# Discrete Gamma (Factorial) Function and Its Series in Terms of a Generalized Difference Operator

Authors: G. Britto Antony Xavier; V. Chandrasekar; S. U. Vasanthakumar; B. Govindan
Journal: Advances in Numerical Analysis
 (2012)

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## Abstract

The recent theory and applications of difference operatorΔL introduced in (M. Maria Susai Manuel et al., 2012) are enriched and extended with a useful tool Δℓ-m[kℓ(n)e-k]||(m-1)ℓ+jk for finding the values of various series of discrete gamma functions in number theory. Illustrative examples show the effectiveness of the obtained results in finding the values of various gamma series.


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## Body

## 1. Introduction

The fractional calculus involving gamma function is a generalization of differential calculus, allowing to define derivatives of real or complex order [1, 2]. It is a mathematical subject that has proved to be very useful in applied fields such as economics, engineering, and physics [3–7]. In 1989, Miller and Ross introduced the discrete analogue of the Riemann-Liouville fractional derivative and proved some properties of the fractional difference operator [8]. In the general fractional h-difference Riemann-Liouville operator mentioned in [9, 10], the presence of the h parameter is particularly interesting from the numerical point of view, because when h tends to zero the solutions of the fractional difference equations can be seen as approximations to the solutions of corresponding Riemann-Liouville fractional differential equation [9, 11]. On the other hand, fractional h sum of order m≥1 ((Δh-mf)(t) Definition 2.8 of [9]) is very useful to derive many interesting results in a different way in number theory such as the sum of the mth partial sums on nth powers of arithmetic, arithmetic-geometric progressions, and products of n consecutive terms of arithmetic progression using Δℓ-mu(k) [12].We observed that no results in number theory using definition 2.8 of [9] had been derived. In this paper, we use Definition 2.8 of [9] in a different way and define discrete gamma factorial function to obtain summation formulas of certain series on gamma function and gamma factorial function in number theory by getting closed and summation form of Δℓ-mu(k)∥(m-1)ℓ+jk, (here we replace f(t) by u(k), h by ℓ, and ν by m on the notations used in [9]).


## 2. Preliminaries

Before stating and proving our results, we present some notations, basic definitions, and preliminary results which will be useful for further subsequent discussions. Letℓ>0,k∈[0,∞),j=k-[k/ℓ]ℓ where [k/ℓ] denotes the integer part of k/ℓ,ℕℓ(j)={j,ℓ+j,2ℓ+j,…} and ℕ1(j)=ℕ(j). Throughout this paper, cj is a constant for all k∈ℕℓ(j) and for any positive integer m, we denote Δℓ-mu(k)∥(m-1)ℓ+jk = Δℓ-1(⋯Δℓ-1(Δℓ-1u(k)|jk)|ℓ+jk⋯)|(m-1)ℓ+jk, where Δℓ-1u(k)|jk = u1(k) = Δℓ-1u(k)-Δℓ-1u(j),Δℓ-1(Δℓ-1u(k)|jk)|ℓ+jk = u2(k) = Δℓ-1u1(k)-Δℓ-1u1(ℓ+j), and so on.Definition 2.1 (see [13]).
For a real valued functionu(k), the generalized difference operator Δℓ and its inverse are, respectively, defined as
(2.1)Δℓu(k)=u(k+ℓ)-u(k),k∈[0,∞),ℓ∈(0,∞),(2.2)ifΔℓv(k)=u(k),thenv(k)=Δℓ-1u(k)+cj.Definition 2.2 (see [10]).
Fork,n∈(0,∞), the ℓ-factorial function is defined by
(2.3)kℓ(n)=ℓnΓ(k/ℓ+1)Γ(k/ℓ+1-n),
where Γ is the Euler gamma function and k1(n)=k(n).Remark 2.3.
Whenn∈ℕ(1), (2.3), and its Δℓ difference become
(2.4)kℓ(n)=∏t=0n-1‍(k-tℓ),Δℓkℓ(n)=(nℓ)kℓ(n-1).Lemma 2.4 (see [13]).
LetSrn’s be the Stirling numbers of second kind. Then,
(2.5)kn=∑r=1n‍Srnℓn-rkℓ(r).Theorem 2.5 (see [13]).
Letu(k),k∈[0,∞) be real valued function. Then for k∈[ℓ,∞),
(2.6)Δℓ-1u(k)|jk=∑r=1[k/ℓ]‍u(k-rℓ).Lemma 2.6.
Letv(k) and w(k) be two real valued functions. Then,
(2.7)Δℓ-1[v(k)w(k)]=v(k)Δℓ-1w(k)-Δℓ-1[Δℓ-1w(k+ℓ)Δℓv(k)].Proof.
From (2.1), we find
(2.8)Δℓ[v(k)z(k)]=z(k+ℓ)Δℓv(k)+v(k)Δℓz(k).
Applying (2.2) in (2.8), we obtain
(2.9)Δℓ-1[v(k)Δℓz(k)]=v(k)z(k)-Δℓ-1[z(k+ℓ)Δℓv(k)].
The proof follows by taking w(k)=Δℓz(k) in (2.9).


## 3. Main Results

In this section, we use the following notations:Lm-1={1,2,…,m-1},0(Lm-1) = {ϕ}, ϕ is an empty set, 1(Lm-1)={{1},{2},…,{m-1}}, 2(Lm-1)={{1,2},{1,3},…,{1,m-1},{2,3},…,{2,m-1},…,{m-2,m-1}}. In general, t(Lm-1) = set of all subsets of size t from the set Lm-1 such that if {m1,m2,…,mt}∈t(Lm-1), then m1<m2<⋯<mt, (m-1)(Lm-1)={{1,2,…,m-1}},℘(Lm-1)=⋃t=0m-1t(Lm-1), power set of Lm-1, ∑t=1m-1f(t)=0 for m≤1, and ∏i=2tf(i)=1 for t≤1.Lemma 3.1 (see [13]).
Letk∈[ℓ,∞). Then,
(3.1)Δℓ-1e-k|jk=e-ke-ℓ-1-e-je-ℓ-1.Lemma 3.2 (see [13]).
Letn be any nonnegative integer. Then,
(3.2)Δℓ-1kℓ(n)|jk=kℓ(n+1)ℓ(n+1)-jℓ(n+1)ℓ(n+1).
In particular, when ℓ=1, k=(r-1)∈ℕ={1,2,3,…}, j=0, then (3.2) becomes
(3.3)Δ-1k(n)|0r-1=(r-1)(n+1)(n+1).Remark 3.3.
For any constantc, since 1=kℓ(0), by (3.2) and linearity of Δℓ-1,
(3.4)Δℓ-1c|jk=cΔℓ-1kℓ(0)|jk=c[kℓ(1)1!ℓ-jℓ(1)1!ℓ].Lemma 3.4.
Letn∈ℕ(0) and r∈ℕ(3). Then,
(3.5)Δ-1k(n)|0r-1=(r-2)(n)+(r-3)(n)+⋯+(1)(n)=(r-1)(n+1)(n+1).Proof.
The proof follows by takingℓ=1, u(k)=k(n) in (2.6) and (3.2).Theorem 3.5.
Letm be a positive integer, ℓ∈(0,∞), and k∈[mℓ,∞). Then,
(3.6)Δℓ-mu(k)∥(m-1)ℓ+jk=∑r=m[k/ℓ]‍(r-1)(m-1)(m-1)!u(k-rℓ).Proof.
TakingΔℓ-1 on (2.6), and applying (2.6) for Δℓ-1u(k-rℓ), we get
(3.7)Δℓ-1(Δℓ-1u(k)|jk)|ℓ+jk=∑r=1[k/ℓ]‍Δℓ-1u(k-rℓ)=∑r=1[k/ℓ]‍∑s=1[(k-rℓ)/ℓ]‍u(k-rℓ-sℓ).
From the notation given this section and ordering the terms u(k-rℓ), we find
(3.8)Δℓ-2u(k)∥ℓ+jk=∑r=2[k/ℓ]‍(r-1)(1)1!u(k-rℓ).
Again, taking Δℓ-1 on (3.8), by (2.6) for Δℓ-1u(k-rℓ), we arrive
(3.9)Δℓ-3u(k)∥2ℓ+jk=∑r=3[k/ℓ]‍[(r-2)(1)1!+(r-3)(1)1!+⋯+(1)(1)1!]u(k-rℓ),
which yields by (3.5),
(3.10)Δℓ-3u(k)∥2ℓ+jk=∑r=3[k/ℓ]‍(r-1)(2)2!u(k-rℓ).
Now, (3.6) will be obtained by continuing this process and using (3.5).Theorem 3.6.
Letm∈ℕ(2), ℓ∈(0,∞), and k∈[mℓ,∞). Then,
(3.11)Δℓ-mu(k)∥(m-1)ℓ+jk=Δℓ-mu(k)|(m-1)ℓ+jk+∑t=1m-1‍∑{m1,…,mt}∈t(Lm-1)‍(-1)t×(Δℓ-m1u((m1-1)ℓ+j))kℓ(m-mt)(m-mt)!ℓm-mt×∏i=2t((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1|(m-1)ℓ+jk.Proof.
Applying the limitj to k on Δℓ-1u(k), we write
(3.12)Δℓ-1u(k)|jk=Δℓ-1u(k)-Δℓ-1u(j),
where Δℓ-1u(j) is constant and Δℓ-1u(k) is a function of k. Taking Δℓ-1 on (3.12), by (3.4) and applying the limit ℓ+j to k, we obtain
(3.13)Δℓ-2u(k)∥ℓ+jk=Δℓ-2u(k)|ℓ+jk-Δℓ-1u(j)kℓ(1)1!ℓ|ℓ+jk,
which can be expressed as
(3.14)Δℓ-2u(k)∥ℓ+jk=Δℓ-2u(k)|ℓ+jk+(-1)tΔℓ-m1u((m1-1)ℓ+j)kℓ(2-mt)(2-mt)!ℓ2-mt|ℓ+jk,
where t=1 and {m1=mt=1}∈1(L1) and is same as
(3.15)Δℓ-2u(k)∥ℓ+jk=Δℓ-2u(k)-Δℓ-2u(ℓ+j)+(-1)tkℓ(2-mt)(2-mt)!ℓ2-mtΔℓ-m1u((m1-1)ℓ+j)-(-1)t(ℓ+j)ℓ(2-mt)(2-mt)!ℓ2-mtΔℓ-m1u((m1-1)ℓ+j),
where all the terms except Δℓ-2u(k), kℓ(2-mt) are constants.
Again takingΔℓ-1 on (3.15), by (3.2) and (3.4), we arrive
(3.16)Δℓ-3u(k)∥2ℓ+jk=Δℓ-3u(k)|2ℓ+jk-Δℓ-2u(ℓ+j)kℓ(1)1!ℓ|2ℓ+jk+(-1)1Δℓ-m1u((m1-1)ℓ+j)kℓ(3-mt)(3-mt)!ℓ3-mt|2ℓ+jk+(-1)2Δℓ-m1u((m1-1)ℓ+j)kℓ(1)1!ℓ(ℓ+j)ℓ(2-mt)(2-mt)!ℓ2-mt|2ℓ+jk,
which is the same as
(3.17)Δℓ-3u(k)∥2ℓ+jk=Δℓ-3u(k)|2ℓ+jk+∑t=12‍∑{mt=1},{mt=2}∈1(L2){m1=1,mt=2}∈2(L2)‍(-1)tΔℓ-m1u((m1-1)ℓ+j)×kℓ(3-mt)(3-mt)!ℓ3-mt((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1|2ℓ+jk,
where i=2. In the same way, we find
(3.18)Δℓ-4u(k)∥3ℓ+jk=Δℓ-4u(k)|3ℓ+jk+∑t=13‍∑{mt}∈1(L3),{m1,mt}∈2(L3){m1,m2,mt}∈3(L3)‍(-1)tΔℓ-m1u((m1-1)ℓ+j)×kℓ(4-mt)(4-mt)!ℓ4-mt∏i=23‍((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1|3ℓ+jk,
which can be expressed as the following:(3.19)Δℓ-4u(k)∥3ℓ+jk=Δℓ-4u(k)|3ℓ+jk+∑t=13‍∑{m1,…,mt}∈t(L3)‍(-1)tΔℓ-m1u((m1-1)ℓ+j)×kℓ(4-mt)(4-mt)!ℓ4-mt∏i=23‍((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1|3ℓ+jk.
The proof completes by continuing this process.Theorem 3.7 (partial summation formula).
Letm∈ℕ(2). Then,
(3.20)∑r=m[k/ℓ]‍(r-1)(m-1)(m-1)!u(k-rℓ)=Δℓ-mu(k)|(m-1)ℓ+jk+∑t=1m-1‍∑{m1,…,mt}∈t(Lm-1)‍(-1)t×(Δℓ-m1u((m1-1)ℓ+j))kℓ(m-mt)(m-mt)!ℓm-mt×∏i=2t‍((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1|(m-1)ℓ+jk.Proof.
The proof follows by equating (3.6) and (3.11).Corollary 3.8.
Letm∈ℕ(2), ℓ∈(0,∞), and k∈[mℓ,∞). Then,
(3.21)∑r=m[k/ℓ]‍(r-1)(m-1)(m-1)!e(k-rℓ)(k-rℓ)ℓ(n)=∑r=1n+1‍(-1)r-1n(r-1)(r-1)![(m+r-2)(r-1)ℓr-1(e-ℓ-1)r+m-1×kℓ(n+1-r)e(k+(r-1)ℓ)+∑t=1m-1‍∑{m1,…,mt}∈t(Lm-1)‍×(-1)t(m1+r-2)(r-1)ℓr-1(e-ℓ-1)r+m1-1×((m1-1)ℓ+j)ℓ(n+1-r)kℓ(m-mt)(m-mt)!ℓm-mte((m1-1)ℓ+j+(r-1)ℓ)×∏i=2t‍((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1](m-1)ℓ+jk.Proof.
Takingv(k)=kℓ(n), w(k)=e-k, u(k)=v(k)w(k) in (2.7), by (2.4), (3.1), and (3.2), we get
(3.22)Δℓ-1u(k)=kℓ(n)e-ke-ℓ-1-Δℓ-1(e-k+ℓe-ℓ-1(nℓ)kℓ(n-1)).
By (3.1) and applying (2.7) for kℓ(n-1)e-k, kℓ(n-2)e-k,…,kℓ(1)e-k, we arrive
(3.23)Δℓ-1u(k)=∑r=1n+1‍(-1)r-1n(r-1)(r-1)!(r-1)(r-1)ℓr-1(e-ℓ-1)rkℓ(n+1-r)e(k+(r-1)ℓ).
Taking Δℓ-1 on and applying (3.23) for (m-1) times, we arrive
(3.24)Δℓ-mu(k)=∑r=1n+1‍(-1)r-1n(r-1)(r-1)!(m+r-2)(r-1)ℓr-1(e-ℓ-1)r+m-1kℓ(n+1-r)e(k+(r-1)ℓ).
The proof follows by taking u(k)=kℓ(n)e-k in Theorem 3.7.The following example illustrates Corollary3.8.Example 3.9.
Consider the case whenm=4 and n=2. In this case, L3={1,2,3}, 1(L3)={{1},{2},{3}}, 2(L3)={{1,2},{1,3},{2,3}}, 3(L3)={{1,2,3}} and (3.21) becomes
(3.25)∑r=4[k/ℓ]‍(r-1)(3)(k-rℓ)ℓ(2)(3)!e(k-rℓ)=∑r=13‍(-1)r-12(r-1)(r-1)![(2+r)(r-1)ℓr-1kℓ(1+r)(e-ℓ-1)r+3e(k+(r-1)ℓ)+∑t=13‍∑{m1,…,mt}∈t(L3)‍(-1)t(m1+r-2)(r-1)ℓr-1(e-ℓ-1)r+m1-1×((m1-1)ℓ+j)ℓ(1+r)kℓ(4-mt)(4-mt)!ℓ4-mte((m1-1)ℓ+j+(r-1)ℓ)∏i=2t‍((mi-1)ℓ+j)ℓ(mi-mi-1)(mi-mi-1)!ℓmi-mi-1]3ℓ+jk.
The double summation expression of (3.25) will be obtained by adding the sums corresponds to 1(L3):
(3.26)(r-1)(r-1)ℓr-1(e-ℓ-1)re(j+(r-1)ℓ)(j)ℓ(1+r)kℓ(3)3!ℓ3+(r)(r-1)ℓr-1(e-ℓ-1)r+1e(ℓ+j+(r-1)ℓ)(ℓ+j)ℓ(1+r)kℓ(2)2!ℓ2+(r+1)(r-1)ℓr-1(e-ℓ-1)r+2e(2ℓ+j+(r-1)ℓ)(2ℓ+j)ℓ(1+r)kℓ(1)ℓ,
corresponds to 2(L3):
(3.27)(r-1)(r-1)ℓr-1(e-ℓ-1)re(j+(r-1)ℓ)(j)ℓ(1+r)kℓ(2)2!ℓ2(ℓ+j)ℓ(1)ℓ+(r-1)(r-1)ℓr-1(e-ℓ-1)re(j+(r-1)ℓ)(j)ℓ(1+r)×kℓ(1)ℓ(2ℓ+j)ℓ(2)2!ℓ2+(r)(r-1)ℓr-1(e-ℓ-1)r+1e(ℓ+j+(r-1)ℓ)(ℓ+j)ℓ(1+r)kℓ(1)ℓ(2ℓ+j)ℓ(2)2!ℓ2,
and to 3(L3):
(3.28)(r-1)(r-1)ℓr-1(e-ℓ-1)re(j+(r-1)ℓ)(j)ℓ(1+r)kℓ(1)ℓ(ℓ+j)ℓ(1)ℓ(2ℓ+j)ℓ(1)ℓ.Corollary 3.10.
Letℓ∈(0,∞) and k∈[ℓ,∞). Then,
(3.29)Δℓ-1[kℓ(n)e-k]|jk=∑r=1[k/ℓ]‍(k-rℓ)ℓ(n)e(k-rℓ)=∑r=1n+1‍-n(r-1)ℓr-1kℓ(n+1-r)(1-e-ℓ)re(k+(r-1)ℓ)|jk.Proof.
Sincekℓ(n)e-k=0 as k→∞ and 0ℓ(r)=0 if r≠0 and 0ℓ(r)=1 if r=0, the upper limit of (3.23) for k→∞ will be zero and lower limit for j=0, gives (3.29).


## 4. Discrete Gamma—Factorial Function

First we derive infinite series formula usingΔℓ-1, which induces the definition of discrete gamma factorial function.Theorem 4.1.
Letk∈[0,∞) and limk→∞Δℓ-1u(k)=0. Then,
(4.1)Δℓ-1u(k)|k∞=∑r=0∞‍u(k+rℓ).Proof.
From (2.6), and expressing its terms in reverse order, we find
(4.2)Δℓ-1u(k)|jk+Δℓ-1u(k)|k∞=u(j)+u(ℓ+j)+⋯+u(k-ℓ)+u(k)+u(k+ℓ)+⋯+u(∞),
which is the same as
(4.3)Δℓ-1u(k)|j∞=∑r=0∞‍u(j+rℓ).
Now (4.1) follows by given condition Δℓ-1u(∞)=0 and then replacing j by k.Definition 4.2.
Forn∈(-∞,∞)-{0} and k∈[0,∞), the discrete k-gamma factorial function is defined as
(4.4)kΓℓ((n))=ℓΔℓ-1[kℓ(n-1)e-k]k∞=ℓ∑r=0∞‍(k+rℓ)ℓ(n-1)e-(k+rℓ),
and the discrete k-gamma function is defined as
(4.5)kΓℓ(n)=ℓΔℓ-1[kn-1e-k]k∞=ℓ∑r=0∞‍(k+rℓ)n-1e-(k+rℓ).
In particular,  when k=0, (4.4) and (4.5) becomes
(4.6)Γℓ((n))=ℓΔℓ-1[kℓ(n-1)e-k]k=0∞=ℓ∑r=1∞‍(rℓ)ℓ(n-1)e-rℓ,(4.7)Γℓ(n)=ℓΔℓ-1[kn-1e-k]k=0∞=ℓ∑r=1∞‍(rℓ)n-1e-rℓ
which can be called as the discrete gamma factorial function and the discrete gamma function, respectively.Theorem 4.3.
Letℓ∈(0,∞) and n∈ℕ. Then,
(4.8)kΓℓ((n+1))=ℓ[∑r=0n‍(n)(r)ℓrkℓ(n-r)e-(k+rℓ)(1-e-ℓ)r+1]=ℓ∑r=0∞‍(k+rℓ)ℓ(n)e-(k+rℓ).Proof.
The proof follows by takingj=k, k=∞ in (3.29) and multiplying it by ℓ.Theorem 4.4.
Letℓ∈(0,∞) and n∈ℕ(1). Then,
(4.9)Γℓ((n+1))=ℓe-ℓ1-e-ℓnΓℓ((n))=n!ℓn+1e-nℓ(1-e-ℓ)n+1.Proof.
From (2.7), (3.1), and (4.6), we get
(4.10)Γℓ((n+1))=ℓ{kℓ(n)e-ke-ℓ-1-Δℓ-1{e-(k+ℓ)e-ℓ-1(nℓ)kℓ(n-1)}}k=0∞.
Since kℓ(n)e-k=0 for k=0 and k=∞, (4.10) gives first part of (4.9) by (4.6). Now second part of (4.9) will be obtained by applying first part of (4.9) again and again and using the identity Γℓ((1))=ℓ/(1-e-ℓ).Theorem 4.5.
LetSrn be as given in (2.5) and n∈ℕ(1). Then,
(4.11)Δℓ-m[kne-k]=∑r=1n‍Srnℓn-rΔℓ-m[kℓ(r)e-k],
which gives
(4.12)Γℓ(n+1)=∑r=1n‍Srnℓn-rΓℓ((r+1))=ℓe-ℓ1-e-ℓ∑r=1n‍Srnℓn-rrΓℓ((r)).Proof.
The proof of (4.11) follows by (2.5) and linearity of Δℓ-1. Now, (4.12) will be obtained by taking m=1 in (4.11) and using (4.6), (4.7), and (4.9).Theorem 4.6.
Letℓ∈(0,∞) and n∈ℕ(1). Then,
(4.13)limℓ→0Γℓ(n)=limℓ→0Γℓ((n))=(n-1)!=Γ(n).Proof.
Asℓ→0, all the terms except the last term of (4.12) will be zero. Also, since limk→∞ℓe-ℓ/(1-e-ℓ)=1, by (4.9), we arrive
(4.14)limℓ→0Γℓ(n)=limℓ→0(n-1)Γℓ((n+1))=limℓ→0Γℓ((n)).
Now, (4.13) follows by replacing n by n-1 and taking limit ℓ→0 on (4.9).Theorem 4.7.
Letk∈[ℓ,∞) and j=k-[k/ℓ]ℓ. Then,
(4.15)∑r=1[k/ℓ]‍(kℓ-r)Γ(kℓ-r+1)=Γ(kℓ+1)-Γ(jℓ+1).Proof.
From (2.1) and (2.2), we have
(4.16)ΔℓΓ(kℓ+1)=kℓΓ(kℓ+1),
which yields
(4.17)Δℓ-1[kℓΓ(kℓ+1)]jk=Γ(kℓ+1).
Now, the proof follows from (2.6) and (4.17).Theorem 4.8.
Letk∈[ℓ,∞) and j=k-[k/ℓ]ℓ. Then,
(4.18)∑r=1[k/ℓ]‍(k/ℓ)-r(k/ℓ)-r+11Γ((k/ℓ)-r+1)=1Γ((j/ℓ)+1)-1Γ((k/ℓ)+1).Proof.
From (2.1) and (2.2), we find
(4.19)Δℓ[1Γ((k/ℓ)+1)]=-(k/ℓ)((k/ℓ)+1)Γ((k/ℓ)+1),
which yields by (2.2),
(4.20)-Δℓ-1[(k/ℓ)((k/ℓ)+1)Γ((k/ℓ)+1)]jk=1Γ((k/ℓ)+1)-1Γ((j/ℓ)+1).
Now, the proof follows from (2.6) and (4.20).Corollary 4.9.
Letk∈[ℓ,∞)andj=k-[k/ℓ]ℓ. Then,
(4.21)∑r=0∞‍(k/ℓ)+r(k/ℓ)+r+11Γ((k/ℓ)+r+1)=1Γ((k/ℓ)+1).Proof.
The proof follows from (4.1) and (4.20).Corollary 4.10.
(i) Whenk=0,
(4.22)∑r=0∞‍rr+11Γ(r+1)=1.
(ii) Whenk=1 and ℓ=0.5,
(4.23)∑r=0∞‍2+r3+r1Γ(3+r)=1Γ3=12!.
(iii) Whenk=nℓ, n∈N(1),
(4.24)∑r=0∞‍n+rn+1+r1Γ(n+1+r)=1Γ(n+1)=1n!.



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*Source: 780646-2012-11-13.xml*