Harmonic Series of Convergence Ratio “1/4" with Cubic Central Binomial Coefficients

Abstract

We examine a useful hypergeometric transformation formula by means of the coefficient extraction method. A large class of “binomial/harmonic series” (of convergence ratio “$1/4$”) containing the cubic central binomial coefficients and harmonic numbers is systematically investigated. Numerous closed summation formulae are established, including a remarkable series about harmonic numbers of the third order.

1. Introduction and Outline

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

$${\mathbf{H}}_n^{\langle m\rangle }\left(\lambda \right):=\sum _{k=0}^{n-1}{\displaystyle \frac{1}{{(\lambda +k)}^m}}\mathrm{and}{\overline{\mathbf{H}}}_n^{\langle m\rangle }\left(\lambda \right):=\sum _{k=0}^{n-1}{\displaystyle \frac{{(-1)}^k}{{(\lambda +k)}^m}}.$$

In case $\lambda ={\displaystyle \frac{1}{2}}$, they become the odd harmonic numbers

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

When $m=1$ and/or $\lambda =1$, they will be suppressed from these notations. There are two almost obvious, but useful equalities:

$${\mathbf{H}}_{2n}^{\langle m\rangle }={\mathbf{O}}_n^{\langle m\rangle }+2^{-m}{\mathbf{H}}_n^{\langle m\rangle }\mathrm{and}{\overline{\mathbf{H}}}_{2n}^{\langle m\rangle }={\mathbf{O}}_n^{\langle m\rangle }-2^{-m}{\mathbf{H}}_n^{\langle m\rangle }.$$

1.1. Coefficient Extraction

For an indeterminate x and $n\in {\mathbb{N}}_0$, the Pochhammer symbol reads as

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

Let $\left[x^m\right]\phi \left(x\right)$ stand for the coefficient of $x^m$ in the formal power series $\phi \left(x\right)$. Then, we have

$${\mathbf{H}}_n\left(\lambda \right)=\left[x\right]{\displaystyle \frac{{(\lambda +x)}_n}{{\left(\lambda \right)}_n}}=\left[x\right]{\displaystyle \frac{{\left(\lambda \right)}_n}{{(\lambda -x)}_n}}.$$

In general, it is not hard to show, by the generating function approach, the following formulae:

$$\left[x^m\right]{\displaystyle \frac{{(\lambda -x)}_n}{{\left(\lambda \right)}_n}}={\mathbf{Y}}_m\left\{-{\mathbf{H}}_n^{\langle k\rangle }\left(\lambda \right)\right\}\mathrm{and}\left[x^m\right]{\displaystyle \frac{{\left(\lambda \right)}_n}{{(\lambda -x)}_n}}={\mathbf{Y}}_m\left\{{\mathbf{H}}_n^{\langle k\rangle }\left(\lambda \right)\right\},$$

(1)

where the Bell polynomials (cf. [1], §3.3) are expressed explicitly by the multiple sum

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

(2)

with $\sigma \left(m\right)$ being the set of all m-partitions represented by m-tuples $(i_1,i_2,\dots ,i_m)\in {\mathbb{N}}_0^m$ subject to the linear condition ${\displaystyle \sum _{k=1}^m}k{i}_k=m$.

1.2. $\Gamma$-Function Expansion

Recall that the Pochhammer symbol can also be expressed in terms of the $\Gamma$-function

$${\left(x\right)}_n={\displaystyle \frac{\Gamma (x+n)}{\Gamma \left(x\right)}}\mathrm{with}\Gamma \left(x\right):={\int }_0^{\infty }{\tau }^{x-1}{e}^{-\tau }d\tau \mathrm{for}\Re \left(x\right)>0.$$

For the sake of brevity, the $\Gamma$-function quotient will be abbreviated to

$$\Gamma \left[\begin{array}{cccc}\alpha ,& \beta ,& \dots ,& \gamma \\ A,& B,& \dots ,& C\end{array}\right]:={\displaystyle \frac{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)\dots \Gamma \left(\gamma \right)}{\Gamma \left(A\right)\Gamma \left(B\right)\dots \Gamma \left(C\right)}}.$$

Denote the Euler constant by ${\displaystyle \gamma :=\underset{n\to \infty }{lim}\left({\mathbf{H}}_n-lnn\right)}$. Then, the logarithmic differentiation of the $\Gamma$-function results in the digamma function (cf. Rainville [2], §9)

$$\psi \left(z\right):={\displaystyle \frac{d}{dz}}ln\Gamma \left(z\right)={\displaystyle \frac{{\Gamma }^{\prime }\left(z\right)}{\Gamma \left(z\right)}}=-\gamma +\sum _{n=0}^{\infty }{\displaystyle \frac{z-1}{(n+1)(n+z)}}.$$

For a real number $\lambda \notin \mathbb{Z}\setminus \mathbb{N}$, we can extract the coefficients

$$\left[x\right]{\displaystyle \frac{\Gamma (\lambda -x)}{\Gamma \left(\lambda \right)}}=-\psi \left(\lambda \right)\mathrm{and}\left[x^2\right]{\displaystyle \frac{\Gamma (\lambda -x)}{\Gamma \left(\lambda \right)}}={\displaystyle \frac{{\psi }^2\left(\lambda \right)+{\psi }^{\prime }\left(\lambda \right)}{2}}$$

from the exponential expression

$${\displaystyle \frac{\Gamma (\lambda -x)}{\Gamma \left(\lambda \right)}}=exp\left\{-x\psi \left(\lambda \right)+\sum _{k=2}^{\infty }{\displaystyle \frac{x^k}{k}}{\zeta }_k\left(\lambda \right)\right\}.$$

(3)

Henceforth, the Riemann and Hurwitz zeta functions are defined respectively by

$$\zeta \left(m\right):=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^m}}\mathrm{and}{\zeta }_m\left(z\right):={\displaystyle \frac{{(-1)}^m}{(m-1)!}}{\mathcal{D}}_z^{m-1}\psi \left(z\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{1}{{(n+z)}^m}}.$$

In addition, we shall utilize G to denote the Catalan constant

$$G:=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n}{{(2n+1)}^2}}\approx 0.91596559417722.$$

1.3. Hypergeometric Series Transformations

Following Bailey [3], the classical hypergeometric series is defined by

$${}_{p}F_q\left[\begin{array}{c}{a}_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q\end{array}|x\right]=\sum _{n=0}^{\infty }{\displaystyle \frac{x^n}{n!}}{\displaystyle \frac{{\left(a_1\right)}_n{\left(a_2\right)}_n\dots {\left(a_p\right)}_n}{{\left(b_1\right)}_n{\left(b_2\right)}_n\dots {\left(b_q\right)}_n}}.$$

Among numerous transformations in the literature (see for example [4,5,6]), the following formula due to Thomae is fundamental (cf. Bailey [3], §3.2; where $\sigma =b+d-a-c-e$):

$$\begin{array}{cc}\hfill {}_{3}F_2\left[\begin{array}{c}\hfill a,c,e\\ \hfill b,d\end{array}|1\right]=& {}_{3}F_2\left[\begin{array}{c}\hfill \sigma ,b-a,d-a\\ \hfill c+\sigma ,e+\sigma \end{array}|1\right]\Gamma \left[\begin{array}{c}\sigma ,b,d\\ a,c+\sigma ,e+\sigma \end{array}\right].\hfill \end{array}$$

(4)

In 2011, the second author discovered, among several transformation formulae, the following useful one, which can be restated as in the lemma below.

Lemma 1

(Chu [5], Theorem 2.7). For four indeterminates $\{a,b,c,d\}$ subject to the condition $\Re (c+d-a-b)>1$, we have the transformation formula

$$(c+d-a-b-1)\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_k{\left(b\right)}_k}{{\left(c\right)}_k{\left(d\right)}_k}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\nabla }_n(a,b,c,d)}{c+d-a-b+2n}}{\displaystyle \frac{{(c-a)}_n{(c-b)}_n{(d-a)}_n{(d-b)}_n}{{\left(c\right)}_n{\left(d\right)}_n{(c+d-a-b)}_{2n}}},$$

where ${\nabla }_n(a,b,c,d)$ is a quadratic polynomial defined by

$${\nabla }_n(a,b,c,d)=(c+d-a-b+2n)(c-1+n)+(d-a+n)(d-b+n).$$

By introducing a free variable x, the aim of this paper is to consider, via the coefficient extraction method (cf. [7,8,9]), the above transformation under the following four parameter settings:

$$\begin{array}{cc}\hfill \mathrm{U}:& a\to {\displaystyle \frac{1}{2}}+ax,b\to {\displaystyle \frac{1}{2}}+bx,c\to 1+cx,d\to 1+dx;\hfill \\ \hfill \mathrm{V}:& a\to ax-{\displaystyle \frac{1}{2}},b\to bx-{\displaystyle \frac{1}{2}},c\to cx,d\to dx;\hfill \\ \hfill \mathrm{W}:& a\to {\displaystyle \frac{1}{2}}+ax,b\to bx-{\displaystyle \frac{1}{2}},c\to cx,d\to 1+dx;\hfill \\ \hfill \Omega :& a\to ax-{\displaystyle \frac{1}{2}},b\to bx-{\displaystyle \frac{1}{2}},c\to 1+cx,d\to 1+dx.\hfill \end{array}$$

Numerous closed formulae will be established for the series (of convergent ratio “${\displaystyle \frac{1}{4}}$”) containing both cubic central binomial coefficients and harmonic numbers, including one prominently conjectured formula by Sun [10] (Conjecture 33 and Equation (99)):

$$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{H}}_n^{\langle 3\rangle }+64{\mathbf{O}}_n^{\langle 3\rangle }\right\}={\displaystyle \frac{200\zeta \left(3\right)}{\pi }}-64G.$$

Our results will be distributed from Section 2 to Section 5, devoted to dealing with the series under the four parameter settings $\{\mathrm{U},\mathrm{V},\mathrm{W},\Omega \}$, respectively. During the course of computing harmonic series, four crucial k-sums emerged. In order not to interrupt the route of investigation, their evaluations are put together in Section 6, where further related problems are briefly discussed.

There exist various summation formulae containing harmonic numbers (cf. [11,12,13,14,15,16,17,18,19,20]). Many of them involve binomial coefficients (cf. [17,21,22,23,24]) and squared central binomial coefficients (cf. [9,25,26,27,28]). To the best of the author’s knowledge, almost all of the identities (except for a few cited explicitly) presented in this paper for the series concerning cubic central binomial coefficients are not only new, but also challenging. They may serve as a reference source to readers for further investigation. In order to ensure accuracy, all the displayed equations throughout the paper have been tested numerically using the inbuilt commands of computer algebra system Wolfram Mathematica (version 11).

2. Series Under Parameter Setting $\mathrm{U}$

Under the parameter setting “$\mathrm{U}$”, the transformation in Lemma 1 becomes

$$\begin{array}{cc}\hfill x(c+d-a-b)& \times {}_{3}F_2\left[\begin{array}{ccc}1,& \frac{1}{2}+ax,& \frac{1}{2}+bx\\ & 1+cx,& 1+dx\end{array}|1\right]=\sum _{n=0}^{\infty }{\displaystyle \frac{{\nabla }_n\left(\frac{1}{2}+ax,\frac{1}{2}+bx,1+cx,1+dx\right)}{1+(c+d-a-b)x+2n}}\hfill \\ & \times {\displaystyle \frac{{(\frac{1}{2}+cx-ax)}_n{(\frac{1}{2}+cx-bx)}_n{(\frac{1}{2}+dx-ax)}_n{(\frac{1}{2}+dx-bx)}_n}{{(1+cx)}_n{(1+dx)}_n{(1+cx+dx-ax-bx)}_{2n}}}.\hfill \end{array}$$

The ${}_{3}F_2$-series on the left diverges when $x=0$. By making use of Thomae’s transformation (4), we can express it as

$${}_{3}F_2\left[\begin{array}{c}cx+dx-ax-bx,cx,dx\\ \frac{1}{2}+cx+dx-ax,\frac{1}{2}+cx+dx-bx\end{array}|1\right]\times {\displaystyle \frac{\Gamma (1+cx)\Gamma (1+dx)\Gamma (cx+dx-ax-bx)}{\Gamma (\frac{1}{2}+cx+dx-ax)\Gamma (\frac{1}{2}+cx+dx-bx)}}.$$

Therefore, we have established the following transformation formula involving a free variable “x” and four independent indeterminates $\{a,b,c,d\}$.

Theorem 1.

Let ${\nabla }_n$ be as in Lemma 1. The following transformation formula holds:

$$\begin{array}{c}{}_{3}F_2\left[\begin{array}{c}cx+dx-ax-bx,cx,dx\\ \frac{1}{2}+cx+dx-ax,\frac{1}{2}+cx+dx-bx\end{array}|1\right]{\displaystyle \frac{\Gamma (1+cx)\Gamma (1+dx)\Gamma (1+cx+dx-ax-bx)}{\Gamma (\frac{1}{2}+cx+dx-ax)\Gamma (\frac{1}{2}+cx+dx-bx)}}\hfill \\ =\sum _{n=0}^{\infty }{\displaystyle \frac{{\nabla }_n\left(\frac{1}{2}+ax,\frac{1}{2}+bx,1+cx,1+dx\right)}{1+(c+d-a-b)x+2n}}{\displaystyle \frac{{(\frac{1}{2}+cx-ax)}_n{(\frac{1}{2}+cx-bx)}_n{(\frac{1}{2}+dx-ax)}_n{(\frac{1}{2}+dx-bx)}_n}{{(1+cx)}_n{(1+dx)}_n{(1+cx+dx-ax-bx)}_{2n}}}.\hfill \end{array}$$

In order to derive identities for binomial/harmonic series from this theorem, we often come across the following remarkable binomial sum, which corresponds to the coefficient of $x^2$ in the above ${}_{3}F_2$-series:

$$\tilde{\mathbf{u}}:=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{k^3{\left(\frac{1}{2}\right)}_k^2}}=8\pi G-14\zeta \left(3\right).$$

(5)

Gosper [29] discovered it by an iterative use of Abel’s lemma on summation by parts. Joakim Petersson found another proof (reproduced in [30]) by differentiating the parametric integral ingeniously:

$$\tilde{\mathbf{u}}={\int }_0^1{\int }_0^1{\displaystyle \frac{-ln(1-xy)}{xy\sqrt{(1-x)(1-y)}}}dxdy.$$

We shall present new alternative proofs for it in Section 6.1.

Both sides of the equality displayed in Theorem 1 are convergent series around $x=0$. By analytic continuation, we can expand, via Equations (1)–(3), the two sides into power series in x. Denote by ${\mathrm{U}}_m(a,b,c,d)$ the equality formed by extracting the coefficients of $x^m$ across the equation in Theorem 1. By assigning particular values for parameters $\{a,b,c,d\}$ and then making simplifications, we can establish several identities for series with harmonic numbers.

To showcase, we illustrate, for $m=1$, how to carry out the procedure just described. In this case, the equation corresponding to ${\mathrm{U}}_1(a,b,c,d)$ reads as

$$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{(a+b-3c-3d)\left(2-(1+6n){\mathbf{H}}_n\right)+6(1+6n)(a+b-c-d){\mathbf{O}}_n\right\}\hfill \\ & ={\displaystyle \frac{16ln2}{\pi }}(a+b-2c-2d).\hfill \end{array}$$

Then, it is trivial for us to write down the two formulae labeled by ${\mathrm{U}}_1(1,1,1,1)$ and ${\mathrm{U}}_1(3,3,1,1)$ under the respective parameter specifications.

Analogously, we can examine the series ${\mathrm{U}}_3(x+y,2x-y,x-y,y)$. After some routine simplifications, it can be written explicitly as shown below:

$$\begin{gathered} \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{\begin{array}{c}36x^2{\mathbf{O}}_n^3+{\mathbf{H}}_n^{\langle 3\rangle }\left(y^2-xy\right)+8{\mathbf{O}}_n^{\langle 3\rangle }\left(3x^2-8xy+8y^2\right)\\ +6{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n\left(x^2-xy+y^2\right)-12{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\left(5x^2-8xy+8y^2\right)\end{array}\right\} \\ ={\displaystyle \frac{4}{3\pi }}\left\{x^2\left(4{ln}^{3}2+{\pi }^{2}ln2\right)+(xy-y^2)\left(48\pi G-108\zeta \left(3\right)+2{\pi }^{2}ln2\right)\right\}. \end{gathered}$$

(6)

From this equation, we can deduce three interesting identities.

• Firstly, comparing the coefficient of $x^2$ in (6) gives rise to the summation formula below:

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{4{\mathbf{O}}_n^{\langle 3\rangle }+{\mathbf{O}}_n\left({\mathbf{H}}_n^{\langle 2\rangle }+6{\mathbf{O}}_n^2-10{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}={\displaystyle \frac{2\pi ln2}{9}}+{\displaystyle \frac{8{ln}^{3}2}{9\pi }}.$$
• Secondly, by extracting the coefficient of $y^2$ across (6), we establish the identity:

  $$\begin{gathered} \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{H}}_n^{\langle 3\rangle }+64{\mathbf{O}}_n^{\langle 3\rangle }+6{\mathbf{O}}_n\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\} \\ ={\displaystyle \frac{32\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{8\pi ln2}{3}}-{\displaystyle \frac{8\tilde{\mathbf{u}}}{\pi }}={\displaystyle \frac{144\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{8\pi ln2}{3}}-64G. \end{gathered}$$

  (7)
  Recall the known equality due to the authors: (cf. [31], Equation (3))

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{O}}_n\left(16{\mathbf{O}}_n^{\langle 2\rangle }-{\mathbf{H}}_n^{\langle 2\rangle }\right)\right\}={\displaystyle \frac{28\zeta \left(3\right)}{3\pi }}+{\displaystyle \frac{4\pi ln2}{9}}.$$
  By adding six times of it to (7), we arrive at the following remarkable identity conjectured by Sun [10] (Equation (99) and Conjecture 33):

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{H}}_n^{\langle 3\rangle }+64{\mathbf{O}}_n^{\langle 3\rangle }\right\}={\displaystyle \frac{88\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{8\tilde{\mathbf{u}}}{\pi }}={\displaystyle \frac{200\zeta \left(3\right)}{\pi }}-64G.$$

  (8)
• Thirdly, let $x^2-xy+y^2=0$, which annihilates ${\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n$ in the summand of (6). In accordance with one of the solutions of algebraic equation $x=2\mathbf{i}$ and $y=\mathbf{i}+\sqrt{3}$, we find from (6) the following formula:

  $$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{H}}_n^{\langle 3\rangle }+40{\mathbf{O}}_n^{\langle 3\rangle }-36{\mathbf{O}}_n({\mathbf{O}}_n^2+{\mathbf{O}}_n^{\langle 2\rangle })\right\}\hfill \\ ={\displaystyle \frac{32\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{16{ln}^{3}2}{3\pi }}-4\pi ln2-{\displaystyle \frac{8\tilde{\mathbf{u}}}{\pi }}={\displaystyle \frac{144\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{16{ln}^{3}2}{3\pi }}-4\pi ln2-64G.\hfill \end{array}$$

  (9)
  Then, it can be shown alternatively that (8) follows also by relating (9) to yet another known identity (cf. Li and Chu [31], Equation (4)):

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{2{\mathbf{O}}_n^{\langle 3\rangle }+3{\mathbf{O}}_n^3+3{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{14\zeta \left(3\right)}{3\pi }}+{\displaystyle \frac{4{ln}^{3}2}{9\pi }}+{\displaystyle \frac{\pi ln2}{3}}.$$

By carrying out the same procedure just demonstrated, further binomial/harmonic series identities can be derived from Theorem 1. The following representatives are recorded as examples, where most of them seem new except for a few series (reproduced for integrity) that were previously evaluated by the authors in [31]. Theoretically, there are infinite choices of “$\{a,b,c,d\}$” values that lead to binomial/harmonic sums. Our selection of identities to include (here and next sections) are based on three criteria: representative sums, elegantly closed formulae, and non-routine outcome results.

• $\begin{array}{|c|}\hline {\mathrm{U}}_0(a,b,c,d)\mathrm{Constant}\mathrm{term}\mathrm{identity}\\ \hline\end{array}$ Ramanujan [32] (cf. Guillera [33])

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)={\displaystyle \frac{4}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_1(1,1,1,1)\\ \hline\end{array}$ (cf. [31], Equation (15))

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{2-(1+6n){\mathbf{H}}_n\right\}={\displaystyle \frac{8ln2}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_1(3,3,1,1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n){\mathbf{O}}_n={\displaystyle \frac{4ln2}{3\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_2(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{8}{1+2n}}+(1+6n){\mathbf{H}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{8\pi }{3}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_2(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{1}{2+4n}}+(1+6n){\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{\pi }{4}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_2(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$ (cf. [31], Equation (12))

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{4}{1+2n}}+4{\mathbf{H}}_n-(1+6n){\mathbf{H}}_n^2\right\}=2\pi -{\displaystyle \frac{16{ln}^{2}2}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_2(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{-1}{6+12n}}+(1+6n){\mathbf{O}}_n^2\right\}={\displaystyle \frac{\pi }{36}}+{\displaystyle \frac{4{ln}^{2}2}{9\pi }}.$$
• $\begin{array}{|c|}\hline \left[y^2\right]{\mathrm{U}}_3(1+y,1-y,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{{\mathbf{H}}_n}{1+2n}}-4{\mathbf{O}}_n^{\langle 2\rangle }+2(1+6n){\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{7\zeta \left(3\right)}{\pi }}-\pi ln2.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(1,2,1,0)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{4{\mathbf{O}}_n^{\langle 3\rangle }+6{\mathbf{O}}_n^3+{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n-10{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{2\pi }{9}}ln2+{\displaystyle \frac{8{ln}^{3}2}{9\pi }}.$$
• $\begin{array}{|c|}\hline \left[y^2\right]{\mathrm{U}}_3(3+y,3-y,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{1}{{(1+2n)}^2}}-{\displaystyle \frac{3{\mathbf{O}}_n}{1+2n}}+(1+6n)\left(4{\mathbf{O}}_n^{\langle 3\rangle }-6{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}={\displaystyle \frac{7\zeta \left(3\right)}{2\pi }}-{\displaystyle \frac{\pi ln2}{2}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{12{\mathbf{H}}_n}{1+2n}}+6{\mathbf{H}}_n^2+(1+6n)\left({\mathbf{H}}_n^{\langle 3\rangle }-{\mathbf{H}}_n^3\right)\right\}={\displaystyle \frac{60\zeta \left(3\right)}{\pi }}+{\displaystyle \frac{32{ln}^{3}2}{\pi }}-12\pi ln2.$$
• $\begin{array}{|c|}\hline \mathrm{Linear}\mathrm{combination:}2\left(11\right)–\left(10\right)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{8{\mathbf{H}}_n}{1+2n}}-2{\mathbf{H}}_n^{\langle 2\rangle }+(1+6n)\left({\mathbf{H}}_n^{\langle 3\rangle }+{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }\right)\right\}={\displaystyle \frac{32\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{16\pi ln2}{3}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(1,-\mathbf{i},-\mathbf{i},1)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{48{\mathbf{O}}_n^{\langle 2\rangle }+6{\mathbf{H}}_n^2+(1+6n)\left({\mathbf{H}}_n^{\langle 3\rangle }-{\mathbf{H}}_n^3-24{\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}={\displaystyle \frac{32{ln}^{3}2}{\pi }}-{\displaystyle \frac{24\zeta \left(3\right)}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(1,1,1,1)\\ \hline\end{array}$

  $$\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{6{\mathbf{H}}_n^{\langle 2\rangle }+12{\mathbf{H}}_n^2-(1+6n)\left({\mathbf{H}}_n^{\langle 3\rangle }+2{\mathbf{H}}_n^3+3{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }\right)\right\}\hfill \\ ={\displaystyle \frac{24\zeta \left(3\right)}{\pi }}+{\displaystyle \frac{64{ln}^{3}2}{\pi }}-8\pi ln2.\hfill \end{array}$$

  (10)
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(2,4-2\sqrt{3},3-\sqrt{3}+\mathbf{i}(1-\sqrt{3}),3-\sqrt{3}-\mathbf{i}(1-\sqrt{3}))\\ \hline\end{array}$

  $$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{4{\mathbf{H}}_n}{1+2n}}+2{\mathbf{H}}_n^{\langle 2\rangle }+6{\mathbf{H}}_n^2-(1+6n)\left({\mathbf{H}}_n^3+{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }\right)\right\}\hfill \\ ={\displaystyle \frac{28\zeta \left(3\right)}{\pi }}+{\displaystyle \frac{32{ln}^{3}2}{\pi }}-{\displaystyle \frac{20\pi }{3}}ln2.\hfill \end{array}$$

  (11)
• $\begin{array}{|c|}\hline \left[y^2\right]{\mathrm{U}}_3(y,6-y,1+\mathbf{i}y,1-\mathbf{i}y)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{16}{{(1+2n)}^2}}-{\displaystyle \frac{48{\mathbf{O}}_n}{1+2n}}-(1+6n)\left({\mathbf{H}}_n^{\langle 3\rangle }+6{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n\right)\right\}\hfill \\ & =64G-{\displaystyle \frac{16\pi ln2}{3}}-{\displaystyle \frac{88\zeta \left(3\right)}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(11,7-6\mathbf{i}\sqrt{2},0,6-2\mathbf{i}\sqrt{2})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{1}{{(1+2n)}^2}}-{\displaystyle \frac{3{\mathbf{O}}_n}{1+2n}}-(1+6n)\left(6{\mathbf{O}}_n^3+{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n-4{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}\hfill \\ & ={\displaystyle \frac{7\zeta \left(3\right)}{2\pi }}-{\displaystyle \frac{13\pi }{18}}ln2-{\displaystyle \frac{8{ln}^{3}2}{9\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(19,17-6\mathbf{i}\sqrt{2},5-4\mathbf{i}\sqrt{2},7+2\mathbf{i}\sqrt{2})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{H}}_n^{\langle 3\rangle }-96{\mathbf{O}}_n^3-10{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n-64{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right\}\hfill \\ & ={\displaystyle \frac{144\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{56\pi }{9}}ln2-{\displaystyle \frac{128{ln}^{3}2}{9\pi }}-64G.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(21,15-6\mathbf{i}\sqrt{6},3-4\mathbf{i}\sqrt{6},9+2\mathbf{i}\sqrt{6})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{5{\mathbf{H}}_n^{\langle 3\rangle }+128{\mathbf{O}}_n^{\langle 3\rangle }-288{\mathbf{O}}_n^3-18{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n\right\}\hfill \\ & ={\displaystyle \frac{720\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{128{ln}^{3}2}{3\pi }}-24\pi ln2-320G.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_3(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{6}{{(1+2n)}^2}}-{\displaystyle \frac{18{\mathbf{O}}_n}{1+2n}}-(1+6n)\left({\mathbf{H}}_n^{\langle 3\rangle }+16{\mathbf{O}}_n^{\langle 3\rangle }-36{\mathbf{O}}_n^3\right)\right\}\hfill \\ & =64G+\pi ln2+{\displaystyle \frac{16{ln}^{3}2}{3\pi }}-{\displaystyle \frac{123\zeta \left(3\right)}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_4(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{2{\mathbf{O}}_n^{\langle 2\rangle }}{1+2n}}-(1+6n)\left({\mathbf{O}}_n^{\langle 4\rangle }-2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\right\}={\displaystyle \frac{{\pi }^3}{96}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_4(1,-1,-1,1)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}(1+6n)\left\{{\mathbf{H}}_n^{\langle 4\rangle }-256{\mathbf{O}}_n^{\langle 4\rangle }+{({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle })}^2\right\}=-{\displaystyle \frac{8{\pi }^3}{45}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_4(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (cf. [31], Equation (20))

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{16{\mathbf{H}}_n^{\langle 2\rangle }}{1+2n}}+(1+6n)\left({\mathbf{H}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2+128{\mathbf{O}}_n^{\langle 4\rangle }\right)\right\}={\displaystyle \frac{52}{45}}{\pi }^3.$$
• $\begin{array}{|c|}\hline \left[y^2\right]{\mathrm{U}}_4(y,-y,1,-1)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{1+2n}}+2(1+6n)\left({\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 2\rangle }+12{\mathbf{O}}_n^{\langle 4\rangle }-8{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\right\}={\displaystyle \frac{{\pi }^3}{12}}.$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_4(0,0,1,-1)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }}{1+2n}}+{\displaystyle \frac{(1+6n)}{8}}{\left({\mathbf{H}}_n^{\langle 4\rangle }-32{\mathbf{O}}_n^{\langle 4\rangle }+({\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\}={\displaystyle \frac{7{\pi }^3}{360}}.$$
• $\begin{array}{|c|}\hline \left[x^2\right]{\mathrm{U}}_4(1+x,1-x,1,1)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }+2{\mathbf{H}}_n^2}{1+2n}}-16{\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }+2(1+6n)\left({\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 2\rangle }+2{\mathbf{H}}_n^2{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}\hfill \\ & =4\pi {ln}^{2}2+{\displaystyle \frac{{\pi }^3}{3}}-{\displaystyle \frac{56ln2}{\pi }}\zeta \left(3\right).\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_4(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{{\displaystyle \frac{24{\mathbf{H}}_n^2}{1+2n}}+8{\mathbf{H}}_n^3-8{\mathbf{H}}_n^{\langle 3\rangle }+(1+6n)\left(3{\mathbf{H}}_n^{\langle 4\rangle }-96{\mathbf{O}}_n^{\langle 4\rangle }-{\mathbf{H}}_n^4+4{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 3\rangle }\right)\right\}\hfill \\ & ={\displaystyle \frac{17{\pi }^3}{15}}-{\displaystyle \frac{64{ln}^{4}2}{\pi }}-{\displaystyle \frac{480ln2}{\pi }}\zeta \left(3\right)+48\pi {ln}^{2}2.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{U}}_4(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{\begin{array}{c}{\displaystyle \frac{4}{{(1+2n)}^3}}-{\displaystyle \frac{12{\mathbf{O}}_n}{{(1+2n)}^2}}+{\displaystyle \frac{18{\mathbf{O}}_n^2}{1+2n}}\\ -(1+6n)\left(17{\mathbf{O}}_n^{\langle 4\rangle }+18{\mathbf{O}}_n^4-2{\mathbf{H}}_n^{\langle 3\rangle }{\mathbf{O}}_n-32{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 3\rangle }\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{8\widehat{\mathbf{u}}}{3\pi }}-{\displaystyle \frac{128Gln2}{3}}-{\displaystyle \frac{\pi {ln}^{2}2}{3}}-{\displaystyle \frac{8{ln}^{4}2}{9\pi }}-{\displaystyle \frac{17{\pi }^3}{288}}+{\displaystyle \frac{82ln2}{\pi }}\zeta \left(3\right),\hfill \end{array}$$

  where $\widehat{\mathbf{u}}$ denotes the k-sum below, that we are unable to evaluate it in closed form:

  $$\widehat{\mathbf{u}}:=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{k^3{\left(\frac{1}{2}\right)}_k^2}}\left\{{\displaystyle \frac{1}{k}}+2{\overline{\mathbf{H}}}_{2k}\right\}.$$

  (12)
• $\begin{array}{|c|}\hline \left[x^4\right]{\mathrm{U}}_5(x,2-x,0,2)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}{{256}^n}}\left\{\left(2-(1+6n){\mathbf{H}}_n\right)\left(2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{O}}_n^{\langle 4\rangle }\right)-{\displaystyle \frac{2{\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }}{1+2n}}\right\}\hfill \\ & ={\displaystyle \frac{31\zeta \left(5\right)}{4\pi }}+{\displaystyle \frac{{\pi }^{3}ln2}{48}}-{\displaystyle \frac{7\pi }{8}}\zeta \left(3\right).\hfill \end{array}$$
  Observe that ${\left({\displaystyle \frac{1}{2}}\right)}_k={\displaystyle \frac{(k+1)!}{4^k}}{C}_k$, where $C_k$ is the Catalan number. Thus, series containing ${\left({\displaystyle \frac{1}{2}}\right)}_k$ can be rewritten in terms of Catalan numbers or central binomial coefficients.

3. Series Under Parameter Setting $\mathrm{V}$

Performing the parameter replacement “$\mathrm{V}$” in Lemma 1

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

and then applying the Thoame transformation Formula (4)

$$\begin{array}{cc}\hfill {}_{3}F_2\left[\begin{array}{ccc}1,& ax-{\displaystyle \frac{1}{2}},\hfill & bx-{\displaystyle \frac{1}{2}}\hfill \\ & cx,\hfill & dx\hfill \end{array}|1\right]={}_{3}F_2\left[\begin{array}{c}cx+dx-ax-bx,cx-1,dx-1\\ cx+dx-ax-{\displaystyle \frac{1}{2}},cx+dx-bx-{\displaystyle \frac{1}{2}}\end{array}|1\right]& \\ \hfill \times \Gamma \left[\begin{array}{c}cx+dx-ax-bx,cx,dx\\ cx+dx-ax-{\displaystyle \frac{1}{2}},cx+dx-bx-{\displaystyle \frac{1}{2}}\end{array}\right]& ,\hfill \end{array}$$

we can state the resulting expression as in the following theorem.

Theorem 2.

Let ${\nabla }_n$ be as in Lemma 1. The following transformation formula holds:

$$\begin{array}{cc}& \Gamma \left[\begin{array}{c}1+cx+dx-ax-bx,1+cx,1+dx\\ cx+dx-ax-{\displaystyle \frac{1}{2}},cx+dx-bx-{\displaystyle \frac{1}{2}}\end{array}\right]{}_{3}F_2\left[\begin{array}{c}cx+dx-ax-bx,cx-1,dx-1\\ cx+dx-ax-{\displaystyle \frac{1}{2}},cx+dx-bx-{\displaystyle \frac{1}{2}}\end{array}|1\right]\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\nabla }_n\left(ax-\frac{1}{2},bx-\frac{1}{2},cx,dx\right){\left[\frac{1}{2}+cx-ax,\frac{1}{2}+cx-bx,\frac{1}{2}+dx-ax,\frac{1}{2}+dx-bx\right]}_n}{{(1+cx+dx-ax-bx)}_{2n+1}{(1+cx)}_{n-1}{(1+dx)}_{n-1}}}.\hfill \end{array}$$

Let $T_k$ stand for the summation term indexed by k in the above ${}_{3}F_2$-series. By putting two initial terms aside, we can express this series as

$$T_0+T_1+T_2\times {}_{4}F_3\left[\begin{array}{c}\hfill 1,2+cx+dx-ax-bx,1+cx,1+dx\\ \hfill 3,{\displaystyle \frac{3}{2}}+cx+dx-ax,{\displaystyle \frac{3}{2}}+cx+dx-bx\end{array}|1\right]$$

In order to derive identities for binomial/harmonic series from Theorem 2, it is necessary to determine the constant term of the ${}_{4}F_3$-series. It is given by the following formula and will be confirmed in Section 6.2:

$$\tilde{\mathbf{v}}:=\sum _{k=0}^{\infty }{\displaystyle \frac{{(k!)}^2}{(2+k){\left(\frac{3}{2}\right)}_k^2}}={\displaystyle \frac{\pi -1}{4}}+{\displaystyle \frac{\pi G}{2}}-{\displaystyle \frac{7\zeta \left(3\right)}{8}}.$$

(13)

Denote by ${\mathrm{V}}_m(a,b,c,d)$ the resulting formula by equating the coefficients of $x^m$ across the equation displayed in Theorem 2. By specifying particular values for parameters $\{a,b,c,d\}$ and then making simplifications, we derive the following binomial/harmonic series identities.

• $\begin{array}{|c|}\hline {\mathrm{V}}_0(a,b,c,d)\mathrm{Constant}\mathrm{term}\mathrm{identity}\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{n^2(2n-1)\right\}={\displaystyle \frac{1}{3\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_1(1,1,1,1)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{n(8n-3)-3n^2(2n-1){\mathbf{H}}_n\right\}={\displaystyle \frac{2}{\pi }}\left\{ln2-1\right\}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_1(3,3,1,1)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{n(2n-1)-6n^2(2n-1){\mathbf{O}}_n\right\}={\displaystyle \frac{-2}{3\pi }}(3+ln2).$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_2(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{3-4n^2}{2n+1}}+3n^2(2n-1){\mathbf{H}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{2\pi }{3}}+{\displaystyle \frac{4}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_2(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{n^2}{2n+1}}+6n^2(2n-1){\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{\pi }{8}}+{\displaystyle \frac{1}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_2(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{16n^2+4n-3}{2n+1}}-2n(8n-3){\mathbf{H}}_n+3n^2(2n-1){\mathbf{H}}_n^2\right\}\hfill \\ & ={\displaystyle \frac{4{ln}^{2}2}{\pi }}-{\displaystyle \frac{8ln2}{\pi }}-{\displaystyle \frac{\pi }{2}}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_2(3,3,1+\mathbf{i}\sqrt{3},d=1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{3n^2-1}{6(2n+1)}}-n(2n-1){\mathbf{O}}_n+3n^2(2n-1){\mathbf{O}}_n^2\right\}\hfill \\ & ={\displaystyle \frac{1}{2\pi }}+{\displaystyle \frac{\pi }{144}}+{\displaystyle \frac{{ln}^{2}2}{9\pi }}+{\displaystyle \frac{2ln2}{3\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline \left[x^2\right]{\mathrm{V}}_3(1+x,1-x,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{n}{2n+1}}-{\displaystyle \frac{n^2}{2n+1}}{\mathbf{H}}_n+2n(8n-3){\mathbf{O}}_n^{\langle 2\rangle }\\ -6n^2(2n-1){\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }\end{array}\right\}={\displaystyle \frac{2ln2}{\pi }}-{\displaystyle \frac{7\zeta \left(3\right)}{4\pi }}+{\displaystyle \frac{\pi }{4}}ln2-{\displaystyle \frac{\pi }{4}}.$$
• $\begin{array}{|c|}\hline \left[x^2\right]{\mathrm{V}}_3(3+x,3-x,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}\hfill {\displaystyle \frac{n(1+4n)}{{(2n+1)}^2}}-{\displaystyle \frac{6n^2}{2n+1}}{\mathbf{O}}_n+6n(2n-1){\mathbf{O}}_n^{\langle 2\rangle }\\ \hfill +12n^2(2n-1)\left(2{\mathbf{O}}_n^{\langle 3\rangle }-3{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)\end{array}\right\}={\displaystyle \frac{7\zeta \left(3\right)}{4\pi }}-{\displaystyle \frac{2ln2}{\pi }}-{\displaystyle \frac{3\pi }{4}}-{\displaystyle \frac{\pi }{4}}ln2.$$
• $\begin{array}{|c|}\hline \left[y\right]{\mathrm{V}}_3(y,6-y,1+y\mathbf{i},1-y\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}\hfill {\displaystyle \frac{n(1+4n)}{{(2n+1)}^2}}-{\displaystyle \frac{6n^2}{2n+1}}{\mathbf{O}}_n+6n(2n-1){\mathbf{O}}_n^{\langle 2\rangle }\\ \hfill +12n^2(2n-1)\left(2{\mathbf{O}}_n^{\langle 3\rangle }-3{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)\end{array}\right\}={\displaystyle \frac{7\zeta \left(3\right)}{4\pi }}-{\displaystyle \frac{2ln2}{\pi }}-{\displaystyle \frac{3\pi }{4}}-{\displaystyle \frac{\pi }{4}}ln2.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_3(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{2}{2n+1}}+n^2(2n-1)({\mathbf{H}}_n^{\langle 3\rangle }-{\mathbf{H}}_n^3)\\ +{\displaystyle \frac{3-4n-16n^2}{2n+1}}{\mathbf{H}}_n+n(8n-3){\mathbf{H}}_n^2\end{array}\right\}=\pi -\pi ln2+{\displaystyle \frac{5\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{8{ln}^{2}2}{\pi }}+{\displaystyle \frac{8{ln}^{3}2}{3\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_3(1,2,1,0)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}6n^2(2n-1)(4{\mathbf{O}}_n^{\langle 3\rangle }+6{\mathbf{O}}_n^3+{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n-10{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle })\\ -n(2n-1)\left({\mathbf{H}}_n^{\langle 2\rangle }-10{\mathbf{O}}_n^{\langle 2\rangle }+18{\mathbf{O}}_n^2\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{\pi }{3}}+{\displaystyle \frac{4{ln}^{3}2}{9\pi }}+{\displaystyle \frac{4{ln}^{2}2}{\pi }}+{\displaystyle \frac{16ln2}{3\pi }}+{\displaystyle \frac{\pi ln2}{9}}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_3(1,1,1,1)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}\hfill 2-(14n-3){\mathbf{H}}_n+n(8n-3)\left({\mathbf{H}}_n^{\langle 2\rangle }+2{\mathbf{H}}_n^2\right)\\ \hfill -n^2(2n-1)({\mathbf{H}}_n^{\langle 3\rangle }+2{\mathbf{H}}_n^3+3{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle })\end{array}\right\}\hfill \\ & ={\displaystyle \frac{2\zeta \left(3\right)}{\pi }}+{\displaystyle \frac{2\pi }{3}}+{\displaystyle \frac{16{ln}^{3}2}{3\pi }}-{\displaystyle \frac{16{ln}^{2}2}{\pi }}+{\displaystyle \frac{8ln2}{\pi }}-{\displaystyle \frac{2\pi }{3}}ln2.\hfill \end{array}$$
• $\begin{array}{|c|}\hline \left[y^2\right]{\mathrm{V}}_3(y,6-y,1+y\mathbf{i},1-y\mathbf{i})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}\hfill {\displaystyle \frac{8n(1+4n)}{3{(2n+1)}^2}}-{\displaystyle \frac{2(3-4n^2)}{2n+1}}{\mathbf{O}}_n+n(2n-1){\mathbf{H}}_n^{\langle 2\rangle }\\ \hfill -n^2(2n-1){\mathbf{H}}_n^{\langle 3\rangle }-6n^2(2n-1){\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n\end{array}\right\}\hfill \\ & ={\displaystyle \frac{8}{3}}-{\displaystyle \frac{4\pi }{3}}-{\displaystyle \frac{8ln2}{3\pi }}-{\displaystyle \frac{4\pi }{9}}ln2+{\displaystyle \frac{16G}{3}}-{\displaystyle \frac{22\zeta \left(3\right)}{3\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_3(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{n(1+4n)}{{(2n+1)}^2}}+{\displaystyle \frac{6(3n^2-1)}{2n+1}}{\mathbf{O}}_n-18n(2n-1){\mathbf{O}}_n^2\\ -n^2(2n-1)\left({\mathbf{H}}_n^{\langle 3\rangle }+16{\mathbf{O}}_n^{\langle 3\rangle }-36{\mathbf{O}}_n^3\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{8}{3}}+{\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{16G}{3}}-{\displaystyle \frac{41\zeta \left(3\right)}{4\pi }}+{\displaystyle \frac{4{ln}^{3}2}{9\pi }}+{\displaystyle \frac{4{ln}^{2}2}{\pi }}+{\displaystyle \frac{\pi ln2}{12}}+{\displaystyle \frac{6ln2}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_4(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{2n^2}{2n+1}}{\mathbf{O}}_n^{\langle 2\rangle }+3n^2(2n-1)\left(2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{O}}_n^{\langle 4\rangle }\right)\right\}={\displaystyle \frac{\pi }{8}}+{\displaystyle \frac{{\pi }^3}{384}}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_4(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{-16}{1+2n}}-{\displaystyle \frac{2(4n^2-3)}{2n+1}}{\mathbf{H}}_n^{\langle 2\rangle }\\ +3n^2(2n-1)\left({\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2+{\mathbf{H}}_n^{\langle 4\rangle }+128{\mathbf{O}}_n^{\langle 4\rangle }\right)\end{array}\right\}={\displaystyle \frac{16\pi }{3}}+{\displaystyle \frac{13{\pi }^3}{45}}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_4(1,-1,1,-1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}2(2n-1)\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)\\ -n^2(2n-1)\left({({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{H}}_n^{\langle 2\rangle })}^2+{\mathbf{H}}_n^{\langle 4\rangle }-256{\mathbf{O}}_n^{\langle 4\rangle }\right)\end{array}\right\}={\displaystyle \frac{2{\pi }^3}{135}}-{\displaystyle \frac{8\pi }{9}}.$$
• $\begin{array}{|c|}\hline \left[x^2{y}^2\right]{\mathrm{V}}_4(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{-1}{2n-1}}+{\displaystyle \frac{n^2{\mathbf{H}}_n^{\langle 2\rangle }}{2n+1}}-6(2n-1){\mathbf{O}}_n^{\langle 2\rangle }\\ +6n^2(2n-1)\left({\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 2\rangle }-8{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2+12{\mathbf{O}}_n^{\langle 4\rangle }\right)\end{array}\right\}={\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{{\pi }^3}{48}}.$$
• $\begin{array}{|c|}\hline {\mathrm{V}}_4(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{24(1+{\mathbf{H}}_n)}{1+2n}}+{\displaystyle \frac{6(3-4n-16n^2)}{2n+1}}{\mathbf{H}}_n^2+4n(8n-3)\left(({\mathbf{H}}_n^3-{\mathbf{H}}_n^{\langle 3\rangle }\right)\\ -3n^2(2n-1)\left({\mathbf{H}}_n^4-3{\mathbf{H}}_n^{\langle 4\rangle }+96{\mathbf{O}}_n^{\langle 4\rangle }-4{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 3\rangle }\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{120\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{120\zeta \left(3\right)ln2}{\pi }}+{\displaystyle \frac{17{\pi }^3}{60}}-{\displaystyle \frac{16{ln}^{4}2}{\pi }}+{\displaystyle \frac{64{ln}^{3}2}{\pi }}+12\pi {ln}^{2}2-24\pi ln2.\hfill \end{array}$$
• $\begin{array}{|c|}\hline \left[y^4\right]{\mathrm{V}}_5(y,1-y,0,1)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{2n{\mathbf{O}}_n^{\langle 2\rangle }}{2n+1}}-n(8n-3)\left({\mathbf{O}}_n^{\langle 4\rangle }-2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\\ -{\displaystyle \frac{2n^2{\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }}{2n+1}}+3n^2(2n-1){\mathbf{H}}_n\left({\mathbf{O}}_n^{\langle 4\rangle }-2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{\pi }{4}}ln2-{\displaystyle \frac{{\pi }^3}{192}}+{\displaystyle \frac{{\pi }^3}{192}}ln2-{\displaystyle \frac{7\pi }{32}}\zeta \left(3\right)-{\displaystyle \frac{7\zeta \left(3\right)}{4\pi }}+{\displaystyle \frac{31\zeta \left(5\right)}{16\pi }}.\hfill \end{array}$$

4. Series Under Parameter Setting $\mathrm{W}$

Making the parameter substitution “$\mathrm{W}$” in Lemma 1

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

and then invoking the Thomae transformation Formula (4)

$$\begin{array}{cc}\hfill {}_{3}F_2\left[\begin{array}{ccc}1,& \hfill {\displaystyle \frac{1}{2}}+ax,& bx-{\displaystyle \frac{1}{2}}\hfill \\ & \hfill cx,& dx+1\hfill \end{array}|1\right]={}_{3}F_2\left[\begin{array}{c}cx+dx-ax-bx,cx-1,dx\\ cx+dx-ax-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+cx+dx-bx\end{array}|1\right]& \\ \hfill \times \Gamma \left[\begin{array}{c}cx+dx-ax-bx,cx,1+dx\\ cx+dx-ax-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+cx+dx-bx\end{array}\right]& ,\hfill \end{array}$$

we can express the resulting transformation as in the following theorem.

Theorem 3.

Let ${\nabla }_n$ be as in Lemma 1. The following transformation formula holds:

$$\begin{array}{cc}& \Gamma \left[\begin{array}{c}1+cx+dx-ax-bx,1+cx,1+dx\\ cx+dx-ax-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+cx+dx-bx\end{array}\right]{}_{3}F_2\left[\begin{array}{c}cx+dx-ax-bx,cx-1,dx\\ cx+dx-ax-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+cx+dx-bx\end{array}|1\right]\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\nabla }_n\left(\frac{1}{2}+ax,bx-\frac{1}{2},cx,1+dx\right){\left[cx-ax-\frac{1}{2},\frac{1}{2}+cx-bx,\frac{1}{2}+dx-ax,\frac{3}{2}+dx-bx\right]}_n}{{(1+cx+dx-ax-bx)}_{2n+1}{(1+cx)}_{n-1}{(1+dx)}_n}}.\hfill \end{array}$$

For the above ${}_{3}F_2$-series, denote by $T_k$ its general term. Taking out the first term, we can reformulate it as

$$T_0+T_1\times {}_{4}F_3\left[\begin{array}{c}1,1+cx+dx-ax-bx,cx,1+dx\\ 2,{\displaystyle \frac{1}{2}}+cx+dx-ax,{\displaystyle \frac{3}{2}}+cx+dx-bx\end{array}|1\right].$$

Then, the coefficient of x in this ${}_{4}F_3$-series will play a crucial role in the subsequent computations. It can be stated equivalently as shown below and validated in Section 6.3:

$$\tilde{\mathbf{w}}:=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{{\left(\frac{3}{2}\right)}_k^2}}\left\{{\displaystyle \frac{2k+1}{k(k+1)}}\right\}=3-\pi +2\pi G-{\displaystyle \frac{7\zeta \left(3\right)}{2}}.$$

(14)

Denote by ${\mathrm{W}}_m(a,b,c,d)$ the resulting formula by equating the coefficients of $x^m$ across the above displayed equation. By specifying particular values for parameters $\{a,b,c,d\}$, and then making simplifications, we establish from Theorem 3 the following identities of binomial/harmonic series.

• $\begin{array}{|c|}\hline {\mathrm{W}}_0(a,b,c,d)\mathrm{Constant}\mathrm{term}\mathrm{identity}\\ \hline\end{array}$

  $$\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{n^2(1+2n)(6n-1)}{2n-1}}\right\}={\displaystyle \frac{2}{\pi }}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_1(1,1,1,1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{(1+2n)(10n-1)}{2n-1}}-{\displaystyle \frac{2n(1+2n)(6n-1)}{2n-1}}{\mathbf{H}}_n\right\}={\displaystyle \frac{4}{\pi }}\left\{4ln2-1\right\}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_1(0,3,0,1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{2n+{\displaystyle \frac{n(1+2n)(6n-1)}{2n-1}}{\mathbf{O}}_n\right\}={\displaystyle \frac{2}{3\pi }}\left\{3+ln2\right\}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}(\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l} \mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t})$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1+2n+8n^2}{(1-2n)(2n+1)}}+{\displaystyle \frac{n(1+2n)(6n-1)}{4(2n-1)}}{\mathbf{H}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{\pi }{3}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}(\mathrm{I}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y} \mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t})$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1-6n+12n^2+24n^3}{{(2n-1)}^3}}\right\}=0.$$
  As an anonymous reviewer observed, this identity follows simply by telescoping

  $$\sum _{n=0}^m\left\{T_n-T_{n-1}\right\}=T_m\mathrm{for}{T}_n:={\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{n(1-6n-4n^2-40n^3)}{{(2n-1)}^3(1+2n)}}+{\displaystyle \frac{2n(1+2n)(6n-1)}{2n-1}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{\pi }{4}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$ (Real part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{2(1+4n-12n^2-64n^3)}{{(2n-1)}^2(1+2n)}}\\ +{\displaystyle \frac{(1-6n-20n^2+88n^3)}{{(2n-1)}^2}}{\mathbf{H}}_n-{\displaystyle \frac{n(1+2n)(6n-1)}{2n-1}}{\mathbf{H}}_n^2\end{array}\right\}=\pi -{\displaystyle \frac{8}{\pi }}{ln}^{2}2.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{2(1-6n-4n^2+56n^3)}{{(2n-1)}^3}}+{\displaystyle \frac{(1-2n-4n^2-24n^3)}{{(2n-1)}^2}}{\mathbf{H}}_n\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$ (Real part)

  $$\begin{array}{c}\hfill \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{(5-23n+18n^2+60n^3-24n^4)}{9{(2n-1)}^3(1+2n)}}\\ -{\displaystyle \frac{(6n-1)(4n^2+8n-1)}{3{(2n-1)}^2}}{\mathbf{O}}_n+{\displaystyle \frac{2n(1+2n)(6n-1)}{2n-1}}{\mathbf{O}}_n^2\end{array}\right\}={\displaystyle \frac{\pi }{36}}+{\displaystyle \frac{4ln2}{3\pi }}+{\displaystyle \frac{4{ln}^{2}2}{9\pi }}.\end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_2(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{(6n-1)(1+12n^2)}{3{(2n-1)}^3}}-{\displaystyle \frac{(24n^3+4n^2+2n-1)}{{(2n-1)}^2}}{\mathbf{O}}_n\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{4n^3}{{(2n-1)}^4}}-{\displaystyle \frac{n(1-4n+12n^2)}{{(2n-1)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}=-{\displaystyle \frac{\pi }{16}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (Real part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{8n(4n^2+4n-1)}{{(2n-1)}^4}}+{\displaystyle \frac{n(1-4n+12n^2)}{{(2n-1)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{2\pi }{3}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1-12n+32n^2-48n^3-80n^4}{{(2n-1)}^4(1+2n)}}-{\displaystyle \frac{1-2n-4n^2-24n^3}{8{(2n-1)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }\right\}=0.$$
• Linear combination: $\begin{array}{|c|}\hline {\mathrm{W}}_3(0,0,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ with the above equation

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1-8n+24n^2-128n^3-176n^4}{2{(2n-1)}^4(1+2n)}}-{\displaystyle \frac{1-2n-4n^2-24n^3}{{(2n-1)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(1,-1,-1,1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{n(1+2n)(6n-1)}{{(2n-1)}^4}}+{\displaystyle \frac{(1+12n^2)(6n-1)}{64{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}={\displaystyle \frac{-\pi }{48}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(x-\mathbf{i},\mathbf{i}-x,x,-x)\\ \hline\end{array}$

  $$\begin{array}{cc}\hfill \left[x^1\right]& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1}{2n-1}}-{\displaystyle \frac{2(1+10n-52n^2+120n^3)}{{(2n-1)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}={\displaystyle \frac{-3\pi }{2}},\hfill \\ \hfill \left[x^2\right]& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{2(4n^2+8n-1)}{{(2n-1)}^2(2n+1)}}-{\displaystyle \frac{n(1-4n+12n^2)}{{(2n-1)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }\\ +{\displaystyle \frac{4(1+6n-36n^2+72n^3)}{{(2n-1)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }\end{array}\right\}={\displaystyle \frac{4\pi }{3}}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(y,-y,\mathbf{i}-y,y-\mathbf{i})\\ \hline\end{array}$

  $$\begin{array}{cc}\hfill \left[y^1\right]& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{(3-56n+136n^2+224n^3+176n^4)}{{(2n-1)}^4(1+2n)}}\\ -{\displaystyle \frac{(3-10n+4n^2-120n^3)}{4{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }\right)\end{array}\right\}={\displaystyle \frac{\pi }{6}},\hfill \\ \hfill \left[y^2\right]& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{(1-28n+64n^2+240n^3+240n^4)}{{(2n-1)}^4(1+2n)}}\\ -{\displaystyle \frac{(3-14n+20n^2-168n^3)}{8{(2n-1)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }+{\displaystyle \frac{4(1-4n+4n^2-48n^3)}{{(2n-1)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }\end{array}\right\}={\displaystyle \frac{\pi }{6}}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(1,1,1,1)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{6(4-32n^2+64n^4)}{{(2n-1)}^3(1+2n)}}{\mathbf{H}}_n-{\displaystyle \frac{(1-10n-4n^2+40n^3)}{{(2n-1)}^2}}\left(3{\mathbf{H}}_n^{\langle 2\rangle }+6{\mathbf{H}}_n^2\right)\\ +{\displaystyle \frac{n(1+2n)(6n-1)}{2n-1}}\left(2{\mathbf{H}}_n^{\langle 3\rangle }+4{\mathbf{H}}_n^3+6{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }\right)\end{array}\right\}\hfill \\ & =8\pi ln2-4\pi +{\displaystyle \frac{96{ln}^{2}2}{\pi }}-{\displaystyle \frac{64{ln}^{3}2}{\pi }}-{\displaystyle \frac{24\zeta \left(3\right)}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$ (Real part)

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{(1+8n^2-96n^3+16n^4)}{{(2n-1)}^4(1+2n)}}+{\displaystyle \frac{(1+4n-12n^2-64n^3)}{{(2n-1)}^2(1+2n)}}{\mathbf{H}}_n\\ +{\displaystyle \frac{(1-6n-20n^2+88n^3)}{4{(2n-1)}^2}}{\mathbf{H}}_n^2+{\displaystyle \frac{n(1+2n)(6n-1)}{6(2n-1)}}\left({\mathbf{H}}_n^{\langle 3\rangle }-{\mathbf{H}}_n^3\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{5\zeta \left(3\right)}{\pi }}+{\displaystyle \frac{8{ln}^{3}2}{3\pi }}-\pi ln2.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(2,0,1+\mathbf{i},1-\mathbf{i})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{(6n-1)(1-10n+12n^2+40n^3)}{{(2n-1)}^4(1+2n)}}\\ -{\displaystyle \frac{(1-6n-4n^2+56n^3)}{{(2n-1)}^3}}{\mathbf{H}}_n-{\displaystyle \frac{(1-2n-4n^2-24n^3)}{4{(2n-1)}^2}}{\mathbf{H}}_n^2\end{array}\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{1-8n+24n^2-128n^3-176n^4}{{(2n-1)}^4(1+2n)}}\\ +{\displaystyle \frac{4(6n-1)(1+12n^2)}{{(2n-1)}^3}}{\mathbf{O}}_n-{\displaystyle \frac{6(24n^3+4n^2+2n-1)}{{(2n-1)}^2}}{\mathbf{O}}_n^2\end{array}\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_3(3,3,1+\mathbf{i}\sqrt{3},1-\mathbf{i}\sqrt{3})\\ \hline\end{array}$ (Real part)

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{19-44n-152n^2+144n^3-1296n^4-1536n^5}{{(2n-1)}^4{(1+2n)}^2}}-{\displaystyle \frac{18(2-9n+6n^2+4n^3-56n^4)}{{(2n-1)}^3(1+2n)}}{\mathbf{O}}_n\\ +{\displaystyle \frac{18(1-8n+20n^2)}{{(2n-1)}^2}}{\mathbf{O}}_n^2+{\displaystyle \frac{n(1+2n)(6n-1)}{2n-1}}\left({\mathbf{H}}_n^{\langle 3\rangle }+16{\mathbf{O}}_n^{\langle 3\rangle }-36{\mathbf{O}}_n^3\right)\end{array}\right\}\hfill \\ & =16-32G-{\displaystyle \frac{3\pi }{4}}-{\displaystyle \frac{\pi }{2}}ln2-{\displaystyle \frac{12{ln}^{2}2}{\pi }}-{\displaystyle \frac{8{ln}^{3}2}{3\pi }}+{\displaystyle \frac{123\zeta \left(3\right)}{2\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_4(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{32n}{{(2n-1)}^3}}-{\displaystyle \frac{(1-6n+12n^2+24n^3)}{{(2n-1)}^3}}{\mathbf{H}}_n^{\langle 2\rangle }\right\}=0.$$
• $\begin{array}{|c|}\hline \left[x^{3}y\right]{\mathrm{W}}_4(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{32n^3}{{(2n-1)}^5}}-{\displaystyle \frac{(1-6n+12n^2+24n^3)}{{(2n-1)}^3}}{\mathbf{O}}_n^{\langle 2\rangle }\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_4(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}\frac{64n^3}{{(2n-1)}^5}-\frac{8n(40n^3+4n^2+6n-1)}{{(2n-1)}^3(2n+1)}{\mathbf{O}}_n^{\langle 2\rangle }\\ +\frac{4n(1+2n)(6n-1)}{(2n-1)}\left(2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{O}}_n^{\langle 4\rangle }\right)\end{array}\right\}={\displaystyle \frac{{\pi }^3}{48}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_4(1,-1,1,-1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}\frac{16({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle })}{2n-1}-\frac{n(1+2n)(6n-1)}{2n-1}\hfill \\ \times \left({\mathbf{H}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-256{\mathbf{O}}_n^{\langle 4\rangle }\right)\hfill \end{array}\right\}={\displaystyle \frac{4{\pi }^3}{45}}.$$
• $\begin{array}{|c|}\hline \left[x^4\right]{\mathrm{W}}_4(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{16n^3}{{(2n-1)}^5}}+{\displaystyle \frac{2n(1-6n-4n^2-40n^3)}{{(2n-1)}^3(1+2n)}}{\mathbf{O}}_n^{\langle 2\rangle }\\ -{\displaystyle \frac{n(1+2n)(6n-1)}{2n-1}}\left({\mathbf{O}}_n^{\langle 4\rangle }-2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}={\displaystyle \frac{{\pi }^3}{192}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_4(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (Real part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{n(1+2n)(6n-1)}{8(2n-1)}}\left({\mathbf{H}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2+128{\mathbf{O}}_n^{\langle 4\rangle }\right)\\ -{\displaystyle \frac{256n^3}{{(2n-1)}^5}}-{\displaystyle \frac{(1+2n+8n^2)}{(2n+1)(2n-1)}}{\mathbf{H}}_n^{\langle 2\rangle }\end{array}\right\}={\displaystyle \frac{13{\pi }^3}{180}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_4(1,-1,-1,1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{64n(1+2n)(6n-1)}{{(2n-1)}^5}}+{\displaystyle \frac{4n(1+2n)(6n-1)}{{(2n-1)}^3}}\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)\\ +{\displaystyle \frac{n(1+2n)(6n-1)}{8(2n-1)}}\left({\mathbf{H}}_n^{\langle 4\rangle }+{({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle })}^2-256{\mathbf{O}}_n^{\langle 4\rangle }\right)\end{array}\right\}={\displaystyle \frac{-{\pi }^3}{90}}.$$
• $\begin{array}{|c|}\hline \left[x^2{y}^2\right]{\mathrm{W}}_4(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{8n(1-4n-20n^2}{{(2n-1)}^5}}+{\displaystyle \frac{n(1-6n-4n^2-40n^3)}{2{(2n-1)}^3(1+2n)}}{\mathbf{H}}_n^{\langle 2\rangle }-{\displaystyle \frac{4(1-2n-4n^2-24n^3)}{{(2n-1)}^3}}{\mathbf{O}}_n^{\langle 2\rangle }\\ +{\displaystyle \frac{n(1+2n)(6n-1)}{(2n-1)}}\left({\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 2\rangle }+12{\mathbf{O}}_n^{\langle 4\rangle }-8{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}={\displaystyle \frac{{\pi }^3}{48}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_4(1,1,1,1)\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{n(1+2n)(6n-1)}{24(2n-1)}}\left(3{\mathbf{H}}_n^{\langle 4\rangle }+3{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2+8{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 3\rangle }+12{\mathbf{H}}_n^2{\mathbf{H}}_n^{\langle 2\rangle }+4{\mathbf{H}}_n^4\right)\\ +{\displaystyle \frac{2n+1}{2n-1}}\left({\mathbf{H}}_n^{\langle 2\rangle }+2{\mathbf{H}}_n^2\right)-{\displaystyle \frac{(1+2n)(10n-1)}{6(2n-1)}}\left({\mathbf{H}}_n^{\langle 3\rangle }+3{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }+2{\mathbf{H}}_n^3\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{4\pi }{3}}ln2-{\displaystyle \frac{4\pi }{3}}{ln}^{2}2-{\displaystyle \frac{{\pi }^3}{90}}-{\displaystyle \frac{32{ln}^{3}2}{3\pi }}+{\displaystyle \frac{16{ln}^{4}2}{3\pi }}-{\displaystyle \frac{4\zeta \left(3\right)}{\pi }}+{\displaystyle \frac{8\zeta \left(3\right)ln2}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_5(1,-1,1,-1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3(1+6n)}{4^n{(n!)}^3}}\left\{256{\mathbf{O}}_n^{\langle 4\rangle }-{\mathbf{H}}_n^{\langle 4\rangle }-{\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right\}={\displaystyle \frac{8{\pi }^3}{45}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_5(1,-1,0,0)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{8n^3}{{(2n-1)}^6}}-{\displaystyle \frac{16n^3{\mathbf{O}}_n^{\langle 2\rangle }}{{(2n-1)}^4}}+{\displaystyle \frac{n(1-4n+12n^2)}{{(2n-1)}^2}}\left(2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{O}}_n^{\langle 4\rangle }\right)\right\}={\displaystyle \frac{{\pi }^3}{384}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_5(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (Imaginary part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{1-2n-4n^2-24n^3}{16{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 4\rangle }+128{\mathbf{O}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2\right)\\ -{\displaystyle \frac{32n(1-4n-4n^2)}{{(2n-1)}^6}}-{\displaystyle \frac{1-12n+32n^2-48n^3-80n^4){\mathbf{H}}_n^{\langle 2\rangle }}{{(2n-1)}^4(2n+1)}}\end{array}\right\}=0.$$
• $\begin{array}{|c|}\hline \left[x^{4}y\right]{\mathrm{W}}_5(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{80n^3}{{(2n-1)}^6}}+{\displaystyle \frac{(1-2n-4n^2-24n^3)}{2{(2n-1)}^2}}{\left({\mathbf{O}}_n^{\langle 4\rangle }-2({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2)\\ +{\displaystyle \frac{(1-8n+24n^2-128n^3-176n^4)}{{(2n-1)}^4(1+2n)}}{\mathbf{O}}_n^{\langle 2\rangle }\end{array}\right\}=0.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_5(1,-1,\mathbf{i},-\mathbf{i})\\ \hline\end{array}$ (Real part)

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{n(1-4n+12n^2)}{2{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2+128{\mathbf{O}}_n^{\langle 4\rangle }\right)\\ -{\displaystyle \frac{64n(1-4n+12n^2)}{{(2n-1)}^6}}-{\displaystyle \frac{8n(1-4n-4n^2){\mathbf{H}}_n^{\langle 2\rangle }}{{(2n-1)}^4}}\end{array}\right\}={\displaystyle \frac{13{\pi }^3}{90}}.$$
• $\begin{array}{|c|}\hline {\mathrm{W}}_5(1,-1,-1,1)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{256n(1+2n)(6n-1)}{{(2n-1)}^6}}+{\displaystyle \frac{16n(1+2n)(6n-1)}{{(2n-1)}^4}}\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)\\ +{\displaystyle \frac{(1+12n^2)(6n-1)}{8{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 4\rangle }-256{\mathbf{O}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}={\displaystyle \frac{-{\pi }^3}{45}}.$$
• $\begin{array}{|c|}\hline \left[x^3{y}^2\right]{\mathrm{W}}_5(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{4n(1-4n-36n^2)}{{(2n-1)}^6}}-{\displaystyle \frac{4n^3{\mathbf{H}}_n^{\langle 2\rangle }}{{(2n-1)}^4}}+{\displaystyle \frac{8n(1+2n)(6n-1)}{{(2n-1)}^4}}{\mathbf{O}}_n^{\langle 2\rangle }\\ +{\displaystyle \frac{n(1-4n+12n^2)}{{(2n-1)}^2}}\left(12{\mathbf{O}}_n^{\langle 4\rangle }+{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 2\rangle }-8{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}={\displaystyle \frac{{\pi }^3}{96}}.$$
• $\begin{array}{|c|}\hline \left[x{y}^4\right]{\mathrm{W}}_5(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{n(1-4n+12n^2)}{{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 4\rangle }-32{\mathbf{O}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\\ -{\displaystyle \frac{64n(3-12n-8n^2)}{{(2n-1)}^6}}-{\displaystyle \frac{16n(1-4n-8n^2)}{{(2n-1)}^4}}\left({\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }\right)\end{array}\right\}={\displaystyle \frac{7{\pi }^3}{180}}.$$
• $\begin{array}{|c|}\hline \left[y^5\right]{\mathrm{W}}_5(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{32n(-1+4n)}{{(2n-1)}^6}}+{\displaystyle \frac{(1-16n+40n^2+32n^3+16n^4)}{{(2n-1)}^4(2n+1)}}\left({\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }\right)\\ -{\displaystyle \frac{(1-2n-4n^2-24n^3)}{8{(2n-1)}^2}}\left({\mathbf{H}}_n^{\langle 4\rangle }-32{\mathbf{O}}_n^{\langle 4\rangle }+{\left({\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}=0.$$
• $\begin{array}{|c|}\hline \left[x^2{y}^3\right]{\mathrm{W}}_5(x,-x,y,-y)\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{4n(3-12n-28n^2)}{{(2n-1)}^6}}-{\displaystyle \frac{(1-2n-4n^2-24n^3)}{4{(2n-1)}^2}}\left(12{\mathbf{O}}_n^{\langle 4\rangle }+{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n^{\langle 2\rangle }-8{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\\ +{\displaystyle \frac{(1-8n+24n^2-128n^3-176n^4)}{8{(2n-1)}^4(1+2n)}}{\mathbf{H}}_n^{\langle 2\rangle }+{\displaystyle \frac{8n(1+2n)(6n-1)}{{(2n-1)}^4}}{\mathbf{O}}_n^{\langle 2\rangle }\end{array}\right\}=0.$$

5. Series Under Parameter Setting $\Omega$

Under the parameter setting “$\Omega$” in Lemma 1

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

we can reformulate the resulting equation as in the theorem below, where the third expression follows by Thomae’s transformation (4).

Theorem 4.

Let ${\nabla }_n$ be as in Lemma 1. We have the transformation formulae $\begin{array}{|c|}\hline \varphi =\Psi =\Phi \\ \hline\end{array}$, where

$$\begin{gathered} \begin{array}{cc}\hfill \varphi & =\sum _{k=0}^{\infty }{\displaystyle \frac{{(ax-\frac{1}{2})}_k{(bx-\frac{1}{2})}_k}{{(1+cx)}_k{(1+dx)}_k}},\hfill \\ \hfill \Phi & ={}_{3}F_2\left[\begin{array}{c}2+cx+dx-ax-bx,cx,dx\\ {\displaystyle \frac{3}{2}}+cx+dx-ax,{\displaystyle \frac{3}{2}}+cx+dx-bx\end{array}|1\right] \\ \times \Gamma \left[\begin{array}{c}2+cx+dx-ax-bx,1+cx,1+dx\\ {\displaystyle \frac{3}{2}}+cx+dx-ax,{\displaystyle \frac{3}{2}}+cx+dx-bx\end{array}\right],\hfill \\ \hfill \Psi & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\nabla }_n\left(ax-\frac{1}{2},bx-\frac{1}{2},1+cx,1+dx\right)}{{(2+cx+dx-ax-bx)}_{2n+2}}}\hfill \\ & \times {\displaystyle \frac{{\left[\frac{3}{2}+cx-ax,\frac{3}{2}+cx-bx,\frac{3}{2}+dx-ax,\frac{3}{2}+dx-bx\right]}_n}{{(1+cx)}_n{(1+dx)}_n}}.\hfill \end{array} \end{gathered}$$

For the above ${}_{3}F_2$-series, the coefficient of x leads to the following k-sum,

$$\tilde{\mathit{\omega }}:=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{{\left(\frac{3}{2}\right)}_k^2}}\left\{{\displaystyle \frac{k+1}{k^2}}\right\}=3\pi -8,$$

(15)

that is useful in the subsequent evaluation and its proof will be given in Section 6.4.

Denote by ${\varphi }_m,{\Phi }_m$, and ${\Psi }_m$ the triplet coefficients of $x^m$ extracted from $\varphi ,\Phi$, and $\Psi$, respectively. By specifying particular values for parameters $\{a,b,c,d\}$, and then making simplifications, we establish from Theorem 4 the following identities of binomial/harmonic series.

• $\begin{array}{|c|}\hline {\Phi }_0={\Psi }_0\mathrm{Constant}\mathrm{term}\mathrm{identity}\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1+2n}{1+n}}\right\}={\displaystyle \frac{32}{3\pi }}.$$
  By considering $\begin{array}{|c|}\hline {\varphi }_0={\Phi }_0\\ \hline\end{array}$, we immediately deduce the following formula as a byproduct:

  $$\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2{(2k-1)}^2}}={\displaystyle \frac{4}{\pi }},$$

  which is a special case of Gauss’ summation formula for ${}_{2}F_1$-series (cf. Bailey [3], §1.3).
  There exist two analogous series in terms of the Catalan constant:

  $$\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2{(2k-1)}^3}}={\displaystyle \frac{4G-6}{\pi }}\mathrm{and}\sum _{k=0}^{\infty }{\displaystyle \frac{k{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2{(2k-1)}^3}}={\displaystyle \frac{2G-1}{\pi }}.$$
• $\begin{array}{|c|}\hline {\Phi }_1={\Psi }_1:a=b=c=d=1\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{3(1+2n)}{1+n}}{\mathbf{H}}_n-{\displaystyle \frac{2}{1+n}}\right\}={\displaystyle \frac{64}{\pi }}\left\{1-ln2\right\}.$$
• $\begin{array}{|c|}\hline {\Phi }_1={\Psi }_1:a=b=3,c=d=1\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{6(1+2n)}{1+n}}{\mathbf{O}}_{n+1}-{\displaystyle \frac{(1+2n)(7+6n)}{{(1+n)}^2}}\right\}={\displaystyle \frac{64ln2}{3\pi }}.$$
  Under the above two parameter settings, we have the system of linear equations:

  $$\begin{array}{cc}\begin{array}{|c|}\hline {\varphi }_1={\Phi }_1:a=b=c=d=1\\ \hline\end{array}\hfill & {\displaystyle \frac{4-8G}{\pi }}-\dot{\mathit{u}}+2\dot{\mathit{v}}={\displaystyle \frac{8(ln2-1)}{\pi }},\hfill \\ \begin{array}{|c|}\hline {\varphi }_1={\Phi }_1:a=b=3,c=d=1\\ \hline\end{array}\hfill & {\displaystyle \frac{8G-4}{\pi }}+{\displaystyle \frac{\dot{\mathit{u}}}{3}}-2\dot{\mathit{v}}={\displaystyle \frac{8ln2}{3\pi }}.\hfill \end{array}$$
  The solutions for $\dot{\mathit{u}}$ and $\dot{\mathit{v}}$ recover the two formulae below:
• Wang–Chu [9]

  $$\begin{array}{cc}\hfill \dot{\mathit{u}}& :=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2{\mathbf{H}}_k}{{(k!)}^2{(2k-1)}^2}}={\displaystyle \frac{12-16ln2}{\pi }},\hfill \end{array}$$
• Wang–Chu [34]

  $$\begin{array}{cc}\hfill \dot{\mathit{v}}& :=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2{\mathbf{O}}_k}{{(k!)}^2{(2k-1)}^2}}={\displaystyle \frac{4G-4ln2}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\Phi }_2={\Psi }_2:a=1,b=-1,c=\mathbf{i},d=-\mathbf{i}\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1}{(1+n)(3+2n)}}+{\displaystyle \frac{3(1+2n)}{8(1+n)}}{\mathbf{H}}_n^{\langle 2\rangle }\right\}\hfill \\ & ={\displaystyle \frac{8\pi }{3}}-{\displaystyle \frac{16}{\pi }}-\begin{array}{|c|}\hline {\displaystyle \frac{4\tilde{\mathit{\omega }}}{\pi }}\\ \hline\end{array}={\displaystyle \frac{8\pi }{3}}+{\displaystyle \frac{16}{\pi }}-12.\hfill \end{array}$$
• $\begin{array}{|c|}\hline {\Phi }_2={\Psi }_2:a=1,b=-1,c=d=0\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{17+48n+24n^2}{2(1+n)(3+2n)}}-{\displaystyle \frac{3(1+2n)}{1+n}}{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right\}={\displaystyle \frac{16}{\pi }}-2\pi .$$
  According to the above two parameter settings, the system of linear equations reads as

  $$\begin{array}{cccc}& \begin{array}{|c|}\hline {\varphi }_2={\Phi }_2:a=1,b=-1,c=-d=\mathbf{i}\\ \hline\end{array}\hfill & \hfill & {\displaystyle \frac{8\pi }{3}}-12=\ddot{\mathit{u}}+4\ddot{\mathit{v}}-4\mathcal{F}\hfill \\ & \begin{array}{|c|}\hline {\varphi }_2={\Phi }_2:a=1,b=-1,c=d=0\\ \hline\end{array}\hfill & & {\displaystyle \frac{16}{\pi }}-\pi =2\mathcal{F}-2\ddot{\mathit{v}}.\hfill \end{array}$$
  Resolving this linear system yields the following formulae:

  $$\begin{array}{cc}\hfill \ddot{\mathit{u}}& :=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2}}{\displaystyle \frac{{\mathbf{H}}_k^{\langle 2\rangle }}{{(2k-1)}^2}}={\displaystyle \frac{2\pi }{3}}+{\displaystyle \frac{32}{\pi }}-12,\hfill \end{array}$$

  (16)

  $$\begin{array}{cc}\hfill \ddot{\mathit{v}}& :=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2}}{\displaystyle \frac{{\mathbf{O}}_k^{\langle 2\rangle }}{{(2k-1)}^2}}={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{8}{\pi }}+\mathcal{F};\hfill \end{array}$$

  (17)

  where the first formula is due to Wang–Chu [9] and

  $$\begin{gathered} \mathcal{F}=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}^2}{{16}^k{(2k-1)}^4}}={}_{4}F_3\left[\begin{array}{cccc}{\displaystyle \frac{-1}{2}},& {\displaystyle \frac{-1}{2}},& {\displaystyle \frac{-1}{2}},& {\displaystyle \frac{-1}{2}}\\ \\ & 1,& {\displaystyle \frac{1}{2}},& {\displaystyle \frac{1}{2}}\end{array}|1\right] \\ ={\displaystyle \frac{3{\pi }^2}{8}}+{\displaystyle \frac{{ln}^{2}2}{2}}-{\displaystyle \frac{16}{\pi }}\Im \left\{{\mathrm{Li}}_3\left(\frac{1+\mathbf{i}}{2}\right)\right\}+{\displaystyle \frac{8(1-G)}{\pi }}. \end{gathered}$$

  (18)
  The last identity is due to Cantarini and D’Aurizio [35], who also found a companion formula

  $$\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}^2}{{16}^k{(2k+1)}^2}}={}_{4}F_3\left[\begin{array}{cccc}{\displaystyle \frac{1}{2}},& {\displaystyle \frac{1}{2}},& {\displaystyle \frac{1}{2}},& {\displaystyle \frac{1}{2}}\\ \\ & 1,& {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]={\displaystyle \frac{3{\pi }^2}{8}}+{\displaystyle \frac{{ln}^{2}2}{2}}-{\displaystyle \frac{16}{\pi }}\Im \left\{{\mathrm{Li}}_3\left(\frac{1+\mathbf{i}}{2}\right)\right\}.$$
  Analogously, we can produce three further series as shown below:

  •

  $\begin{array}{|c|}\hline {\varphi }_2={\Phi }_2:a=0,b=0,c=1,d=\mathbf{i}\\ \hline\end{array}$

  $$\begin{array}{cc}{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2}}{\displaystyle \frac{{\mathbf{H}}_k^2}{{(2k-1)}^2}}& ={\displaystyle \frac{4\tilde{\mathit{\omega }}}{\pi }}+{\displaystyle \frac{64}{\pi }}-{\displaystyle \frac{10\pi }{3}}+{\displaystyle \frac{64{ln}^{2}2}{\pi }}-{\displaystyle \frac{96ln2}{\pi }}\\ & =12+{\displaystyle \frac{32}{\pi }}-{\displaystyle \frac{10\pi }{3}}+{\displaystyle \frac{64{ln}^{2}2}{\pi }}-{\displaystyle \frac{96ln2}{\pi }}.\end{array}$$

  (19)

  •

  $\begin{array}{|c|}\hline {\varphi }_2={\Phi }_2:a=1,b=\mathbf{i},c=0,d=0\\ \hline\end{array}$

  $${\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{4{ln}^{2}2}{\pi }}-{\displaystyle \frac{4ln2}{\pi }}=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2}}\left\{{\displaystyle \frac{4k^2}{{(2k-1)}^4}}-{\displaystyle \frac{4k{\mathbf{O}}_k}{{(2k-1)}^3}}+{\displaystyle \frac{{\mathbf{O}}_k^2}{{(2k-1)}^2}}\right\}.$$

  •

  $\begin{array}{|c|}\hline \left[xy\right]{\varphi }_2=\left[xy\right]{\Phi }_2:a=x,b=\mathbf{i}x,c=y,d=\mathbf{i}y\\ \hline\end{array}$

  $${\displaystyle \frac{2\pi }{3}}-{\displaystyle \frac{12}{\pi }}-{\displaystyle \frac{16{ln}^{2}2}{\pi }}+{\displaystyle \frac{20ln2}{\pi }}=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2}}\left\{{\displaystyle \frac{2k{\mathbf{H}}_k}{{(2k-1)}^3}}-{\displaystyle \frac{{\mathbf{H}}_k{\mathbf{O}}_k}{{(2k-1)}^2}}\right\}.$$

  Unfortunately, these two formulae are far from sufficiency for determining values of the following “4” series in closed form (except for those in Equations (16)–(19)).

  $$\overline{\mathit{\omega }}:=\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2}{{(k!)}^2}}\left\{{\displaystyle \frac{k\times \{{\mathbf{H}}_k,{\mathbf{O}}_k\}}{{(2k-1)}^3}},{\displaystyle \frac{\{{\mathbf{O}}_k^2,{\mathbf{H}}_k{\mathbf{O}}_k\}}{{(2k-1)}^2}}\right\}.$$

  (20)
• $\begin{array}{|c|}\hline {\Phi }_2={\Psi }_2:a=2,b=0,c=1+\mathbf{i},d=1-\mathbf{i}\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{1}{(1+n)(3+2n)}}+{\displaystyle \frac{{\mathbf{H}}_n}{1+n}}-{\displaystyle \frac{3(1+2n)}{4(1+n)}}{\mathbf{H}}_n^2\right\}=4\pi -24+{\displaystyle \frac{64ln2}{\pi }}-{\displaystyle \frac{32{ln}^{2}2}{\pi }}.$$
• $\begin{array}{|c|}\hline {\Phi }_2={\Psi }_2:a=3,b=3,c=1+\mathbf{i}\sqrt{3},d=1-\mathbf{i}\sqrt{3}\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{74+324n+489n^2+312n^3+72n^4}{2{(1+n)}^3(3+2n)}}\\ -{\displaystyle \frac{3(1+2n)(7+6n)}{{(1+n)}^2}}{\mathbf{O}}_{n+1}+{\displaystyle \frac{9(1+2n)}{1+n}}{\mathbf{O}}_{n+1}^2\end{array}\right\}=16-{\displaystyle \frac{48}{\pi }}+{\displaystyle \frac{2\pi }{3}}+{\displaystyle \frac{32{ln}^{2}2}{3\pi }}.$$
• $\begin{array}{|c|}\hline {\Phi }_3={\Psi }_3:a=2,b=0,c=1+\mathbf{i},d=1-\mathbf{i}\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{2{\mathbf{H}}_n}{(1+n)(3+2n)}}+{\displaystyle \frac{(1+2n)}{2(1+n)}}{\mathbf{H}}_n^{\langle 3\rangle }+{\displaystyle \frac{{\mathbf{H}}_n^2}{1+n}}-{\displaystyle \frac{(1+2n)}{2(1+n)}}{\mathbf{H}}_n^3\right\}\hfill \\ & =16\pi -16\pi ln2+96ln2-{\displaystyle \frac{256}{\pi }}+{\displaystyle \frac{80\zeta \left(3\right)}{\pi }}-{\displaystyle \frac{128{ln}^{2}2}{\pi }}+{\displaystyle \frac{128{ln}^{3}2}{3\pi }}-{\displaystyle \frac{16\widehat{\mathit{\omega }}}{\pi }},\hfill \end{array}$$

  where $\widehat{\mathit{\omega }}$ denotes the k-sum below that unlikely admits a closed form value:

  $$\widehat{\mathit{\omega }}:=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{{\left(\frac{3}{2}\right)}_k^2}}\left\{{\displaystyle \frac{k+1}{k^3}}\right\}\left(1+2k{\overline{\mathbf{H}}}_{2k+1}\right).$$

  (21)
  In order to be able to evaluate the remaining series in closed form, the additional parameter restrictions $cd=0$ and/or $a+b=2c+2d$ will be imposed for simplicity.
• $\begin{array}{|c|}\hline \left[x^2\right]{\varphi }_3=\left[x^2\right]{\Psi }_3:a\to 1+x,b\to 1-x,c\to 0,d\to 2\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{{\displaystyle \frac{4}{1+n}}-{\displaystyle \frac{(17+48n+24n^2)}{(1+n)(3+2n)}}{\mathbf{H}}_n-{\displaystyle \frac{4}{1+n}}{\mathbf{O}}_{n+1}^{\langle 2\rangle }+{\displaystyle \frac{6(1+2n)}{1+n}}{\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right\}\hfill \\ & =8\pi -8\pi ln2-{\displaystyle \frac{128}{\pi }}+{\displaystyle \frac{64ln2}{\pi }}+{\displaystyle \frac{56\zeta \left(3\right)}{\pi }}.\hfill \end{array}$$
• $\begin{array}{|c|}\hline \left[x^2\right]{\varphi }_3=\left[x^2\right]{\Psi }_3:a\to 1+x,b\to 1-x,c\to 0,d\to 1\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{607+2570n+3700n^2+2224n^3+480n^4}{4{(1+n)}^2{(3+2n)}^2}}-{\displaystyle \frac{17+48n+24n^2}{4(1+n)(3+2n)}}\left({\mathbf{H}}_n+6{\mathbf{O}}_{n+1}\right)\\ -{\displaystyle \frac{23+62n+36n^2}{2{(1+n)}^2}}{\mathbf{O}}_{n+1}^{\langle 2\rangle }+{\displaystyle \frac{3(1+2n)}{2(1+n)}}\left({\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }-4{\mathbf{O}}_{n+1}^{\langle 3\rangle }+6{\mathbf{O}}_{n+1}{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right)\end{array}\right\}=2\pi -{\displaystyle \frac{16}{\pi }}.$$
• $\begin{array}{|c|}\hline \left[x^2\right]{\varphi }_3=\left[x^2\right]{\Psi }_3:a\to 1+x,b\to 2-x,c\to 0,d\to 1\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{571+2486n+3636n^2+2208n^3+480n^4}{6{(1+n)}^2{(3+2n)}^2}}-{\displaystyle \frac{17+48n+24n^2}{(1+n)(3+2n)}}{\mathbf{O}}_{n+1}\\ -{\displaystyle \frac{(1+2n)(7+6n)}{{(1+n)}^2}}{\mathbf{O}}_{n+1}^{\langle 2\rangle }+{\displaystyle \frac{2(1+2n)}{1+n}}\left(3{\mathbf{O}}_{n+1}{\mathbf{O}}_{n+1}^{\langle 2\rangle }-2{\mathbf{O}}_{n+1}^{\langle 3\rangle }\right)\end{array}\right\}\hfill \\ & ={\displaystyle \frac{4\pi }{3}}ln2+{\displaystyle \frac{32}{3\pi }}-{\displaystyle \frac{32}{3\pi }}ln2-{\displaystyle \frac{28}{3\pi }}\zeta \left(3\right).\hfill \end{array}$$
• $\begin{array}{|c|}\hline \left[x^4\right]{\Phi }_4=\left[x^4\right]{\Psi }_4:a\to 1+x,b\to 1-x,c\to 0,d\to 2\\ \hline\end{array}$

  $$\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{(25+72n+36n^2)}{(1+n)(3+2n)}}-{\displaystyle \frac{2(17+48n+24n^2)}{(1+n)(3+2n)}}{\mathbf{O}}_{n+1}^{\langle 2\rangle }\\ -{\displaystyle \frac{3(1+2n)}{1+n}}\left({\mathbf{O}}_{n+1}^{\langle 4\rangle }-2{\left({\mathbf{O}}_{n+1}^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}={\displaystyle \frac{32}{\pi }}+{\displaystyle \frac{{\pi }^3}{12}}-4\pi .$$
• $\begin{array}{|c|}\hline \left[x^4\right]{\Phi }_5=\left[x^4\right]{\Psi }_5:a\to 1+x,b\to 1-x,c\to 0,d\to 2\\ \hline\end{array}$

  $$\begin{array}{cc}& \sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\frac{3}{2}\right)}_n^3}{4^n{(n!)}^3}}\left\{\begin{array}{c}{\displaystyle \frac{6}{1+n}}-{\displaystyle \frac{(25+72n+36n^2)}{(1+n)(3+2n)}}{\mathbf{H}}_n-{\displaystyle \frac{8{\mathbf{O}}_{n+1}^{\langle 2\rangle }}{1+n}}+{\displaystyle \frac{2(17+48n+24n^2)}{(1+n)(3+2n)}}{\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }\\ -{\displaystyle \frac{2}{1+n}}\left({\mathbf{O}}_{n+1}^{\langle 4\rangle }-2{\left({\mathbf{O}}_{n+1}^{\langle 2\rangle }\right)}^2\right)+{\displaystyle \frac{3(1+2n){\mathbf{H}}_n}{1+n}}\left({\mathbf{O}}_{n+1}^{\langle 4\rangle }-2{\left({\mathbf{O}}_{n+1}^{\langle 2\rangle }\right)}^2\right)\end{array}\right\}\hfill \\ & =16\pi -8\pi ln2-7\pi \zeta \left(3\right)-{\displaystyle \frac{192}{\pi }}+{\displaystyle \frac{64ln2}{\pi }}-{\displaystyle \frac{{\pi }^3}{6}}+{\displaystyle \frac{{\pi }^3}{6}}ln2+{\displaystyle \frac{56}{\pi }}\zeta \left(3\right)+{\displaystyle \frac{62}{\pi }}\zeta \left(5\right).\hfill \end{array}$$

6. Evaluation of Four $\mathbf{k}$-Sums

In order to establish identities for the series involving harmonic numbers, we have utilized the closed form values of four k-sums $\tilde{\mathbf{u}},\tilde{\mathbf{v}},\tilde{\mathbf{w}}$, and $\tilde{\mathit{\omega }}$ in the precedent sections. Their proofs are produced below.

6.1. Proof of (5)

First recalling the beta integral

$${\displaystyle \frac{(k-1)!}{{\left(\frac{1}{2}\right)}_k}}={\displaystyle \frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(k\right)}{\Gamma (\frac{1}{2}+k)}}=2{\int }_0^1{\displaystyle \frac{y^{2k-1}}{\sqrt{1-y^2}}}dy$$

and then making use of the known ${\Omega }_{0,0}$-sum (cf. [6], Corollary 10), we can express

$$\begin{array}{cc}\hfill \tilde{\mathbf{u}}& ={\int }_0^1{\displaystyle \frac{2dy}{y\sqrt{1-y^2}}}\sum _{k=1}^{\infty }{\displaystyle \frac{(k-1){!}^2}{k!{\left(\frac{1}{2}\right)}_k}}{y}^{2k}\hfill \\ & ={\int }_0^1{\displaystyle \frac{-2dy}{y\sqrt{1-y^2}}}\left[x^2\right]cos(2xarcsiny)\hfill \\ & =4{\int }_0^1{\displaystyle \frac{{arcsin}^2\left(y\right)dy}{y\sqrt{1-y^2}}}.\hfill \end{array}$$

Under the change of variables $y\to sin\theta$, the above integral becomes

$$\tilde{\mathbf{u}}={\int }_0^{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{4{\theta }^2}{sin\theta }}d\theta =8\pi G-14\zeta \left(3\right).$$

This integral value can be found in [36] (§5.5), where more improper integrals were evaluated in closed form.

The following alternative proof is based on a double integral. First, it is routine to write the series in terms of the beta integral

$$\begin{array}{cc}\hfill \tilde{\mathbf{u}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{\Gamma }^2\left(\frac{1}{2}\right){\Gamma }^2\left(n\right)}{n{\Gamma }^2(\frac{1}{2}+n)}}& =\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n}}{\int }_0^1{\int }_0^1{\displaystyle \frac{{\left(xy\right)}^{n-1}}{\sqrt{(1-x)(1-y)}}}dxdy\hfill \\ & ={\int }_0^1{\int }_0^1{\displaystyle \frac{-ln(1-xy)}{xy\sqrt{(1-x)(1-y)}}}dxdy.\hfill \end{array}$$

Evaluating this integral by Mathematica (Wolfram, version 11) directly confirms the claimed result. However, a human proof would be expected for this curious identity. Joakim Petersson may be the first to find this integral value by differentiating the parametric integral (see [30]). Here, we present another proof. Observing the following easily verified equalities

$$\begin{array}{cc}& \int {\displaystyle \frac{dx}{x\sqrt{1-x}}}=ln\left({\displaystyle \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}}\right)+C,\hfill \\ & \underset{x\to 0}{lim}ln(1-xy)ln\left({\displaystyle \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}}\right)=0,\hfill \\ & \underset{x\to 1}{lim}ln(1-xy)ln\left({\displaystyle \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}}\right)=0;\hfill \end{array}$$

and then applying the integration by parts, we have an alternative expression

$$\begin{array}{cc}\hfill \tilde{\mathbf{u}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{\Gamma }^2\left(\frac{1}{2}\right){\Gamma }^2\left(n\right)}{n{\Gamma }^2(\frac{1}{2}+n)}}& ={\int }_0^1{\int }_0^1{\displaystyle \frac{ln\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)}{(1-xy)\sqrt{1-y}}}dxdy.\hfill \end{array}$$

Under the change of variables $\begin{array}{|c|}\hline x\to {\displaystyle \frac{4T}{{(1+T)}^2}}\rightleftharpoons T\to \left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)\\ \hline\end{array}$, the above integral becomes

$$\tilde{\mathbf{u}}={\int }_0^1{\int }_0^1{\displaystyle \frac{4(T-1)lnT}{{(1+T)}^3-4T(1+T)y}}{\displaystyle \frac{dTdy}{\sqrt{1-y}}}.$$

Keeping in mind that

$${\int }_0^1{\displaystyle \frac{zlnT}{1+Tz}}dT={\mathrm{Li}}_2\left(-z\right)$$

and then decomposing the rational function into partial fractions

$${\displaystyle \frac{4(T-1)}{{(1+T)}^3-4T(1+T)y}}={\displaystyle \frac{4u(1+u)}{(1+Tu)(1-u)(u-v)}}+{\displaystyle \frac{4v(1+v)}{(1+Tv)(1-v)(v-u)}}-{\displaystyle \frac{8}{(1+T)(1-u)(1-v)}},$$

where $\begin{array}{|c|}\hline u,v=1-2y\pm 2\sqrt{y^2-y}\\ \hline\end{array}$, we reduce, after substitutions and simplifications, the double integral to a single one:

$$\tilde{\mathbf{u}}={\int }_0^{1}dy\left\{{\displaystyle \frac{{\pi }^2}{6y\sqrt{1-y}}}+{\displaystyle \frac{{\mathrm{Li}}_2\left(2y-1+2\sqrt{y(y-1)}\right)}{y\sqrt{1-y}}}+{\displaystyle \frac{{\mathrm{Li}}_2\left(2y-1-2\sqrt{y(y-1)}\right)}{y\sqrt{1-y}}}\right\}.$$

Luckily enough, these two dilogarithm functions with reciprocal arguments can be unified by

$${\mathrm{Li}}_2\left(2y-1+2\sqrt{y(y-1)}\right)+{\mathrm{Li}}_2\left(2y-1-2\sqrt{y(y-1)}\right)=-{\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{1}{2}}{ln}^2\left(1-2y+2\sqrt{y(y-1)}\right)$$

that enables us to simplify further

$$\tilde{\mathbf{u}}=-{\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{{ln}^2\left(1-2y+2\sqrt{y(y-1)}\right)}{y\sqrt{1-y}}}dy.$$

Finally, making another change of variables $\begin{array}{|c|}\hline y\to {sin}^2\theta \rightleftharpoons \theta \to arcsin\sqrt{y}\\ \hline\end{array}$, we return surprisingly to the same simpler integral expression as demonstrated before:

$$\tilde{\mathbf{u}}={\int }_0^{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{4{\theta }^2}{sin\theta }}d\theta =8\pi G-14\zeta \left(3\right).$$

6.2. Proof of (13)

We remark that the combined difference “$\left(5\right)-16\times \left(13\right)$” results in an interesting formula, for which a direct and elementary proof would be desirable:

$$\tilde{\mathbf{u}}-16\tilde{\mathbf{v}}=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{k^3{\left(\frac{1}{2}\right)}_k^2}}\left\{{\displaystyle \frac{1-3k}{k+1}}\right\}=4-4\pi .$$

(22)

To prove (13), it is enough to show (22). For the sequence ${\mathcal{T}}_k$ defined below, compute its difference

$${\mathcal{T}}_k={\displaystyle \frac{{(k!)}^2}{(k+1){\left(\frac{1}{2}\right)}_k^2}}:{\mathcal{T}}_{k-1}-{\mathcal{T}}_k={\displaystyle \frac{{(k!)}^2}{k^3{\left(\frac{1}{2}\right)}_k^2}}\left\{{\displaystyle \frac{1-3k}{4(k+1)}}\right\}.$$

Then, we can evaluate the series by telescoping

$$\begin{array}{cc}\hfill \tilde{\mathbf{u}}-16\tilde{\mathbf{v}}& =4\underset{m\to \infty }{lim}\sum _{k=1}^m\left({\mathcal{T}}_{k-1}-{\mathcal{T}}_k\right)=4-4\underset{m\to \infty }{lim}{\mathcal{T}}_m\hfill \\ & =4-4\pi \underset{m\to \infty }{lim}\Gamma \left[\begin{array}{c}1+m,1+m,1+m\\ \\ 2+m,{\displaystyle \frac{1}{2}}+m,{\displaystyle \frac{1}{2}}+m\end{array}\right]=4-4\pi .\hfill \end{array}$$

The formula in (13) can be shown similarly as (5). Rewriting the series in terms of the beta integral

$$\begin{array}{cc}\hfill \tilde{\mathbf{v}}& =\sum _{k=0}^{\infty }{\displaystyle \frac{{(k!)}^2}{(2+k){\left(\frac{3}{2}\right)}_k^2}}=\sum _{k=0}^{\infty }{\displaystyle \frac{(k+1)!k!}{2(k+2)!{\left(\frac{3}{2}\right)}_k}}\mathrm{Beta}\left(k+1,{\displaystyle \frac{1}{2}}\right)\hfill \\ & =\sum _{k=0}^{\infty }\left[x^0\right]{\displaystyle \frac{{(1-x)}_k{(2+x)}_k}{(k+2)!{\left(\frac{3}{2}\right)}_k}}{\int }_0^1{\displaystyle \frac{y^{2k+1}}{\sqrt{1-y^2}}}dy,\hfill \end{array}$$

we can proceed with

$$\begin{array}{cc}\hfill \tilde{\mathbf{v}}& =\left[x^0\right]\sum _{j=2}^{\infty }{\displaystyle \frac{{(1-x)}_{j-2}{(2+x)}_{j-2}}{j!{\left(\frac{3}{2}\right)}_{j-2}}}{\int }_0^1{\displaystyle \frac{y^{2j-3}}{\sqrt{1-y^2}}}dy\begin{array}{|c|}\hline k\to j-2\\ \hline\end{array}\hfill \\ & ={\int }_0^1{\displaystyle \frac{y^{-3}dy}{4\sqrt{1-y^2}}}\left[x^2\right]{\displaystyle \frac{1}{{(1+x)}^2}}\left\{1+2x(1+x)y^2-{}_{2}F_1\left[\begin{array}{c}\hfill x,-x-1\\ \hfill -{\displaystyle \frac{1}{2}}\end{array}|y^2\right]\right\}.\hfill \end{array}$$

Evaluating the ${}_{2}F_1$-series by [6] (${\Omega }_{-1,-1}(x,y)$)

$${}_{2}F_1\left[\begin{array}{c}\hfill x,-x-1\\ \hfill -{\displaystyle \frac{1}{2}}\end{array}|y^2\right]=\sqrt{1-y^2}cos[(1+2x)arcsiny]+(1+2x)ysin[(1+2x)arcsiny]$$

and then determining the coefficient

$$\begin{array}{cc}\hfill \left[x^2\right]& {\displaystyle \frac{1}{{(1+x)}^2}}\left\{1+2x(1+x)y^2-{}_{2}F_1\left[\begin{array}{c}\hfill x,-x-1\\ \hfill -{\displaystyle \frac{1}{2}}\end{array}|y^2\right]\right\}\hfill \\ \hfill =& 2\left\{y^2-2y\sqrt{1-y^2}arcsiny+{arcsin}^{2}y\right\},\hfill \end{array}$$

we obtain the following integral expression

$$\begin{array}{cc}\hfill \tilde{\mathbf{v}}& ={\displaystyle \frac{1}{2}}{\int }_0^1{\displaystyle \frac{y^{-3}dy}{\sqrt{1-y^2}}}\left\{y^2-2y\sqrt{1-y^2}arcsiny+{arcsin}^{2}y\right\}\hfill \\ & ={\displaystyle \frac{1}{2}}{\int }_0^{{\displaystyle \frac{\pi }{2}}}\left\{{\displaystyle \frac{{\theta }^2}{{sin}^3\theta }}-{\displaystyle \frac{2\theta cos\theta }{{sin}^2\theta }}+{\displaystyle \frac{1}{{sin}^2\theta }}\right\}d\theta \begin{array}{|c|}\hline y\to sin\theta \\ \hline\end{array}\hfill \\ & ={\displaystyle \frac{\pi -1}{4}}+{\displaystyle \frac{\pi G}{2}}-{\displaystyle \frac{7\zeta \left(3\right)}{8}}.\begin{array}{|c|}\hline \mathrm{Mathematica}(\mathrm{Version}11)\\ \hline\end{array}\hfill \end{array}$$

6.3. Proof of (14)

Notice that linear combination “$4\times \left(13\right)-\left(14\right)$” yields the following identity:

$$4\tilde{\mathbf{v}}-\tilde{\mathbf{w}}=\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{{\left(\frac{1}{2}\right)}_k^2}}\times \left\{{\displaystyle \frac{1}{k^2(2k+1)}}\right\}=2\pi -4.$$

(23)

To prove (14), it suffices to show (23). Write the k-sum in terms of hypergeometric series

$$\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{{\left(\frac{1}{2}\right)}_k^2}}\times \left\{{\displaystyle \frac{1}{k^2(2k+1)}}\right\}=\left[x^2\right]\left\{1-{}_{3}F_2\left[\begin{array}{ccc}1,& x,& -x\\ & {\displaystyle \frac{1}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]\right\}.$$

According to the linear relation

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

we can reformulate the series

$$\begin{array}{cc}\hfill {}_{3}F_2\left[\begin{array}{ccc}1,& x,& -x\\ & {\displaystyle \frac{1}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]& ={\displaystyle \frac{2x+1}{2}}{}_{3}F_2\left[\begin{array}{ccc}1,& 1+x,& -x\\ & {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]+{\displaystyle \frac{1-2x}{2}}{}_{3}F_2\left[\begin{array}{ccc}1,& 1-x,& x\\ & {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]\hfill \\ & ={\displaystyle \frac{2x+1}{2}}\times {\displaystyle \frac{1+sin\left(\pi x\right)}{{(1+2x)}^2}}+{\displaystyle \frac{1-2x}{2}}\times {\displaystyle \frac{1-sin\left(\pi x\right)}{{(1-2x)}^2}},\hfill \end{array}$$

where both ${}_{3}F_2$-series have been evaluated by Whipple’s theorem (cf. Bailey [3], §3.4).

Finally, the claimed value follows by extracting the coefficient of $x^2$:

$$4\tilde{\mathbf{v}}-\tilde{\mathbf{w}}=-2\left[x^2\right]\left\{{\displaystyle \frac{2x+1}{2}}\times {\displaystyle \frac{1+sin\left(\pi x\right)}{{(1+2x)}^2}}\right\}=2\pi -4.$$

Analogous to (5) and (13), we record the following proof for (14). Reformulating the series in terms of the beta integral

$$\begin{array}{cc}\hfill \tilde{\mathbf{w}}& =\sum _{k=1}^{\infty }{\displaystyle \frac{{(k!)}^2}{{\left(\frac{3}{2}\right)}_k^2}}\times \left\{{\displaystyle \frac{2k+1}{k(k+1)}}\right\}\hfill \\ & =\sum _{k=1}^{\infty }{\displaystyle \frac{k!(k-1)!}{2(k+1)!{\left(\frac{1}{2}\right)}_k}}\mathrm{Beta}\left(k+1,{\displaystyle \frac{1}{2}}\right)\hfill \\ & =\sum _{k=1}^{\infty }\left[x\right]{\displaystyle \frac{{\left(x\right)}_k{(1-x)}_k}{(k+1)!{\left(\frac{1}{2}\right)}_k}}{\int }_0^1{\displaystyle \frac{y^{2k+1}}{\sqrt{1-y^2}}}dy,\hfill \end{array}$$

we can proceed with

$$\begin{array}{cc}\hfill \tilde{\mathbf{w}}& =\left[x\right]\sum _{j=2}^{\infty }{\displaystyle \frac{{\left(x\right)}_{j-1}{(1-x)}_{j-1}}{j!{\left(\frac{1}{2}\right)}_{j-1}}}{\int }_0^1{\displaystyle \frac{y^{2j-1}}{\sqrt{1-y^2}}}dy\begin{array}{|c|}\hline k\to j-1\\ \hline\end{array}\hfill \\ & ={\int }_0^1{\displaystyle \frac{y^{-1}dy}{2\sqrt{1-y^2}}}\left[x^2\right]{\displaystyle \frac{1}{1-x}}\left\{1-2x(1-x)y^2-{}_{2}F_1\left[\begin{array}{c}\hfill x-1,-x\\ \hfill -{\displaystyle \frac{1}{2}}\end{array}|y^2\right]\right\}.\hfill \end{array}$$

Evaluating the ${}_{2}F_1$-series by [6] (${\Omega }_{-1,-1}(x,y)$)

$${}_{2}F_1\left[\begin{array}{c}\hfill x-1,-x\\ \hfill -{\displaystyle \frac{1}{2}}\end{array}|y^2\right]=\sqrt{1-y^2}cos[(1-2x)arcsiny]+(1-2x)ysin[(1-2x)arcsiny]$$

and then determining the coefficient

$$\begin{array}{cc}\hfill \left[x^2\right]& {\displaystyle \frac{1}{1-x}}\left\{1-2x(1-x)y^2-{}_{2}F_1\left[\begin{array}{c}\hfill x-1,-x\\ \hfill -{\displaystyle \frac{1}{2}}\end{array}|y^2\right]\right\}\hfill \\ \hfill =& 2\left\{y^2-2y\sqrt{1-y^2}arcsiny+{arcsin}^{2}y\right\},\hfill \end{array}$$

we obtain the following integral expression:

$$\begin{array}{cc}\hfill \tilde{\mathbf{w}}& ={\int }_0^1{\displaystyle \frac{y^{-1}dy}{\sqrt{1-y^2}}}\left\{y^2-2y\sqrt{1-y^2}arcsiny+{arcsin}^{2}y\right\}\hfill \\ & ={\int }_0^{{\displaystyle \frac{\pi }{2}}}\left\{sin\theta -2\theta cos\theta +{\displaystyle \frac{{\theta }^2}{sin\theta }}\right\}d\theta \begin{array}{|c|}\hline y\to sin\theta \\ \hline\end{array}\hfill \\ & =3-\pi +2\pi G-{\displaystyle \frac{7\zeta \left(3\right)}{2}}\begin{array}{|c|}\hline \mathrm{Follow}\mathrm{the}\mathrm{proof}\mathrm{for} \left(5\right)\\ \hline\end{array}\hfill \end{array}$$

6.4. Proof of (15)

Express the k-sum in terms of hypergeometric series

$$\tilde{\mathit{\omega }}=\left[x^2\right]\left\{1-{}_{3}F_2\left[\begin{array}{ccc}2,& x,& -x\\ & {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]\right\}.$$

According to the linear relation

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

we can manipulate the series

$$\begin{array}{cc}\hfill {}_{3}F_2\left[\begin{array}{ccc}2,& x,& -x\\ & {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]& ={\displaystyle \frac{x+1}{2}}{}_{3}F_2\left[\begin{array}{ccc}1,& 1+x,& -x\\ & {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]+{\displaystyle \frac{1-x}{2}}{}_{3}F_2\left[\begin{array}{ccc}1,& 1-x,& x\\ & {\displaystyle \frac{3}{2}},& {\displaystyle \frac{3}{2}}\end{array}|1\right]\hfill \\ & ={\displaystyle \frac{x+1}{2}}\times {\displaystyle \frac{1+sin\left(\pi x\right)}{{(1+2x)}^2}}+{\displaystyle \frac{1-x}{2}}\times {\displaystyle \frac{1-sin\left(\pi x\right)}{{(1-2x)}^2}},\hfill \end{array}$$

where both ${}_{3}F_2$-series have been evaluated by Whipple’s theorem (cf. Bailey [3], §3.4). Finally, the claimed value follows by extracting the coefficient of $x^2$:

$$\tilde{\mathit{\omega }}=-2\left[x^2\right]\left\{{\displaystyle \frac{x+1}{2}}\times {\displaystyle \frac{1+sin\left(\pi x\right)}{{(1+2x)}^2}}\right\}=3\pi -8.$$

7. Conclusions and Further Problems

Under four parameter settings $\{\mathrm{U},\mathrm{V},\mathrm{W},\Omega \}$, the hypergeomeric transformation in Lemma 1 has efficiently been utilized to derive numerous formulae for binomial/harmonic series via the coefficient extraction method. However, the associated k-sums emerged during the course in realizing this process is not routine as commented on in (12) and (21), and shown by the four sums evaluated in Section 6.

There exist different k-sums where the factorial quotients in the summands are replaced by their reciprocals as in (20). The authors fail to evaluate these k-sums in closed form, which prevents us from finding more analytic values for the series containing harmonic numbers of higher order. The interested readers are encouraged to make further attempts.