Sums of Reciprocals of Triple Binomial Coefficients

Abstract

We investigate the integral representation of infinite sums involving the reciprocals of triple binomial coefficients. We also recover some wellknown properties of 𝜁(3) and extend the range of results given by other authors.

1. Introduction

In this paper, we investigate the summation of the reciprocal of triple products of combinatorial coefficients. In particular, we develop integral representations for^∞_𝑛=01(^{𝑗𝑎𝑛+𝑗})(^{𝑘𝑏𝑛+𝑘})(^{𝑙𝑐𝑛+𝑙}),^∞_𝑛=0(^{𝑛𝑛+𝑚−1})(^{𝑗𝑎𝑛+𝑗})(^{𝑘𝑏𝑛+𝑘})(^{𝑙𝑐𝑛+𝑙}),(1.1)and their alternating series counterparts.

For the representation of sums of reciprocals of single and double binomial coefficients, one may refer to some results in the papers [1–3], see also [4].

For designated cases of the parameter values (𝑎,𝑏,𝑐,𝑗,𝑘,𝑙,𝑚), various particular sums may be expressed in terms of 𝜁(2) and 𝜁(3). For many interesting properties of the Zeta function, the reader is referred to [5].

The representation of sums in terms of integrals is extremely useful because it allows one to estimate bounds on the sums in cases they cannot be written in closed form. Convexity properties for sums may also be investigated.

Apéry's [6], see also Beukers [7], proof of the irrationality of 𝜁(3) uses an elementary and quite complicated construction of the approximants 𝛼_𝑛/𝛽_𝑛∈𝑄 to this number based on a recurrence relation. The integral representation¹⁰𝑥1−𝑥𝑦1−𝑦𝑧(1−𝑧)^𝑛1−(1−𝑥𝑦)𝑧^𝑛+1𝑑𝑥𝑑𝑦𝑑𝑧=2𝛽_𝑛𝜁(3)−2𝛼_𝑛(1.2)for the sequence {𝛼_𝑛,𝛽_𝑛} was proposed.

It is important to note that other integral representations of 𝜁(3) are available in terms of both single and double integrals. Guillera and Sondow [8] list a number of them including the classical results¹⁰−ln(𝑥𝑦)1−𝑥𝑦𝑑𝑥𝑑𝑦=2𝜁(3),₁₀ln(2−𝑥𝑦)1−𝑥𝑦𝑑𝑥𝑑𝑦=58𝜁(3).(1.3)In a recent paper, Muzaffar [9] also obtained some results of the combinatorial type^∞_𝑛=0(^𝑛2𝑛)(^12𝑛+1)(^{12𝑛+𝑘+1})(^{2𝑛+2𝑘𝑛+𝑘})=𝛼^𝑘𝜋²+𝛽_𝑘(1.4)by utilising the power series expansion of (sin⁻¹𝑥)^𝑞, and (𝛼_𝑘,𝛽_𝑘) are constants depending on 𝑘≥0. In this paper, we complement and extend some of the results given by Muzaffar.

There are some identities in the literature involving reciprocals of triple products of combinatorial coefficients, one prominent identity is the Dougall identity, see [10] or [11],1+2^∞_𝑛=1(−1)^𝑛(^𝑎𝑛)(^𝑏𝑛)(^𝑐𝑛)(^{𝑎𝑛+𝑎})(^{𝑏𝑛+𝑏})(^{𝑐𝑛+𝑐})=(^{𝑏𝑎+𝑏+𝑐})(^{𝑏𝑎+𝑏})(^{𝑏𝑏+𝑐})(1.5)for 𝑅(𝑎+𝑏+𝑐)>−1.

2. The Main Results

In this section, we develop integral identities for reciprocals of triple products of binomial coefficients.

Theorem 2.1. For 𝑎,𝑏, and 𝑐 positive real numbers and 𝑗,𝑘,𝑙≥0, then 𝑋𝑌𝑍=𝑥^𝑎𝑦^𝑏𝑧^𝑐.(2.9)𝑆𝑎,𝑏,𝑐,𝑗,𝑘,𝑙=^∞_𝑛=01=𝑗𝑘𝑙(^{𝑗𝑎𝑛+𝑗}_∞)(^{𝑘𝑏𝑛+𝑘})(^{𝑙𝑐𝑛+𝑙})_𝑛=0Γ𝑎𝑛+1Γ𝑗Γ𝑏𝑛+1Γ𝑘Γ𝑐𝑛+1Γ𝑙=𝑗𝑘𝑙_∞Γ𝑎𝑛+𝑗+1Γ𝑏𝑛+𝑘+1Γ𝑐𝑛+𝑙+1_𝑛=0𝐵𝑎𝑛,𝑗+1𝐵𝑏𝑛,𝑘+1𝐵𝑐𝑛,𝑙+1,(2.10)Γ(⋅)𝐵(⋅,⋅)and similarly 𝑆𝑎,𝑏,𝑐,𝑗,𝑘,𝑙=𝑗𝑘𝑙^∞_𝑛=0^1𝑥=01−𝑥^𝑗−1=𝑗𝑘𝑙_1𝑥=0𝑥^𝑎𝑛𝑑𝑥^1𝑦=01−𝑦^𝑘−1𝑦^𝑏𝑛𝑑𝑦^1𝑧=01−𝑧^𝑙−1𝑧^𝑐𝑛𝑑𝑧^1𝑦=0^1𝑧=01−𝑥^𝑗−11−𝑦^𝑘−11−𝑧^∞𝑙−1_𝑛=0𝑥^𝑎𝑦^𝑏𝑧^𝑐^𝑛𝑑𝑥𝑑𝑦𝑑𝑧(2.11)𝑆(𝑎,𝑏,𝑐,𝑗,𝑘,𝑙)=𝑗𝑘𝑙¹⁰1−𝑥^𝑗−11−𝑦^𝑘−11−𝑧^𝑙−11−𝑋𝑌𝑍𝑑𝑥𝑑𝑦𝑑𝑧,(2.12)=𝑆(𝑎,𝑏,𝑐,𝑗,𝑘,𝑙)_∞_𝑛=0𝑎𝑏𝑐𝑛³Γ𝑎𝑛Γ𝑗+1Γ𝑏𝑛Γ𝑘+1Γ𝑐𝑛Γ𝑙+1=Γ𝑎𝑛+𝑗+1Γ𝑏𝑛+𝑘+1Γ𝑐𝑛+𝑙+1_∞_𝑛=0=1+𝑎𝑏𝑐₁₀𝑎𝑏𝑐𝑛³𝐵𝑗+1,𝑎𝑛𝐵𝑘+1,𝑏𝑛𝐵𝑘+1,𝑐𝑛1−𝑥^𝑗1−𝑦^𝑘1−𝑧^𝑙𝑥𝑦𝑧^∞_𝑛=1=1+𝑎𝑏𝑐₁₀𝑛³𝑥^𝑎𝑦^𝑏𝑧^𝑐^𝑛𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥^𝑗1−𝑦^𝑘1−𝑧^𝑙𝑥𝑦𝑧1−𝑋𝑌𝑍⁴𝑋𝑌𝑍𝑋𝑌𝑍²+4𝑋𝑌𝑍+1𝑑𝑥𝑑𝑦𝑑𝑧(2.13)_{𝑗+𝑘+𝑙+1}𝐹_{𝑗+𝑘+𝑙}[𝑎+11,𝑎1,𝑎2,…,𝑎𝑗,1𝑏,2𝑏,…,𝑘𝑏,1𝑐,2𝑐,…,𝑐𝑙=𝑎,𝑎+𝑎2,…,𝑎+𝑎𝑗,𝑏+𝑏1,𝑏+𝑏2,…,𝑏+𝑏𝑘,𝑐+𝑐1,𝑐+𝑐2,…,𝑐+𝑐𝑙|1]_{𝑎+𝑏+𝑐+1}𝐹_{𝑎+𝑏+𝑐}[𝑗+11,1,1,1,𝑎1,𝑎2,…,𝑎−𝑎1,1𝑏,2𝑏,…,𝑏−𝑏1,1𝑐,2𝑐,…,𝑐−𝑐1𝑎,𝑗+𝑎2,…,𝑗+𝑎𝑎,𝑘+𝑏1,𝑘+𝑏2,…,𝑘+𝑏𝑏,𝑙+𝑐1,𝑙+𝑐2,…,𝑙+𝑐𝑐|1],_{𝑗+𝑘+𝑙+1}𝐹_{𝑗+𝑘+𝑙}[𝑎+11,𝑎1,𝑎2,…,𝑎𝑗,1𝑏,2𝑏,…,𝑘𝑏,1𝑐,2𝑐,…,𝑐𝑙=𝑎,𝑎+𝑎2,…,𝑎+𝑎𝑗,𝑏+𝑏1,𝑏+𝑏2,…,𝑏+𝑏𝑘,𝑐+𝑐1,𝑐+𝑐2,…,𝑐+𝑐𝑙|−1]_{𝑎+𝑏+𝑐+1}𝐹_{𝑎+𝑏+𝑐}[𝑗+11,1,1,1,𝑎1,𝑎2,…,𝑎−𝑎1,1𝑏,2𝑏,…,𝑏−𝑏1,1𝑐,2𝑐,…,𝑐−𝑐1𝑎,𝑗+𝑎2,…,𝑗+𝑎𝑎,𝑘+𝑏1,𝑘+𝑏2,…,𝑘+𝑏𝑏,𝑙+𝑐1,𝑙+𝑐2,…,𝑙+𝑐𝑐|−1](2.14)where 𝑆1,1,1,1,1,1=^∞_𝑛=01𝑛+1³=𝜁3=¹⁰𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥𝑦𝑧=⁴=1+₁₀𝐹₃[12,1,2,1,2,1|1]1−𝑥1−𝑦(1−𝑧)(𝑥𝑦𝑧)²+4𝑥𝑦𝑧+11−𝑥𝑦𝑧⁴𝑑𝑥𝑑𝑦𝑑𝑧.(3.1)

Proof. Consider (2.1):𝜁(3)where (−1)^𝑛𝑛!𝜁𝑛+1=^1𝑥=0ln𝑥^𝑛1−𝑥𝑑𝑥=^1𝑥=0ln1−𝑥^𝑛𝑥𝑑𝑥.(3.2) is the classical Gamma function and =𝑆2,2,2,1,1,1_∞_𝑛=012𝑛+1³=78𝜁3=^𝜋/4𝑥=0lncos(𝑥)lnsin(𝑥)cos(𝑥)sin(𝑥)𝑑𝑥=¹⁰𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥²𝑦²𝑧²=1+8¹⁰1−𝑥1−𝑦(1−𝑧)𝑥𝑦𝑧(𝑥𝑦𝑧)⁴+4(𝑥𝑦𝑧)²+11−𝑥²𝑦²𝑧²⁴=𝑑𝑥𝑑𝑦𝑑𝑧₄𝐹₃[213,12,12,12,32,23|1],(3.3) is the Beta function. It holds that 𝜁3=−5^{ln(𝜙2)𝑥=0}𝑥ln2sinh𝑥2𝑑𝑥,(3.4)by an allowable change of integral and sum, and hence we have 𝜙which is the result (2.2).
To prove identity (2.3), consider (2.1) and expand as follows:(1+√5)/2which is the result (2.3). The results (2.6) and (2.7) may be obtained in a similar fashion and therefore will not be pursued here.

The hypergeometric representation (2.4) and (2.8) can be obtained by the consideration of the ratio of successive terms (2.1) and (2.5), respectively.

We may also note that from known properties of the hypergeometric function, we may write, from (2.4) and (2.8),𝑆4,2,3,𝑗,𝑘,𝑙=^∞_𝑛=01=_∞(^{𝑗4𝑛+𝑗})(^{𝑘2𝑛+𝑘})(^{𝑙3𝑛+𝑙})_𝑛=0𝑗!𝑘!𝑙!=𝑗𝑘𝑙∏^𝑗𝑟=1₁₀4𝑛+𝑟∏^𝑘𝑟=12𝑛+𝑟∏^𝑙𝑟=13𝑛+𝑟(1−𝑥)^𝑗−1(1−𝑦)^𝑘−1(1−𝑧)^𝑙−1𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥⁴𝑦²𝑧³=₁₀𝐹₉[𝑗+11,1,1,1,34,23,21,12,13,41=𝛼₁4,𝑗+42,𝑗+43𝑗+44,𝑘+21,𝑘+22,𝑙+31,𝑙+32,𝑙+33|1]+𝛼₂𝜋+𝛼₃𝜁2+𝛼₄ln(2)+𝛼₅ln(3)+𝛼₆𝜁(3).(3.5)Examples
Example 3.1. It holds that 𝑗=5,𝑘=6
Other integral representations of 𝑙=6 do exist, some of which are as follows.
Finch [12] gave the expression𝛼₁=49576279909317⋅11⋅7²⋅3²⋅2⁴,𝛼₂=167⋅2²⁵17⋅13⋅11⋅7²⋅3²−23⋅5⋅3¹⁶√3𝛼₃17⋅13⋅11⋅7²⋅2⁴,=1709⋅5⋅27,𝛼⁴=−755357⋅2¹⁹17⋅13⋅11⋅7²⋅3²,𝛼₅=43⋅23⋅5⋅3¹⁶17⋅13⋅11⋅7²⋅2⁴,𝛼₆=5³⋅3⋅2².(3.6)
Lord [13] posed the problem to show that𝑇1,1,1,1,1,1=^∞_𝑛=0−1^𝑛𝑛+1³=3=4𝜁(3)₁₀𝑑𝑥𝑑𝑦𝑑𝑧1+𝑥𝑦𝑧=⁴=1−₁₀𝐹₃[12,1,2,1,2,1|−1]1−𝑥1−𝑦(1−𝑧)(𝑥𝑦𝑧)²−4𝑥𝑦𝑧+11+𝑥𝑦𝑧⁴𝑑𝑥𝑑𝑦𝑑𝑧.(3.7)the last three expressions are directly from (2.2), (2.3), and (2.4), respectively.
Nan-Yue and Williams [14] also gave𝑇2/=3,2/3,5/6,2,4,3_∞_𝑛=0−1^𝑛=_∞(^22𝑛/3+2)(^42𝑛/3+4)(^35𝑛/6+3)_𝑛=0−1^𝑛2⁵3¹¹=24𝑛₁₀+3²(𝑛+6)2𝑛+3²(2𝑛+9)(5𝑛+6)(5𝑛+12)(5𝑛+18)1−𝑥1−𝑦³1−𝑧²𝑑𝑥𝑑𝑦𝑑𝑧1+𝑥^2/3𝑦^2/3𝑧^5/6=₁₀𝐹₉[15,32,32,3,3,92,6,65,152,158=720,65263,4⋅,146,61291,151,157,253,7|−1]13⋅11⋅7⋅5⋅3⋅2+3³⋅2⁴𝜁(2)+3²⋅2¹⁰7𝐺+1091⋅15⋅⋅73²⋅2⁹ln2−2⁸⋅3⁴11⋅7²+113⋅5₆𝜋⋅√511⋅7²⋅2²−271⋅5⁵11⋅7²⋅2³ln𝛼−113⋅5⁶⋅√511⋅7²⋅2²+271⋅5⁵−511⋅7²⋅2³ln₄𝜙⋅3⋅√5⋅𝜙𝛼√511⋅2+37⋅5⁴⋅√5⋅𝛼𝜙√57²𝜋,(3.8)where 𝐺 = golden ratio = 𝜙.
Example 3.2. It holds that 𝛼For (√5−1)/2., and 𝑎,𝑏,𝑐, we have the values 𝑚
Example 3.3. For the alternating case,𝑗,𝑘,𝑙≥0
Example 3.4. It holds that 𝑗+𝑘+𝑙≥𝑚,where 𝑋𝑌𝑍 is Catalan's constant, 𝑝_𝛼=𝑝𝑝+1⋯𝑝+𝛼−1=Γ𝑝+𝛼Γ𝑝(3.19) is the golden ratio, and 𝑅𝑎,=𝑏,𝑐,𝑗,𝑘,𝑙,𝑚_∞_𝑛=0−1^𝑛(^{𝑛𝑛+𝑚−1})=_∞(^{𝑗𝑎𝑛+𝑗})(^{𝑘𝑏𝑛+𝑘})(^{𝑙𝑐𝑛+𝑙})_𝑛=0−1^𝑛𝑛+𝑚𝑛−1𝑎𝑏𝑐𝑛³Γ𝑎𝑛Γ𝑗+1Γ𝑏𝑛Γ𝑘+1Γ𝑐𝑛Γ𝑙+1=Γ𝑎𝑛+𝑗+1Γ𝑏𝑛+𝑘+1Γ𝑐𝑛+𝑙+1_∞_𝑛=0−1^𝑛𝑎𝑏𝑐𝑛³=1+𝑎𝑏𝑐_∞𝑛+𝑚𝑛−1𝐵𝑎𝑛,𝑗+1𝐵𝑏𝑛,𝑘+1𝐵𝑐𝑛,𝑙+1_𝑛=1−1^𝑛𝑛³𝑛+𝑚𝑛−1¹⁰1−𝑥^𝑗−1𝑥^𝑎𝑛𝑑𝑥¹⁰1−𝑦^𝑘−1𝑦^𝑏𝑛𝑑𝑦¹⁰1−𝑧^𝑙−1𝑧^𝑐𝑛𝑑𝑧.(3.20) = silver ratio = =1+𝑎𝑏𝑐₁₀𝑅𝑎,𝑏,𝑐,𝑗,𝑘,𝑙,𝑚1−𝑥^𝑗1−𝑦^𝑘1−𝑧^𝑙𝑥𝑦𝑧^∞_𝑛=1(−1)^𝑛𝑛+𝑚𝑚−−11𝑛³𝑥^𝑎=1−𝑚𝑎𝑏𝑐₁₀𝑦^𝑏𝑧^𝑐^𝑛𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥^𝑗1−𝑦^𝑘1−𝑧^𝑙𝑋𝑌𝑍𝑋𝑌𝑍²−(3𝑚+1)𝑋𝑌𝑍+1𝑥𝑦𝑧1+𝑋𝑌𝑍^𝑚+3𝑑𝑥𝑑𝑦𝑑𝑧(3.21)

Now consider the following theorem, which is a generalisation of Theorem 2.1.

Theorem 3.5. For =𝑗𝑘𝑙_∞𝑅𝑎,𝑏,𝑗,𝑐,𝑘,𝑙,𝑚=_𝑛=0𝑗−𝑘𝑙1_∞^𝑛𝑛+𝑚𝑛−1ΓΓ𝑎𝑛𝑎𝑛++1𝑗Γ+𝑗1ΓΓ𝑏𝑏𝑛𝑛++1𝑘+Γ1𝑘ΓΓ𝑐𝑐𝑛𝑛++𝑙1+Γ1𝑙=𝑗𝑘𝑙_∞_𝑛=0−1^𝑛𝑛+𝑚𝑛−1𝐵𝑎𝑛+1,𝑗𝐵𝑏𝑛+1,𝑘𝐵𝑐𝑛+1,𝑙_𝑛=0(−1)^𝑛𝑛+𝑚𝑛−1¹⁰=𝑗𝑘𝑙₁₀1−𝑥^𝑗−1𝑥^𝑎𝑛𝑑𝑥¹⁰1−𝑦^𝑘−1𝑦^𝑏𝑛𝑑𝑦¹⁰1−𝑧^𝑙−1𝑧^𝑐𝑛𝑑𝑧1−𝑥^𝑗−11−𝑦^𝑘−11−𝑧^∞𝑙−1_𝑛=0(−1)^𝑛𝑛+𝑚𝑛−1𝑥^𝑎𝑦^𝑏𝑧^𝑐^𝑛𝑑𝑥𝑑𝑦𝑑𝑧(3.22), and 𝑅𝑎,𝑏,𝑗,𝑐,𝑘,𝑙,𝑚=𝑗𝑘𝑙¹⁰1−𝑥^𝑗−11−𝑦^𝑘−11−𝑧^𝑙−11+𝑋𝑌𝑍^𝑚𝑑𝑥𝑑𝑦𝑑𝑧(3.23) positive real numbers and 𝑚=1, with =𝑄4,3,2,5,3,6,11_∞_𝑛=0(^𝑛𝑛+10)=90₁₀(^54𝑛+5)(^33𝑛+3)(^62𝑛+6)1−𝑥⁴1−𝑦²1−𝑧⁵1−𝑥⁴𝑦³𝑧²¹¹=1+264₁₀𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥⁵1−𝑦³1−𝑧⁶𝑥³𝑦²𝑧𝑥⁸𝑦⁶𝑧⁴+34𝑥⁴𝑦³𝑧²+1𝑑𝑥𝑑𝑦𝑑𝑧1−𝑥⁴𝑦³𝑧²¹⁴=₁₀𝐹₉[114,,17,1,1,34,23,21,12,13,41=3413⋅43⋅5⋅32,49,2,2,53,47,32,43|1]2¹⁴𝜁2+1931⋅5092¹²−90379√37⋅3⋅2⁵𝜋−1745⋅93²⋅⋅1223₁₁1−3798807797⋅3⋅2¹⁰ln2+22567⋅3²7⋅2⁵ln3.(4.1) then 𝑅1/4,1/6,1/2,5,3,7,14∶=^∞_𝑛=0−1^𝑛(^𝑛𝑛+13)=_∞(^5𝑛/4+5)(^3𝑛/6+3)(^7𝑛/2+7)_𝑛=0(−1)^𝑛5!3!7!𝑛+1₁₃13!𝑛/4+1₅𝑛/6+1₃𝑛/2+1₇=105¹⁰1−𝑥⁴1−𝑦²1−𝑧⁶1+𝑥^1/4𝑦^1/6𝑧^1/2¹⁴=1+7𝑑𝑥𝑑𝑦𝑑𝑧24¹⁰1−𝑥⁵1−𝑦³1−𝑧⁷×𝑥^1/2𝑦^1/3𝑧+43𝑥^1/4𝑦^1/6𝑧^1/2+1𝑥^3/4𝑦^5/6𝑧^1/21−𝑥^1/4𝑦^1/6𝑧^1/2¹⁷=𝑑𝑥𝑑𝑦𝑑𝑧₁₆=7𝐹⋅₁₅5[2⋅,334₃,,54,,56,,76,,78,,98,,91,01,11,21,31,21,31,21,31,41,51,41,71,61,91,82,120|−1]⋅2⁴13𝜁2−16273⋅15⋅⋅121³⋅3.(4.2)𝜁(3)^∞_𝑛=0𝑛^𝑠(^{𝑛𝑛+𝑚−1})(^{𝑗𝑎𝑛+𝑗})(^{𝑘𝑏𝑛+𝑘})(^{𝑙𝑐𝑛+𝑙}),^∞_𝑛=0(^{𝑛𝑛+𝑚−1})(^{𝑎𝑛+𝑗𝑏𝑛})(^{𝑐𝑛+𝑘𝑑𝑛})(^{𝑝𝑛+𝑗𝑞𝑛}).(5.1)𝜁𝜁𝜁𝜁(sin⁻¹𝑥)^𝑞log(2sinh(𝜃/2))log(2sin(𝜃/2))where is given by (2.9) and is Pochhammer's symbol.

Proof. Consider (3.14):By an allowable change of integral and sum, we havewhich is the result (3.17).
To arrive at the result (3.16), considerby an allowable change of sum and integral, hencewhich is the result (3.16).

The hypergeometric representations (3.13) and (3.18) can be obtained by the consideration of the ratio of successive terms (3.9) and (3.14), respectively.

In the case when Theorem 3.5 reduces to Theorem 2.1.

Examples
Example 4.1. It holds that
Example 4.2. It holds that

3. Conclusion

We have provided triple integral identities for sums of the reciprocal of triple binomial coefficients. In doing so, we have recovered the standard representation for and have generalised and extended some results published previously by other authors.

In another forum, we will extend our results to consider binomial coefficients of the form

Acknowledgment

This paper was completed while the author was a Visiting Professor at the Dipartimento di Sistemi e Informatica, Universita di Firenze. I wish to express my sincere thanks to Professor Sprugnoli for his hospitality.