Odd Euler Sums and Harmonic Series with Cubic Central Binomial Coefficients in Denominators

Abstract

By means of the coefficient extraction method, we examine a transformation of a classical hypergeometric series. Three classes of infinite series (of convergence rate “$1/4$”) with harmonic numbers in numerators and cubic central binomial coefficients in denominators are expressed in terms of odd Euler sums. Several new closed formulae are established.

1. Introduction and Motivation

Let $\mathbb{N}$ be the set of natural numbers with ${\mathbb{N}}_0=\left\{0\right\}\cup \mathbb{N}$. For $m,n\in \mathbb{N}$, define the harmonic numbers of order m, the odd harmonic numbers of order m, and the skew-harmonic numbers of order m, respectively, by

$${\mathbf{H}}_n^{\langle m\rangle }:=\sum _{k=1}^n{\displaystyle \frac{1}{k^m}},{\mathbf{O}}_n^{\langle m\rangle }:=\sum _{k=1}^n{\displaystyle \frac{1}{{(2k-1)}^m}},\mathrm{and}{\overline{\mathbf{H}}}_n^{\langle m\rangle }:=\sum _{k=1}^n{\displaystyle \frac{{(-1)}^{k-1}}{k^m}}.$$

When $m=1$, it will be omitted from these notations. They satisfy the following almost trivial but useful relations:

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

The infinite sums concerning harmonic numbers have wide applications in combinatorial analysis and number theory (cf. [1,2,3]). The aim of this paper is to investigate, by applying the coefficient extraction method (cf. [4,5,6]) to a hypergeometric series transformation (cf. [7]), the infinite series (of convergent rate “${\displaystyle \frac{1}{4}}$”) involving harmonic numbers in numerators and the cubic central binomial coefficients in denominators.

Compared with existing works in the literature (cf. [8,9,10]) on the series with squared central binomial coefficients, this may be considered as a significant progress in searching closed-form formulae for the harmonic series of similar binomial structures.

The rest of the paper is divided into six sections. The “coefficient extraction” method is sketched in Section 2, where a known (and crucial) hypergeometric series transformation for subsequent applications will be reproduced. Section 3, Section 4 and Section 5 constitute the central part of the paper, which is devoted to evaluating three classes of harmonic series under different parameter settings. Several series of convergence rate “1/4” will be expressed as odd Euler sums (cf. [11,12,13]) and then evaluated in closed form in terms of the Riemann zeta function, the Catalan constant (cf. [14,15,16]) and the polylogarithmic function (cf. [17,18]), which are defined, respectively, as follows:

$$\begin{array}{cc}& \zeta \left(m\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^m}}\mathrm{where}m>1,\hfill \\ & G=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}}{{(2n-1)}^2}}\approx 0.915965594,\hfill \\ & {\mathrm{Li}}_m\left(z\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{z^n}{n^m}},\mathrm{where}m\ge 1\mathrm{and}\left|z\right|<1.\hfill \end{array}$$

To realize the closed-form formulae just mentioned, it is indispensable to evaluate the associated k-sums. In order not to disrupt the principal objective (of finding exact values of n-series), we put these k-sums together in Section 6, where they are expressed as odd Euler sums and evaluated explicitly in closed form. Finally, the paper will conclude in Section 7, where further variants of the harmonic series are exemplified and possible working prospects are briefly discussed.

It should be pointed out that all the harmonic series presented in this paper are new. In particular, they refine a few similar series appearing in the author’s previous work [4]. In order to ensure accuracy, all the displayed equations throughout the paper have been checked numerically using appropriately devised Mathematica (version 11) commands.

2. Coefficient Extraction and Hypergeometric Transformation

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

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

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

$$\begin{gathered} {\mathbf{H}}_n=\left[x\right]{\displaystyle \frac{n!}{{(1-x)}_n}},{\mathbf{O}}_n=\left[x\right]{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n}{2{(\frac{1}{2}-x)}_n}}; \\ {\mathbf{H}}_n=\left[x\right]{\displaystyle \frac{{(1+x)}_n}{n!}},{\mathbf{O}}_n=\left[x\right]{\displaystyle \frac{{(\frac{1}{2}+x)}_n}{2{\left(\frac{1}{2}\right)}_n}}. \end{gathered}$$

(1)

It is not hard to show (cf. [4]), by the generating function approach, the following general formulae that play the key role in the “coefficient extraction” approach:

$$\begin{array}{l}\left[x^m\right]{\displaystyle \frac{n!}{{(1-x)}_n}}={\mathbf{Y}}_m\left({\mathbf{H}}_n^{\langle k\rangle }\right),\left[x^m\right]{\displaystyle \frac{{\left(\frac{1}{2}\right)}_n}{{(\frac{1}{2}-x)}_n}}={\mathbf{Y}}_m\left(2^k{\mathbf{O}}_n^{\langle k\rangle }\right);\\ \left[x^m\right]{\displaystyle \frac{{(1-x)}_n}{n!}}={\mathbf{Y}}_m\left(-{\mathbf{H}}_n^{\langle k\rangle }\right),\left[x^m\right]{\displaystyle \frac{{(\frac{1}{2}-x)}_n}{{\left(\frac{1}{2}\right)}_n}}={\mathbf{Y}}_m\left(-2^k{\mathbf{O}}_n^{\langle k\rangle }\right).\end{array}$$

(2)

where the Bell polynomials (cf. [19], §3.3) are expressed explicitly by the multiple sum

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

running over the m-partition set $\sigma \left(m\right)$ represented by m-tuples $(i_1,i_2,\dots ,i_m)\in {\mathbb{N}}_0^m$ subject to the linear condition ${\displaystyle \sum _{k=1}^m}k{i}_k=m$. The six initial polynomials are recorded as below:

$$\begin{array}{cc}\hfill {\mathbf{Y}}_0\left(X_k\right)& =1,\hfill \\ \hfill {\mathbf{Y}}_1\left(X_k\right)& =X_1,\hfill \\ \hfill {\mathbf{Y}}_2\left(X_k\right)& ={\displaystyle \frac{1}{2}}\left\{X_1^2+X_2\right\},\hfill \\ \hfill {\mathbf{Y}}_3\left(X_k\right)& ={\displaystyle \frac{1}{6}}\left\{X_1^3+3X_2{X}_1+2X_3\right\},\hfill \\ \hfill {\mathbf{Y}}_4\left(X_k\right)& ={\displaystyle \frac{1}{24}}\left\{X_1^4+6X_2{X}_1^2+8X_3{X}_1+3X_2^2+6X_4\right\},\hfill \\ \hfill {\mathbf{Y}}_5\left(X_k\right)& ={\displaystyle \frac{1}{120}}\left\{X_1^5+10X_2{X}_1^3+20X_3{X}_1^2+15X_2^2{X}_1+30X_4{X}_1+20X_2{X}_3+24X_5\right\}.\hfill \end{array}$$

To determine analytical values of a harmonic series via the “coefficient extraction” method, the following hypergeometric transformation formula discovered by the second author [7] (Theorem 2.7) will be crucial. For four indeterminate parameters $\{a,b,c,d\}$ subject to the condition $\Re (c+d-a-b)>1$, we have the transformation formula

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

(3)

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

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

(4)

According to the three parameter specifications

$$\begin{array}{cc}\hfill \mathrm{Parameter}\mathrm{setting}\mathrm{A}:& a\to {\displaystyle \frac{1}{2}}+ax,b\to {\displaystyle \frac{1}{2}}+bx,c\to {\displaystyle \frac{3}{2}}+cx,d\to {\displaystyle \frac{3}{2}}+dx;\hfill \\ \hfill \mathrm{Parameter}\mathrm{setting}\mathrm{B}:& a\to ax-{\displaystyle \frac{1}{2}},b\to bx-{\displaystyle \frac{1}{2}},c\to {\displaystyle \frac{1}{2}}+cx,d\to {\displaystyle \frac{1}{2}}+dx;\hfill \\ \hfill \mathrm{Parameter}\mathrm{setting}\mathrm{C}:& a\to {\displaystyle \frac{1}{2}}+ax,b\to bx-{\displaystyle \frac{1}{2}},c\to {\displaystyle \frac{1}{2}}+cx,d\to {\displaystyle \frac{3}{2}}+dx;\hfill \end{array}$$

we shall establish numerous closed-form formulae for harmonic series with cubic central binomial coefficients in denominators. Seven representative series are highlighted in advance as follows:

$$\begin{array}{cccc}& \mathrm{Equation}\left(6\right):\hfill & & {\displaystyle \frac{7\zeta \left(3\right)}{3}}-{\displaystyle \frac{{\pi }^2}{3}}ln2=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3{(1+2n)}^3}}\left\{(2+3n){\mathbf{H}}_n\right\},\hfill \\ & \mathrm{Equation}\left(15\right):\hfill & & {\displaystyle \frac{{\pi }^4}{48}}+{\pi }^2=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{1}{1+n}}+6n{\mathbf{H}}_n^{\langle 2\rangle }\right\},\hfill \\ & \mathrm{Equation}\left(22\right):\hfill & & {\displaystyle \frac{7\zeta \left(3\right)}{2}}+{\displaystyle \frac{{\pi }^2}{4}}-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+n}{{(1+2n)}^3}}+{\displaystyle \frac{2(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{O}}_n\right\},\hfill \\ & \mathrm{Equation}\left(24\right):\hfill & & {\displaystyle \frac{{\pi }^4}{16}}-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1-2n^2}{(1+n){(1+2n)}^4}}+{\displaystyle \frac{2(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }\right\},\hfill \\ & \mathrm{Equation}\left(11\right):\hfill & & {\displaystyle \frac{{\pi }^6}{240}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n(2+3n)}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3{(1+2n)}^3}}\left\{{\mathbf{O}}_{n+1}^{\langle 4\rangle }-{\mathbf{H}}_n^{\langle 4\rangle }+{({\mathbf{H}}_n^{\langle 2\rangle }-{\mathbf{O}}_{n+1}^{\langle 2\rangle })}^2\right\},\hfill \\ & \mathrm{Equation}\left(19\right):\hfill & & {\displaystyle \frac{{\pi }^6}{1440}}+{\displaystyle \frac{{\pi }^4}{12}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{2{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}+3n\left(2{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{H}}_n^{\langle 4\rangle }\right)\right\},\hfill \\ & \mathrm{Equation}\left(10\right):\hfill & & {\displaystyle \frac{7{\pi }^2}{8}}\zeta \left(3\right)-{\displaystyle \frac{31}{4}}\zeta \left(5\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3{(1+2n)}^3}}\{{\displaystyle \frac{{\mathbf{O}}_{n+1}}{n+1}}-{\mathbf{H}}_n^{\langle 2\rangle }+2(2+3n){\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_{n+1}\}.\hfill \end{array}$$

3. Series Under Parameter Setting A

In this case, the resulting Equation (3) can be reformulated as in the theorem below.

Theorem 1.

Let ${\rho }_n(a,b,c,d)$ be as in (4). The following transformation formula holds:

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

Observe that both sides of the above equality are analytic functions in the neighborhood of $x=0$ and therefore can be expanded into Maclaurin series. Denote by “${\Omega }_m:={\Omega }_m(a,b,c,d)$” the resulting equation formed by the coefficients of $x^m$ extracted across the equality in Theorem 1. Then, we can establish numerous identities for the harmonic series of convergence rate “1/4”. For instance, by making use of (1), we can determine the constant terms (corresponding to $x^0$) and extract coefficients of x. Thus, we obtain the following equalities

$$\begin{array}{cc}\hfill {\Omega }_0(a,b,c,d)& =\sum _{n=0}^{\infty }{\displaystyle \frac{{16}^n(2+3n)}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3{(2n+1)}^3}}=\sum _{k=0}^{\infty }{\displaystyle \frac{2}{{(1+2k)}^2}}={\displaystyle \frac{{\pi }^2}{4}},\hfill \\ \hfill {\Omega }_1(a,b,c,d)& =\sum _{n=0}^{\infty }{\displaystyle \frac{{16}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3{(2n+1)}^3}}\left\{\begin{array}{c}\hfill (a+b-3c-3d)+3(a+b-c-d)(2+3n){\mathbf{H}}_n\\ \hfill -2(a+b-3c-3d)(2+3n){\mathbf{O}}_{n+1}\end{array}\right\}\hfill \\ & =\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{8(a+b)}{{(1+2k)}^3}}-{\displaystyle \frac{8(a+b-c-d){\mathbf{O}}_{k+1}}{{(1+2k)}^2}}\right\}\hfill \\ & ={\displaystyle \frac{7}{2}}(a+b+c+d)\zeta \left(3\right)-(a+b-c-d){\pi }^{2}ln2;\hfill \end{array}$$

(5)

where the k-sums in the penultimate line have been evaluated in Section 6.1. Then, by assigning particular values for $\{a,b,c,d\}$, we find the three initial formulae displayed below. Further identities can be shown analogously

• Constant term identity

  $${\displaystyle \frac{{\pi }^2}{4}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{16}^n(2+3n)}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3{(2n+1)}^3}}$$
• Ω₁: a = b = c = d = 1

  $$-{\displaystyle \frac{7\zeta \left(3\right)}{2}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{1-2(2+3n){\mathbf{O}}_{n+1}\right\}.$$
• Ω₁: a = b = 3, c = d = 1

  $${\displaystyle \frac{7\zeta \left(3\right)}{3}}-{\displaystyle \frac{{\pi }^2}{3}}ln2=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{(2+3n){\mathbf{H}}_n\right\}.$$

  (6)
• Ω₂: a = 1, b = −1, c = d = 0

  $${\displaystyle \frac{{\pi }^4}{48}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{n+1}}+2(2+3n){\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$
• Ω₂: a = −b = 1, c = −d = i

  $${\displaystyle \frac{{\pi }^4}{16}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{n+1}}+2(2+3n){\mathbf{O}}_{n+1}^{\langle 2\rangle }\right\}.$$
• Ω₂: a = 2, b = 0, c = 1 + i, d = 1 − i

  $$-{\displaystyle \frac{{\pi }^4}{16}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{n+1}}+4{\mathbf{O}}_{n+1}-4(2+3n){\mathbf{O}}_{n+1}^2\right\}.$$

  (7)
• Ω₂: a = b = 3, $c=1+\mathbf{i}\sqrt{3},d=1-\mathbf{i}\sqrt{3}$

  $${\displaystyle \frac{533{\pi }^4}{2160}}-{\displaystyle \frac{8{\pi }^2}{9}}{ln}^{2}2-{\displaystyle \frac{16{ln}^{4}2}{9}}-{\displaystyle \frac{128}{3}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{n+1}}-6(2+3n){\mathbf{H}}_n^2\right\}.$$

  (8)
• Ω₂: a = x, b = 2x + y, c = y, d = x

  $$\begin{array}{cccc}& \left[y^2\right]\hfill & & \sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\{2{\mathbf{O}}_{n+1}+(2+3n)\left({\mathbf{H}}_n^{\langle 2\rangle }-2{\mathbf{O}}_{n+1}^{\langle 2\rangle }-2{\mathbf{O}}_{n+1}^2\right\}=-{\displaystyle \frac{{\pi }^4}{12}},\hfill \\ & \left[xy\right]\hfill & & \sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{n+1}}+3{\mathbf{H}}_n-6(2+3n){\mathbf{H}}_n{\mathbf{O}}_{n+1}\right\}\hfill \\ & & & ={\displaystyle \frac{{\pi }^4}{45}}+{\displaystyle \frac{2{\pi }^2}{3}}{ln}^{2}2-{\displaystyle \frac{2{ln}^{4}2}{3}}-16{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right),\hfill \\ & \left[x^2\right]\hfill & & \sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{(2+3n)\left(5{\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_{n+1}^{\langle 2\rangle }-9{\mathbf{H}}_n^2\right)\right\}\hfill \\ & & & ={\displaystyle \frac{31{\pi }^4}{180}}-{\displaystyle \frac{4{\pi }^2}{3}}{ln}^{2}2-{\displaystyle \frac{8{ln}^{4}2}{3}}-64{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$

  (9)
• Linear combinations of (7), (8) and (9)

  $$\begin{array}{cc}\hfill {\displaystyle \frac{16{\pi }^2}{9}}{ln}^{2}2-{\displaystyle \frac{17{\pi }^4}{432}}-{\displaystyle \frac{40{ln}^{4}2}{9}}-{\displaystyle \frac{320}{3}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{11}{n+1}}+24{\mathbf{H}}_{2n+1}-24(2+3n){\mathbf{H}}_{2n+1}^2\right\},\hfill \\ \hfill {\displaystyle \frac{8{ln}^{4}2}{27}}-{\displaystyle \frac{32{\pi }^2}{27}}{ln}^{2}2-{\displaystyle \frac{469{\pi }^4}{6840}}+{\displaystyle \frac{64}{9}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{n+1}}+8{\overline{\mathbf{H}}}_{2n+1}-8(2+3n){\overline{\mathbf{H}}}_{2n+1}^2\right\}.\hfill \end{array}$$
• Ω₃: a = b = c = d = 1

  $$\begin{array}{cc}& {\displaystyle \frac{93}{4}}\zeta \left(5\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{2{\mathbf{O}}_{n+1}^{\langle 3\rangle }+4{\mathbf{O}}_{n+1}^3+6{\mathbf{O}}_{n+1}{\mathbf{O}}_{n+1}^{\langle 2\rangle }-{\displaystyle \frac{3{\mathbf{O}}_{n+1}^{\langle 2\rangle }}{2+3n}}-{\displaystyle \frac{6{\mathbf{O}}_{n+1}^2}{2+3n}}\right\}.\hfill \end{array}$$
• Ω₃: a = 2, b = 0, c = 1 + i, d = 1 − i

  $${\displaystyle \frac{21{\pi }^2}{8}}\zeta \left(3\right)-{\displaystyle \frac{93}{4}}\zeta \left(5\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{3{\mathbf{O}}_{n+1}}{n+1}}+6{\mathbf{O}}_{n+1}^2+4(2+3n)({\mathbf{O}}_{n+1}^{\langle 3\rangle }-{\mathbf{O}}_{n+1}^3)\right\}.$$
• [a²d] Ω₃: b → d − a, c → 0

  $${\displaystyle \frac{7{\pi }^2}{8}}\zeta \left(3\right)-{\displaystyle \frac{31}{4}}\zeta \left(5\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{{\mathbf{O}}_{n+1}}{n+1}}-{\mathbf{H}}_n^{\langle 2\rangle }+2(2+3n){\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_{n+1}\right\}.$$

  (10)
• [a] Ω₃: $b\to 3+3\mathbf{i}\sqrt{3}-a,c=2,d=\mathbf{i}\sqrt{3}-1$

  $$\begin{array}{cc}\hfill {\displaystyle \frac{341}{8}}\zeta \left(5\right)& +{\displaystyle \frac{{\pi }^4}{12}}ln2-{\displaystyle \frac{17{\pi }^2}{4}}\zeta \left(3\right)\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{{(n+1)}^2}}-{\displaystyle \frac{3{\mathbf{H}}_n}{n+1}}+2(2+3n)(2{\mathbf{H}}_n^{\langle 3\rangle }-3{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle })\right\}.\hfill \end{array}$$
• [a²x] Ω₃: b → 3x − a, c → ai, d → x − ai

  $$\begin{array}{cc}\hfill {\displaystyle \frac{93}{8}}\zeta \left(5\right)& -{\displaystyle \frac{7{\pi }^2}{2}}\zeta \left(3\right)+{\displaystyle \frac{{\pi }^4}{4}}ln2\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{1}{{(n+1)}^2}}-{\displaystyle \frac{3{\mathbf{H}}_n}{n+1}}-2(2+3n)\left(2{\mathbf{O}}_{n+1}^{\langle 3\rangle }+3{\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right)\right\}.\hfill \end{array}$$
• Ω₃: $a=2\sqrt{3},b=3\mathbf{i}+\sqrt{3},c=2\mathbf{i},d=\sqrt{3}-\mathbf{i}$

  $$\begin{array}{cc}& {\displaystyle \frac{341}{2}}\zeta \left(5\right)+{\displaystyle \frac{32{ln}^{5}2}{15}}+{\displaystyle \frac{16{\pi }^2}{9}}{ln}^{3}2-{\displaystyle \frac{61{\pi }^4}{45}}ln2-256{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+6{\pi }^2\zeta \left(3\right)\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{10{\mathbf{H}}_n^{\langle 3\rangle }+16{\mathbf{O}}_{n+1}^{\langle 3\rangle }-9{\mathbf{H}}_n^3-9{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }\right\}.\hfill \end{array}$$
• Ω₃: a = b = 3, $c=1+\mathbf{i}\sqrt{3},d=1-\mathbf{i}\sqrt{3}$

  $$\begin{array}{cc}& 512{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{533{\pi }^4}{180}}ln2-{\displaystyle \frac{99{\pi }^2}{4}}\zeta \left(3\right)-{\displaystyle \frac{1705}{8}}\zeta \left(5\right)-{\displaystyle \frac{64{ln}^{5}2}{15}}-{\displaystyle \frac{32{\pi }^2}{9}}{ln}^{3}2\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{3}{{(n+1)}^2}}-{\displaystyle \frac{9{\mathbf{H}}_n}{n+1}}-(2+3n)(8{\mathbf{H}}_n^{\langle 3\rangle }-18{\mathbf{H}}_n^3+32{\mathbf{O}}_{n+1}^{\langle 3\rangle })\right\}.\hfill \end{array}$$
• Ω₃: $a=5+4\mathbf{i}\sqrt{2},b=7+2\mathbf{i}\sqrt{2},c=3,d=1+2\mathbf{i}\sqrt{2}$

  $$\begin{array}{cc}& 31\zeta \left(5\right)+{\displaystyle \frac{32{ln}^{5}2}{45}}+{\displaystyle \frac{16{\pi }^2}{27}}{ln}^{3}2-{\displaystyle \frac{169{\pi }^4}{540}}ln2-{\displaystyle \frac{256}{3}}{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{21{\pi }^2}{8}}\zeta \left(3\right)\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{2{\mathbf{O}}_{n+1}^{\langle 3\rangle }-3{\mathbf{H}}_n^3+2{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }-5{\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right\}.\hfill \end{array}$$
• Ω₃: $a=3+4\mathbf{i}\sqrt{6},b=9+2\mathbf{i}\sqrt{6},c=5,d=2\mathbf{i}\sqrt{6}-1$

  $$\begin{array}{cc}& 124\zeta \left(5\right)+{\displaystyle \frac{32{ln}^{5}2}{15}}+{\displaystyle \frac{16{\pi }^2}{9}}{ln}^{3}2-{\displaystyle \frac{199{\pi }^4}{180}}ln2-256{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{57{\pi }^2}{8}}\zeta \left(3\right)\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{4{\mathbf{H}}_n^{\langle 3\rangle }+10{\mathbf{O}}_{n+1}^{\langle 3\rangle }-9{\mathbf{H}}_n^3-9{\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right\}.\hfill \end{array}$$
• Ω₃: a = c = 1, b = 2, d = 0

  $$\begin{array}{cc}& {\displaystyle \frac{256}{3}}{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{31{\pi }^4}{135}}ln2-{\displaystyle \frac{31}{2}}\zeta \left(5\right)-{\displaystyle \frac{32{ln}^{5}2}{45}}-{\displaystyle \frac{16{\pi }^2}{27}}{ln}^{3}2-3{\pi }^2\zeta \left(3\right)\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{2{\mathbf{H}}_n^{\langle 3\rangle }+3{\mathbf{H}}_n^3-5{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }+8{\mathbf{H}}_n{\mathbf{O}}_{n+1}^{\langle 2\rangle }\right\}.\hfill \end{array}$$
• Ω₄: a = d = 1, b = c = −1

  $${\displaystyle \frac{{\pi }^6}{240}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3(2+3n)}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\mathbf{O}}_{n+1}^{\langle 4\rangle }-{\mathbf{H}}_n^{\langle 4\rangle }+{({\mathbf{H}}_n^{\langle 2\rangle }-{\mathbf{O}}_{n+1}^{\langle 2\rangle })}^2\right\}.$$

  (11)
• Ω₄: a = 1, b = −1, c = d = 0

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\pi }^6}{1440}}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{2{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}+(2+3n)\left(2{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{H}}_n^{\langle 4\rangle }\right)\right\}\hfill \\ & ={\displaystyle \frac{{\pi }^6}{30}}-16\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{2{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(1+2k)}^4}}+{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 4\rangle }}{{(1+2k)}^2}}-{\displaystyle \frac{{\left({\mathbf{O}}_{k+1}^{\langle 2\rangle }\right)}^2}{{(1+2k)}^2}}\right\}.\hfill \end{array}$$

  (12)
• Ω₄: a = 1, b = −1, c = −d = i

  $$\begin{array}{cc}\hfill {\displaystyle \frac{19{\pi }^6}{1440}}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(n!)}^3}{4^n{\left(\frac{3}{2}\right)}_n^3}}\left\{{\displaystyle \frac{2{\mathbf{O}}_{n+1}^{\langle 2\rangle }}{n+1}}+(2+3n)\left({\mathbf{H}}_n^{\langle 4\rangle }+2{\mathbf{O}}_{n+1}^{\langle 4\rangle }+2{\left({\mathbf{O}}_{n+1}^{\langle 2\rangle }\right)}^2\right)\right\}\hfill \\ & ={\displaystyle \frac{{\pi }^6}{120}}-16\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(1+2k)}^4}}-{\displaystyle \frac{{\left({\mathbf{O}}_{k+1}^{\langle 2\rangle }\right)}^2}{{(1+2k)}^2}}\right\}.\hfill \end{array}$$

  (13)
  The k-sums in (12) and (13) have been evaluated by the formulae in Section 6.4.

4. Series Under Parameter Setting B

Performing parameter replacements in (3)

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

we can reformulate the resulting equation as in the theorem below.

Theorem 2.

Let ${\rho }_n(a,b,c,d)$ be as in (4). The following transformation formula holds:

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

By means of the coefficient extraction method, we first construct the equation “${\Omega }_m:={\Omega }_m(a,b,c,d)$” by comparing the coefficients of $x^m$ across the equality displayed in the above theorem. Then, specifying the parameters $\{a,b,c,d\}$ with particular values, we derive the following identities.

• Constant term identity

  $${\displaystyle \frac{2}{3}}+{\displaystyle \frac{{\pi }^2}{12}}=\sum _{n=0}^{\infty }{\displaystyle \frac{n{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}.$$
• Ω₁: a = b = c = d = 1

  $${\displaystyle \frac{{\pi }^2}{2}}+{\displaystyle \frac{7\zeta \left(3\right)}{2}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n(6n{\mathbf{O}}_n-1)}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}.$$
• Ω₁: a = b = 3, c = d = 1

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\pi }^2}{3}}ln2-{\displaystyle \frac{7\zeta \left(3\right)}{3}}-{\displaystyle \frac{{\pi }^2}{2}}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{2n}{(1+2n)}}-3n{\mathbf{H}}_n\right\}\hfill \\ & =\sum _{k=0}^{\infty }{\displaystyle \frac{8{\mathbf{O}}_{k+1}}{3{(2k+1)}^2}}-{\displaystyle \frac{7\zeta \left(3\right)}{2}}-{\displaystyle \frac{{\pi }^2}{2}}.\hfill \end{array}$$

  (14)
  The above k-sum is explicitly evaluated in Section 6.1.
• Ω₂: a = 1, b = −1, c = d = 0

  $$\begin{array}{cc}\hfill {\pi }^2+{\displaystyle \frac{{\pi }^4}{48}}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{1}{1+n}}+6n{\mathbf{H}}_n^{\langle 2\rangle }\right\}\hfill \\ & =\sum _{k=0}^{\infty }{\displaystyle \frac{8{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^2}}+32\sum _{k=0}^{\infty }{\displaystyle \frac{k(k+1)}{{(2k+1)}^4}}.\hfill \end{array}$$

  (15)
  Here, the two k-sums are evaluated by the formulae provided in Section 6.2.
• Ω₂: a = −b = 1, c = −d = i

  $${\displaystyle \frac{{\pi }^2}{2}}+{\displaystyle \frac{{\pi }^4}{16}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{1}{1+n}}+6n{\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$
• Ω₂: a = 2, b = 0, c = 1 + i, d = 1 − i

  $$-{\displaystyle \frac{{\pi }^4}{16}}-14\zeta \left(3\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{1}{n+1}}+4{\mathbf{O}}_n-12n{\mathbf{O}}_n^2\right\}.$$

  (16)
• Ω₂: a = b = 3 , $c=1+\mathbf{i}\sqrt{3},d=1-\mathbf{i}\sqrt{3}$

  $$\begin{array}{cc}\hfill & {\displaystyle \frac{533{\pi }^4}{4320}}+4{\pi }^{2}ln2-{\displaystyle \frac{64}{3}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-28\zeta \left(3\right)-{\displaystyle \frac{3{\pi }^2}{2}}-{\displaystyle \frac{8{ln}^{4}2}{9}}-{\displaystyle \frac{4{\pi }^2}{9}}{ln}^{2}2\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{1-12n-12n^2}{2(1+n){(1+2n)}^2}}+{\displaystyle \frac{12n}{1+2n}}{\mathbf{H}}_n-9n{\mathbf{H}}_n^2\right\}\hfill \\ & =\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{64(k+1){\mathbf{O}}_{k+1}}{{(2k+1)}^3}}+{\displaystyle \frac{44{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{3{(2k+1)}^2}}-{\displaystyle \frac{32{\mathbf{O}}_{k+1}^2}{3{(2k+1)}^2}}\right\}-48\sum _{k=0}^{\infty }{\displaystyle \frac{(k+1)(k+2)}{{(2k+1)}^4}};\hfill \end{array}$$

  (17)

  where

  $$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\displaystyle \frac{(k+1)(k+2)}{{(2k+1)}^4}}={\displaystyle \frac{{\pi }^4}{128}}+{\displaystyle \frac{{\pi }^2}{32}}+{\displaystyle \frac{7}{8}}\zeta \left(3\right).\hfill \end{array}$$
  These k-sums are evaluated by the formulae given in Section 6.1 and Section 6.2.
• Ω₂: a = x, b = 2x + y, c = y, d = x

  $$\begin{array}{cccc}& \left[xy\right]\hfill & \hfill & 16{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{{\pi }^4}{45}}+{\displaystyle \frac{2{ln}^{4}2}{3}}-{\displaystyle \frac{2{\pi }^2}{3}}{ln}^{2}2+{\pi }^2-2{\pi }^{2}ln2+35\zeta \left(3\right)\hfill \\ & & & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{1}{(1+n)(1+2n)}}-3{\mathbf{H}}_n-{\displaystyle \frac{12n{\mathbf{O}}_n}{1+2n}}+18{\mathbf{H}}_n{\mathbf{O}}_n\right\},\hfill \\ & \left[x^2\right]\hfill & \hfill & {\displaystyle \frac{16}{3}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{31{\pi }^4}{2160}}+{\displaystyle \frac{2{ln}^{4}2}{9}}+{\displaystyle \frac{2{\pi }^2}{9}}{ln}^{2}2+{\displaystyle \frac{{\pi }^2}{3}}-{\pi }^{2}ln2+7\zeta \left(3\right)\hfill \\ & & & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{2n}{{(1+2n)}^3}}-{\displaystyle \frac{3n{\mathbf{H}}_n}{{(1+2n)}^2}}-{\displaystyle \frac{n}{4(1+2n)}}\left(5{\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }-9{\mathbf{H}}_n^2\right)\right\},\hfill \\ & \left[y^2\right]\hfill & \hfill & -{\displaystyle \frac{7}{6}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^4}{72}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{{\mathbf{O}}_n}{3+6n}}+{\displaystyle \frac{n}{2(1+2n)}}\left({\mathbf{H}}_n^{\langle 2\rangle }-2{\mathbf{O}}_n^{\langle 2\rangle }-2{\mathbf{O}}_n^2\right)\right\}.\hfill \end{array}$$

  (18)
• Ω₃: a = 2, b = 0, c = 1 + i, d = 1 − i

  $$\begin{array}{cc}\hfill {\displaystyle \frac{7{\pi }^2}{8}}\zeta \left(3\right)& -{\displaystyle \frac{31}{4}}\zeta \left(5\right)-{\displaystyle \frac{{\pi }^4}{16}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{{\mathbf{O}}_n}{n+1}}+2{\mathbf{O}}_n^2+4n({\mathbf{O}}_n^{\langle 3\rangle }-{\mathbf{O}}_n^3)\right\}\hfill \\ & =8\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^2}}+{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^3}}+{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 3\rangle }}{{(2k+1)}^2}}\right\}-16\sum _{k=0}^{\infty }{\displaystyle \frac{2(k+1)}{{(2k+1)}^5}};\hfill \end{array}$$

  where

  $$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\displaystyle \frac{k+1}{{(2k+1)}^5}}={\displaystyle \frac{{\pi }^4}{192}}+{\displaystyle \frac{31}{64}}\zeta \left(5\right).\hfill \end{array}$$
  These k-sums are evaluated by the formulae displayed in Section 6.2 and Section 6.3.
• [a] Ω₃: b → 2 − a, c → 1 + (a − 1)i, d → 1 − (a − 1)i

  $${\displaystyle \frac{7{\pi }^2}{8}}\zeta \left(3\right)+{\displaystyle \frac{{\pi }^4}{16}}+{\displaystyle \frac{7}{2}}\zeta \left(3\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{{\mathbf{O}}_n}{n+1}}-{\mathbf{O}}_n^{\langle 2\rangle }+6n({\mathbf{O}}_n^{\langle 3\rangle }+{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle })\right\}.$$
• [a] Ω₃: b → 2 − a, c = 1 + i, d = 1 − i

  $${\displaystyle \frac{7{\pi }^2}{8}}\zeta \left(3\right)-{\displaystyle \frac{31}{4}}\zeta \left(5\right)+{\displaystyle \frac{{\pi }^4}{48}}+7\zeta \left(3\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{{\mathbf{O}}_n}{n+1}}-{\mathbf{H}}_n^{\langle 2\rangle }+6n{\mathbf{O}}_n{\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$
• [a] Ω₃: b → 3 − a, c = 0, d = 1

  $$\begin{array}{cc}& {\displaystyle \frac{341}{48}}\zeta \left(5\right)-{\displaystyle \frac{14}{3}}\zeta \left(3\right)-{\displaystyle \frac{17{\pi }^2}{24}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^4}{48}}+{\displaystyle \frac{2{\pi }^2}{3}}ln2+{\displaystyle \frac{{\pi }^4}{72}}ln2\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{3+4n}{6{(1+n)}^2(1+2n)}}-{\displaystyle \frac{{\mathbf{H}}_n}{2(n+1)}}+{\displaystyle \frac{2n{\mathbf{H}}_n^{\langle 2\rangle }}{1+2n}}+n(2{\mathbf{H}}_n^{\langle 3\rangle }-3{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle })\right\}.\hfill \end{array}$$
• Ω₃: b → 3 − a, c → ai, d → 1 − ai

  $$\begin{array}{cc}& {\displaystyle \frac{31}{16}}\zeta \left(5\right)-{\displaystyle \frac{7}{3}}\zeta \left(3\right)-{\displaystyle \frac{7{\pi }^2}{12}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^4}{16}}+{\displaystyle \frac{{\pi }^2}{3}}ln2+{\displaystyle \frac{{\pi }^4}{24}}ln2\hfill \\ & =\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{3+4n}{6{(1+n)}^2(1+2n)}}-{\displaystyle \frac{{\mathbf{H}}_n}{2(n+1)}}+{\displaystyle \frac{2n{\mathbf{O}}_n^{\langle 2\rangle }}{1+2n}}-n\left(2{\mathbf{O}}_n^{\langle 3\rangle }+3{\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)\right\}.\hfill \end{array}$$
• Ω₃: a = b = c = d = 1

  $$\begin{array}{cc}& {\displaystyle \frac{7}{2}}\zeta \left(3\right)+{\displaystyle \frac{31}{4}}\zeta \left(5\right)+{\displaystyle \frac{{\pi }^4}{8}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{2n\left({\mathbf{O}}_n^{\langle 3\rangle }+2{\mathbf{O}}_n^3+3{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }\right)-{\mathbf{O}}_n^{\langle 2\rangle }-2{\mathbf{O}}_n^2\right\}.\hfill \end{array}$$
• Ω₄: a = d = 1, b = c = −1

  $${\displaystyle \frac{{\pi }^6}{720}}-{\displaystyle \frac{{\pi }^4}{72}}=\sum _{n=0}^{\infty }{\displaystyle \frac{n{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\mathbf{O}}_n^{\langle 4\rangle }-{\mathbf{H}}_n^{\langle 4\rangle }+{({\mathbf{H}}_n^{\langle 2\rangle }-{\mathbf{O}}_n^{\langle 2\rangle })}^2\right\}.$$
• Ω₄: a = 1, b = −1, c = d = 0

  $$\begin{array}{cc}& {\displaystyle \frac{{\pi }^6}{1440}}+{\displaystyle \frac{{\pi }^4}{12}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{2{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}+3n\left(2{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{H}}_n^{\langle 4\rangle }\right)\right\}\hfill \\ & =16\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{2{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^2}}-{\displaystyle \frac{2{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^4}}+{\displaystyle \frac{{\left({\mathbf{O}}_{k+1}^{\langle 2\rangle }\right)}^2}{{(2k+1)}^2}}-{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 4\rangle }}{{(2k+1)}^2}}-{\displaystyle \frac{8k(k+1)}{{(2k+1)}^6}}\right\}.\hfill \end{array}$$

  (19)
• Ω₄: a = 1, b = −1, c = −d = i

  $$\begin{array}{cc}\hfill {\displaystyle \frac{19{\pi }^6}{1440}}& +{\displaystyle \frac{{\pi }^4}{8}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3(1+2n)}}\left\{{\displaystyle \frac{2{\mathbf{O}}_n^{\langle 2\rangle }}{n+1}}+3n\left({\mathbf{H}}_n^{\langle 4\rangle }+2{\mathbf{O}}_n^{\langle 4\rangle }+2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\right\}\hfill \\ & =16\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^2}}-{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^4}}+{\displaystyle \frac{{\left({\mathbf{O}}_{k+1}^{\langle 2\rangle }\right)}^2}{{(2k+1)}^2}}-{\displaystyle \frac{2k(k+1)}{{(2k+1)}^6}}\right\}.\hfill \end{array}$$

  (20)
  The k-sums appearing in the last two identities are evaluated by employing the formulae given in Section 6.2 and Section 6.4.

5. Series Under Parameter Setting C

Performing parameter replacements in (3)

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

we can reformulate the resulting equation as in the theorem below.

Theorem 3.

Let ${\rho }_n(a,b,c,d)$ be as in (4). The following transformation formula holds:

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

According to this theorem, we can form the equation “${\Omega }_m:={\Omega }_m(a,b,c,d)$” by extracting the coefficients of $x^m$. Then by specializing the parameters $\{a,b,c,d\}$ to particular values, we can prove the following infinite series formulae.

• Constant term identity

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\pi }^2}{4}}& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{(1+n)(1+3n)}{n{(1+2n)}^2}}\right\}\hfill \\ & ={\displaystyle \frac{(d-b){\pi }^2}{4(a-c)}}+{\displaystyle \frac{4(a+b-c-d)}{(a-c)}}\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k}{(2k+1)(2k-1)}}.\hfill \end{array}$$

  (21)
  The above k-sum is evaluated by one formula in Section 6.5.
• Ω₁: a = −b = c = −d = 1

  $${\displaystyle \frac{{\pi }^2}{4}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{(2+3n)}{{(1+2n)}^3}}\right\}.$$
• Ω₁: a = b = c = d = 1

  $${\displaystyle \frac{7\zeta \left(3\right)}{2}}+{\displaystyle \frac{{\pi }^2}{4}}-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+n}{{(1+2n)}^3}}+{\displaystyle \frac{2(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{O}}_n\right\}.$$

  (22)
• Ω₁: a = b = 3, c = d = 1

  $$\begin{array}{c}{\displaystyle \frac{{\pi }^2}{3}}ln2-{\displaystyle \frac{7\zeta \left(3\right)}{3}}-{\displaystyle \frac{{\pi }^2}{4}}+{\displaystyle \frac{2}{3}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{(1+3n)(1+3n+n^2)}{3n^2{(1+2n)}^3}}-{\displaystyle \frac{(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{H}}_n\right\}\hfill \\ ={\displaystyle \frac{7\zeta \left(3\right)}{3}}+{\displaystyle \frac{{\pi }^2}{4}}+{\displaystyle \frac{2}{3}}-{\displaystyle \frac{4}{3}}\sum _{k=1}^{\infty }\left\{{\displaystyle \frac{6{\mathbf{O}}_k}{(2k+1)(2k-1)}}-{\displaystyle \frac{{\mathbf{O}}_k}{{(2k+1)}^2}}+{\displaystyle \frac{3{\mathbf{O}}_k}{{(2k-1)}^2}}\right\}\hfill \\ +{\displaystyle \frac{16}{3}}\sum _{k=1}^{\infty }\left\{{\displaystyle \frac{{\mathbf{O}}_k^2}{(2k+1)(2k-1)}}-{\displaystyle \frac{{\mathbf{O}}_k^{\langle 2\rangle }}{(2k+1)(2k-1)}}\right\}.\end{array}$$

  (23)
  These k-sums are evaluated by the formulae given in Section 6.1 and Section 6.5.
• Imaginary part of Ω₂: a = −b = 1, c = −d = i

  $$2=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+6n+12n^2+6n^3}{n^3{(1+2n)}^3}}\right\}.$$
• Real part of Ω₂: a = −b = 1, c = −d = i

  $${\displaystyle \frac{{\pi }^4}{16}}-1=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{2(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{O}}_n^{\langle 2\rangle }+{\displaystyle \frac{1-2n^2}{(1+n){(1+2n)}^4}}\right\}.$$

  (24)
• Ω₂: a = 1, b = −1, c = d = 0

  $${\displaystyle \frac{{\pi }^4}{48}}+1=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{2(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }-{\displaystyle \frac{1+5n+8n^2+5n^3}{n^3(1+n){(1+2n)}^2}}\right\}.$$
• Real part of Ω₂: a = 2, b = 0, c = 1 + i, d = 1 − i

  $${\displaystyle \frac{{\pi }^4}{64}}+{\displaystyle \frac{1}{4}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{O}}_n^2-{\displaystyle \frac{(1+6n+12n^2+10n^3)}{2n^2{(1+2n)}^3}}{\mathbf{O}}_n-{\displaystyle \frac{1}{4(1+n){(1+2n)}^3}}\right\}.$$
• Real part of Ω₂: a = b = 3, $c=1+\mathbf{i}\sqrt{3},d=1-\mathbf{i}\sqrt{3}$

  $$\begin{array}{cc}& {\displaystyle \frac{17}{18}}-{\displaystyle \frac{533{\pi }^4}{12960}}-{\displaystyle \frac{2}{3}}{\pi }^{2}ln2+{\displaystyle \frac{64}{9}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{14}{3}}\zeta \left(3\right)+{\displaystyle \frac{8{ln}^{4}2}{27}}+{\displaystyle \frac{4{\pi }^2}{27}}{ln}^{2}2\hfill \\ & =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{H}}_n^2-{\displaystyle \frac{2(1+3n)(1+3n+n^2)}{3n^2{(1+2n)}^3}}{\mathbf{H}}_n+{\displaystyle \frac{1+7n+18n^2+3n^3-6n^4}{18n^3(1+n){(1+2n)}^3}}\right\}.\hfill \end{array}$$
• Ω₂: a = x, b = 2x + y, c = y, d = x

  $$\begin{array}{cc}\hfill \left[xy\right]& {\displaystyle \frac{1}{2}}+{\displaystyle \frac{{\pi }^4}{90}}-{\displaystyle \frac{{ln}^{4}2}{3}}+{\displaystyle \frac{{\pi }^2}{3}}{ln}^{2}2-8{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)+{\pi }^{2}ln2-14\zeta \left(3\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\hfill \\ & \times \left\{{\displaystyle \frac{2+10n+16n^2+7n^3}{2n^3(1+n){(1+2n)}^2}}-{\displaystyle \frac{3(1+4n+5n^2){\mathbf{H}}_n}{2n^2{(1+2n)}^2}}+{\displaystyle \frac{(1+4n){\mathbf{O}}_n}{n^2{(1+2n)}^2}}-{\displaystyle \frac{3(1+n)(1+3n)}{n{(1+2n)}^2}}{\mathbf{H}}_n{\mathbf{O}}_n\right\},\hfill \\ \hfill [x^2]& 32{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{31{\pi }^4}{360}}+{\displaystyle \frac{4{ln}^{4}2}{3}}+{\displaystyle \frac{2{\pi }^2}{3}}{ln}^{2}2-4{\pi }^{2}ln2+28\zeta \left(3\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\hfill \\ & \times \left\{{\displaystyle \frac{(1+4n+2n^2)(1+4n+6n^2)}{n^3{(1+2n)}^4}}-{\displaystyle \frac{3(1+4n){\mathbf{H}}_n}{n^2{(1+2n)}^2}}-{\displaystyle \frac{(1+n)(1+3n)}{2n{(1+2n)}^2}}\left(5{\mathbf{H}}_n^{\langle 2\rangle }-8{\mathbf{O}}_n^{\langle 2\rangle }-9{\mathbf{H}}_n^2\right)\right\},\hfill \\ \hfill \left[y^2\right]& {\displaystyle \frac{7}{2}}\zeta \left(3\right)+{\displaystyle \frac{{\pi }^4}{24}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1}{2n^3}}+{\displaystyle \frac{(1+4n+5n^2){\mathbf{O}}_n}{n^2{(1+2n)}^2}}-{\displaystyle \frac{(1+n)(1+3n)}{2n{(1+2n)}^2}}\left({\mathbf{H}}_n^{\langle 2\rangle }-2{\mathbf{O}}_n^{\langle 2\rangle }-2{\mathbf{O}}_n^2\right)\right\}.\hfill \end{array}$$
• Imaginary part of Ω₄: a = 1, b = −1, c = −d = i

  $$2=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+6n+12n^2+6n^3}{n^3{(1+2n)}^3}}{\mathbf{O}}_n^{\langle 2\rangle }-{\displaystyle \frac{2}{{(1+2n)}^5}}\right\}.$$
• Real part of Ω₄: a = 1, b = −1, c = −d = i

  $$2-{\displaystyle \frac{19{\pi }^6}{1440}}=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1}{n^5}}-{\displaystyle \frac{2(1-2n^2){\mathbf{O}}_{n+1}^{\langle 2\rangle }}{(1+n){(1+2n)}^4}}-{\displaystyle \frac{(1+n)(1+3n)}{n{(1+2n)}^2}}\left({\mathbf{H}}_n^{\langle 4\rangle }+2{\mathbf{O}}_n^{\langle 4\rangle }+2{\left({\mathbf{O}}_n^{\langle 2\rangle }\right)}^2\right)\right\}.$$
• Ω₄: a = 1, b = −1, c = d = 0

  $$\begin{array}{cc}\hfill {\displaystyle \frac{{\pi }^6}{1440}}& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1}{n^5}}-{\displaystyle \frac{2(1+5n+8n^2+5n^3)}{n^3(1+n){(1+2n)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }+{\displaystyle \frac{(1+n)(1+3n)}{n{(1+2n)}^2}}\left(2{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{H}}_n^{\langle 4\rangle }\right)\right\}\hfill \\ & =32\sum _{k=1}^{\infty }\left\{{\displaystyle \frac{{\left({\mathbf{O}}_k^{\langle 2\rangle }\right)}^2}{(2k+1)(2k-1)}}-{\displaystyle \frac{{\mathbf{O}}_k^{\langle 4\rangle }}{(2k+1)(2k-1)}}-{\displaystyle \frac{{\mathbf{O}}_k^{\langle 2\rangle }}{{(2k-1)}^4}}\right\}\hfill \\ & +16\sum _{k=1}^{\infty }\left\{{\displaystyle \frac{{\left({\mathbf{O}}_k^{\langle 2\rangle }\right)}^2}{{(2k-1)}^2}}-{\displaystyle \frac{{\mathbf{O}}_k^{\langle 4\rangle }}{{(2k-1)}^2}}-{\displaystyle \frac{2{\mathbf{O}}_k^{\langle 2\rangle }}{{(2k-1)}^3}}\right\}+31\zeta \left(5\right)+{\displaystyle \frac{{\pi }^6}{30}}.\hfill \end{array}$$

  (25)
  These k-sums are evaluated by the formulae in Section 6.3, Section 6.4 and Section 6.5.

6. $\mathit{K}$-Sums (Odd Euler–Sums)

During the course of determining closed-form values for the infinite series exhibited in the previous sections, it is necessary to find the exact values of the k-sums in the form

$$\sum _{k=0}^{\infty }{(\pm 1)}^k{\displaystyle \frac{\nabla \left(k\right)}{{(2k+1)}^{\lambda }}},\mathrm{where}\nabla \left(k\right)\mathrm{is}\mathrm{a}\mathrm{combined}\mathrm{product}\mathrm{of}\mathrm{harmonic}\mathrm{numbers}{\mathbf{H}}_k^{\langle *\rangle }\mathrm{and}{\mathbf{O}}_k^{\langle *\rangle }.$$

For this k-sum, its depth equals $\lambda$ plus the order of $\nabla \left(k\right)$. To evaluate these k-sums, we shall convert them into the odd Euler sums, which is defined, for $\tilde{m}=(m_1,m_2,\cdots ,m_{\mathcal{\ell }})\in {\mathbb{N}}^{\mathcal{\ell }}$ with $m_1>1$, by the following multifold sum

$$T\left[\tilde{m}\right]=\sum _{k_1>k_2>\cdots >k_{\mathcal{\ell }}\ge 0}\prod _{i=1}^{\mathcal{\ell }}{\displaystyle \frac{1}{{(2k_i+1)}^{m_i}}}.$$

For instance, we have four k-sums of depth 2:

$$\begin{array}{cccc}& \sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^2}}={\displaystyle \frac{{\pi }^2}{8}},\hfill & \hfill & \sum _{k=1}^{\infty }{(-1)}^k{\displaystyle \frac{{\mathbf{H}}_k}{2k+1}}=G-{\displaystyle \frac{\pi ln2}{2}},\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k}{{(2k+1)}^2}}=G;\hfill & & \sum _{k=1}^{\infty }{(-1)}^k{\displaystyle \frac{{\mathbf{O}}_k}{2k+1}}={\displaystyle \frac{\pi ln2}{8}}-{\displaystyle \frac{G}{2}};\hfill \end{array}$$

where the two formulae on the right column are equivalent to those appearing in Olaikhan [20] (§4.1.19 and §4.1.20). Hoffman [11] made a systematic investigation on odd Euler sums and found numerous remarkable formulae. They are utilized to evaluate k-sums as below. We are going to display the resulting formulae of k-sums according to their depth and present a few detailed proofs as exemplification.

6.1. K-Sums of Depth 3

These values can be deduced from Olaikhan [20] (§4.1.17 and §4.5.5) and have been utilized in computing (5), (14) and (23).

$$\begin{array}{cccc}& \sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^3}}={\displaystyle \frac{7\zeta \left(3\right)}{8}},\hfill & \hfill & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}}{{(2k+1)}^2}}={\displaystyle \frac{{\pi }^2}{8}}ln2+{\displaystyle \frac{7\zeta \left(3\right)}{16}},\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k}{{(2k+1)}^2}}={\displaystyle \frac{7\zeta \left(3\right)}{4}}-{\displaystyle \frac{{\pi }^2}{4}}ln2,\hfill & \hfill & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k}{{(2k+1)}^2}}={\displaystyle \frac{{\pi }^2}{8}}ln2-{\displaystyle \frac{7\zeta \left(3\right)}{16}}.\hfill \end{array}$$

6.2. K-Sums of Depth 4

The formulae have been used in deriving (15), (17), (19) and (20).

$$\begin{array}{cccc}& \sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^4}}={\displaystyle \frac{{\pi }^4}{96}},\hfill & & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}}{{(2k+1)}^3}}={\displaystyle \frac{83{\pi }^4}{5760}}+{\displaystyle \frac{{\pi }^2}{24}}{ln}^{2}2-{\displaystyle \frac{{ln}^{4}2}{24}}-{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right),\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^2}}={\displaystyle \frac{5{\pi }^4}{384}},\hfill & & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^2}{{(2k+1)}^2}}={\displaystyle \frac{83{\pi }^4}{5760}}+{\displaystyle \frac{{\pi }^2}{6}}{ln}^{2}2-{\displaystyle \frac{{ln}^{4}2}{24}}-{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$

The two series on the right column are justified by first rewriting

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}}{{(2k+1)}^3}}& =\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^3}}\left\{{\displaystyle \frac{1}{2k+1}}+{\mathbf{O}}_k\right\}=T\left[4\right]+T[3,1],\hfill \\ \hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^2}{{(2k+1)}^2}}& =\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^2}}\left\{{\displaystyle \frac{1}{{(2k+1)}^2}}+{\displaystyle \frac{2{\mathbf{O}}_k}{2k+1}}+{\mathbf{O}}_k^2\right\}\hfill \\ & =T\left[4\right]+T[2,2]+2T[3,1]+2T[2,1,1];\hfill \end{array}$$

and then applying the values of odd Euler sums collected by Hoffman [11].

Alternatively, they can be reformulated by

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}}{{(2k+1)}^3}}& =T\left[4\right]+{\displaystyle \frac{1}{2}}\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k}+{\overline{\mathbf{H}}}_{2k}}{{(2k+1)}^3}},\hfill \\ \hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^2}{{(2k+1)}^2}}& =T\left[4\right]+\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k}+{\overline{\mathbf{H}}}_{2k}}{{(2k+1)}^3}}+\sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^2}{{(2k+1)}^2}};\hfill \end{array}$$

and then evaluated by the following summation formulae that are deduced by combining the related values appearing in the compendium by Olaikhan [20]:

$$\begin{array}{cc}& \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k}}{{(2k+1)}^3}}={\displaystyle \frac{17{\pi }^4}{1440}}-{\displaystyle \frac{7ln2}{8}}\zeta \left(3\right)-{\displaystyle \frac{{ln}^{4}2}{24}}+{\displaystyle \frac{{\pi }^2}{24}}{ln}^{2}2-{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right),\hfill \\ & \sum _{k=1}^{\infty }{\displaystyle \frac{{\overline{\mathbf{H}}}_{2k}}{{(2k+1)}^3}}={\displaystyle \frac{{\pi }^2}{24}}{ln}^{2}2-{\displaystyle \frac{11{\pi }^4}{2880}}-{\displaystyle \frac{{ln}^{4}2}{24}}+{\displaystyle \frac{7\zeta \left(3\right)}{8}}ln2-{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right);\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k}^{\langle 2\rangle }}{{(2k+1)}^2}}={\displaystyle \frac{{ln}^{4}2}{12}}-{\displaystyle \frac{53{\pi }^4}{2880}}-{\displaystyle \frac{{\pi }^2}{12}}{ln}^{2}2+{\displaystyle \frac{7}{4}}\zeta \left(3\right)ln2+2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right),\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\overline{\mathbf{H}}}_{2k}^{\langle 2\rangle }}{{(2k+1)}^2}}={\displaystyle \frac{17{\pi }^4}{720}}-{\displaystyle \frac{{ln}^{4}2}{12}}+{\displaystyle \frac{{\pi }^2}{12}}{ln}^{2}2-{\displaystyle \frac{7}{4}}\zeta \left(3\right)ln2-2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right);\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k}^2}{{(2k+1)}^2}}={\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{7\zeta \left(3\right)}{8}}ln2+{\displaystyle \frac{11{\pi }^4}{2880}}+{\displaystyle \frac{{ln}^{4}2}{24}}-{\displaystyle \frac{{\pi }^2}{24}}{ln}^{2}2,\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\overline{\mathbf{H}}}_{2k}^2}{{(2k+1)}^2}}=5{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{7\zeta \left(3\right)}{8}}ln2-{\displaystyle \frac{19{\pi }^4}{576}}+{\displaystyle \frac{5{ln}^{4}2}{24}}+{\displaystyle \frac{7{\pi }^2}{24}}{ln}^{2}2;\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^2}{{(2k+1)}^2}}={\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{23{\pi }^4}{5760}}+{\displaystyle \frac{{ln}^{4}2}{24}}+{\displaystyle \frac{{\pi }^2}{12}}{ln}^{2}2,\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k^2}{{(2k+1)}^2}}=8{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{61{\pi }^4}{1440}}+{\displaystyle \frac{{ln}^{4}2}{3}}+{\displaystyle \frac{{\pi }^2}{6}}{ln}^{2}2.\hfill \end{array}$$

6.3. K-Sums of Depth 5

These sums are more complex and therefore cumbersome to handle.

• The easiest one:

  $$\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^5}}=T\left[5\right]={\displaystyle \frac{31}{32}}\zeta \left(5\right).$$
• Combination of [20] (§4.1.14 and §4.1.17):

  $$\sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}}{{(2k+1)}^4}}={\displaystyle \frac{{\pi }^4}{96}}ln2-{\displaystyle \frac{{\pi }^2}{64}}\zeta \left(3\right)+{\displaystyle \frac{31\zeta \left(5\right)}{64}}.$$
• Combinations of [20] (§4.2.5 and §4.3.13) with [20](§4.5.11):

  $$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k+1}^{\langle 2\rangle }}{{(2k+1)}^3}}={\displaystyle \frac{29{\pi }^2}{96}}\zeta \left(3\right)-{\displaystyle \frac{155}{64}}\zeta \left(5\right),\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k^{\langle 2\rangle }}{{(2k+1)}^3}}={\displaystyle \frac{49{\pi }^2}{48}}\zeta \left(3\right)-{\displaystyle \frac{93}{8}}\zeta \left(5\right);\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^3}}={\displaystyle \frac{3{\pi }^2}{64}}\zeta \left(3\right)+{\displaystyle \frac{31}{64}}\zeta \left(5\right).\hfill \end{array}$$
  The last formula has been used in computing (25).
• Combinations of [20] (§4.2.6 and §4.3.16) with [20] (§4.5.12):

  $$\begin{array}{cc}& \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k+1}^{\langle 3\rangle }}{{(2k+1)}^2}}={\displaystyle \frac{155}{64}}\zeta \left(5\right)-{\displaystyle \frac{5{\pi }^2}{48}}\zeta \left(3\right),\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k^{\langle 3\rangle }}{{(2k+1)}^2}}={\displaystyle \frac{31}{2}}\zeta \left(5\right)-{\displaystyle \frac{4{\pi }^2}{3}}\zeta \left(3\right);\hfill \\ & \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 3\rangle }}{{(2k+1)}^2}}={\displaystyle \frac{{\pi }^2}{16}}\zeta \left(3\right)+{\displaystyle \frac{31}{64}}\zeta \left(5\right).\hfill \end{array}$$
• Combination of [20] (§4.2.7 and §4.3.14):

  $$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k+1}^2}{{(2k+1)}^3}}& =2{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+2ln2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{19{\pi }^2}{96}}\zeta \left(3\right)\hfill \\ & +{\displaystyle \frac{93}{64}}\zeta \left(5\right)+{\displaystyle \frac{7{ln}^{2}2}{8}}\zeta \left(3\right)+{\displaystyle \frac{{ln}^{5}2}{15}}-{\displaystyle \frac{{\pi }^2}{18}}{ln}^{3}2.\hfill \end{array}$$
• Combination of [20] (§4.2.9 and §4.3.18):

  $$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{H}}_{2k+1}^3}{{(2k+1)}^2}}& =3{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+3ln2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{11{\pi }^2}{192}}\zeta \left(3\right)\hfill \\ & +{\displaystyle \frac{31}{8}}\zeta \left(5\right)+{\displaystyle \frac{21{ln}^{2}2}{16}}\zeta \left(3\right)+{\displaystyle \frac{{ln}^{5}2}{10}}-{\displaystyle \frac{{\pi }^2}{12}}{ln}^{3}2.\hfill \end{array}$$
  Furthermore, we have three more involved k-sums of depth 5.
• Combination of [20] (§4.2.17 and §4.3.17):

  $$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }& {\displaystyle \frac{{\mathbf{O}}_{k+1}{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(2k+1)}^2}}=\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{1}{{(2k+1)}^5}}+{\displaystyle \frac{{\mathbf{O}}_k}{{(2k+1)}^4}}+{\displaystyle \frac{{\mathbf{O}}_k^{\langle 2\rangle }}{{(2k+1)}^3}}+{\displaystyle \frac{{\mathbf{O}}_k{\mathbf{O}}_k^{\langle 2\rangle }}{{(2k+1)}^2}}\right\}\hfill \\ \hfill =& T\left[5\right]+T[4,1]+\left(T[3,2]+T[2,3]\right)+\left(T[2,2,1]+T[2,1,2]\right)\hfill \\ \hfill =& T\left[5\right]+\left\{T\left[4\right]ln2-{\displaystyle \frac{1}{2}}T\left[5\right]-{\displaystyle \frac{1}{7}}T\left[2\right]T\left[3\right]\right\}+\{T[2]T[3]-T[5]\}\hfill \\ & +\left\{{\displaystyle \frac{7}{8}}T\left[5\right]-{\displaystyle \frac{5}{7}}T\left[2\right]T\left[3\right]+{\displaystyle \frac{1}{4}}T\left[4\right]ln2\right\}\hfill \\ \hfill =& {\displaystyle \frac{93}{256}}\zeta \left(5\right)+{\displaystyle \frac{{\pi }^2}{64}}\zeta \left(3\right)+{\displaystyle \frac{5{\pi }^4}{384}}ln2.\hfill \end{array}$$
• The k-sum involving ${\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)$:

  $$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }& {\displaystyle \frac{{\mathbf{O}}_{k+1}^2}{{(2k+1)}^3}}=\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{1}{{(2k+1)}^5}}+{\displaystyle \frac{2{\mathbf{O}}_k}{{(2k+1)}^4}}+{\displaystyle \frac{{\mathbf{O}}_k^2}{{(2k+1)}^3}}\right\}\hfill \\ \hfill =& T\left[5\right]+2T[3,1,1])+(2T[4,1]+T[3,2])\hfill \\ \hfill =& T\left[5\right]+2\left\{T\left[4\right]ln2-{\displaystyle \frac{1}{2}}T\left[5\right]-{\displaystyle \frac{1}{7}}T\left[2\right]T\left[3\right]\right\}+\left\{{\displaystyle \frac{3}{7}}T\left[2\right]T\left[3\right]-{\displaystyle \frac{1}{2}}T\left[5\right]\right\}\hfill \\ & +2\left\{{\displaystyle \frac{5}{21}}T\left[2\right]T\left[3\right]-{\displaystyle \frac{23}{248}}T\left[5\right]-{\displaystyle \frac{37}{60}}T\left[4\right]ln2+{\displaystyle \frac{{ln}^{2}2}{2}}T\left[3\right]-{\displaystyle \frac{ln2}{2}}\zeta (\overline{3},1)-{\displaystyle \frac{1}{2}}\zeta (\overline{3},1,1)\right\}\hfill \\ \hfill =& 2{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{83{\pi }^4}{2880}}ln2+{\displaystyle \frac{{\pi }^2}{36}}{ln}^{3}2-{\displaystyle \frac{{\pi }^2}{64}}\zeta \left(3\right)-{\displaystyle \frac{217}{128}}\zeta \left(5\right)-{\displaystyle \frac{{ln}^{5}2}{60}}.\hfill \end{array}$$
• Another k-sum involving ${\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)$:

  $$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }& {\displaystyle \frac{{\mathbf{O}}_{k+1}^3}{{(2k+1)}^2}}=\sum _{k=0}^{\infty }\left\{{\displaystyle \frac{1}{{(2k+1)}^5}}+{\displaystyle \frac{3{\mathbf{O}}_k}{{(2k+1)}^4}}+{\displaystyle \frac{3{\mathbf{O}}_k^2}{{(2k+1)}^3}}+{\displaystyle \frac{{\mathbf{O}}_k^3}{{(2k+1)}^2}}\right\}\hfill \\ \hfill =& T\left[5\right]+6T[3,1,1]+3\left(T\right[4,1]+T[3,2\left]\right)+3\left(T\right[2,2,1]+T[2,1,2\left]\right)+T[2,3]+6T[2,1,1,1]\hfill \\ \hfill =& T\left[5\right]+6\left\{{\displaystyle \frac{5}{21}}T\left[2\right]T\left[3\right]-{\displaystyle \frac{37}{60}}T\left[4\right]ln2+{\displaystyle \frac{{ln}^{2}2}{2}}T\left[3\right]-{\displaystyle \frac{23}{248}}T\left[5\right]-{\displaystyle \frac{ln2}{2}}\zeta (\overline{3},1)-{\displaystyle \frac{1}{2}}\zeta (\overline{3},1,1)\right\}\hfill \\ & +3\left\{T\left[4\right]ln2-{\displaystyle \frac{1}{2}}T\left[5\right]-{\displaystyle \frac{1}{7}}T\left[2\right]T\left[3\right]\right\}+3\left\{{\displaystyle \frac{3}{7}}T\left[2\right]T\left[3\right]-{\displaystyle \frac{1}{2}}T\left[5\right]\right\}\hfill \\ & +3\left\{{\displaystyle \frac{7}{8}}T\left[5\right]-{\displaystyle \frac{5}{7}}T\left[2\right]T\left[3\right]+{\displaystyle \frac{1}{4}}T\left[4\right]ln2\right\}+\left\{{\displaystyle \frac{4}{7}}T\left[2\right]T\left[3\right]-{\displaystyle \frac{1}{2}}T\left[5\right]\right\}\hfill \\ & +6\left\{{\displaystyle \frac{2}{21}}T\left[2\right]T\left[3\right]+{\displaystyle \frac{11}{60}}T\left[4\right]ln2-{\displaystyle \frac{1}{4}}T\left[3\right]{ln}^{2}2+{\displaystyle \frac{1}{6}}T\left[2\right]{ln}^{3}2-{\displaystyle \frac{35}{248}}T\left[5\right]+{\displaystyle \frac{ln2}{4}}\zeta (\overline{3},1)+{\displaystyle \frac{1}{4}}\zeta (\overline{3},1,1)\right\}\hfill \\ \hfill =& 3{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{83{\pi }^4}{1920}}ln2+{\displaystyle \frac{{\pi }^2}{6}}{ln}^{3}2+{\displaystyle \frac{{\pi }^2}{64}}\zeta \left(3\right)-{\displaystyle \frac{713}{256}}\zeta \left(5\right)-{\displaystyle \frac{{ln}^{5}2}{40}}.\hfill \end{array}$$
  In the last two k-sums, we have made use of the values of two alternating Euler sums:

  $$\begin{array}{cc}\hfill \zeta (\overline{3},1)& =\sum _{n>k\ge 1}{\displaystyle \frac{{(-1)}^n}{n^{3}k}}=2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{{\pi }^4}{48}}+{\displaystyle \frac{7ln2}{4}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^2}{12}}{ln}^{2}2+{\displaystyle \frac{{ln}^{4}2}{12}};\hfill \\ \hfill \zeta (\overline{3},1,1)& =\sum _{n>k>j\ge 1}{\displaystyle \frac{{(-1)}^n}{n^{3}kj}}=\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{2k^3}}\left\{{\mathbf{H}}_{k-1}^2-{\mathbf{H}}_{k-1}^{\langle 2\rangle }\right\}=\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{2k^3}}\left\{{\mathbf{H}}_k^2-{\mathbf{H}}_k^{\langle 2\rangle }+{\displaystyle \frac{2}{k^2}}-{\displaystyle \frac{2{\mathbf{H}}_k}{k}}\right\}\hfill \\ & ={\displaystyle \frac{33}{32}}\zeta \left(5\right)+{\displaystyle \frac{{\pi }^2}{12}}\zeta \left(3\right)-{\displaystyle \frac{7{ln}^{2}2}{8}}\zeta \left(3\right)+{\displaystyle \frac{{\pi }^2}{18}}{ln}^{3}2-{\displaystyle \frac{{ln}^{5}2}{15}}-2ln2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-2{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$
  Their evaluation depends upon the following values of Euler sums:
  Olaihkan [20] (§4.2.5)$\begin{array}{cc}& \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k^{\langle 2\rangle }}{k^3}}={\displaystyle \frac{{\pi }^2}{2}}\zeta \left(3\right)-{\displaystyle \frac{9}{2}}\zeta \left(5\right),\hfill \end{array}$
  Olaihkan [20] (§4.2.7)$\begin{array}{cc}& \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{H}}_k^2}{k^3}}={\displaystyle \frac{7}{2}}\zeta \left(5\right)-{\displaystyle \frac{{\pi }^2}{6}}\zeta \left(3\right);\hfill \end{array}$
  Olaihkan [20] (§4.3.13)$\begin{array}{cc}& \sum _{k=1}^{\infty }{(-1)}^k{\displaystyle \frac{{\mathbf{H}}_k^{\langle 2\rangle }}{k^3}}={\displaystyle \frac{11}{32}}\zeta \left(5\right)-{\displaystyle \frac{5{\pi }^2}{48}}\zeta \left(3\right),\hfill \end{array}$
  Olaihkan [20] (§4.3.12)$\begin{array}{cc}& \sum _{k=1}^{\infty }{(-1)}^k{\displaystyle \frac{{\mathbf{H}}_k}{k^4}}={\displaystyle \frac{{\pi }^2}{12}}\zeta \left(3\right)-{\displaystyle \frac{59}{32}}\zeta \left(5\right),\hfill \end{array}$
  Olaihkan [20] (§4.3.7)$\begin{array}{cc}& \sum _{k=1}^{\infty }{(-1)}^k{\displaystyle \frac{{\mathbf{H}}_k}{k^3}}=2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{11{\pi }^4}{360}}+{\displaystyle \frac{7ln2}{4}}\zeta \left(3\right)-{\displaystyle \frac{{\pi }^2}{12}}{ln}^{2}2+{\displaystyle \frac{{ln}^{4}2}{12}},\hfill \end{array}$
  Olaihkan [20] (§4.3.14)$\begin{array}{cc}& \sum _{k=1}^{\infty }{(-1)}^k{\displaystyle \frac{{\mathbf{H}}_k^2}{k^3}}=\left\{\begin{array}{c}\hfill {\displaystyle \frac{19}{32}}\zeta \left(5\right)+{\displaystyle \frac{11{\pi }^2}{48}}\zeta \left(3\right)-{\displaystyle \frac{7{ln}^{2}2}{4}}\zeta \left(3\right)+{\displaystyle \frac{{\pi }^2}{9}}{ln}^{3}2\\ \hfill -{\displaystyle \frac{2{ln}^{5}2}{15}}-4ln2{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right)-4{\mathrm{Li}}_5\left({\displaystyle \frac{1}{2}}\right)\end{array}\right\}.\hfill \end{array}$

6.4. K-Sums of Depth 6

They are employed in the derivation of (12), (13), (19), (20) and (25).

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(2k+1)}^6}}& =T\left[6\right]={\displaystyle \frac{{\pi }^6}{960}},\hfill \\ \hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 2\rangle }}{{(1+2k)}^4}}& ={\displaystyle \frac{6T\left[6\right]+T^2\left(3\right)}{7}}={\displaystyle \frac{{\pi }^6}{1120}}+{\displaystyle \frac{7{\zeta }^2\left(3\right)}{64}},\hfill \\ \hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\mathbf{O}}_{k+1}^{\langle 4\rangle }}{{(1+2k)}^2}}& ={\displaystyle \frac{39T\left[6\right]-4T^2\left(3\right)}{28}}={\displaystyle \frac{13{\pi }^6}{8960}}-{\displaystyle \frac{7{\zeta }^2\left(3\right)}{64}},\hfill \\ \hfill \sum _{k=0}^{\infty }{\displaystyle \frac{{\left({\mathbf{O}}_{k+1}^{\langle 2\rangle }\right)}^2}{{(1+2k)}^2}}& =T\left[6\right]+T[2,4]+2T[4,2]+2T[2,2,2]\hfill \\ & ={\displaystyle \frac{25}{24}}T\left[6\right]+{\displaystyle \frac{11T\left[6\right]-4T^2\left(3\right)}{28}}+{\displaystyle \frac{2T^2\left(3\right)-2T\left[6\right]}{7}}\hfill \\ & ={\displaystyle \frac{193{\pi }^6}{161280}}+{\displaystyle \frac{7{\zeta }^2\left(3\right)}{64}}.\hfill \end{array}$$

6.5. Series Rearrangements

In addition, the k-sums with ${(4k^2-1)}^{\lambda }$ in denominators were encountered in Section 5. They can be treated by the series rearrangements.

• k-sum of depth 2: The value has been used in computing (21) and (23).

  $$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k}{(2k-1)(2k+1)}}& =\sum _{k\ge i\ge 1}{\displaystyle \frac{1}{(2i-1)(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^2}}={\displaystyle \frac{{\pi }^2}{16}}.\hfill \end{array}$$
• k-sums of depth 3:

  $$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^{\langle 2\rangle }}{(2k-1)(2k+1)}}& =\sum _{k\ge i\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^3}}={\displaystyle \frac{7}{16}}\zeta \left(3\right),\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^2}{(2k-1)(2k+1)}}& =\sum _{k\ge \max \{i,j\}\ge 1}{\displaystyle \frac{1}{(2i-1)(2j-1)(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i,j\ge 1}{\displaystyle \frac{1}{(2i-1)(2j-1)\left(2\max \right\{i,j\}-1)}}\hfill \\ & =\sum _{i>j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2(2j-1)}}+{\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^3}}\hfill \\ & =T[2,1]+{\displaystyle \frac{1}{2}}T\left[3\right]={\displaystyle \frac{{\pi }^2}{8}}ln2.\hfill \end{array}$$
• k-sums of depth 4:

  $$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^{\langle 3\rangle }}{(2k-1)(2k+1)}}& =\sum _{k\ge i\ge 1}{\displaystyle \frac{1}{{(2i-1)}^3(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^4}}={\displaystyle \frac{{\pi }^4}{192}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^3}{(2k-1)(2k+1)}}& =\sum _{k\ge \max \{i,j,r\}\ge 1}{\displaystyle \frac{1}{(2i-1)(2j-1)(2r-1)(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i,j,r\ge 1}{\displaystyle \frac{1}{(2i-1)(2j-1)(2r-1)\left(2\max \right\{i,j,r\}-1)}}\hfill \\ & =3\sum _{i>j>r\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2(2j-1)(2r-1)}}+{\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^4}}\hfill \\ & +{\displaystyle \frac{3}{2}}\sum _{i>j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2{(2j-1)}^2}}+{\displaystyle \frac{3}{2}}\sum _{i>j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^3(2j-1)}}\hfill \\ & =3T[2,1,1]+{\displaystyle \frac{1}{2}}T\left[4\right]+{\displaystyle \frac{3}{2}}(T[2,2]+T[3,1])\hfill \\ & ={\displaystyle \frac{{\pi }^4}{192}}+{\displaystyle \frac{3{\pi }^2}{16}}{ln}^{2}2,\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k{\mathbf{O}}_k^{\langle 2\rangle }}{(2k-1)(2k+1)}}& =\sum _{k\ge \max \{i,j\}\ge 1}{\displaystyle \frac{1}{(2i-1){(2j-1)}^2(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i,j\ge 1}{\displaystyle \frac{1}{(2i-1){(2j-1)}^2(2\max \{i,j\}-1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i>j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2{(2j-1)}^2}}+{\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^4}}\hfill \\ & +{\displaystyle \frac{1}{2}}\sum _{i>j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^3(2j-1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}T\left[4\right]+{\displaystyle \frac{1}{2}}(T[2,2]+T[3,1])\hfill \\ & ={\displaystyle \frac{49{\pi }^4}{5760}}+{\displaystyle \frac{{\pi }^2}{48}}{ln}^{2}2-{\displaystyle \frac{{ln}^{4}2}{48}}-{\displaystyle \frac{1}{2}}{\mathrm{Li}}_4\left({\displaystyle \frac{1}{2}}\right).\hfill \end{array}$$
• k-sums of depth 5: the values have been utilized in computing (25)

  $$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\mathbf{O}}_k^{\langle 4\rangle }}{(2k-1)(2k+1)}}& =\sum _{k\ge i\ge 1}{\displaystyle \frac{1}{{(2i-1)}^4(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^5}}={\displaystyle \frac{31\zeta \left(5\right)}{64}},\hfill \\ \hfill \sum _{k=1}^{\infty }{\displaystyle \frac{{\left({\mathbf{O}}_k^{\langle 2\rangle }\right)}^2}{(2k-1)(2k+1)}}& =\sum _{k\ge \max \{i,j\}\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2{(2j-1)}^2(2k-1)(2k+1)}}\hfill \\ & ={\displaystyle \frac{1}{2}}\sum _{i,j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^2{(2j-1)}^2(2\max \{i,j\}-1)}}\hfill \\ & =\sum _{i>j\ge 1}{\displaystyle \frac{1}{{(2i-1)}^3{(2j-1)}^2}}+{\displaystyle \frac{1}{2}}\sum _{i=1}^{\infty }{\displaystyle \frac{1}{{(2i-1)}^5}}\hfill \\ & =T[3,2]+{\displaystyle \frac{1}{2}}T\left[5\right]={\displaystyle \frac{3{\pi }^2}{64}}\zeta \left(3\right).\hfill \end{array}$$

7. Conclusions and Further Prospects

By employing the “coefficient extraction” method, we have systematically examined a harmonic series of convergence rate “$1/4$” with cubic central binomial coefficients in denominators. They correspond to three parameter settings “$A,B$ and C” in (3), wherein the weight polynomial ${\rho }_n(a,b,c,d)$ in (4) is factorizable into a linear functions.

If considering the different parameter replacements in (3)

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

then by carrying out the same “coefficient extraction” procedure, we can show the following identities for the series weighted by quadratic or cubic polynomials.

$$\begin{array}{cc}\hfill 2& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+6n+12n^2+6n^3}{n^3{(1+2n)}^3}}{\mathbf{O}}_n-{\displaystyle \frac{2}{{(1+2n)}^4}}\right\},\hfill \\ \hfill 0& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+6n+12n^2+6n^3}{n^3{(1+2n)}^3}}{\mathbf{H}}_n^{\langle 2\rangle }-{\displaystyle \frac{1}{n^5}}\right\},\hfill \\ \hfill 2& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+6n+12n^2+6n^3}{n^3{(1+2n)}^3}}{\mathbf{O}}_n^2-{\displaystyle \frac{4{\mathbf{O}}_n}{{(1+2n)}^4}}-{\displaystyle \frac{2}{{(1+2n)}^5}}\right\}.\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\pi }^2}{2}}-2& =\sum _{n=1}^{\infty }{\displaystyle \frac{{16}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}\right\},\hfill \\ \hfill {\displaystyle \frac{7}{2}}\zeta \left(3\right)-1& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}{\mathbf{O}}_n+{\displaystyle \frac{1}{{(1+2n)}^3}}\right\},\hfill \\ \hfill {\displaystyle \frac{{\pi }^4}{48}}& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }-{\displaystyle \frac{1}{2n^4}}\right\},\hfill \\ \hfill {\displaystyle \frac{{\pi }^4}{16}}-2& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}{\mathbf{O}}_{n+1}^{\langle 2\rangle }+{\displaystyle \frac{1+4n+2n^2}{2n^4{(1+2n)}^2}}\right\},\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\pi }^4}{32}}& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}{\mathbf{O}}_n^2+{\displaystyle \frac{2{\mathbf{O}}_n}{{(1+2n)}^3}}-{\displaystyle \frac{1}{4n^4}}\right\},\hfill \\ \hfill {\displaystyle \frac{{\pi }^6}{720}}& =\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}\left(2{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-{\mathbf{H}}_n^{\langle 4\rangle }\right)-{\displaystyle \frac{2{\mathbf{H}}_n^{\langle 2\rangle }}{n^4}}+{\displaystyle \frac{1}{n^6}}\right\},\hfill \\ \hfill {\displaystyle \frac{14\zeta \left(3\right)}{3}}& -{\displaystyle \frac{2{\pi }^2}{3}}ln2=\sum _{n=1}^{\infty }{\displaystyle \frac{{16}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}{\mathbf{H}}_n-{\displaystyle \frac{1}{n^3}}\right\},\hfill \\ \hfill {\displaystyle \frac{7{\pi }^2}{8}}\zeta \left(3\right)& -{\displaystyle \frac{31}{4}}\zeta \left(5\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\left(16\right)}^n}{{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3}}\left\{{\displaystyle \frac{1+4n+6n^2}{n^2{(1+2n)}^2}}{\mathbf{H}}_n^{\langle 2\rangle }{\mathbf{O}}_n+{\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{{(1+2n)}^3}}-{\displaystyle \frac{{\mathbf{O}}_n}{2n^4}}\right\}.\hfill \end{array}$$

Finally, observing that all four parameter settings for (3) examined in this paper are characterized by half integers for $\{a,b,c,d\}$, different parameter substitutions (with integer shifts) would lead to more summation formulae for complementary and/or generalized harmonic series. They may be worth further exploration.