Series of Convergence Rate −1/4 Containing Harmonic Numbers

Abstract

Two general transformations for hypergeometric series are examined by means of the coefficient extraction method. Several interesting closed formulae are shown for infinite series containing harmonic numbers and binomial/multinomial coefficients. Among them, three conjectured identities due to Z.-W. Sun are also confirmed.

1. Introduction and Outline

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

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

In general for $n\in \mathbb{Z}$, we can express it as the $\Gamma$-function ratio

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

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

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

The $\Gamma$-function is one of the important special functions (cf. Rainville [1] §8). The following two power series expansions (cf. [2]) will be very useful in this paper.

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

(1)

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

(2)

where ${\sigma }_k$ and ${\tau }_k$ are defined respectively by

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

with the Euler–Mascheroni constant and the Riemann zeta function being, respectively, given by

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

Throughout the paper, we shall utilize, for $x\in \mathbb{R}$ and $n,\lambda \in {\mathbb{N}}_0$, the following notations for harmonic numbers:

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

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

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

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

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

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

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

(3)

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

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

(4)

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

There are numerous infinite series identities containing central binomial coefficients (cf. [5,6,7,8,9]). Recently, the “coefficient extraction” method just displayed has been successfully applied in evaluating, in closed form, infinite series involving both harmonic numbers and central binomial coefficients (see for example [2,3,10,11]). In this paper, the same approach will be employed further to examine the two useful hypergeometric series transformation formulae from [12]. They express the series of unit argument in terms of faster convergent ones, i.e., the series of convergent rate “$\frac{-1}{4}$” (the limit of term ratio $T_{n+1}/T_n$ as $n\to \infty$, where $T_n$ denotes the general term of the series). In order to facilitate subsequent applications, they are reproduced below and will be referred to shortly as $\Phi (a;b,c,d,e)$ and $\Psi (a;b,c,d,e)$.

For five complex parameters $\{a,b,c,d,e\}$ subject to $\Re (1+2a-b-c-d-e)>0$, the first transformation (cf. [12] Theorem 9) can be stated as

$$\begin{array}{cc}\sum _{k=0}^{\infty }(a+2k){\left[\begin{array}{c}b,c,d,e\\ 1+a-b,1+a-c,1+a-d,1+a-e\end{array}\right]}_k\hfill & \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n\times {\mathrm{U}}_n(a,b,c,d,e)}{{(1+a-e)}_{2n+1}}\hfill & \Phi (a;b,c,d,e)\\ \times \frac{{\left(b\right)}_n{\left(c\right)}_n{\left(d\right)}_n{(1+a-b-e)}_n{(1+a-c-e)}_n{(1+a-d-e)}_n}{{(1+a-b)}_n{(1+a-c)}_n{(1+a-d)}_n{(1+2a-b-c-d-e)}_{n+1}},\hfill & \end{array}$$

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

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

Under the same condition, the second (in symmetric form) transformation (cf. [12] Theorem 10) reads as

$$\begin{array}{cc}\sum _{k=0}^{\infty }(a+2k){\left[\begin{array}{c}b,c,d,e\\ 1+a-b,1+a-c,1+a-d,1+a-e\end{array}\right]}_k\hfill & \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n\times {\mathrm{V}}_n(a,b,c,d,e)}{{(1+2a-b-c-d-e)}_{2n+2}}\hfill & \Psi (a;b,c,d,e)\\ \times \frac{\left\{\begin{array}{c}{(1+a-b-c)}_n{(1+a-b-d)}_n{(1+a-b-e)}_n\\ {(1+a-c-d)}_n{(1+a-c-e)}_n{(1+a-d-e)}_n\end{array}\right\}}{{(1+a-b)}_n{(1+a-c)}_n{(1+a-d)}_n{(1+a-e)}_n},\hfill & \end{array}$$

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

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

In the next section, four main cases of $\Phi (a;b,c,d,e)$ will be systematically explored that will result in a number of closed formulae for infinite series of convergence rate “$\frac{-1}{4}$” involving harmonic numbers and binomial/multinomial coefficients. Then we shall carry on similar investigations for $\Psi (a;b,c,d,e)$ in Section 3. Among numerous infinite series identities established in this paper, we highlight the following five representatives as examples:

$$\begin{array}{cc}\hfill \mathrm{Equation}\left(5\right)& \frac{{\pi }^4}{24}=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{4-5{(2n+1)}^2{\mathbf{O}}_n^{\langle 2\rangle }}{{(-16)}^n{(2n+1)}^4}.\hfill \\ \hfill \mathrm{Equation}\left(6\right)& \frac{{\pi }^2\zeta \left(3\right)}{2}=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{6+5{(2n+1)}^3{\mathbf{O}}_n^{\langle 3\rangle }}{{(-16)}^n{(2n+1)}^5}.\hfill \\ \hfill \mathrm{Equation}\left(7\right)& \frac{7{\pi }^6}{7200}=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{1+5{(2n+1)}^4{\mathbf{O}}_n^{\langle 4\rangle }}{{(-16)}^n{(2n+1)}^6}.\hfill \\ \hfill \mathrm{Equation}\left(8\right)& 2\zeta \left(5\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^5\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{9-5n^2{\mathbf{H}}_n^{\langle 2\rangle }\right\}.\hfill \\ \hfill \mathrm{Equation}\left(9\right)& 2\zeta {\left(3\right)}^2=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^6\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{1+5n^3{\mathbf{H}}_n^{\langle 3\rangle }\right\}.\hfill \end{array}$$

Among them, the last formula was proposed as a conjecture by Sun [13] (Equation 4.5), who raised also the second and the third identities as open problems in the monograph [14] (Conjectures 10.71 and 10.74).

2. Identities from Transformation $\Phi (\mathit{a};\mathit{b},\mathit{c},\mathit{d},\mathit{e})$

This section will be devoted to presenting several infinite series evaluations mainly by employing transformation $\Phi (a;b,c,d,e)$. For five complex parameters $\{a,b,c,d,e\}$ in $\Phi (a;b,c,d,e)$, letting $e=a$ and assuming the condition $\Re (1+a-b-c-d)>0$, we derive, by making use of Dougall’s summation theorem for the well–poised ${}_5{F}_4$-series (cf. Bailey [15] §4.4), the summation formula as in the following lemma.

Lemma 1.

Let $\{a,b,c,d\}$ be four complex parameters such that the series below is well–defined. Then there holds

$$\begin{array}{cc}& \Gamma \left[\begin{array}{c}1+a-b,1+a-c,1+a-d,1+a-b-c-d\\ a,1+a-b-c,1+a-b-d,1+a-c-d\end{array}\right]\hfill \\ & =\sum _{n=0}^{\infty }{(-1)}^n\frac{{\left(b\right)}_n{\left(c\right)}_n{\left(d\right)}_n{\mathrm{U}}_n(a,b,c,d)}{(2n+1)!{(1+a-b-c-d)}_{n+1}}\hfill \\ & \times \frac{{(1-b)}_n{(1-c)}_n{(1-d)}_n}{{(1+a-b)}_n{(1+a-c)}_n{(1+a-d)}_n},\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\mathrm{U}}_n(a,b,c,d)& =(1+2n)(a-b+n)(1+a-c-d+2n)\hfill \\ & +(b+n)(1-c+n)(1-d+n).\hfill \end{array}$$

It should be pointed out that the parameter restriction $\Re (1+a-b-c-d)>0$ has been removed by analytic continuation for the above formula. In fact, it is valid in the whole complex space of dimension 4 except for the hyperplanes determined by

$$\left\{b-a\in \mathbb{N}\right\}\cup \left\{c-a\in \mathbb{N}\right\}\cup \left\{d-a\in \mathbb{N}\right\}\cup \left\{b+c+d-a\in \mathbb{N}\right\}.$$

2.1. Series with $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$

Performing the parameter replacements

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

we derive from Lemma 1 the summation formula as in the theorem below.

Theorem 2.

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

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

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

$$\begin{array}{cc}\hfill {\mathrm{P}}_n(a,b,c,d;x)& =(1+2n)(\frac{1}{2}+ax-bx+n)(1+ax-cx-dx+2n)\hfill \\ & +(\frac{1}{2}+bx+n)(\frac{1}{2}-cx+n)(\frac{1}{2}-dx+n).\hfill \end{array}$$

Observe that both sides of the above equation are analytic functions of x in the neighborhood of $x=0$. They can be expanded into power series in accordance with (1)–(4). Then by comparing the coefficients ${\mathrm{A}}_m(a,b,c,d)$ of $x^m$ across, we can show the following infinite series identities.

• A₀(a, b, c, d)

  $$\frac{{\pi }^2}{10}=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n{(2n+1)}^2}.$$
• A₂(1, 2, 0, 0) Dilcher and Vignat [16]

  $$\frac{{\pi }^4}{24}=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{4-5{(2n+1)}^2{\mathbf{O}}_n^{\langle 2\rangle }}{{(-16)}^n{(2n+1)}^4}.$$

  (5)
• A₃(3, 2, 2, 2) Conjecture 10.71 by Sun [14]

  $$\frac{{\pi }^2}{2}\zeta \left(3\right)=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{6+5{(2n+1)}^3{\mathbf{O}}_n^{\langle 3\rangle }}{{(-16)}^n{(2n+1)}^5}.$$

  (6)
• A₄(2 + $\sqrt{-2}$, 2, 2, 2$\sqrt{-2}$) Conjecture 10.74 by Sun [14]

  $$\frac{7{\pi }^6}{7200}=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{1+5{(2n+1)}^4{\mathbf{O}}_n^{\langle 4\rangle }}{{(-16)}^n{(2n+1)}^6}.$$

  (7)
• A₈(2 + $\sqrt{-2}$, 2, 2, 2$\sqrt{-2}$)

  $$\frac{181{\pi }^{10}}{9072000}=\sum _{n=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\frac{2+10{(2n+1)}^4{\mathbf{O}}_n^{\langle 4\rangle }-5{(2n+1)}^8\left\{3{\mathbf{O}}_n^{\langle 8\rangle }-5{\left({\mathbf{O}}_n^{\langle 4\rangle }\right)}^2\right\}}{{(-16)}^n{(2n+1)}^{10}}.$$

2.2. Series with ${\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^{-1}$

We begin with the following three easily derived identities by extracting initial coefficients of x directly from combined equations of $\Phi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]$ and $\Psi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]$.

• $\boxed{\left[x^0\right]\Phi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]}$

  $$\frac{2{\pi }^2}{3}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right){(1+2n)}^2}\frac{(7+10n)}{(1+n)}.$$
• $\boxed{3\left[x^1\right]\Psi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]-4\left[x^1\right]\Phi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]}$

  $$12\zeta \left(3\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{13+20n+4n^2}{{(n+1)}^2{(2n+1)}^3}-\frac{4(7+10n){\mathbf{H}}_n}{(n+1){(2n+1)}^2}\right\}.$$
• $\boxed{4\left[x^1\right]\Phi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]+\left[x^1\right]\Psi [\frac{3}{2}+2x,1+x,1+x,\frac{1}{2}+x,\frac{1}{2}+x]}$

  $$60\zeta \left(3\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{77+188n+116n^2}{{(n+1)}^2{(2n+1)}^3}+\frac{8(7+10n){\mathbf{O}}_n}{(n+1){(2n+1)}^2}\right\}.$$

More series of a similar type can be evaluated by applying Lemma 1. In fact, performing further the parameter replacements

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

the summation formula in Lemma 1 becomes the following one:

Theorem 3.

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

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

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

$$\begin{array}{cc}\hfill {\mathrm{P}}_n(a,b,c,d;x)& =(1+2n)(ax-bx+n)(\frac{1}{2}+ax-cx-dx+2n)\hfill \\ & +(bx+n)(\frac{1}{2}-cx+n)(1-dx+n).\hfill \end{array}$$

According to this theorem, for each fixed nonnegative integer m, we can first write down the equation corresponding to the coefficient ${\mathrm{B}}_m(a,b,c,d)$ of $x^m$ (across the above equation). Based on this equation, we can then construct a system of refined linear equations by comparing coefficients of monomials $a^i{b}^j{c}^k{d}^{\ell }$ (subject to the restriction $i+j+k+\ell =m$). By resolving the system of linear equations, we establish the following identities. This procedure will be denominated as “Resolving linear system" formed by ${\mathrm{B}}_m(a,b,c,d)$.

• B₁(a,b,c,d)

  $$1=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}(2+5n)}{n(2n+1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}.$$
• Resolving linear system B₂(a,b,c,d)

  $$0=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{2}{n^2}-\frac{(2+5n){\mathbf{H}}_n}{n(2n+1)}\right\}.$$
• Resolving linear system B₃(a,b,c,d)

  $$0=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{2}{n^3}-\frac{2+5n}{n(2n+1)}{\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$
• Resolving linear system B₂(a,b,c,d)

  $$1=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{1}{{(2n+1)}^2}+\frac{(2+5n){\mathbf{O}}_n}{n(2n+1)}\right\}.$$
• Resolving linear system B₃(a,b,c,d)

  $$1=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{1}{{(1+2n)}^3}+\frac{2+5n}{n(1+2n)}{\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$
• Resolving linear system B₂(a,b,c,d)

  $$\frac{{\pi }^2}{3}=2-\sum _{n=1}^{\infty }\frac{{(-1)}^n(3+12n+10n^2)}{n^2{(2n+1)}^2\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}.$$
• Resolving linear system B₂(a,b,c,d)

  $$\frac{{\pi }^2}{3}=2+\sum _{n=1}^{\infty }\frac{{(-1)}^n\left\{(1+4n+6n^2)-2n(1+2n)(2+5n){\mathbf{H}}_n\right\}}{n^2{(2n+1)}^2\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}.$$
• Resolving linear system B₃(a,b,c,d)

  $$\frac{7}{2}\zeta \left(3\right)=3+\sum _{n=1}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{3n^2}{n^2{(2n+1)}^3}-\frac{3+12n+10n^2}{n^2{(2n+1)}^2}{\mathbf{O}}_n\right\}.$$
• Resolving linear system B₃(a,b,c,d)

  $$\zeta \left(3\right)=2+\sum _{n=1}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{3+12n+10n^2}{n^2{(2n+1)}^2}{\mathbf{H}}_n-\frac{1+6n+12n^2+6n^3}{n^3{(2n+1)}^3}\right\}.$$
• Resolving linear system B₄(a,b,c,d)

  $$\frac{{\pi }^4}{45}=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{\frac{3+12n+10n^2}{n^2{(2n+1)}^2}{\mathbf{H}}_n^{\langle 2\rangle }-\frac{7}{n^4}\right\}.$$

2.3. Series with $\left(\genfrac{}{}{0pt}{}{4n}{n,n,n,n}\right)$

Performing the parameter replacements

$$a=\frac{1}{2}+ax,b=\frac{1}{4}+bx,c=\frac{1}{2}+cx,d=\frac{3}{4}+dx,$$

we derive from Lemma 1 the following summation formula.

Theorem 4.

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

$$\begin{array}{cc}\hfill \Gamma & \left[\begin{array}{c}\frac{1}{4}+ax-bx,1+ax-cx,\frac{3}{4}+ax-dx,1+ax-bx-cx-dx\\ \\ \frac{1}{2}+ax,\frac{3}{4}+ax-bx-cx,\frac{1}{2}+ax-bx-dx,\frac{1}{4}+ax-cx-dx\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{(-1)}^n\frac{{\mathrm{P}}_n(a,b,c,d;x)\times {(\frac{1}{2}+cx)}_n{(\frac{1}{2}-cx)}_n}{(2n+1)!{(1+ax-bx-cx-dx)}_n}\hfill \\ & \times \frac{{(\frac{1}{4}+bx)}_n{(\frac{3}{4}+dx)}_n{(\frac{3}{4}-bx)}_n{(\frac{1}{4}-dx)}_n}{{(\frac{1}{4}+ax-bx)}_{n+1}{(1+ax-cx)}_n{(\frac{3}{4}+ax-dx)}_n},\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\mathrm{P}}_n(a,b,c,d;x)& =(1+2n)(\frac{1}{4}+ax-bx+n)(\frac{1}{4}+ax-cx-dx+2n)\hfill \\ & +(\frac{1}{4}+bx+n)(\frac{1}{2}-cx+n)(\frac{1}{4}-dx+n).\hfill \end{array}$$

Analogously, several infinite series identities can be deduced by extracting the coefficients ${\mathrm{C}}_m(a,b,c,d)$ of $x^m$ from the above equation. Five of them are highlighted as examples.

• C₀(1,1,1,1): Ramanujan [17]

  $$\frac{8}{\pi }=\sum _{n=0}^{\infty }\frac{\left(4n\right)!(3+20n)}{n{!}^4{(-1024)}^n}.$$
• C₁(1,1,1,1)

  $$\frac{24ln2}{\pi }=\sum _{n=0}^{\infty }\frac{\left(4n\right)!}{n{!}^4{(-1024)}^n}\left\{\frac{5+12n}{2n+1}-(3+20n){\mathbf{H}}_n\right\}.$$
• C₁(2,1,1,1)

  $$\frac{8ln2}{\pi }=\sum _{n=0}^{\infty }\frac{\left(4n\right)!}{n{!}^4{(-1024)}^n}\left\{\frac{3+8n}{2n+1}+2(3+20n){\mathbf{O}}_{2n}\right\}.$$
• C₂(1,1,0,1)

  $$\frac{8\pi }{3}=\sum _{n=0}^{\infty }\frac{\left(4n\right)!}{n{!}^4{(-1024)}^n}\left\{\frac{16}{4n+1}-(3+20n)\left({\mathbf{H}}_n^{\langle 2\rangle }-16{\mathbf{O}}_{2n}^{\langle 2\rangle }\right)\right\}.$$
• C₂(0,0,1,0)

  $$\frac{20\pi }{6}-\frac{72{ln}^{2}2}{\pi }=\sum _{n=0}^{\infty }\frac{\left(4n\right)!(3+20n)}{n{!}^4{(-1024)}^n}\left\{\frac{2(5+12n){\mathbf{H}}_n}{(2n+1)(3+20n)}-\left({\mathbf{H}}_n^2-2{\overline{\mathbf{H}}}_{2n}^{\langle 2\rangle }\right)\right\}.$$

2.4. Series with $\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)$

Performing the parameter replacements

$$a=1+ax,b=\frac{1}{4}+bx,c=\frac{1}{2}+cx,d=\frac{3}{4}+dx,$$

we derive from Lemma 1 the following summation formula.

Theorem 5.

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

$$\begin{array}{cc}\hfill \Gamma & \left[\begin{array}{c}\frac{1}{2}+ax-bx-cx-dx,\frac{3}{4}+ax-bx,\frac{1}{2}+ax-cx,\frac{1}{4}+ax-dx\\ 1+ax,\frac{1}{4}+ax-bx-cx,1+ax-bx-dx,\frac{3}{4}+ax-cx-dx\end{array}\right]\hfill \\ \hfill =& \sum _{n=0}^{\infty }{(-1)}^n\frac{{\mathrm{P}}_n(a,b,c,d;x)\times {(\frac{1}{4}+bx)}_n{(\frac{1}{2}+cx)}_n{(\frac{3}{4}+dx)}_n}{(2n+1)!{(\frac{1}{2}+ax-bx-cx-dx)}_{n+1}}\hfill \\ & \times \frac{{(\frac{3}{4}-bx)}_n{(\frac{1}{2}-cx)}_n{(\frac{1}{4}-dx)}_n(\frac{1}{4}+ax-bx-cx)}{{(\frac{3}{4}+ax-bx)}_{n+1}{(\frac{1}{2}+ax-cx)}_{n+1}{(\frac{1}{4}+ax-dx)}_{n+1}},\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\mathrm{P}}_n(a,b,c,d;x)& =(1+2n)(\frac{3}{4}+ax-bx+n)(\frac{3}{4}+ax-cx-dx+2n)\hfill \\ & +(\frac{1}{4}+bx+n)(\frac{1}{2}-cx+n)(\frac{1}{4}-dx+n).\hfill \end{array}$$

From this formula, we can derive several infinite series identities by extracting the coefficients ${\mathrm{D}}_m(a,b,c,d)$ of $x^m$. Some of them are recorded as follows.

• D₀(a,b,c,d)

  $$\pi =\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)(19+80n+80n^2)}{{(-64)}^n(1+2n){(1+4n)}_3}.$$
• D₁(0,0,1,0)

  $$\pi ln2=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)(1+4n)}{{(-64)}^n{(1+2n)}^2(3+4n)}\left\{\frac{13+20n}{2+4n}-\frac{19+80n+80n^2}{{(1+4n)}^2}{\mathbf{O}}_n\right\}.$$
• D₁(1,0,1,0)

  $$6\pi ln2=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)(1+4n)}{{(-64)}^n(3+4n)}\left\{\frac{(5+8n)}{{(1+2n)}^3(3+4n)}+\frac{2(19+80n+80n^2)}{{(1+2n)}^2{(1+4n)}^2}{\mathbf{O}}_{2n+1}\right\}.$$
• D₂(1,1,0,1)

  $$\frac{{\pi }^3}{12}=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)}{{(-64)}^n}\frac{(19+80n+80n^2)({\mathbf{O}}_{n+1}^{\langle 2\rangle }-4{\mathbf{O}}_{2n}^{\langle 2\rangle })-4}{(1+2n){(1+4n)}_3}.$$
• D₂(0,0,1,0)

  $$\frac{{\pi }^3-4\pi {ln}^{2}2}{8}=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{4n}{2n}\right)(1+4n)}{{(-64)}^n(3+4n)}\left\{\frac{19+28n}{2{(1+2n)}^4}+\frac{13+20n}{{(1+2n)}^3}{\mathbf{O}}_n-\frac{19+80n+80n^2}{{(1+2n)}^2{(1+4n)}^2}{\mathbf{O}}_n^2\right\}.$$

3. Identities from Transformation $\Psi (\mathit{a};\mathit{b},\mathit{c},\mathit{d},\mathit{e})$

Analogously, in this section we are going to show further evaluations of infinite series involving harmonic numbers and binomial/multinomial coefficients mainly by applying the transformation $\Psi (a;b,c,d,e)$. For five complex parameters $\{a,b,c,d,e\}$ in $\Psi (a;b,c,d,e)$, by letting $e=a$ and assuming the condition $\Re (1+a-b-c-d)>0$, we derive, again by making use of Dougall’s summation theorem for the well–poised ${}_5{F}_4$-series (cf. Bailey [15] §4.4), the following summation formula.

Lemma 6.

Let $\{a,b,c,d\}$ be four complex parameters such that the series below is well–defined. Then there holds

$$\begin{array}{cc}& \Gamma \left[\begin{array}{c}1+a-b,1+a-c,1+a-d,1+a-b-c-d\\ a,1+a-b-c,1+a-b-d,1+a-c-d\end{array}\right]\hfill \\ & =\sum _{n=0}^{\infty }{(-1)}^n\frac{{(1-b)}_n{(1-c)}_n{(1-d)}_n}{{(1+a-b-c-d)}_{2n+2}}{\mathrm{V}}_n(a,b,c,d)\hfill \\ & \times \frac{{(1+a-b-c)}_n{(1+a-b-d)}_n{(1+a-c-d)}_n}{n!{(1+a-b)}_n{(1+a-c)}_n{(1+a-d)}_n},\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {\mathrm{V}}_n(a,b,c,d)& =n(1+2a-b-c-d+2n)(2+a-b-c-d+2n)\hfill \\ & +(1+a-b-c+n)(1+a-b-d+n)(1+a-c-d+n).\hfill \end{array}$$

We remark that the parameter restriction $\Re (1+a-b-c-d)>0$ has been removed by analytic continuation for the above formula. In fact, it is valid in the whole complex space of dimension 4 except for the hyperplanes determined by

$$\left\{b-a\in \mathbb{N}\right\}\cup \left\{c-a\in \mathbb{N}\right\}\cup \left\{d-a\in \mathbb{N}\right\}\cup \left\{b+c+d-a\in \mathbb{N}\right\}.$$

3.1. Series with $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$ Again

Performing further the parameter replacements

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

the summation formula in in Lemma 6 becomes the following one:

Theorem 7.

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

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

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

$$\begin{array}{cc}\hfill {\mathrm{Q}}_n(a,b,c,d;x)=n(1+ax-bx-cx-dx+2n)(1+2ax-bx-cx-dx+2n)& \\ \hfill +(\frac{1}{2}+ax-bx-cx+n)(1+ax-bx-dx+n)(\frac{1}{2}+ax-cx-dx+n)& .\hfill \end{array}$$

According to the equation corresponding to the coefficient ${\mathcal{A}}_m(a,b,c,d)$ of $x^m$ across the equation in Theorem 7, we can directly show the first three identities below by assigning specific numeric values to $\{a,b,c,d\}$. Instead, the remaining series are evaluated, as done in Section 2.2, by constructing and then resolving linear systems of equations obtained by refining the equation corresponding to the coefficient ${\mathcal{A}}_m(a,b,c,d)$ for each fixed nonnegative integer m.

• 𝒜₁(a,b,c,d)

  $$4ln2=3+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)(1+5n)}{n(2n+1){(-16)}^n}.$$
• 𝒜₂(3,0,−1,0)

  $$\frac{7{\pi }^2-24{ln}^{2}2}{6}=10+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{1+4n+24n^2}{2n^2{(2n+1)}^2}+\frac{(1+5n){\mathbf{H}}_n}{n(2n+1)}\right\}.$$
• 𝒜₂(1,0,1,0)

  $$\frac{{\pi }^2+24{ln}^{2}2}{12}=2+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{4n^2-4n-1}{4n^2{(2n+1)}^2}+\frac{(1+5n){\mathbf{O}}_n}{n(2n+1)}\right\}.$$
• Resolving linear system 𝒜₃(a,b,c,d)

  $${\pi }^{2}ln2=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{7}{{(2n+1)}^3}+\frac{10{\mathbf{O}}_n}{{(2n+1)}^2}\right\}.$$
• Resolving linear system 𝒜₃(a,b,c,d)

  $$28\zeta \left(3\right)-{\pi }^{2}ln2=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{27}{{(2n+1)}^3}+\frac{5{\mathbf{H}}_n}{{(2n+1)}^2}\right\}.$$
  Linear combinations of the above two series result in the next two series.

  $$\begin{array}{cc}& 14\zeta \left(3\right)=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{17}{{(2n+1)}^3}+\frac{5{\mathbf{H}}_{2n}}{{(2n+1)}^2}\right\}.\hfill \\ & 14\zeta \left(3\right)-{\pi }^{2}ln2=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{10}{{(2n+1)}^3}-\frac{5{\overline{\mathbf{H}}}_{2n}}{{(2n+1)}^2}\right\}.\hfill \end{array}$$
• Resolving linear system 𝒜₃(a,b,c,d)

  $$30\zeta \left(3\right)=36+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{36}{{(2n+1)}^3}-\frac{(1+5n){\mathbf{H}}_n^{\langle 2\rangle }}{n(2n+1)}\right\}.$$
• Resolving linear system 𝒜₃(a,b,c,d)

  $$\frac{7\zeta \left(3\right)}{2}=4+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{4}{{(2n+1)}^3}-\frac{(1+5n){\mathbf{O}}_n^{\langle 2\rangle }}{n(2n+1)}\right\}.$$
• Resolving linear system 𝒜₄(a,b,c,d)

  $$\frac{7{\pi }^4}{10}=68+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{68}{{(2n+1)}^4}-\frac{5{\mathbf{H}}_n^{\langle 2\rangle }}{{(2n+1)}^2}-\frac{(1+5n){\mathbf{H}}_n^{\langle 3\rangle }}{n(2n+1)}\right\}.$$
• Resolving linear system 𝒜₄(a,b,c,d)

  $$\frac{3{\pi }^4}{16}=18+\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{18}{{(2n+1)}^4}-\frac{5{\mathbf{O}}_n^{\langle 2\rangle }}{{(2n+1)}^2}-\frac{(1+5n){\mathbf{O}}_n^{\langle 3\rangle }}{n(2n+1)}\right\}.$$
• Resolving linear system 𝒜₄(a,b,c,d)

  $$\frac{5{\pi }^4}{24}+{\pi }^2{ln}^{2}2=\sum _{n=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{25}{{(2n+1)}^4}+\frac{14{\mathbf{O}}_n}{{(2n+1)}^3}+\frac{10\left({\mathbf{O}}_n^2-2{\mathbf{O}}_n^{\langle 2\rangle }\right)}{{(2n+1)}^2}\right\}.$$
• Resolving linear system 𝒜₄(a,b,c,d)

  $$\frac{{\pi }^4}{8}-14\zeta \left(3\right)ln2=\sum _{n=1}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}{{(-16)}^n}\left\{\frac{8{\mathbf{H}}_n}{{(2n+1)}^3}-\frac{{\mathbf{O}}_n^{\langle 2\rangle }}{n^2}-\frac{2(1+5n){\mathbf{H}}_n{\mathbf{O}}_n^{\langle 2\rangle }}{n(2n+1)}\right\}.$$

3.2. Series with ${\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^{-1}$ Again

Performing further the parameter replacements

$$a=1+ax,b=1+bx,c=1+cx,d=1+dx,$$

the summation formula in Lemma 6 becomes the following one:

Theorem 8.

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

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

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

$$\begin{array}{cc}\hfill {\mathrm{Q}}_n(a,b,c,d;x)& =n(ax-bx-cx-dx+2n)(2ax-bx-cx-dx+2n)\hfill \\ & +(ax-bx-cx+n)(ax-bx-dx+n)(ax-cx-dx+n).\hfill \end{array}$$

According to (1)–(4), the coefficients ${\mathcal{B}}_m(a,b,c,d)$ of $x^m$ can be extracted across the above equation, which leads to the following interesting infinite series identities.

• $ℬ$₃(a,b,c,d) Apéry series (cf. [3,9,16,18])

  $$\frac{2}{5}\zeta \left(3\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^3\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}.$$
• $ℬ$₄(1,1,−1,1) Conjectured by Sun [13] (Equation 3.13) and proved by Chu [3]

  $$\frac{{\pi }^4}{30}=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^4\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{10n{\mathbf{H}}_n-3\right\}.$$
• $ℬ$₄(3,−1,1,−1)

  $$\frac{7{\pi }^4}{30}=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^4\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{27+20n{\mathbf{O}}_n\right\}.$$
  By linear combinations, we deduce two further equivalent series.

  $$\begin{array}{cc}& \frac{{\pi }^4}{15}=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^4\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{6+5n{\mathbf{H}}_{2n}\right\}.\hfill \\ & \frac{{\pi }^4}{50}=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^4\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{3+2{\overline{\mathbf{H}}}_{2n}\right\}.\hfill \end{array}$$
• $ℬ$₅(0,1,−2,1)

  $$2\zeta \left(5\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^5\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{9-5n^2{\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$

  (8)
  This series was first evaluated by Koecher [19] (see also [3,7,16,18]). By Mathematica, we detected (but failed to prove) the following counterpart series

  $$\frac{56{\pi }^2}{3}\zeta \left(3\right)-166\zeta \left(5\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^5\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{119-20n^2{\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$
  The above two evaluations may serve as refinements of a conjectured formula made by Sun [20] (Conjecture 2.1).
• $ℬ$₅(3,2,−1,2) [3] (Example 3.6)

  $$4\zeta \left(5\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^5\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{5-6n{\mathbf{H}}_n+10n^2{\mathbf{H}}_n^2\right\}.$$
• $ℬ$₆(0,1,−2,1) Conjectured by Sun [13] (Equation 4.5) and verified by Chu [3]
• $$2\zeta {\left(3\right)}^2=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^6\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{1+5n^3{\mathbf{H}}_n^{\langle 3\rangle }\right\}.$$

  (9)
• $ℬ$₇(0,1,−2,1) Chu [3] (Example 3.16)

  $$4\zeta \left(7\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^7\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{26-18n^2{\mathbf{H}}_n^{\langle 2\rangle }+5n^4{\left({\mathbf{H}}_n^{\langle 2\rangle }\right)}^2-5n^4{\mathbf{H}}_n^{\langle 4\rangle }\right\}.$$
• $ℬ$₈(0,1,−2,1)

  $$20\zeta \left(3\right)\zeta \left(5\right)=\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^8\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}\left\{1+(9-5n^2{\mathbf{H}}_n^{\langle 2\rangle })(1+5n^3{\mathbf{H}}_n^{\langle 3\rangle })+25n^5{\mathbf{H}}_n^{\langle 5\rangle }\right\}.$$

3.3. Series with ${\left(\genfrac{}{}{0pt}{}{4n}{2n,n,n}\right)}^{-1}$

As done in Section 2.2, we can directly deduce the following two identities from transformation $\Phi [1+2x,\frac{1}{2}+x,\frac{1}{2}+x,\frac{1}{2}+x,\frac{1}{2}+x]$.

• $\boxed{\left[x^0\right]\Phi [1+2x,\frac{1}{2}+x,\frac{1}{2}+x,\frac{1}{2}+x,\frac{1}{2}+x]}$

  $$\frac{7\zeta \left(3\right)}{16}=\sum _{n=0}^{\infty }\frac{{(-16)}^n\left(2n\right)!n{!}^2}{(4+4n)!(2n+1)}\left\{13+32n+20n^2\right\}.$$
• $\boxed{\left[x^1\right]\Phi [1+2x,\frac{1}{2}+x,\frac{1}{2}+x,\frac{1}{2}+x,\frac{1}{2}+x]}$

  $$\frac{{\pi }^4}{16}=\sum _{n=0}^{\infty }\frac{{(-16)}^n\left(2n\right)!n{!}^2}{(3+4n)!}\left\{\frac{4(5+6n)}{{(2n+1)}^2}+\frac{(13+32n+20n^2)}{(n+1)(2n+1)}{\mathbf{O}}_{2n+2}\right\}.$$

Next under the parameter setting

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

the summation formula in Lemma 6 becomes the following one:

Theorem 9.

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

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

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

$$\begin{array}{cc}\hfill {\mathrm{Q}}_n(a,b,c,d;x)=n(\frac{3}{2}+ax-bx-cx-dx+2n)(\frac{3}{2}+2ax-bx-cx-dx+2n)& \\ \hfill +(1+ax-bx-cx+n)(1+ax-bx-dx+n)(1+ax+cx-dx+n)& .\hfill \end{array}$$

Analogously, a number of infinite series identities can be derived by extracting the coefficients ${\mathcal{C}}_m(a,b,c,d)$ of $x^m$ from this equation. Some remarkable ones are displayed as follows:

• $𝒞$₀(a,b,c,d)

  $$\frac{{\pi }^2}{16}=\sum _{n=0}^{\infty }\frac{\left(2n\right)!n{!}^2{(-16)}^n}{(3+4n)!}\left\{4+5n\right\}.$$
• $𝒞$₁(2,0,1,1)

  $$\frac{{\pi }^{2}ln2}{4}=\sum _{n=0}^{\infty }\frac{\left(2n\right)!n{!}^2{(-16)}^n}{(3+4n)!}\left\{\frac{12+14n}{1+2n}-(4+5n)({\mathbf{H}}_n-6{\mathbf{O}}_n)\right\}.$$
• $𝒞$₁(2,1,1,1)

  $$\frac{{\pi }^{2}ln2}{8}=\sum _{n=0}^{\infty }\frac{\left(2n\right)!n{!}^2{(-16)}^n}{(3+4n)!}\left\{\frac{12+15n}{1+2n}+(4+5n)(6{\mathbf{O}}_n-{\mathbf{O}}_{2n+2})\right\}.$$
• $𝒞$₂(0,−2,1,1)

  $$\frac{{\pi }^4}{48}=\sum _{n=0}^{\infty }\frac{\left(2n\right)!n{!}^2{(-16)}^n}{(3+4n)!}\left\{\frac{12+16n}{{(1+2n)}^2}-(4+5n){\mathbf{H}}_n^{\langle 2\rangle }\right\}.$$
• $𝒞$₃(0,−2,1,1)

  $$\frac{{\pi }^2\zeta \left(3\right)}{2}=\sum _{n=0}^{\infty }\frac{\left(2n\right)!n{!}^2{(-16)}^n}{(3+4n)!}\left\{\frac{12(3+4n)}{{(1+2n)}^3}+(4+5n){\mathbf{H}}_n^{\langle 3\rangle }\right\}.$$

3.4. Series with $(\genfrac{}{}{0pt}{}{2n}{n})/(\genfrac{}{}{0pt}{}{4n}{2n})$

First, we state two identities implied in the power series expansions across transformation $\Phi [2+2x,1+x,1+x,1+x,\frac{3}{2}+x]$.

• $\boxed{\left[x^0\right]\Phi [2+2x,1+x,1+x,1+x,\frac{3}{2}+x]}$

  $$\frac{{\pi }^2}{12}=\sum _{n=0}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(3+4n)!n{!}^2}\frac{5+14n+10n^2}{{(1+n)}^2}.$$
• $\boxed{\left[x^1\right]\Phi [2+2x,1+x,1+x,1+x,\frac{3}{2}+x]}$

  $$\frac{\zeta \left(3\right)}{2}=\sum _{n=0}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(3+4n)!n{!}^2}\left\{\frac{5+14n+10n^2}{{(1+n)}^2}{\mathbf{O}}_{2n+1}-\frac{{(2n+1)}^3(4n+5)}{{(1+n)}^3(4n+3)}\right\}.$$

Then under the parameter setting

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

the summation formula in Lemma 6 becomes the following one:

Theorem 10.

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

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

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

$$\begin{array}{cc}\hfill {\mathrm{Q}}_n(a,b,c,d;x)& =n(\frac{1}{2}+ax-bx-cx-dx+2n)(\frac{1}{2}+2ax-bx-cx-dx+2n)\hfill \\ & +(\frac{1}{2}+ax-bx-cx+n)(ax-bx-dx+n)(\frac{1}{2}+ax-cx-dx+n).\hfill \end{array}$$

From the last formula, we can deduce several infinite series identities by extracting the coefficients ${\mathcal{D}}_m(a,b,c,d)$ of $x^m$. Some of them are recorded below as examples.

• $𝒟$₂(a,b,c,d)

  $$\frac{{\pi }^2}{12}=1+\sum _{n=1}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(2+4n)!n{!}^2}\left\{\frac{1+6n+10n^2}{n^2}\right\}.$$
• $𝒟$₃(0,1,0,1)

  $$\frac{3}{4}\zeta \left(3\right)=1-\sum _{n=1}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(2+4n)!n{!}^2}\left\{\frac{1+8n+20n^2+12n^3}{2(2n+1)n^3}-\frac{1+6n+10n^2}{n^2}{\mathbf{H}}_n\right\}.$$
• $𝒟$₃(2,−1,4,−1)

  $$\frac{5}{8}\zeta \left(3\right)=\frac{1}{2}-\sum _{n=1}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(2+4n)!n{!}^2}\left\{\frac{1+8n+20n^2+14n^3}{2(2n+1)n^3}+\frac{1+6n+10n^2}{n^2}{\mathbf{O}}_n\right\}.$$
• $𝒟$₃(2,−1,2,−1)

  $$\frac{\zeta \left(3\right)}{2}=1+\sum _{n=1}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(2+4n)!n{!}^2}\left\{\frac{1+5n+6n^2}{n^3}+\frac{1+6n+10n^2}{n^2}{\mathbf{O}}_{2n+1}\right\}.$$
• $𝒟$₄(0,−1,0,1)

  $$\frac{{\pi }^4}{720}=\sum _{n=1}^{\infty }\frac{{(-1)}^n\left(2n\right){!}^3}{(2+4n)!n{!}^2}\left\{\frac{1+6n+8n^2}{4n^4}-\frac{1+6n+10n^2}{n^2}{\mathbf{O}}_n^{\langle 2\rangle }\right\}.$$