Further Formulae for Harmonic Series with Convergence Rate “−1/4”

Abstract

By applying the “coefficient extraction method” to the symmetric transformation of hypergeometric series due to Chu and Zhang (2014), an overview is presented systematically for a large class of infinite series of convergence rate “$-1/4$” concerning harmonic numbers. Numerous closed formulae in terms of mathematical constants (such as $\pi$, $ln2$ and the Riemann zeta values) are established. They may serve as a reference source for readers in their further investigations.

1. Introduction and Outline

Among many useful functions in mathematics and physics, the $\mathrm{\Gamma }$-function plays an important role (cf. Rainville [1], §8):

$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\infty }{y}^{x-1}{e}^{-y}\mathrm{d}y\mathrm{for}\Re \left(x\right)>0.$$

The two Maclaurin series expansions (cf. [2]) below will be extensively utilized in this paper:

$$\begin{array}{cc}\hfill \mathrm{\Gamma }(1-x)& =exp\left\{\sum _{k\ge 1}{\displaystyle \frac{{\sigma }_k}{k}}{x}^k\right\},\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill \mathrm{\Gamma }(\frac{1}{2}-x)& =\sqrt{\pi }exp\left\{\sum _{k\ge 1}{\displaystyle \frac{{\tau }_k}{k}}{x}^k\right\};\hfill \end{array}$$

(2)

where ${\sigma }_k$ and ${\tau }_k$ are explicitly given by

$$\begin{array}{cccccccccc}& {\sigma }_1=\gamma \hfill & & \mathrm{and}\hfill & & {\sigma }_m=\zeta \left(m\right)\hfill & & \mathrm{for}\hfill & & m\ge 2;\hfill \\ & {\tau }_1=\gamma +2ln2\hfill & & \mathrm{and}\hfill & & {\tau }_m=(2^m-1)\zeta \left(m\right)\hfill & & \mathrm{for}\hfill & & m\ge 2;\hfill \end{array}$$

with the Euler–Mascheroni constant and the Riemann zeta function being denoted by

$$\gamma =\underset{n\to \infty }{lim}\left({\mathbf{H}}_n-lnn\right)\mathrm{and}\zeta \left(\lambda \right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^{\lambda }}}\mathrm{for}\Re \left(\lambda \right)>1.$$

For $n\in {\mathbb{N}}_0$ and an indeterminate x, the shifted factorials are usually defined by

$${\left(x\right)}_0=1\mathrm{and}{\left(x\right)}_n=x(x+1)\cdots (x+n-1)\mathrm{with}n\in \mathbb{N}.$$

In general, we can express it as the $\mathrm{\Gamma }$-function ratio

$${\left(x\right)}_n={\displaystyle \frac{\mathrm{\Gamma }(x+n)}{\mathrm{\Gamma }\left(x\right)}},\mathrm{where}n\in \mathbb{Z}.$$

To reduce lengthy expressions, the following compact notations will be employed:

$$\begin{array}{cc}\hfill {\left[\alpha ,\beta ,\cdots ,\gamma \right]}_n& ={\left(\alpha \right)}_n{\left(\beta \right)}_n\cdots {\left(\gamma \right)}_n,\hfill \\ \hfill {\left[\begin{array}{cccc}\alpha ,& \beta ,& \cdots ,& \gamma \\ A,& B,& \cdots ,& C\end{array}\right]}_n& ={\displaystyle \frac{{\left(\alpha \right)}_n{\left(\beta \right)}_n\cdots {\left(\gamma \right)}_n}{{\left(A\right)}_n{\left(B\right)}_n\cdots {\left(C\right)}_n}},\hfill \\ \hfill \mathrm{\Gamma }\left[\begin{array}{cccc}\alpha ,& \beta ,& \cdots ,& \gamma \\ A,& B,& \cdots ,& C\end{array}\right]& ={\displaystyle \frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)\cdots \mathrm{\Gamma }\left(\gamma \right)}{\mathrm{\Gamma }\left(A\right)\mathrm{\Gamma }\left(B\right)\cdots \mathrm{\Gamma }\left(C\right)}}.\hfill \end{array}$$

For $x\in \mathbb{R}$ and $n,\lambda \in {\mathbb{N}}_0$, define the harmonic numbers by

$$\begin{array}{cccc}& {\mathbf{H}}_n^{\langle \lambda \rangle }\left(x\right)=\sum _{k=0}^{n-1}{\displaystyle \frac{1}{{(x+k)}^{\lambda }}},\hfill & \hfill & {\overline{\mathbf{H}}}_n^{\langle \lambda \rangle }\left(x\right)=\sum _{k=0}^{n-1}{\displaystyle \frac{{(-1)}^k}{{(x+k)}^{\lambda }}};\hfill \\ & {\mathbf{O}}_n^{\langle \lambda \rangle }\left(x\right)=\sum _{k=0}^{n-1}{\displaystyle \frac{1}{{(x+2k)}^{\lambda }}},\hfill & \hfill & {\overline{\mathbf{O}}}_n^{\langle \lambda \rangle }\left(x\right)=\sum _{k=0}^{n-1}{\displaystyle \frac{{(-1)}^k}{{(x+2k)}^{\lambda }}}.\hfill \end{array}$$

When $\lambda =1$ and/or $x=1$, they will be omitted from these notations. We record also the following simple, but useful relations:

$$\begin{array}{cccc}& {\mathbf{H}}_{2n}^{\langle \lambda \rangle }={\mathbf{O}}_n^{\langle \lambda \rangle }+2^{-\lambda }{\mathbf{H}}_n^{\langle \lambda \rangle },\hfill & \hfill & {\mathbf{H}}_n^{\langle \lambda \rangle }\left(\frac{1}{2}\right)=2^{\lambda }{\mathbf{O}}_n^{\langle \lambda \rangle };\hfill \\ & {\overline{\mathbf{H}}}_{2n}^{\langle \lambda \rangle }={\mathbf{O}}_n^{\langle \lambda \rangle }-2^{-\lambda }{\mathbf{H}}_n^{\langle \lambda \rangle },\hfill & \hfill & {\overline{\mathbf{H}}}_n^{\langle \lambda \rangle }\left(\frac{1}{2}\right)=2^{\lambda }{\overline{\mathbf{O}}}_n^{\langle \lambda \rangle }.\hfill \end{array}$$

Let $\left[x^m\right]\varphi \left(x\right)$ stand for the coefficient of $x^m$ in the formal power series $\varphi \left(x\right)$. Then harmonic numbers can be obtained by extracting coefficients:

$$\begin{array}{cccc}& \left[x\right]{\displaystyle \frac{{(1+x)}_n}{n!}}={\mathbf{H}}_n,\hfill & \hfill & \left[x^2\right]{\displaystyle \frac{{(1+x)}_n}{n!}}={\displaystyle \frac{{\mathbf{H}}_n^2-{\mathbf{H}}_n^{\langle 2\rangle }}{2}},\hfill \\ & \left[x\right]{\displaystyle \frac{n!}{{(1-x)}_n}}={\mathbf{H}}_n,\hfill & & \left[x^2\right]{\displaystyle \frac{n!}{{(1-x)}_n}}={\displaystyle \frac{{\mathbf{H}}_n^2+{\mathbf{H}}_n^{\langle 2\rangle }}{2}},\hfill \\ & \left[y\right]{\displaystyle \frac{{(\frac{1}{2}+y)}_n}{{\left(\frac{1}{2}\right)}_n}}=2{\mathbf{O}}_n,\hfill & & \left[y^2\right]{\displaystyle \frac{{(\frac{1}{2}+y)}_n}{{\left(\frac{1}{2}\right)}_n}}=2({\mathbf{O}}_n^2-{\mathbf{O}}_n^{\langle 2\rangle }),\hfill \\ & \left[y\right]{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n}{{(\frac{1}{2}-y)}_n}}=2{\mathbf{O}}_n,\hfill & & \left[y^2\right]{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n}{{(\frac{1}{2}-y)}_n}}=2({\mathbf{O}}_n^2+{\mathbf{O}}_n^{\langle 2\rangle }).\hfill \end{array}$$

By means of the generating function technique, it is not hard to prove that (cf. Chu [3]) in general there hold the relations:

$$\left[x^m\right]{\displaystyle \frac{{(\lambda -x)}_n}{{\left(\lambda \right)}_n}}={\mathbf{Y}}_m(-\mathrm{hh})\mathrm{and}\left[x^m\right]{\displaystyle \frac{{\left(\lambda \right)}_n}{{(\lambda -x)}_n}}={\mathbf{Y}}_m\left(\mathrm{hh}\right).$$

(3)

Here “${\mathrm{hh}}_k$” denotes the higher harmonic number ${\mathrm{hh}}_k:={\mathbf{H}}_n^{\langle k\rangle }\left(\lambda \right)$, and the Bell polynomials ([4], §3.3) are given explicitly by

$${\mathbf{Y}}_m(\pm \mathrm{hh})=\sum _{\begin{array}{c}\omega \left(m\right)\end{array}}\prod _{k=1}^m{\displaystyle \frac{{\left\{\pm {\mathbf{H}}_n^{\langle k\rangle }\left(\lambda \right)\right\}}^{{\ell }_k}}{{\ell }_k!k^{{\ell }_k}}},$$

(4)

where the multifold sum runs over $\omega \left(m\right)$, the set of m-partitions represented by m-tuples of $({\ell }_1,{\ell }_2,\dots ,{\ell }_m)\in {\mathbb{N}}_0^m$ satisfying the condition ${\displaystyle \sum _{k=1}^m}k{\ell }_k=m$.

There exist numerous infinite series about harmonic numbers (cf. [5,6,7,8,9,10]). They have actually become active research topics for their applications in mathematics (particularly number theory, analytic combinators), physics (standing waves in strings), and computer science (algorithmic analysis), just for examples. Recently, the “coefficient extraction” method has been shown to be powerful in evaluating, in closed form, infinite series involving harmonic numbers (see, for example, [2,3,11,12]), which can briefly be sketched as follows:

• For a given hypergeometric series equality (formula or transformation) $\mathrm{\Omega }(a;b,c,d,e)$, reformulate it by identifying a variable “x” and eventual parameters $\{a,b,c,d,e\}$, so that both sides of the resulting equality $\mathrm{\Omega }(a;b,c,d,e|x)$ are analytic in x at $x=0$.
• By applying (1)–(3), extract across $\mathrm{\Omega }(a;b,c,d,e|x)$ the coefficients of $x^m$ for small integer values of m and then denote the resulting equality by ${\mathrm{\Omega }}_m(a;b,c,d,e)$.
• Assigning parameters $\{a,b,c,d,e\}$ to specific numerical values, we find the corresponding infinite series identity for ${\mathrm{\Omega }}_m(a;b,c,d,e)$.

By carrying out this procedure, we shall present an overview in this paper through a useful symmetric transformation of hypergeometric series in [13]. In order to facilitate subsequent applications, we reproduce it below and will refer to it in short as $\mathrm{\Omega }(a;b,c,d,e)$.

For five complex parameters $\{a,b,c,d,e\}$ subject to $\Re (1+2a-b-c-d-e)>0$, define the well-poised sum (whose numerator parameters can be paired off with denominator parameters such that each pair has the same sum; see Bailey [14], §2.5) by

$$\mathrm{\Omega }(a;b,c,d,e):=\sum _{k=0}^{\infty }(a+2k){\left[\begin{array}{c}b,c,d,e\\ 1+a-b,1+a-c,1+a-d,1+a-e\end{array}\right]}_k.$$

(5)

When $a=e$, this series can be evaluated by Dougall’s theorem (cf. Bailey [14], §4.4) as

$$\mathrm{\Omega }(a;b,c,d,a)=\mathrm{\Gamma }\left[\begin{array}{c}1+a-b,1+a-c,1+a-d,1+a-b-c-d\\ a,1+a-b-c,1+a-b-d,1+a-c-d\end{array}\right].$$

(6)

Then, the symmetric transformation discovered by Chu and Zhang [13] (Theorem 10), denominated subsequently by Ω(a;b,c,d,e), is recorded below:

$$\begin{gathered} \mathrm{\Omega }(a;b,c,d,e)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n\times {\mathrm{\Delta }}_n(a,b,c,d,e)}{{(1+2a-b-c-d-e)}_{2n+2}}} \\ \times {\displaystyle \frac{\left\{\begin{array}{c}{(1+a-b-c)}_n{(1+a-b-d)}_n{(1+a-b-e)}_n\\ {(1+a-c-d)}_n{(1+a-c-e)}_n{(1+a-d-e)}_n\end{array}\right\}}{{(1+a-b)}_n{(1+a-c)}_n{(1+a-d)}_n{(1+a-e)}_n}}, \end{gathered}$$

(7)

where the cubic polynomial ${\mathrm{\Delta }}_n(a,b,c,d,e)$ is given by

$$\begin{array}{r}{\mathrm{\Delta }}_n(a,b,c,d,e)=(1+a-b-c+n)(1+a-b-d+n)(1+a-c-d+n)\\ +(a-e+n)(1+2a-b-c-d+2n)(2+2a-b-c-d-e+2n).\end{array}$$

(8)

In the rest of the paper, five classes of harmonic series will be systematically investigated by exclusively making use of well-poised series transformation (7) for $\mathrm{\Omega }(a;b,c,d,e)$. Numerous identities for infinite series of convergence rate “$\frac{-1}{4}$” (i.e., the limit of the term ratio ${\displaystyle \frac{T_{n+1}}{T_n}}$ as $n\to \infty$ for the series ${\sum }_n{T}_n$) containing harmonic numbers will be shown. Among those identities established in this paper, we anticipate the following five representatives as examples:

$$\begin{array}{cc}\hfill \mathrm{Equation}\left(11\right):& \sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{3}{n^2}}+{\displaystyle \frac{2(2+5n)}{n(1+2n)}}{\mathbf{O}}_n\right\}=-{\displaystyle \frac{{\pi }^2}{3}},\hfill \\ \hfill \mathrm{Equation}\left(19\right):& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{3(1+2n)-(2+27n+45n^2){\mathbf{H}}_{3n}\right\}=6,\hfill \\ \hfill \mathrm{Equation}\left(20\right):& \sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+4n)}{n^3}}-{\displaystyle \frac{2(3+10n)}{n}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}=2\zeta \left(3\right),\hfill \\ \hfill \mathrm{Equation}\left(22\right):& \sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{2(16+63n+54n^2)}{3(1+2n)}}+(59+216n+180n^2){\mathbf{O}}_{3n}\right\}=-9\sqrt{3}\pi ,\hfill \\ \hfill \mathrm{Equation}\left(23\right):& \sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{3(3+4n)}{2}}-(8+45n+45n^2){\mathbf{H}}_{3n+1}\right\}=3ln\left({\displaystyle \frac{16}{27}}\right).\hfill \end{array}$$

Nowadays, computer algebra has become indispensable for symbolic calculus. Most closed form formulae presented in this paper are obtained by making use of the inbuilt code of “Wolfram Mathematica” (version 11). We have also verified experimentally all the displayed equations in order to ensure accuracy.

2. Series with Factorial Quotient $\begin{array}{|c|}\hline \frac{\mathit{n}!}{{\frac{\mathbf{1}}{\mathbf{2}}}_{\mathit{n}}}={\mathbf{4}}^{\mathit{n}}/\left(\begin{array}{c}\mathbf{2}\mathit{n}\\ \mathit{n}\end{array}\right)\\ \hline\end{array}$

There are numerous infinite series identities containing central binomial coefficients (cf. [15,16,17]). In this section, we evaluate a number of harmonic series (cf. [18,19,20]) with the central binomial coefficient in denominators. These series are divided into two groups. A few of them have appeared previously in [21] (§2.2 and 3.2) as explicitly indicated.

2.1. Series in Group A

For five parameters $\{a,b,c,d,e\}$ subject to the conditions $\Re (1+2a-b-c-d-e)>0$ and $\Re (a-e)>0$, the following reciprocal identity (cf. Chu and Zhang [13], Theorem 5) holds:

$$\mathrm{\Omega }(a;b,c,d,e)=(a-e){\displaystyle \frac{\mathrm{\Omega }(1+2a-b-c-d;1+a-b-c,1+a-b-d,1+a-c-d,e)}{1+2a-b-c-d-e}}.$$

Now performing the parameter replacements in (7)

$$a\to {\displaystyle \frac{1}{2}}+ax,b\to {\displaystyle \frac{1}{2}}+bx,c\to 1+cx,d\to {\displaystyle \frac{1}{2}}+dx,e\to ex;$$

and then reformulating it by applying the above reciprocal relation for the $\mathrm{\Omega }$-series, we find, after some simplifications, the following transformation formula.

Theorem 1.

For a variable x and five complex parameters $\{a,b,c,d,e\}$, there holds

$$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{(2k+2ax-bx-cx-dx)}{(k+ax-bx-cx)(k+ax-cx-dx)}}\hfill \\ & \times {\left[\begin{array}{c}1+ax-bx-cx,1+ax-cx-dx,ex,{\displaystyle \frac{1}{2}}+ax-bx-dx\\ 1+ax-bx,1+ax-dx,1+2ax-bx-cx-dx-ex,{\displaystyle \frac{1}{2}}+ax-cx\end{array}\right]}_k\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+ax-bx-dx,{\displaystyle \frac{1}{2}}+ax-cx-ex\\ {\displaystyle \frac{1}{2}}+ax-cx,{\displaystyle \frac{1}{2}}+ax-ex\end{array}\right]}_n{\displaystyle \frac{{\mathrm{\Delta }}_n\left(\frac{1}{2}+ax,\frac{1}{2}+bx,1+cx,\frac{1}{2}+dx,ex\right)}{{(1+2ax-bx-cx-dx-ex)}_{2n+1}}}\hfill \\ & \times {\displaystyle \frac{{(-1)}^n{\left[1+ax-bx-cx,1+ax-bx-ex,1+ax-cx-dx,1+ax-dx-ex\right]}_n}{{\left[1+ax-bx,1+ax-dx\right]}_n(n+ax-bx-cx)(n+ax-cx-dx)(\frac{1}{2}+n+ax-ex)}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. Specifying further parameters $\{a,b,c,d,e\}$ with particular values, we can establish from Theorem 1 several infinite series identities. Their derivations are almost mechanical. As examples, we illustrate the three identities concerning the harmonic numbers of the first order.

According to (1)–(4), we write the equality by extracting the coefficient of x across the equation displayed in Theorem 1:

$$\begin{array}{cc}& \begin{array}{r}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\{{\displaystyle \frac{(2a-b-3c-d-3e)(2+5n)}{2n(1+2n)}}{\mathbf{H}}_n-{\displaystyle \frac{(2a-b-3c-d)(1+4n)}{n^2{(1+2n)}^2}}\\ -{\displaystyle \frac{(2a+b-c+d-e)(2+5n)}{n(1+2n)}}{\mathbf{O}}_n-{\displaystyle \frac{(10a-3b-13c-3d-3e)}{{(1+2n)}^2}}\}\end{array}\hfill \\ & =2a+b-c+d-3e+\sum _{k=1}{\displaystyle \frac{2e}{k^2}}=2a+b-c+d-3e+{\displaystyle \frac{{\pi }^{2}e}{3}}.\begin{array}{|c|}\hline {\mathrm{\Omega }}_1(a;b,c,d,e)\\ \hline\end{array}\hfill \end{array}$$

Then it is routine to verify the formulae below for four series labeled by ${\mathrm{\Omega }}_1$ (three of them having summands containing ${\mathbf{H}}_n$ and ${\mathbf{O}}_n$) by specializing the parameters $\{a,b,c,d,e\}$ as explicitly indicated.

• ${\mathrm{\Omega }}_0$ Constant term identity (cf. [21])

  $$-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2+5n}{n(1+2n)}}\right\}.$$

  (9)
• ${\mathrm{\Omega }}_1$: a = 2, b = d = −1, c = e = 1

  $$2-{\displaystyle \frac{{\pi }^2}{3}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{3+12n+10n^2}{n^2{(1+2n)}^2}}\right\}.$$

  (10)
• ${\mathrm{\Omega }}_1$: a = 3, b = −1, c = 0, d = 1, e = 2

  $$-{\displaystyle \frac{{\pi }^2}{3}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{3}{n^2}}+{\displaystyle \frac{2(2+5n)}{n(1+2n)}}{\mathbf{O}}_n\right\}.$$

  (11)
• ${\mathrm{\Omega }}_1$: a = b = d = 2, c = e = 0 (cf. [21])

  $$-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{{(1+2n)}^2}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{O}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$: a = 1, b = d = −1, c = e = 0 (cf. [21])

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^2}}-{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 2, b, d = ±i$\sqrt{7}-1,c,e=1\pm \sqrt{5}$ (Apéry series [22])

  $$-{\displaystyle \frac{2}{5}}\zeta \left(3\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{n^3\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}.$$

  (12)
• ${\mathrm{\Omega }}_2$ a = c = e = 0, b = −d = −1 (cf. [21])

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^3}}-{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$

  (13)
• ${\mathrm{\Omega }}_2$ a = c = −d = 1, b = e = 0 (cf. [21])

  $$-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{{(1+2n)}^3}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$

  (14)
• ${\mathrm{\Omega }}_2$ a = 4, b = −2, c = 3, d = 1, e = 0

  $$-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{{(1+2n)}^3}}+{\displaystyle \frac{2{\mathbf{O}}_n}{{(1+2n)}^2}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{O}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = e = 0, c = 2, b, d = ±i$\sqrt{7}$

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^3}}-{\displaystyle \frac{4{\mathbf{H}}_n}{n^2}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{H}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_3$ a = c = −d = 1, b = e = 0

  $$-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{{(1+2n)}^4}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{O}}_n^{\langle 3\rangle }\right\}.$$
• ${\mathrm{\Omega }}_3$ a = c = e = 0, b = −d = −1

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^4}}-{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{H}}_n^{\langle 3\rangle }\right\}.$$
• ${\mathrm{\Omega }}_4$ a = c = e = 0, b = −d = −1

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^5}}-{\displaystyle \frac{2}{n^3}}{\mathbf{H}}_n^{\langle 2\rangle }+{\displaystyle \frac{(2+5n)}{2n(1+2n)}}\left({\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{H}}_n^{\langle 4\rangle }\right)\right\}.$$

  (15)
• ${\mathrm{\Omega }}_4$ a = −b = c = 1, d = e = 0

  $$-6=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{6}{{(1+2n)}^5}}-{\displaystyle \frac{2{\mathbf{O}}_n^{\langle 2\rangle }}{{(1+2n)}^3}}-{\displaystyle \frac{(2+5n)}{n(1+2n)}}\left({\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-7{\mathbf{O}}_n^{\langle 4\rangle }\right)\right\}.$$

  (16)

  By employing the modified Abel lemma on summation by parts (cf. [13]), we can extend (13), (14), (15) and (16), respectively, to the following general formulae ($m\in \mathbb{N}$):

  $$\begin{array}{cc}\hfill 0& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^{m+1}}}-{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{H}}_n^{\langle m\rangle }\right\},\hfill \\ \hfill -1& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{{(1+2n)}^{m+1}}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\mathbf{O}}_n^{\langle m\rangle }\right\};\hfill \\ \hfill 0& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2}{n^{2m+1}}}-{\displaystyle \frac{2{\mathbf{H}}_n^{\langle m\rangle }}{n^{m+1}}}+{\displaystyle \frac{(2+5n)}{2n(1+2n)}}{\left({\mathbf{H}}_n^{\langle m\rangle }\right)}^2\right\},\hfill \\ \hfill -1& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{{(1+2n)}^{2m+1}}}+{\displaystyle \frac{2{\mathbf{O}}_n^{\langle m\rangle }}{{(1+2n)}^{m+1}}}+{\displaystyle \frac{(2+5n)}{n(1+2n)}}{\left({\mathbf{O}}_n^{\langle m\rangle }\right)}^2\right\}.\hfill \end{array}$$

2.2. Series in Group B

Making the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to ax-{\displaystyle \frac{1}{2}},b\to bx,c\to {\displaystyle \frac{1}{2}}+cx,d\to dx,e\to ax-{\displaystyle \frac{1}{2}};$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 2.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}\hfill \mathrm{\Gamma }& \left[\begin{array}{c}1+ax-bx-cx-dx,1+ax-cx,{\displaystyle \frac{1}{2}}+ax-bx,{\displaystyle \frac{1}{2}}+ax-dx\\ \\ ax-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+ax-bx-dx,1+ax-bx-cx,1+ax-cx-dx\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{{\left[\frac{1}{2}+ax-bx-dx,\frac{1}{2}-cx\right]}_n{\mathrm{\Delta }}_n\left(ax-\frac{1}{2},bx,\frac{1}{2}+cx,dx,ax-\frac{1}{2}\right)}{{\left[\frac{1}{2}+ax-bx,\frac{1}{2}+ax-dx\right]}_n{(1+ax-bx-cx-dx)}_{2n+1}}}\hfill \\ & \times {\displaystyle \frac{{\left[1+ax-bx-cx,1+ax-bx-dx,1-bx,1-dx\right]}_n(n+ax-cx)}{{(1+ax-cx)}_n(n+ax-bx-cx)(n+ax-cx-dx)}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 2 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{1-5n\right\}.$$
• ${\mathrm{\Omega }}_1$: a = 2, b = 0, c = −1, d = 1

  $${\displaystyle \frac{1}{4}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2+3n}{4n}}+(1-5n){\mathbf{O}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$: a = b = 0, c = d = 1

  $$-{\displaystyle \frac{1}{2}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2-3n}{2n}}-(1-5n){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 2, b = 1 − i$\sqrt{3}$, c = −4, d = 1 + i$\sqrt{3}$ (Combined with (10))

  $$-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{(1+n)}{{(1+2n)}^2}}-(1-5n){\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 2, b = 1 − $\sqrt{5}$, c = −4, d = 1 + $\sqrt{5}$ (Combined with (9) and (10))

  $$-8=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2(5+8n)}{n^2(1+2n)}}-5(1-5n){\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, b = 0, c = 1, d = 0

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1+3n}{1+2n}}{\mathbf{O}}_n+(1-5n){\mathbf{O}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 2, c = −2 − 2i, b, d = 2 − i ± $\sqrt{3}$ (The imaginary part combined with (10))

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1}{n^2}}+{\displaystyle \frac{2-3n-12n^2}{2n(1+2n)}}{\mathbf{H}}_n-(1-5n){\mathbf{H}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_3$ a = −b = 1, c = −d = −2 (Combined with (10) and (12)–(14))

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2-4n}{n^3}}+(5n-1){\mathbf{H}}_n^{\langle 3\rangle }\right\}.$$

  (17)
• ${\mathrm{\Omega }}_3$ a = 2, b = 1 + i, c = −4, d = 1 − i (Combined with (10) and (12)–(14))

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1+n}{{(1+2n)}^3}}+(5n-1){\mathbf{O}}_n^{\langle 3\rangle }\right\}.$$

  (18)
• ${\mathrm{\Omega }}_3$ a = 2, b = 1 + i, c = −4, d = 1 − i Combined with (18)

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{1+n}{{(1+2n)}^3}}+{\displaystyle \frac{3+9n}{(1+2n)}}{\mathbf{O}}_n^2+2(1-5n){\mathbf{O}}_n^3\right\}.$$
• ${\mathrm{\Omega }}_3$ a = 2i, c = 2 − 2i, b, d = 1 + 2i ± i$\sqrt{3}$ (Combined with (10), (12), (17) and (18))

  $$\begin{array}{cc}\hfill 0& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\left\{{\displaystyle \frac{2-4n}{n^3}}\left(1-3n{\mathbf{H}}_n+3n^2{\mathbf{H}}_n^2\right)+(5n-1){\mathbf{H}}_n^3\right\},\hfill \\ \hfill -12& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\{{\displaystyle \frac{(41+230n+396n^2+148n^3-128n^4)}{n^3{(1+2n)}^3}}\hfill \\ & -{\displaystyle \frac{6(5+20n+22n^2)}{n^2{(1+2n)}^2}}{\mathbf{H}}_n+{\displaystyle \frac{24(2+3n)}{1+2n}}{\mathbf{H}}_n^2+8(1-5n){\mathbf{H}}_n^3\},\hfill \\ \hfill -12& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}\{{\displaystyle \frac{(17+86n+108n^2-44n^3-128n^4)}{n^3{(1+2n)}^3}}\hfill \\ & +{\displaystyle \frac{6(3+12n+10n^2)}{n^2{(1+2n)}^2}}{\mathbf{H}}_n+{\displaystyle \frac{12(3n-2)}{n}}{\mathbf{H}}_n^2+8(1-5n){\mathbf{H}}_n^3\}.\hfill \end{array}$$

3. Series with Binomial Coefficient $\begin{array}{|c|}\hline \left(\begin{array}{c}\mathbf{3}\mathit{n}\\ \mathit{n}\end{array}\right)={\left(\frac{\mathbf{27}}{\mathbf{4}}\right)}^{\mathit{n}}{\left[\begin{array}{c}\frac{\mathbf{1}}{\mathbf{3}}, \frac{\mathbf{2}}{\mathbf{3}}\\ \mathbf{1}, \frac{\mathbf{1}}{\mathbf{2}}\end{array}\right]}_{\mathit{n}}\\ \hline\end{array}$

This section will be devoted to infinite series with the binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{3n}{n}}\right)$. They do not appear in the literature as frequently as those with the central binomial coefficient.

3.1. Series in Group A

Making the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to ax-1,b\to {\displaystyle \frac{1}{3}}+cx,c\to cx-{\displaystyle \frac{1}{3}},d\to ex,e\to ax-1;$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 3.

For a variable x and three complex parameters $\{a,c,e\}$, there holds

$$\begin{array}{cc}\hfill (ax-1)& \left({\displaystyle \frac{1}{3}}-cx\right)\mathrm{\Gamma }\left[\begin{array}{c}1+ax-ex,1+ax-2cx-ex,{\displaystyle \frac{1}{3}}+ax-cx,{\displaystyle \frac{2}{3}}+ax-cx\\ \\ ax,1+ax-2cx,{\displaystyle \frac{1}{3}}+ax-cx-ex,{\displaystyle \frac{2}{3}}+ax-cx-ex\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\left[\begin{array}{c}1+ax-cx,1+ax-2cx,1-ex\\ 1+ax-cx-ex,1+ax-ex,1-cx\end{array}\right]}_n\hfill \\ & \times {\displaystyle \frac{(n+ax-ex)(n-cx+\frac{1}{3}){\left[1+3ax-3cx-3ex,1-3cx\right]}_{3n}}{(n+ax-2cx)(n+ax-cx-ex-\frac{1}{3})}}\hfill \\ & \times {\displaystyle \frac{(n+ax-cx-\frac{1}{3}){\mathrm{\Delta }}_n\left(ax-1,\frac{1}{3}+cx,cx-\frac{1}{3},ex,ax-1\right)}{n!{(1+ax-2cx-ex)}_{2n+1}{(1+3ax-3cx)}_{3n}}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,c,e\}$, we derive from Theorem 3 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$0=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{n(2-9n^2)\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, c = e = 0

  $$8=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{(1-9n-81n^2)-5n(2-9n^2){\mathbf{H}}_{2n+1}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = c = −e = 1

  $$7=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{{\displaystyle \frac{29-9n-243n^2+81n^3+648n^4}{(1-3n)(1+3n)}}-10n(2-9n^2){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3c = 3e = 3

  $$25=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{{\displaystyle \frac{11-87n-225n^2+513n^3+1134n^4}{(1-3n)(1+3n)}}-30n(2-9n^2){\mathbf{H}}_{3n}\right\}.$$

3.2. Series in Group B

Making the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to ax-1,b\to {\displaystyle \frac{2}{3}}+cx,c\to cx-{\displaystyle \frac{2}{3}},d\to ex,e\to ax-1;$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 4.

For a variable x and three complex parameters $\{a,c,e\}$, there holds

$$\begin{array}{cc}\hfill (ax-1)& \left({\displaystyle \frac{2}{3}}-cx\right)\mathrm{\Gamma }\left[\begin{array}{c}1+ax-ex,1+ax-2cx-ex,{\displaystyle \frac{1}{3}}+ax-cx,{\displaystyle \frac{2}{3}}+ax-cx\\ \\ ax,1+ax-2cx,{\displaystyle \frac{1}{3}}+ax-cx-ex,{\displaystyle \frac{2}{3}}+ax-cx-ex\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\left[\begin{array}{c}1+ax-cx,1+ax-2cx,1-ex\\ 1+ax-cx-ex,1+ax-ex,1-cx\end{array}\right]}_n\hfill \\ & \times {\displaystyle \frac{(n+ax-cx-\frac{2}{3}){\mathrm{\Delta }}_n\left(ax-1,\frac{2}{3}+cx,cx-\frac{2}{3},ex,ax-1\right)}{{(1+ax-2cx-ex)}_{2n+1}{(1+3ax-3cx)}_{3n}}}\hfill \\ & \times {\displaystyle \frac{(n+ax-ex)(n-cx+\frac{2}{3}){\left[1+3ax-3cx-3ex,1-3cx\right]}_{3n}}{n!(n+ax-cx-ex-\frac{2}{3})(n+ax-2cx)}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,c,e\}$, we derive from Theorem 4 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$0=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{n(2+3n)(13-45n^2)\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, c = e = 0

  $$10=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{(2+3n)(4-9n-81n^2)-n(2+3n)(13-45n^2){\mathbf{H}}_{2n+1}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = c = −e = 1

  $$1=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{{\displaystyle \frac{140-36n-729n^2+81n^3+648n^4}{2(2-3n)}}-n(2+3n)(13-45n^2){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3c = 3e = 3

  $${\displaystyle \frac{19}{3}}=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{{\displaystyle \frac{68-264n-657n^2+783n^3+1134n^4}{6(2-3n)}}-n(2+3n)(13-45n^2){\mathbf{H}}_{3n}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, c = e = 0

  $$36=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)(3n+2)}{2n+1}}\left\{90n+2(4-9n-81n^2){\mathbf{H}}_{2n+1}-n(13-45n^2)({\mathbf{H}}_{2n+1}^{\langle 2\rangle }+{\mathbf{H}}_{2n+1}^2)\right\}.$$

3.3. Series in Group C

Performing further the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to ax,b\to {\displaystyle \frac{1}{3}}+cx,c\to {\displaystyle \frac{2}{3}}+cx,d\to ex,e\to ax$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 5.

For a variable x and three complex parameters $\{a,c,e\}$, there holds

$$\begin{array}{cc}\hfill \mathrm{\Gamma }& \left[\begin{array}{c}1+ax-2cx-ex,1+ax-ex,{\displaystyle \frac{1}{3}}+ax-cx,{\displaystyle \frac{2}{3}}+ax-cx\\ ax,1+ax-2cx,{\displaystyle \frac{1}{3}}+ax-ex-cx,{\displaystyle \frac{2}{3}}+ax-ex-cx\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{(1+ax-2cx)}_n{(1-ex)}_n{\mathrm{\Delta }}_n\left(ax,\frac{1}{3}+cx,\frac{2}{3}+cx,ex,ax\right)}{(n+ax-2cx){(1+ax-ex)}_n{(1+ax-2cx-ex)}_{2n+1}}}\hfill \\ & \times {\displaystyle \frac{{(\frac{1}{3}-cx)}_n{(\frac{2}{3}-cx)}_n{(\frac{1}{3}+ax-cx-ex)}_n{(\frac{2}{3}+ax-cx-ex)}_n}{n!{(\frac{1}{3}+ax-cx)}_n{(\frac{2}{3}+ax-cx)}_n}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,c,e\}$, we derive from Theorem 5 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$0=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{2+27n+45n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, c = e = 0

  $$9=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{9(1+4n)-(2+27n+45n^2){\mathbf{H}}_{2n+1}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, c = e = 1

  $$6=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{3(1+2n)-(2+27n+45n^2){\mathbf{H}}_{3n}\right\}.$$

  (19)
• ${\mathrm{\Omega }}_1$ a = c = 1, e = −1

  $$18=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{18(1+2n)-(2+27n+45n^2){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, c = e = 0

  $$0=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{18-18(1+4n){\mathbf{H}}_{2n+1}+(2+27n+45n^2)({\mathbf{H}}_{2n+1}^{\langle 2\rangle }+{\mathbf{H}}_{2n+1}^2)\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 3, c = e = 1

  $$1+9ln3=\sum _{n=1}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{3n}{n}\right)}{2n+1}}\left\{{\displaystyle \frac{1+n}{n}}+12(1+2n){\mathbf{H}}_{3n}-(2+27n+45n^2)({\mathbf{H}}_{3n}^{\langle 2\rangle }+2{\mathbf{H}}_{3n}^2)\right\}.$$

3.4. Series in Group D

Under the parameter replacements

$$a\to ax,b\to bx,c\to {\displaystyle \frac{1}{3}}+cx,d\to cx-{\displaystyle \frac{1}{3}},e\to dx,$$

the equality (7) for $\mathrm{\Omega }(a;b,c,d,e)$ becomes the following transformation formula.

Theorem 6.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}\hfill 3& \sum _{k=0}^{\infty }{\displaystyle \frac{(ax+2k)(\frac{1}{3}+ax-bx-cx)(\frac{1}{3}+ax-cx-dx){\left[bx,dx,\frac{1}{3}+cx,cx-\frac{1}{3}\right]}_k}{{(1+ax-bx)}_k{(1+ax-dx)}_k{(\frac{2}{3}+ax-cx)}_k{(\frac{1}{3}+ax-cx)}_{k+1}}}\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{{(1+ax-cx)}_n{\left[1+3ax-3bx-3cx,1+3ax-3cx-3dx\right]}_{3n+1}}{{(1+3ax-3cx)}_{3n+1}{(1+2ax-bx-2cx-dx)}_{2n+2}}}\hfill \\ & \times {\displaystyle \frac{{(1+ax-2cx)}_n{(1+ax-bx-dx)}_n{\mathrm{\Delta }}_n\left(ax,bx,\frac{1}{3}+cx,cx-\frac{1}{3},dx\right)}{{(1+ax-bx)}_n{(1+ax-dx)}_n{(1+ax-bx-cx)}_n{(1+ax-cx-dx)}_n}}.\hfill \end{array}$$

In particular for $a=d$, the sum with respect to k on the left can be evaluated by Dougall’s Formula (6) for ${}_{5}F_4$-series

$$\left(1-3cx\right)\mathrm{\Gamma }\left[\begin{array}{c}1+ax-bx,1+ax-bx-2cx,{\displaystyle \frac{1}{3}}+ax-cx,{\displaystyle \frac{2}{3}}+ax-cx\\ ax,1+ax-2cx,{\displaystyle \frac{1}{3}}+ax-bx-cx,{\displaystyle \frac{2}{3}}+ax-bx-cx\end{array}\right].$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 6 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$0=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{8+36n+45n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 2, b = d = 1, c = 0

  $$18=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{{\displaystyle \frac{26+99n+81n^2}{1+n}}-(8+36n+45n^2){\mathbf{H}}_{2n+2}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = d = 2, c = 1

  $$54=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{{\displaystyle \frac{26+108n+81n^2}{1+n}}-6(8+36n+45n^2){\mathbf{H}}_{3n+1}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = c = 1, d = −1

  $$18=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{{\displaystyle \frac{10+36n+27n^2}{1+n}}-2(8+36n+45n^2){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_2$ a = c = 0, b = 1, d = −1 (simplified by (24))

  $$54(1-ln3)=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{9-(8+36n+45n^2)(2{\mathbf{H}}_n^{\langle 2\rangle }-9{\mathbf{H}}_{3n+1}^{\langle 2\rangle }\right\}.$$
• ${\mathrm{\Omega }}_2$ a = c = 1, b = d = 0

  $$0=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{(8+36n+45n^2){\mathbf{H}}_n^2-{\displaystyle \frac{10+36n+27n^2}{1+n}}{\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_2$ a = d = 1, b = c = 0

  $$\begin{array}{ll}0={\displaystyle \sum _{n=0}^{\infty }}{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)& \left\{\begin{array}{r}{\displaystyle \frac{18(3+5n)}{(1+n)}}-{\displaystyle \frac{2(26+99n+81n^2)}{(1+n)}}{\mathbf{H}}_{2n+2}\\ +(8+36n+45n^2)({\mathbf{H}}_{2n+2}^2+{\mathbf{H}}_{2n+2}^{\langle 2\rangle })\end{array}\right\}\end{array}.$$
• ${\mathrm{\Omega }}_2$ a = d = 3, b = c = 1

  $$\begin{array}{ll}54(1+ln3)={\displaystyle \sum _{n=0}^{\infty }}{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)& \left\{\begin{array}{r}-9+{\displaystyle \frac{2(26+108n+81n^2)}{(1+n)}}{\mathbf{H}}_{3n+1}\\ -3(8+36n+45n^2)(2{\mathbf{H}}_{3n+1}^2+{\mathbf{H}}_{3n+1}^{\langle 2\rangle })\end{array}\right\}\end{array}.$$
• ${\mathrm{\Omega }}_3$ a = c = 1, b = d = 0

  $$0=\sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{3n+1}{n}}\right)\left\{{\displaystyle \frac{3(10+36n+27n^2)}{1+n}}{\mathbf{H}}_n^2-(8+36n+45n^2)(2{\mathbf{H}}_n^3+{\mathbf{H}}_n^{\langle 3\rangle })\right\}.$$

4. Series with Binomial Quotient $\begin{array}{|c|}\hline \left(\begin{array}{c}\mathbf{2}\mathit{n}\\ \mathit{n}\end{array}\right)/\left(\begin{array}{c}\mathbf{4}\mathit{n}\\ \mathbf{2}\mathit{n}\end{array}\right)={\left(\frac{\mathbf{1}}{\mathbf{4}}\right)}^{\mathit{n}}{\left[\begin{array}{c}\frac{\mathbf{1}}{\mathbf{2}}, \frac{\mathbf{1}}{\mathbf{2}}\\ \frac{\mathbf{1}}{\mathbf{4}}, \frac{\mathbf{3}}{\mathbf{4}}\end{array}\right]}_{\mathit{n}}\\ \hline\end{array}$

The series treated in this section are quite different from those considered in [21] (§3.4) since the harmonic numbers in the summands are tied with different weight factors.

4.1. Series in Group A

Making the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to {\displaystyle \frac{1}{2}}+ax,b\to {\displaystyle \frac{1}{2}}+bx,c\to 1+cx,d\to {\displaystyle \frac{1}{2}}+dx,e\to {\displaystyle \frac{1}{2}}+ax;$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 7.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}\hfill \left({\displaystyle \frac{-1}{cx}}\right)& \mathrm{\Gamma }\left[\begin{array}{c}1+ax-bx,1+ax-dx,{\displaystyle \frac{1}{2}}+ax-cx,{\displaystyle \frac{1}{2}}+ax-bx-cx-dx\\ \\ 1+ax-bx-cx,1+ax-cx-dx,{\displaystyle \frac{1}{2}}+ax,{\displaystyle \frac{1}{2}}+ax-bx-dx\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{(-1)}^n{\left[\begin{array}{c}1+ax-bx-cx,1+ax-cx-dx,1-cx\\ 1+ax-bx,{\displaystyle \frac{1}{2}}+ax-cx,1+ax-dx\end{array}\right]}_n{\displaystyle \frac{{\left[\frac{1}{2}-bx,\frac{1}{2}-dx\right]}_n}{n!(n-cx)}}\hfill \\ & \times {\displaystyle \frac{{(\frac{1}{2}+ax-bx-dx)}_n{\mathrm{\Delta }}_n\left(\frac{1}{2}+ax,\frac{1}{2}+bx,1+cx,\frac{1}{2}+dx,\frac{1}{2}+ax\right)}{(n+ax-bx-cx)(n+ax-cx-dx){(\frac{1}{2}+ax-bx-cx-dx)}_{2n+1}}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 7 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$4-8ln2=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{3+10n}{n}}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = 1, c = 2, d = 0

  $$2{\pi }^2-32{ln}^{2}2=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{7(1+4n)}{n^2}}-{\displaystyle \frac{6(3+10n)}{n}}{\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 2, b = d = 1, c = 0

  $$-{\displaystyle \frac{2{\pi }^2}{3}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+4n)}{n^2}}+{\displaystyle \frac{4(3+10n)}{n}}{\mathbf{O}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = 1, c = 0, d = −1

  $$4-{\displaystyle \frac{2{\pi }^2}{3}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+n)}{n^2}}+{\displaystyle \frac{(3+10n)}{n}}{\mathbf{O}}_{2n+1}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 0, b = 1, c = 0, d = −1

  $$2\zeta \left(3\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+4n)}{n^3}}-{\displaystyle \frac{2(3+10n)}{n}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$

  (20)
• ${\mathrm{\Omega }}_2$ a = b = 1, c = d = 0

  $$-3\zeta \left(3\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+4n)}{4n^3}}+{\displaystyle \frac{(1+4n)}{n^2}}{\mathbf{O}}_n+{\displaystyle \frac{2(3+10n)}{n}}{\mathbf{O}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, b = c = d = 0

  $$8-12\zeta \left(3\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+2n)}{n^3}}+{\displaystyle \frac{2(1+n)}{n^2}}{\mathbf{O}}_{2n+1}+{\displaystyle \frac{(3+10n)}{n}}\left({\mathbf{O}}_{2n+1}^{\langle 2\rangle }+{\mathbf{O}}_{2n+1}^2\right)\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 6, c = 4, b, d = 1 ± i$\sqrt{7}$

  $$4{\pi }^{2}ln2-19\zeta \left(3\right)-{\displaystyle \frac{64}{3}}{ln}^{3}2=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n}}}\left\{{\displaystyle \frac{9}{2n^3}}-{\displaystyle \frac{21}{n^2}}{\mathbf{H}}_n+{\displaystyle \frac{(3+10n)}{n(1+4n)}}\left(2{\mathbf{H}}_n^{\langle 2\rangle }+9{\mathbf{H}}_n^2\right)\right\}.$$
• ${\mathrm{\Omega }}_3$ a = b = 1, c = d = 0

  $$-{\displaystyle \frac{7{\pi }^4}{360}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+4n)}{16n^4}}+{\displaystyle \frac{(1+4n)}{4n^3}}{\mathbf{O}}_n+{\displaystyle \frac{(1+4n)}{2n^2}}{\mathbf{O}}_n^2+{\displaystyle \frac{(3+10n)}{3n}}\left({\mathbf{O}}_n^{\langle 3\rangle }+2{\mathbf{O}}_n^3\right)\right\}.$$
• ${\mathrm{\Omega }}_4$ a = 0, b = 1, c = 0, d = −1

  $${\displaystyle \frac{\zeta \left(5\right)}{2}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{(1+4n)}{4n^5}}-{\displaystyle \frac{(1+4n)}{n^3}}{\mathbf{O}}_n^{\langle 2\rangle }+{\displaystyle \frac{(3+10n)}{n}}\left({\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{O}}_n^{\langle 4\rangle }\right)\right\}.$$

4.2. Series in Group B

Making the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to ax-{\displaystyle \frac{1}{2}},b\to {\displaystyle \frac{1}{2}}+bx,c\to cx,d\to {\displaystyle \frac{1}{2}}+dx,e\to ax-{\displaystyle \frac{1}{2}};$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 8.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}& \mathrm{\Gamma }\left[\begin{array}{c}1+ax-bx,1+ax-dx,{\displaystyle \frac{1}{2}}+ax-cx,{\displaystyle \frac{1}{2}}+ax-bx-cx-dx\\ \\ 1+ax-bx-cx,1+ax-cx-dx,ax-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+ax-bx-dx\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{(-1)}^n{\left[\begin{array}{c}1+ax-bx-cx,1+ax-cx-dx,1-cx\\ 1+ax-bx,{\displaystyle \frac{1}{2}}+ax-cx,1+ax-dx\end{array}\right]}_n{\displaystyle \frac{(n+ax-bx)(n+ax-dx)}{n!(n-\frac{1}{2}+ax-bx-dx)}}\hfill \\ & \times {\displaystyle \frac{{\left[\frac{1}{2}+ax-bx-dx,\frac{1}{2}-bx,\frac{1}{2}-dx\right]}_n{\mathrm{\Delta }}_n\left(ax-\frac{1}{2},\frac{1}{2}+bx,cx,\frac{1}{2}+dx,ax-\frac{1}{2}\right)}{(n+ax-bx-cx)(n+ax-cx-dx){(\frac{1}{2}+ax-bx-cx-dx)}_{2n+1}}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 8 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$-{\displaystyle \frac{1}{2}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{n(1+5n)\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = 1, c = 2, d = 0

  $${\displaystyle \frac{3}{2}}-2ln2=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{2+7n-19n^2}{1-2n}}-3n(1+5n){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 2, b = d = 1, c = 0

  $$-{\displaystyle \frac{1}{2}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{3n^2}{1-2n}}+2n(1+5n){\mathbf{O}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = 1, c = 0, d = −1

  $${\displaystyle \frac{1}{2}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{2n-n(1+5n){\mathbf{O}}_{2n+1}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = c = 0, b = 1, d = −1

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{1+4n(1+5n){\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$
• ${\mathrm{\Omega }}_2$ a = b = 1, c = d = 0

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{3n^2}{1-2n}}{\mathbf{O}}_n+n(1+5n){\mathbf{O}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, b = c = d = 0

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{1-8n{\mathbf{O}}_{2n+1}+2n(1+5n)\left({\mathbf{O}}_{2n+1}^{\langle 2\rangle }+{\mathbf{O}}_{2n+1}^2\right)\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 6, c = 4, b, d = 1 ± i$\sqrt{7}$

  $$\begin{array}{cc}\hfill 12ln2-8{ln}^{2}2+{\displaystyle \frac{{\pi }^2}{2}}-8& =\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\{{\displaystyle \frac{1+6n-22n^2+24n^3}{n{(1-2n)}^2}}\hfill \\ & -{\displaystyle \frac{6(2+7n-19n^2)}{1-2n}}{\mathbf{H}}_n+n(1+5n)\left(2{\mathbf{H}}_n^{\langle 2\rangle }+9{\mathbf{H}}_n^2\right)\}.\hfill \end{array}$$
• ${\mathrm{\Omega }}_3$ a = b = 1, c = d = 0

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{9n^2}{1-2n}}{\mathbf{O}}_n^2+n(1+5n)\left({\mathbf{O}}_n^{\langle 3\rangle }+2{\mathbf{O}}_n^3\right)\right\}.$$
• ${\mathrm{\Omega }}_4$ a = c = 0, b = 1, d = −1

  $$0=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\mathbf{O}}_n^{\langle 2\rangle }+2n(1+5n)\left({\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{O}}_n^{\langle 4\rangle }\right)\right\}.$$

4.3. Series in Group C

Under the parameter replacements

$$a\to {\displaystyle \frac{1}{2}}+ax,b\to {\displaystyle \frac{1}{2}}+bx,c\to {\displaystyle \frac{1}{2}}+cx,d\to dx,e\to {\displaystyle \frac{1}{2}}+ex;$$

the equality in (7) for $\mathrm{\Omega }(a;b,c,d,e)$ becomes the following transformation formula.

Theorem 9.

For a variable x and five complex parameters $\{a,b,c,d,e\}$, there holds

$$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{(\frac{1}{2}+ax+2k){(\frac{1}{2}+bx)}_k{(\frac{1}{2}+cx)}_k{\left(dx\right)}_k{(\frac{1}{2}+ex)}_k}{{(1+ax-bx)}_k{(1+ax-cx)}_k{(\frac{1}{2}+ax-dx)}_{k+1}{(1+ax-ex)}_k}}\hfill \\ \hfill =& \sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left[\frac{1}{2}+ax-bx-cx,\frac{1}{2}+ax-bx-ex,\frac{1}{2}+ax-cx-ex,1+ax-dx-ex\right]}_n}{{(1+ax-bx)}_n{(1+ax-cx)}_n{(1+ax-ex)}_n}}\hfill \\ & \times {\displaystyle \frac{{(1+ax-bx-dx)}_n{(1+ax-cx-dx)}_n{\mathrm{\Delta }}_n\left(\frac{1}{2}+ax,\frac{1}{2}+bx,\frac{1}{2}+cx,dx,\frac{1}{2}+ex\right)}{{(\frac{1}{2}+ax-dx)}_{n+1}{(\frac{1}{2}+2ax-bx-cx-dx-ex)}_{2n+2}}}.\hfill \end{array}$$

In particular for $a=e$, the sum with respect to k on the left can be evaluated by Dougall’s Formula (6) for ${}_{5}F_4$-series

$$\mathrm{\Gamma }\left[\begin{array}{c}1+ax-bx,1+ax-cx,{\displaystyle \frac{1}{2}}+ax-dx,{\displaystyle \frac{1}{2}}+ax-bx-cx-dx\\ {\displaystyle \frac{1}{2}}+ax,{\displaystyle \frac{1}{2}}+ax-bx-cx,1+ax-bx-dx,1+ax-cx-dx\end{array}\right].$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d,e\}$, we derive from Theorem 9 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$1=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{1+5n+5n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = e = 1, c = −1, d = 0

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2){\mathbf{O}}_{2n+2}-(1+2n)\right\}.$$
• ${\mathrm{\Omega }}_1$ a = c = e = 1, b = d = 0

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2){\mathbf{O}}_n+{\displaystyle \frac{1+2n}{4}}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = e = 3, b = 0, c = 1, d = 2

  $${\displaystyle \frac{-4ln2}{3}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2){\mathbf{H}}_n-{\displaystyle \frac{3+4n}{6}}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = d = e = 0, b = 1, c = −1

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2){\mathbf{O}}_n^{\langle 2\rangle }+{\displaystyle \frac{1}{4}}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = c = e = 1, b = 0, d = 0

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2){\mathbf{O}}_n^2+{\displaystyle \frac{1+2n}{2}}{\mathbf{O}}_n+{\displaystyle \frac{1}{4}}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = e = 6, b, c = 1 ± $\sqrt{-7}$, d = 4

  $$16{ln}^{2}2-{\pi }^2=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2)(2{\mathbf{H}}_n^{\langle 2\rangle }+9{\mathbf{H}}_n^2)-3(3+4n){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_2$ a = e = 1, b = c = d = 0

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2)({\mathbf{O}}_{2n+2}^{\langle 2\rangle }+{\mathbf{O}}_{2n+2}^2)-2(1+2n){\mathbf{O}}_{2n+2}+{\displaystyle \frac{1}{2}}\right\}.$$
• Coefficient of “$a{b}^2$” in ${\mathrm{\Omega }}_3$ e→a, c→−b, d→0

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\mathbf{O}}_{2n+2}-4(1+2n){\mathbf{O}}_n^{\langle 2\rangle }+4(1+5n+5n^2){\mathbf{O}}_n^{\langle 2\rangle }{\mathbf{O}}_{2n+2}\right\}.$$
• ${\mathrm{\Omega }}_3$ a = d = 0, b = c = 1, e = −2

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{(1+5n+5n^2){\mathbf{O}}_n^{\langle 3\rangle }+{\displaystyle \frac{1}{4(1+2n)}}\right\}.$$
• ${\mathrm{\Omega }}_4$ a = d = 0, b = c = 1, e = −2

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\mathbf{O}}_n^{\langle 2\rangle }-2(1+5n+5n^2)\left({\mathbf{O}}_n^{\langle 4\rangle }-{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\right\}.$$
• ${\mathrm{\Omega }}_5$ a = d = 0, b = c = 1, e = −2

  $$0=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{\left(\frac{1}{2}\right)}_n^2}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{{\mathbf{O}}_n^{\langle 2\rangle }}{1+2n}}+{\mathbf{O}}_n^{\langle 3\rangle }-4(1+5n+5n^2)({\mathbf{O}}_n^{\langle 5\rangle }-{\mathbf{O}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 3\rangle })\right\}.$$

5. Series with Binomial Quotient $\begin{array}{|c|}\hline \left(\begin{array}{c}\mathbf{3}\mathit{n}\\ \mathit{n}\end{array}\right)/\left(\begin{array}{c}\mathbf{4}\mathit{n}\\ \mathbf{2}\mathit{n}\end{array}\right)={\left(\frac{\mathbf{27}}{\mathbf{64}}\right)}^{\mathit{n}}{\left[\begin{array}{c}\frac{\mathbf{1}}{\mathbf{3}}, \frac{\mathbf{2}}{\mathbf{3}}\\ \frac{\mathbf{1}}{\mathbf{4}}, \frac{\mathbf{3}}{\mathbf{4}}\end{array}\right]}_{\mathit{n}}\\ \hline\end{array}$

In this section, we turn to consider the harmonic series with $\left({\displaystyle \genfrac{}{}{0pt}{}{3n}{n}}\right)/\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right)$ in the summands.

5.1. Series in Group A

Making the parameter replacements in (7) for $\mathrm{\Omega }(a;b,c,d,e)$

$$a\to ax,b\to {\displaystyle \frac{1}{6}}+cx,c\to {\displaystyle \frac{5}{6}}+cx,d\to {\displaystyle \frac{1}{2}}+ex,e\to ax;$$

and then applying Dougall’s summation Formula (6) for the ${}_{5}F_4$-series, we obtain the following transformation formula.

Theorem 10.

For a variable x and three complex parameters $\{a,c,e\}$, there holds

$$\begin{array}{cc}\hfill \left(ax\right)& \mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+ax-ex,{\displaystyle \frac{1}{2}}+ax-2cx-ex,{\displaystyle \frac{1}{6}}+ax-cx,{\displaystyle \frac{5}{6}}+ax-cx\\ 1+ax,1+ax-2cx,{\displaystyle \frac{1}{3}}+ax-cx-ex,{\displaystyle \frac{2}{3}}+ax-cx-ex\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{{(1+3ax-3cx-3ex)}_{3n}{\mathrm{\Delta }}_n\left(ax,\frac{1}{6}+cx,\frac{5}{6}+cx,\frac{1}{2}+ex,ax\right)}{n!(n+ax-cx-ex-\frac{1}{3}){(\frac{1}{2}+ax-2cx-ex)}_{2n+1}}}\hfill \\ & \times {\displaystyle \frac{{(1+ax-2cx)}_{n-1}{(\frac{1}{2}-ex)}_n{(\frac{1}{2}+ax-cx)}_n{(\frac{1}{2}-3cx)}_{3n}}{{(1+ax-cx-ex)}_n{(\frac{1}{2}+ax-ex)}_n{(\frac{1}{2}-cx)}_n{(\frac{1}{2}+3ax-3cx)}_{3n}}}.\hfill \end{array}$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,c,e\}$, we derive from Theorem 10 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$-8=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{-1}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left(13-180n^2\right).$$
• ${\mathrm{\Omega }}_1$ a = 0, c = −e = 1

  $$8=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{-1}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{9(8n^2-1)}{n}}+(13-180n^2)({\mathbf{H}}_n-{\mathbf{O}}_{2n+1})\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 2, c = e = 1

  $$8-12\sqrt{3}\pi =\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{-1}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{18+(13-180n^2)(6{\mathbf{O}}_{3n}-{\mathbf{O}}_{2n+1})\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 0, c = e = 1

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{-1}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+1}}}\left\{{\displaystyle \frac{3-22n+48n^2+36n^3}{n(3n-1)}}+(13-180n^2)({\mathbf{H}}_{3n}-{\mathbf{O}}_{2n+1})\right\}.$$

5.2. Series in Group B

Under the parameter replacements

$$a\to {\displaystyle \frac{1}{2}}+ax,b\to bx,c\to {\displaystyle \frac{1}{3}}+cx,d\to {\displaystyle \frac{2}{3}}+cx,e\to {\displaystyle \frac{1}{2}}+dx,$$

the series in (7) for $\mathrm{\Omega }(a;b,c,d,e)$ gives rise to the following transformation formula.

Theorem 11.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}\hfill 3& \sum _{k=0}^{\infty }{\displaystyle \frac{(\frac{1}{6}+ax-bx-cx)(\frac{1}{2}+ax+2k){(\frac{1}{2}+dx)}_k{(1+3cx)}_{3k}{(\frac{1}{2}+ax-cx)}_k{\left(bx\right)}_k}{(k+\frac{1}{6}+ax-cx){(\frac{1}{2}+ax-bx)}_{k+1}{(1+cx)}_k{(\frac{1}{2}+3ax-3cx)}_{3k}{(1+ax-dx)}_k}}\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{{(1+3ax-3cx-3dx)}_{3n}{\mathrm{\Delta }}_n\left(\frac{1}{2}+ax,bx,\frac{1}{3}+cx,\frac{2}{3}+cx,\frac{1}{2}+dx\right)}{{(1+ax-cx-dx)}_n(n+\frac{1}{6}+ax-cx){(\frac{1}{2}+2ax-bx-2cx-dx)}_{2n+2}}}\hfill \\ & \times {\displaystyle \frac{{(\frac{1}{2}+ax-2cx)}_n{(1+ax-bx-dx)}_n{(\frac{1}{2}+ax-cx)}_n{(\frac{1}{2}+3ax-3bx-3cx)}_{3n+1}}{{(1+ax-dx)}_n{(\frac{1}{2}+ax-bx)}_{n+1}{(\frac{1}{2}+ax-bx-cx)}_n{(\frac{1}{2}+3ax-3cx)}_{3n}}}.\hfill \end{array}$$

In particular for $a=d$, the sum with respect to k on the left can be evaluated by Dougall’s Formula (6) for ${}_{5}F_4$-series

$$\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+ax-bx,{\displaystyle \frac{1}{2}}+ax-bx-2cx,{\displaystyle \frac{1}{6}}+ax-cx,{\displaystyle \frac{5}{6}}+ax-cx\\ {\displaystyle \frac{1}{2}}+ax,{\displaystyle \frac{1}{2}}+ax-2cx,{\displaystyle \frac{1}{6}}+ax-bx-cx,{\displaystyle \frac{5}{6}}+ax-bx-cx\end{array}\right].$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 11 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$-{\displaystyle \frac{32}{15}}=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{7+36n+36n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = d = 1, b = c = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{36(1+2n)}{5}}-(7+36n+36n^2){\mathbf{O}}_{2n+2}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, b = 0, c = 1, d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{4(1+3n)(2+3n)}{5(1+2n)}}+(7+36n+36n^2){\mathbf{O}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, b = c = 0, d = 2

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{36(1+2n)}{5}}+(7+36n+36n^2)\left(3{\mathbf{H}}_{3n}-{\mathbf{H}}_n\right)\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = c = 1, d = 3 (simplified by (25))

  $${\displaystyle \frac{18}{5}}ln3=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{36n^2+18n-1}{15(1+2n)}}+(7+36n+36n^2){\mathbf{H}}_{6n+1}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = d = 1, b = c = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{18-72(1+2n){\mathbf{O}}_{2n+2}+5(7+36n+36n^2)\left({\mathbf{O}}_{2n+2}^{\langle 2\rangle }+{\mathbf{O}}_{2n+2}^2\right)\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, b = 0, c = 1, d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{3}\right)}_n{\left(\frac{2}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{4(1+3n)(2+3n)}{5{(1+2n)}^2}}+{\displaystyle \frac{8(1+3n)(2+3n)}{5(1+2n)}}{\mathbf{O}}_n+(7+36n+36n^2){\mathbf{O}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 2, b = 1, c = 1, d = 1

  $$\begin{array}{cc}\hfill 20& +\sum _{k=1}^{\infty }{\displaystyle \frac{\left(3k\right)!{\left(\frac{1}{2}\right)}_k}{k!{\left(\frac{1}{2}\right)}_{3k}}}\left\{{\displaystyle \frac{2}{{(1+2k)}^2}}+{\displaystyle \frac{18}{{(1+6k)}^2}}-{\displaystyle \frac{1}{k^2}}+{\displaystyle \frac{2(1+4k)\left({\overline{\mathbf{H}}}_{2k}-3{\overline{\mathbf{H}}}_{6k}\right)}{k(1+2k)(1+6k)}}\right\}\hfill \\ & =\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\{{\displaystyle \frac{4(2+3n)}{9{(1+2n)}^2}}-{\displaystyle \frac{19+54n+36n^2}{18(1+2n)(1+3n)}}\left({\mathbf{H}}_n+6{\mathbf{O}}_{3n+1}\right)\hfill \\ & +{\displaystyle \frac{5(7+36n+36n^2)}{72(1+3n)}}\left({\mathbf{H}}_n^{\langle 2\rangle }+36{\mathbf{O}}_{3n+1}^{\langle 2\rangle }+{({\mathbf{H}}_n+6{\mathbf{O}}_{3n+1})}^2\right)\}.\hfill \end{array}$$

  (21)

5.3. Series in Group C

Under the parameter replacements

$$a\to ax,b\to bx-{\displaystyle \frac{1}{2}},c\to {\displaystyle \frac{1}{6}}+cx,d\to {\displaystyle \frac{5}{6}}+cx,e\to dx,$$

the transformation Formula (7) for $\mathrm{\Omega }(a;b,c,d,e)$ becomes the following one.

Theorem 12.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}\hfill 3& \sum _{k=0}^{\infty }{\displaystyle \frac{(\frac{1}{3}+ax-bx-cx)(\frac{1}{2}+ax-bx-dx)(ax+2k){(bx-\frac{1}{2})}_k{(\frac{1}{2}+3cx)}_{3k}{(\frac{1}{2}+ax-cx)}_k{\left(dx\right)}_k}{(ax-2cx){(\frac{1}{2}+ax-bx)}_{k+1}{(\frac{1}{2}+cx)}_k{(\frac{1}{2}+3ax-3cx)}_{3k}{(1+ax-dx)}_k}}\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{{(1+3ax-3bx-3cx)}_{3n+1}{\mathrm{\Delta }}_n\left(ax,bx-\frac{1}{2},\frac{1}{6}+cx,\frac{5}{6}+cx,dx\right)}{(n+ax-2cx){(1+ax-bx-cx)}_n{(\frac{1}{2}+2ax-bx-2cx-dx)}_{2n+2}}}\hfill \\ & \times {\displaystyle \frac{{(1+ax-2cx)}_n{(\frac{1}{2}+ax-bx-dx)}_{n+1}{(\frac{1}{2}+ax-cx)}_n{(\frac{1}{2}+3ax-3cx-3dx)}_{3n}}{{(1+ax-dx)}_n{(\frac{1}{2}+ax-bx)}_{n+1}{(\frac{1}{2}+ax-cx-dx)}_n{(\frac{1}{2}+3ax-3cx)}_{3n}}}.\hfill \end{array}$$

In particular for $a=d$, the sum with respect to k on the left can be evaluated by Dougall’s Formula (6) for ${}_{5}F_4$-series

$$\left({\displaystyle \frac{1}{2}}-bx\right)\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+ax-bx,{\displaystyle \frac{1}{2}}+ax-bx-2cx,{\displaystyle \frac{1}{6}}+ax-cx,{\displaystyle \frac{5}{6}}+ax-cx\\ ax,1+ax-2cx,{\displaystyle \frac{2}{3}}+ax-bx-cx,{\displaystyle \frac{1}{3}}+ax-bx-cx\end{array}\right].$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 12 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$36=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{59+216n+180n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = 1, c = d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{36(1+2n)-(59+216n+180n^2){\mathbf{O}}_{2n+2}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = 4, c = 1, d = 0

  $${\displaystyle \frac{224}{3}}=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{3(3+8n+8n^2)}{n}}-(59+216n+180n^2){\mathbf{H}}_{3n+1}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = c = 1, b = d = 0

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{9(1+4n)(3+4n)}{n}}-(59+216n+180n^2){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = 2, c = 1, d = 2 (simplified by (27)–(29))

  $$-9\sqrt{3}\pi =\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{2(16+63n+54n^2)}{3(1+2n)}}+(59+216n+180n^2){\mathbf{O}}_{3n}\right\}.$$

  (22)
• ${\mathrm{\Omega }}_2$ a = c = 1, b = d = 0

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{9(1+4n)(3+4n)}{n^2}}\left(1-2n{\mathbf{H}}_n\right)+(59+216n+180n^2){\mathbf{H}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = b = 1, c = d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{18-72(1+2n){\mathbf{O}}_{2n+2}+(59+216n+180n^2)({\mathbf{O}}_{2n+2}^{\langle 2\rangle }+{\mathbf{O}}_{2n+2}^2)\right\}.$$
• ${\mathrm{\Omega }}_3$ a = c = 1, b = d = 0

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{2}{3}\right)}_n{\left(\frac{4}{3}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{\begin{array}{c}\hfill {\displaystyle \frac{27(1+4n)(3+4n)}{n^3}}(1-2n{\mathbf{H}}_n+2n^2{\mathbf{H}}_n^2)\\ \hfill -(59+216n+180n^2)({\mathbf{H}}_n^{\langle 3\rangle }+2{\mathbf{H}}_n^3)\end{array}\right\}.$$

6. Series with Binomial Quotient $\begin{array}{|c|}\hline \left(\begin{array}{c}\mathbf{6}\mathit{n}\\ \mathbf{2}\mathit{n}\end{array}\right)/\left(\begin{array}{c}\mathbf{3}\mathit{n}\\ \mathit{n}\end{array}\right)={\left(\frac{\mathbf{27}}{\mathbf{4}}\right)}^{\mathit{n}}{\left[\begin{array}{c}\frac{\mathbf{1}}{\mathbf{6}}, \frac{\mathbf{5}}{\mathbf{6}}\\ \frac{\mathbf{1}}{\mathbf{4}}, \frac{\mathbf{3}}{\mathbf{4}}\end{array}\right]}_{\mathit{n}}\\ \hline\end{array}$

Finally, we are going to examine the series with the above binomial quotient in the title.

6.1. Series in Group A

Under the parameter replacements

$$a\to \frac{1}{2}+ax,b\to bx,c\to \frac{1}{6}+cx,d\to \frac{5}{6}+cx,e\to \frac{1}{2}+dx$$

the transformation Formula (7) for $\mathrm{\Omega }(a;b,c,d,e)$ becomes the following one.

Theorem 13.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{(1+3ax-3bx-3cx)(\frac{1}{2}+ax+2k){(\frac{1}{2}+dx)}_k{(\frac{1}{2}+3cx)}_{3k}{(1+ax-cx)}_k{\left(bx\right)}_k}{{(\frac{1}{2}+ax-bx)}_{k+1}{(\frac{1}{2}+cx)}_k{(1+3ax-3cx)}_{3k+1}{(1+ax-dx)}_k}}\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{{(\frac{1}{2}+3ax-3cx-3dx)}_{3n}{\mathrm{\Delta }}_n\left(\frac{1}{2}+ax,bx,\frac{1}{6}+cx,\frac{5}{6}+cx,\frac{1}{2}+dx\right)}{{(\frac{1}{2}+ax-cx-dx)}_n{(\frac{1}{2}+2ax-bx-2cx-dx)}_{2n+2}}}\hfill \\ & \times {\displaystyle \frac{{(\frac{1}{2}+ax-2cx)}_n{(1+ax-bx-dx)}_n{(1+ax-cx)}_n{(1+3ax-3bx-3cx)}_{3n+1}}{{(1+ax-dx)}_n{(\frac{1}{2}+ax-bx)}_{n+1}{(1+ax-bx-cx)}_n{(1+3ax-3cx)}_{3n+1}}}.\hfill \end{array}$$

In particular for $a=d$, the sum with respect to k on the left can be evaluated by Dougall’s Formula (6) for ${}_{5}F_4$-series

$$\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+ax-bx,{\displaystyle \frac{1}{2}}+ax-bx-2cx,{\displaystyle \frac{1}{3}}+ax-cx,{\displaystyle \frac{2}{3}}+ax-cx\\ {\displaystyle \frac{1}{2}}+ax,{\displaystyle \frac{1}{2}}+ax-2cx,{\displaystyle \frac{1}{3}}+ax-bx-cx,{\displaystyle \frac{2}{3}}+ax-bx-cx\end{array}\right].$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 13 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$-{\displaystyle \frac{5}{3}}=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{8+45n+45n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = d = 1, b = c = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{9(1+2n)-(8+45n+45n^2){\mathbf{O}}_{2n+2}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = 0, c = 1, d = 4

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{23+108n+108n^2}{12(1+2n)}}+(8+45n+45n^2){\mathbf{O}}_{3n}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, b = 0, c = 1, d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{(1+6n)(5+6n)}{4(1+2n)}}+(8+45n+45n^2){\mathbf{O}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 3, b = 2, c = 1, d = 2 (simplified by (26))

  $$3ln\left({\displaystyle \frac{16}{27}}\right)=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{3(3+4n)}{2}}-(8+45n+45n^2){\mathbf{H}}_{3n+1}\right\}.$$

  (23)
• ${\mathrm{\Omega }}_2$ a = 1, b = 0, c = 1, d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{(1+6n)(5+6n)}{4{(1+2n)}^2}}+{\displaystyle \frac{(1+6n)(5+6n)}{2(1+2n)}}{\mathbf{O}}_n+(8+45n+45n^2){\mathbf{O}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = 1, b = c = 0, d = 1

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{1}{6}\right)}_n{\left(\frac{5}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{9-36(1+2n){\mathbf{O}}_{2n+2}+2(8+45n+45n^2)\left({\mathbf{O}}_{2n+2}^{\langle 2\rangle }+{\mathbf{O}}_{2n+2}^2\right)\right\}.$$

6.2. Series in Group B

Under the parameter setting

$$a\to ax,b\to bx-\frac{1}{2},c\to \frac{1}{3}+cx,d\to \frac{2}{3}+cx,e\to dx$$

the transformation Formula (7) for $\mathrm{\Omega }(a;b,c,d,e)$ becomes the following one.

Theorem 14.

For a variable x and four complex parameters $\{a,b,c,d\}$, there holds

$$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{(ax+2k){(1+3cx)}_{3k}}{{(1+3ax-3cx)}_{3k}}}{\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+bx,1+dx,1+ax-cx\\ {\displaystyle \frac{1}{2}}+ax-bx,1+ax-dx,1+cx\end{array}\right]}_k\hfill \\ & \times {\displaystyle \frac{(bx-\frac{1}{2})\left(dx\right)(\frac{1}{2}+3ax-3bx-3cx)(\frac{1}{2}+ax-bx-dx)}{(\frac{1}{2}+ax-bx+k)(k+bx-\frac{1}{2})(k+dx)(ax-2cx)}}\hfill \\ \hfill =& \sum _{n=0}^{\infty }{\left({\displaystyle \frac{-1}{27}}\right)}^n{\displaystyle \frac{{(\frac{1}{2}+3ax-3bx-3cx)}_{3n+1}{\mathrm{\Delta }}_n\left(ax,bx-\frac{1}{2},\frac{1}{3}+cx,\frac{2}{3}+cx,dx\right)}{(n+ax-2cx){(\frac{1}{2}+ax-bx-cx)}_n{(\frac{1}{2}+2ax-bx-2cx-dx)}_{2n+2}}}\hfill \\ & \times {\displaystyle \frac{{(\frac{1}{2}+ax-bx-dx)}_{n+1}{(1+ax-2cx)}_n{(1+ax-cx)}_n{(1+3ax-3cx-3dx)}_{3n}}{{(\frac{1}{2}+ax-bx)}_{n+1}{(1+ax-dx)}_n{(1+ax-cx-dx)}_n{(1+3ax-3cx)}_{3n}}}.\hfill \end{array}$$

In particular for $a=d$, the sum with respect to k on the left can be evaluated by Dougall’s Formula (6) for ${}_{5}F_4$-series

$$3\left({\displaystyle \frac{1}{2}}-bx\right)\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{1}{2}}+ax-bx,{\displaystyle \frac{1}{3}}+ax-cx,{\displaystyle \frac{2}{3}}+ax-cx,{\displaystyle \frac{1}{2}}+ax-bx-2cx\\ \\ ax,{\displaystyle \frac{5}{6}}+ax-bx-cx,{\displaystyle \frac{1}{6}}+ax-bx-cx,1+ax-2cx\end{array}\right].$$

Denote by ${\mathrm{\Omega }}_m$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, we derive from Theorem 14 the following infinite series identities.

• ${\mathrm{\Omega }}_0$ Constant term identity

  $$18=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{31+108n+90n^2\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = 1, c = d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{18(1+2n)-(31+108n+90n^2){\mathbf{O}}_{2n+2}\right\}.$$
• ${\mathrm{\Omega }}_1$ a = c = 1, b = d = 0

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{9(1+4n)(3+4n)}{2n}}-(31+108n+90n^2){\mathbf{H}}_n\right\}.$$
• ${\mathrm{\Omega }}_1$ a = 1, b = 2, c = d = 0

  $$-94=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{18(1+n)-(31+108n+90n^2)\left({\mathbf{O}}_n-3{\mathbf{O}}_{3n+1}\right)\right\}.$$
• ${\mathrm{\Omega }}_1$ a = b = 3, c = d = 1 (simplified by (26) and (30))

  $${\displaystyle \frac{352+27\pi \sqrt{3}}{9}}=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{27+56n-36n^2-72n^3}{12n(1+2n)}}-(31+108n+90n^2){\mathbf{H}}_{6n+1}\right\}.$$
• ${\mathrm{\Omega }}_2$ a = c = 1, b = d = 0

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{{\displaystyle \frac{9(1+4n)(3+4n)}{2n^2}}\left(1-2n{\mathbf{H}}_n\right)+(31+108n+90n^2){\mathbf{H}}_n^2\right\}.$$
• ${\mathrm{\Omega }}_2$ a = b = 1, c = d = 0

  $$0=\sum _{n=0}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{9-36(1+2n){\mathbf{O}}_{2n+2}+(31+108n+90n^2)\left({\mathbf{O}}_{2n+2}^{\langle 2\rangle }+{\mathbf{O}}_{2n+2}^2\right)\right\}.$$
• ${\mathrm{\Omega }}_3$ a = c = 1, b = d = 0

  $$0=\sum _{n=1}^{\infty }{(-1)}^n{\displaystyle \frac{{\left(\frac{5}{6}\right)}_n{\left(\frac{7}{6}\right)}_n}{{\left(\frac{1}{2}\right)}_{2n+2}}}\left\{\begin{array}{c}\hfill {\displaystyle \frac{27(1+4n)(3+4n)}{2n^3}}\left(1-2n{\mathbf{H}}_n+2n^2{\mathbf{H}}_n^2\right)\\ \hfill -(31+108n+90n^2)\left({\mathbf{H}}_n^{\langle 3\rangle }+2{\mathbf{H}}_n^3\right)\end{array}\right\}.$$

7. k-Sums and Further Observations

In the course of deriving closed formulae of harmonic series, we have encountered seven k-sums that need to be evaluated separately. Each k-sum is labeled by the section number where it appeared. The first three of them can be treated by the hypergeometric series approach.

$$\begin{array}{|c|}\hline \hfill \S 3.4\\ \hline\end{array}:3(ln3-1)=\sum _{k=1}^{\infty }{\displaystyle \frac{2}{k(3k-1)(3k+1)}},$$

(24)

$$\begin{array}{|c|}\hline \hfill \S 5.2\\ \hline\end{array}:4-3ln3=\sum _{k=1}^{\infty }{\left[\begin{array}{c}{\displaystyle \frac{1}{3}},{\displaystyle \frac{2}{3}}\\ \\ {\displaystyle \frac{1}{6}},{\displaystyle \frac{5}{6}}\end{array}\right]}_k{\displaystyle \frac{(1+4k)}{k(1+2k)(1+6k)}},$$

(25)

$$\begin{array}{|c|}\hline \hfill \S 6.1\\ \hline\end{array}:1-3ln3+4ln2=\sum _{k=1}^{\infty }{\left[\begin{array}{c}{\displaystyle \frac{1}{6}},{\displaystyle \frac{5}{6}}\\ \\ {\displaystyle \frac{2}{3}},{\displaystyle \frac{4}{3}}\end{array}\right]}_k{\displaystyle \frac{(1+4k)}{k(1+2k)}}.$$

(26)

Proof of

(24). Expressing the sum in terms of well–poised ${}_{5}F_4$-series and then evaluating it by Dixon’s summation theorem (cf. Bailey [14], §3.1), we can show (24) as follows:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{2}{k(3k-1)(3k+1)}}& =\underset{y\to 0}{lim}{\displaystyle \frac{2}{y}}\left\{1-{}_{3}F_2\left[\begin{array}{c}\hfill y,{\displaystyle \frac{1}{3}},y-{\displaystyle \frac{1}{3}}\\ \\ \hfill {\displaystyle \frac{2}{3}}+y,{\displaystyle \frac{4}{3}}\end{array}|1\right]\right\}\hfill \\ & =\underset{y\to 0}{lim}{\displaystyle \frac{2}{y}}\left\{1-\mathrm{\Gamma }\left[\begin{array}{c}1+{\displaystyle \frac{y}{2}},1-{\displaystyle \frac{y}{2}},{\displaystyle \frac{2}{3}}+y,{\displaystyle \frac{4}{3}}\\ \\ 1+y,{\displaystyle \frac{2}{3}}+{\displaystyle \frac{y}{2}},{\displaystyle \frac{4}{3}}-{\displaystyle \frac{y}{2}}\end{array}\right]\right\}\hfill \\ & =-2\left[y\right]\mathrm{\Gamma }\left[\begin{array}{c}1+{\displaystyle \frac{y}{2}},1-{\displaystyle \frac{y}{2}},{\displaystyle \frac{2}{3}}+y,{\displaystyle \frac{4}{3}}\\ \\ 1+y,{\displaystyle \frac{2}{3}}+{\displaystyle \frac{y}{2}},{\displaystyle \frac{4}{3}}-{\displaystyle \frac{y}{2}}\end{array}\right]=3ln3-3. \square \hfill \end{array}$$

Proof of

(25). This can be written as well–poised ${}_{5}F_4$-series and then evaluated by (6) as follows:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\left[\begin{array}{c}{\displaystyle \frac{1}{3}},{\displaystyle \frac{2}{3}}\\ \\ {\displaystyle \frac{1}{6}},{\displaystyle \frac{5}{6}}\end{array}\right]}_k{\displaystyle \frac{(1+4k)}{k(1+2k)(1+6k)}}& =\underset{y\to 0}{lim}{\displaystyle \frac{1}{y}}\left\{-1+{}_{5}F_4\left[\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2}},{\displaystyle \frac{5}{4}},& {\displaystyle \frac{1}{3}},{\displaystyle \frac{2}{3}},& y\\ \\ \hfill {\displaystyle \frac{1}{4}},& {\displaystyle \frac{5}{6}},{\displaystyle \frac{7}{6}},& {\displaystyle \frac{3}{2}}-y\end{array}|1\right]\right\}\hfill \\ & =\underset{y\to 0}{lim}{\displaystyle \frac{1}{y}}\{-1+\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{5}{6}},{\displaystyle \frac{7}{6}},{\displaystyle \frac{1}{2}}-y,{\displaystyle \frac{3}{2}}-y\\ \\ {\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{5}{6}}-y,{\displaystyle \frac{7}{6}}-y\end{array}\right]\hfill \\ & =\left[y\right]\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{5}{6}},{\displaystyle \frac{7}{6}},{\displaystyle \frac{1}{2}}-y,{\displaystyle \frac{3}{2}}-y\\ \\ {\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{5}{6}}-y,{\displaystyle \frac{7}{6}}-y\end{array}\right]=4-3ln3. \square \hfill \end{array}$$

Proof of

(26). This can be confirmed analogously as below:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\left[\begin{array}{c}{\displaystyle \frac{1}{6}},{\displaystyle \frac{5}{6}}\\ \\ {\displaystyle \frac{2}{3}},{\displaystyle \frac{4}{3}}\end{array}\right]}_k{\displaystyle \frac{(1+4k)}{k(1+2k)}}& =\underset{y\to 0}{lim}{\displaystyle \frac{1}{y}}\left\{-1+{}_{5}F_4\left[\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2}},{\displaystyle \frac{5}{4}},& {\displaystyle \frac{1}{6}},{\displaystyle \frac{5}{6}},& y\\ \\ \hfill {\displaystyle \frac{1}{4}},& {\displaystyle \frac{2}{3}},{\displaystyle \frac{4}{3}},& {\displaystyle \frac{3}{2}}-y\end{array}|1\right]\right\}\hfill \\ & =\underset{y\to 0}{lim}{\displaystyle \frac{1}{y}}\{-1+\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{2}{3}},{\displaystyle \frac{4}{3}},{\displaystyle \frac{1}{2}}-y,{\displaystyle \frac{3}{2}}-y\\ \\ {\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{2}{3}}-y,{\displaystyle \frac{4}{3}}-y\end{array}\right]\hfill \\ & =\left[y\right]\mathrm{\Gamma }\left[\begin{array}{c}{\displaystyle \frac{2}{3}},{\displaystyle \frac{4}{3}},{\displaystyle \frac{1}{2}}-y,{\displaystyle \frac{3}{2}}-y\\ \\ {\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{2}{3}}-y,{\displaystyle \frac{4}{3}}-y\end{array}\right]=1-3ln3+4ln2. \square \hfill \end{array}$$

The remaining four k-series are variants of double Euler sums (cf. [23,24,25]) and can be shown by the series rearrangement.

$$\begin{array}{|c|}\hline \hfill \S 5.3\\ \hline\end{array}:\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k}{(2k+1)(2k-1)}}=ln2$$

(27)

$$\begin{array}{|c|}\hline \hfill \S 5.3\\ \hline\end{array}:\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k}{(2k+1)(2k-1)}}={\displaystyle \frac{{\pi }^2}{16}}$$

(28)

$$\begin{array}{|c|}\hline \hfill \S 5.3\\ \hline\end{array}:\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{3k}}{(2k+1)(2k-1)}}={\displaystyle \frac{{\pi }^2}{48}}+{\displaystyle \frac{\pi \sqrt{3}}{8}},$$

(29)

$$\begin{array}{|c|}\hline \hfill \S 6.2\\ \hline\end{array}:\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{3k}}{(2k+1)(2k-1)}}={\displaystyle \frac{\pi \sqrt{3}+2ln2}{6}}.$$

(30)

Proof of

(27). Rewriting the series as a double one, we have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k}{(2k+1)(2k-1)}}& =\sum _{j\le k}{\displaystyle \frac{1}{j(2k+1)(2k-1)}}=\sum _{j=1}^{\infty }{\displaystyle \frac{1}{2j}}\sum _{k=j}^{\infty }\left\{{\displaystyle \frac{1}{2k-1}}-{\displaystyle \frac{1}{2k+1}}\right\}\hfill \\ & =\sum _{j=1}^{\infty }{\displaystyle \frac{1}{2j(2j-1)}}=\sum _{j=1}^{\infty }\left\{{\displaystyle \frac{1}{2j-1}}-{\displaystyle \frac{1}{2j}}\right\}=\sum _{j=1}^{\infty }{\displaystyle \frac{{(-1)}^{j-1}}{j}}=ln2. \square \hfill \end{array}$$

Proof of

(28). This can be validated similarly as follows:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k}{(2k+1)(2k-1)}}& =\sum _{j\le k}{\displaystyle \frac{1}{(2j-1)(2k+1)(2k-1)}}=\sum _{j=1}^{\infty }{\displaystyle \frac{1}{4j-2}}\sum _{k=j}^{\infty }\left\{{\displaystyle \frac{1}{2k-1}}-{\displaystyle \frac{1}{2k+1}}\right\}\hfill \\ & =\sum _{j=1}^{\infty }{\displaystyle \frac{1}{2{(2j-1)}^2}}={\displaystyle \frac{\zeta \left(2\right)}{2}}\left(1-{\displaystyle \frac{1}{4}}\right)={\displaystyle \frac{{\pi }^2}{16}}. \square \hfill \end{array}$$

Proof of

(29). The remaining two k-sums require more elaborations. Denote by $\lceil x\rceil$ the minimum integer $\ge x$. We can prove the formula in the following manner:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{3k}}{(2k+1)(2k-1)}}& =\sum _{j\le 3k}^{\infty }{\displaystyle \frac{1}{(2j-1)(2k+1)(2k-1)}}=\sum _{j=1}^{\infty }{\displaystyle \frac{1}{2(2j-1)(2\lceil \frac{j}{3}\rceil -1)}}\hfill \\ \hfill \begin{array}{|c|}\hline j\to 3i-“0, 1, 2”\\ \hline\end{array}& =\sum _{i=1}^{\infty }\left\{{\displaystyle \frac{1}{2(2i-1)(6i-1)}}+{\displaystyle \frac{1}{2(2i-1)(6i-3)}}+{\displaystyle \frac{1}{2(2i-1)(6i-5)}}\right\}\hfill \\ & =\sum _{i=1}^{\infty }\left\{{\displaystyle \frac{1}{6{(2i-1)}^2}}+{\displaystyle \frac{3}{4(6i-5)}}-{\displaystyle \frac{3}{4(6i-1)}}\right\}\hfill \\ & ={\displaystyle \frac{\zeta \left(2\right)}{6}}\left(1-{\displaystyle \frac{1}{4}}\right)+\underset{m\to \infty }{lim}\sum _{i=-m}^m{\displaystyle \frac{1}{8(\frac{1}{6}-i)}}\hfill \\ & ={\displaystyle \frac{\zeta \left(2\right)}{8}}+{\displaystyle \frac{\pi }{8}}cot\left({\displaystyle \frac{\pi }{6}}\right)={\displaystyle \frac{{\pi }^2}{48}}+{\displaystyle \frac{\pi \sqrt{3}}{8}}. \square \hfill \end{array}$$

Proof of

(30). Analogously, we can show it as below:

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{3k}}{(2k+1)(2k-1)}}& =\sum _{j\le 3k}^{\infty }{\displaystyle \frac{1}{j(2k+1)(2k-1)}}=\sum _{j=1}^{\infty }{\displaystyle \frac{1}{2j(2\lceil \frac{j}{3}\rceil -1)}}\hfill \\ \hfill \begin{array}{|c|}\hline j\to 3i-“0, 1, 2”\\ \hline\end{array}& =\sum _{i=1}^{\infty }\left\{{\displaystyle \frac{1}{2(2i-1)\left(3i\right)}}+{\displaystyle \frac{1}{2(2i-1)(3i-1)}}+{\displaystyle \frac{1}{2(2i-1)(3i-2)}}\right\}\hfill \\ & =\sum _{i=1}^{\infty }\left\{{\displaystyle \frac{1}{6i-3}}-{\displaystyle \frac{1}{6i}}-{\displaystyle \frac{3}{2(3i-1)}}+{\displaystyle \frac{3}{2(3i-2)}}\right\}\hfill \\ & ={\displaystyle \frac{ln2}{3}}+\underset{m\to \infty }{lim}\sum _{i=-m}^m{\displaystyle \frac{1}{2(\frac{1}{3}-i)}}\hfill \\ & ={\displaystyle \frac{ln2}{3}}+{\displaystyle \frac{\pi }{2}}cot\left({\displaystyle \frac{\pi }{3}}\right)={\displaystyle \frac{\pi \sqrt{3}+2ln2}{6}}. \square \hfill \end{array}$$

Furthermore, we come across the following combined k-sums in Equation (21)

$$\sum _{k=1}^{\infty }{\displaystyle \frac{\left(3k\right)!{\left(\frac{1}{2}\right)}_k}{k!{\left(\frac{1}{2}\right)}_{3k}}}\left\{{\displaystyle \frac{1}{k^2}},{\displaystyle \frac{1}{{(1+2k)}^2}},{\displaystyle \frac{1}{{(1+6k)}^2}},{\displaystyle \frac{(1+4k)\left({\overline{\mathbf{H}}}_{2k}-3{\overline{\mathbf{H}}}_{6k}\right)}{k(1+2k)(1+6k)}}\right\}.$$

Their evaluations seem more difficult (even decomposing the summands into small pieces). These obstacles prevent us from determining the exact value (in closed form) of the corresponding n-series displayed in (21). The interested reader is enthusiastically encouraged to make further attempts.