ne 1$, the following reduction formula holds true:

$$\begin{array}{ccc} \hfill & \hfill & {}_{3}F_2\left(\begin{array}{c}b,c+n,d+m\\ c,d\end{array};z\right)\hfill \\ \hfill & =& {\displaystyle \frac{1}{{\left(1-z\right)}^{b+m}}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k{\left(d+m\right)}_k}{{\left(c\right)}_k{\left(d\right)}_k}}{\left({\displaystyle \frac{z}{1-z}}\right)}^k{}_{2}F_1\left(\begin{array}{c}-m,d-b\\ d+k\end{array};z\right).\hfill \end{array}$$

(72)

Proof. 

According to the differentiation Formula (15) and the reduction Formula (59), we have

$${}_{3}F_2\left(\begin{array}{c}b,c+n,d+m\\ c,d\end{array};z\right)={\displaystyle \frac{z^{1-d}}{{\left(d\right)}_m}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(b\right)}_k}{{\left(c\right)}_k}}{\displaystyle \frac{d^m}{d{z}^m}}\left[{\displaystyle \frac{z^{k+m+d-1}}{{\left(1-z\right)}^{b+k}}}\right].$$

Apply the differentiation Formula (68) and take into account the property

$${\displaystyle \frac{{\left(d+k\right)}_m}{{\left(d\right)}_m}}={\displaystyle \frac{{\left(d+m\right)}_k}{{\left(d\right)}_k}},$$

to complete the proof. □

Remark 7.

Note that

$$\underset{z\to 1}{lim}\left|{}_{3}F_2\left(\begin{array}{c}b,c+n,d+m\\ c,d\end{array};z\right)\right|=\infty .$$

Remark 8.

As a consistency test, note that (72) is reduced to (59), taking $b=d$.

7. Conclusions

Throughout this paper, we have systematically used the hypergeometric identity (26) to obtain a set of summation formulas that do not seem to be reported in the existing literature. In order to see how this method works, consider a vector of parameters $\overrightarrow{\alpha }=\left({\alpha }_1,\dots ,{\alpha }_{\ell }\right)$ and adopt the notation $\overrightarrow{\alpha }+k=\left({\alpha }_1+k,\dots ,{\alpha }_{\ell }+k\right)$. If we know the reduction formula of a generalized hypergeometric function of the form

$${}_{p}F_q\left(\begin{array}{c}{a}_1\left(\overrightarrow{\alpha }\right),\dots ,a_p\left(\overrightarrow{\alpha }\right)\\ b_1\left(\overrightarrow{\alpha }\right),\dots ,b_q\left(\overrightarrow{\alpha }\right)\end{array};z\right),$$

in such a way that $\forall i=1,\dots ,p$ and $\forall j=1,\dots ,q$, we have

$$\left\{\begin{array}{c}{a}_i\left(\overrightarrow{\alpha }+k\right)=a_i\left(\overrightarrow{\alpha }\right)+k,\\ b_j\left(\overrightarrow{\alpha }+k\right)=b_j\left(\overrightarrow{\alpha }\right)+k,\end{array}\right.$$

and then, according to (26), we obtain, $\forall n\in {\mathbb{Z}}_{\ge 0}$, a reduction formula for the generalized hypergeometric function:

$${}_{p+1}F_{q+1}\left(\begin{array}{c}{a}_1\left(\overrightarrow{\alpha }\right),\dots ,a_p\left(\overrightarrow{\alpha }\right),c+n\\ b_1\left(\overrightarrow{\alpha }\right),\dots ,b_q\left(\overrightarrow{\alpha }\right),c\end{array};z\right).$$

It is worth noting that during the development of the proofs, we have found some interesting formulas, such as the recursive Formula (31) and the proof of the conjecture given in (54). In addition, in (34) and (42), we have proved two finite sums involving the psi function. Also, with the new reformulation given in (58) of the reduction formula given in the literature, i.e., (55), we were able to obtain two new equivalent reduction formulas in (61) and (62). Finally, for the special case given in (72), which is not included in (61), we have developed a particular proof.