New Generalizations of Two Ramanujan Series for \( 1/\pi \)

1. Introduction

Let ${\left(x\right)}_n={\prod }_{k=0}^{n-1}\left(x+k\right)$, for x a complex number and n a non-negative integer. Our main purpose in this note is to derive the following families of Ramanujan-like series for $1/\pi$:

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k}{k{!}^3}}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3\left(4k+1-2\mathcal{\ell }\right)}{{\prod }_{j=1}^{\mathcal{\ell }}\left(2k-2j+1\right){\prod }_{j=1}^m\left(2k-2j+1\right){\prod }_{j=1}^n\left(2k-2j+1\right){\prod }_{j=1}^{m-\mathcal{\ell }}\left(k+j\right){\prod }_{j=1}^{n-\mathcal{\ell }}\left(k+j\right)}}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{m+n}}{{\left(\frac{1}{2}\right)}_m{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{2}\right)}_{m+n-\mathcal{\ell }}{2}^{\mathcal{\ell }+m+n-1}}}{\displaystyle \frac{1}{\pi }}\hfill \end{array}$$

(1)

and

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}{\displaystyle \frac{{\prod }_{j=1}^{2n}\left(2k+j\right)}{{\prod }_{j=1}^n{\left(k+j\right)}^2\left(2k-2j+1\right)}}\left(6k+1+2n\right)={\displaystyle \frac{{(-1)}^n{2}^{n+2}}{{\left(\frac{1}{2}\right)}_n}}{\displaystyle \frac{1}{\pi }};$$

(2)

valid for ℓ, m, and n non-negative integers.

Identities (1) and (2) are respective generalizations of the Ramanujan series [10]:

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

(3)

and

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

(4)

When $\mathcal{\ell }=0=m$, identity (1) reduces to

$${\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1\right)}{{\prod }_{j=1}^n\left(2k-2j+1\right)\left(k+j\right)}}={\displaystyle \frac{{(-1)}^n}{{\left(\frac{1}{2}\right)}_n^2{2}^{n-1}}}{\displaystyle \frac{1}{\pi }},$$

which was derived by Levrie [9, Theorem 7] using Fourier-Legendre theory. In fact we will see that identities (1) and (2) are particular cases of more general results.

Let $H_j$ be the $j^{\mathrm{th}}$ harmonic number and let $O_j$ be the $j^{\mathrm{th}}$ odd harmonic number. We will also derive the following series involving harmonic and odd harmonic numbers:

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k{\left(\frac{1}{2}\right)}_k^3\left(4k+1\right)\left(2O_{k-n}+H_{k+n}\right)}{k{!}^3{\prod }_{j=1}^n\left(k+j\right)\left(2k-2j+1\right)}}={\displaystyle \frac{{(-1)}^n}{2^{n-2}}}{\displaystyle \frac{1}{{\left(\frac{1}{2}\right)}_n^2}}\left(2O_n-ln2\right){\displaystyle \frac{1}{\pi }}$$

(5)

and

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}{\displaystyle \frac{{\prod }_{j=1}^{2n}\left(2k+j\right)}{{\prod }_{j=1}^n{\left(k+j\right)}^2\left(2k-2j+1\right)}}\left(\left(6k+1+2n\right)\left(2O_{k-n}-2O_{k+n}+H_{k+n}\right)-2\right)\hfill \\ \hfill & ={(-1)}^n{\displaystyle \frac{2^{n+3}}{{\left(\frac{1}{2}\right)}_n}}\left(O_n-ln2\right){\displaystyle \frac{1}{\pi }}.\hfill \end{array}$$

(6)

Identity (5) at $n=0$ reduces to the harmonic number complement of (3):

$${\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}\left(4k+1\right)H_{2k}=-{\displaystyle \frac{2ln2}{\pi }},$$

(7)

derived by Hou et al. [8] and also Campbell [4].

The special case $n=0$ of identity (6), namely,

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}\left(\left(6k+1\right)H_k-2\right)=-{\displaystyle \frac{8ln2}{\pi }},$$

(8)

was derived by Hou et al. [8].

The now famous identity (3) was discovered in 1859 by Bauer [2] via a Fourier-Legendre expansion, and rediscovered in 1914 by Ramanujan who included it in his famous first letter to G. H. Hardy. Early proofs relied on classical techniques like swapping limit operations while later approaches utilized advanced tools like Dougall’s bilateral hypergeometric theorem. For a survey of the Bauer-Ramanujan series (3) and related results and analyses, the reader is referred to the recent article by Campbell and Levrie [3].

Let $\Gamma \left(z\right)$ be Euler’s Gamma function. Generalized binomial coefficients are defined for complex numbers r and s by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{r}{s}}\right)={\displaystyle \frac{\Gamma (r+1)}{\Gamma (s+1)\Gamma (r-s+1)}}.$$

(9)

The extended Pochhammer relation ${\left(x\right)}_y$ is defined for complex numbers x and y that are not non-positive integers and such that $x+y\ne 0$ by

$${\left(x\right)}_y={\displaystyle \frac{\Gamma \left(x+y\right)}{\Gamma \left(x\right)}}=\left({\displaystyle \genfrac{}{}{0pt}{}{x+y-1}{y}}\right)\Gamma \left(y+1\right).$$

(10)

If, in particular, $y=n$ is a non-negative integer, then we have the Pochhammer symbol

$${\left(x\right)}_n={\displaystyle \prod _{k=0}^{n-1}}\left(x+k\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{x+n-1}{n}}\right)n!$$

(11)

Harmonic numbers, $H_j$, and odd harmonic numbers, $O_j$, are defined for non-negative integers j by

$$H_j={\displaystyle \sum _{k=1}^j}{\displaystyle \frac{1}{k}},O_j={\displaystyle \sum _{k=1}^j}{\displaystyle \frac{1}{2k-1}},H_0=0=O_0.$$

(12)

The identity

$$H_{n-1/2}=2O_n-2ln2,$$

(13)

extends harmonic numbers to half-integer arguments and is a consequence of the link between harmonic numbers, odd harmonic numbers and the digamma function.

2. Required Results

Lemma 1. 

If k and x are complex numbers, then

$$\left({\displaystyle \genfrac{}{}{0pt}{}{k}{x+1/2}}\right)={\displaystyle \frac{2}{\pi }}{\displaystyle \frac{1}{\left(2x+1\right)}}{\displaystyle \frac{x!k!}{{\left(\frac{1}{2}\right)}_x{\left(\frac{1}{2}\right)}_k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k-1/2}{x}}\right).$$

(14)

Proof. 

The identity

$$\left({\displaystyle \genfrac{}{}{0pt}{}{a-b}{c}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{a-c}{b}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{a}{c}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{a}{b}}\right)}^{-1},$$

(15)

with $a=k$, $b=1/2$ and $c=k-1/2-x$ gives

$$\left({\displaystyle \genfrac{}{}{0pt}{}{k-1/2}{x}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{x+1/2}{1/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k}{x+1/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1/2}}\right)}^{-1};$$

(16)

and hence (14) since

$$\left({\displaystyle \genfrac{}{}{0pt}{}{x+1/2}{1/2}}\right)={\displaystyle \frac{2x+1}{x!}}{\left({\displaystyle \frac{1}{2}}\right)}_x$$

(17)

and

$$\left({\displaystyle \genfrac{}{}{0pt}{}{k}{1/2}}\right)={\displaystyle \frac{2}{\pi }}{\displaystyle \frac{k!}{{\left(\frac{1}{2}\right)}_k}}.$$

(18)

□

Lemma 2. 

If x is a complex number that is not a negative integer, then

$${\left(-x+{\displaystyle \frac{1}{2}}\right)}_k={\displaystyle \frac{{\left(\frac{1}{2}\right)}_x{\left(\frac{1}{2}\right)}_k}{x!}}{\left({\displaystyle \genfrac{}{}{0pt}{}{k-1/2}{x}}\right)}^{-1}cos\left(\pi x\right)$$

(19)

and

$${\left(x+{\displaystyle \frac{1}{2}}\right)}_k={\left({\displaystyle \frac{1}{2}}\right)}_k\left({\displaystyle \genfrac{}{}{0pt}{}{2k+2x}{2x}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+x}{x}}\right)}^{-1}.$$

(20)

In particular, if n is a non-negative integer, then

$${\left(-n+{\displaystyle \frac{1}{2}}\right)}_k={(-1)}^n{\displaystyle \frac{2^n{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{2}\right)}_k}{{\prod }_{j=1}^n\left(2k-2j+1\right)}}$$

(21)

and

$${\left(n+{\displaystyle \frac{1}{2}}\right)}_k={\displaystyle \frac{1}{2^{2n}}}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k}{{\left(\frac{1}{2}\right)}_n}}{\displaystyle \frac{{\prod }_{j=1}^{2n}\left(2k+j\right)}{{\prod }_{j=1}^n\left(k+j\right)}}.$$

(22)

Proof. 

We prove (19).

$${\left(-x+{\displaystyle \frac{1}{2}}\right)}_k=k!\left({\displaystyle \genfrac{}{}{0pt}{}{k-x-1/2}{k}}\right).$$

(23)

Now, using (15) with $a=k$, $b=x+1/2$ and $c=k$, write

$$\left({\displaystyle \genfrac{}{}{0pt}{}{k-x-1/2}{k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{0}{x+1/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k}{x+1/2}}\right)}^{-1}={\displaystyle \frac{2}{\pi }}{\displaystyle \frac{cos\left(\pi x\right)}{2x+1}}{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{x+1/2}}\right)}^{-1},$$

(24)

since

$$\left({\displaystyle \genfrac{}{}{0pt}{}{0}{s}}\right)=\left\{\begin{array}{cc}{\displaystyle \frac{sin\left(\pi s\right)}{\pi s}},\hfill & \mathrm{if}s\ne 0;\hfill \\ 1,\hfill & \mathrm{if}s=0.\hfill \end{array}\right.$$

(25)

Identity (19) now follows by the use of (14). □

Lemma 3. 

We have

$${\displaystyle \frac{d}{dx}}{\left(x\right)}_y={\left(x\right)}_y\left(H_{x+y-1}-H_{x-1}\right).$$

(26)

3. Generalizations of Ramanujan Series for $1/\mathit{\pi }$

Lemma 4. 

If k is a non-negative integer, then

$$\underset{d\to \infty }{lim}{\displaystyle \frac{{\left(d\right)}_k}{{\left(-d\right)}_k}}={(-1)}^k.$$

(27)

Proof. 

Now

$${\left(-d\right)}_k=k!\left({\displaystyle \genfrac{}{}{0pt}{}{-d+k-1}{k}}\right)={(-1)}^{k}k!\left({\displaystyle \genfrac{}{}{0pt}{}{d}{k}}\right).$$

(28)

But

$${\left(d\right)}_k=k!\left({\displaystyle \genfrac{}{}{0pt}{}{d+k-1}{k}}\right)\sim k!\left({\displaystyle \genfrac{}{}{0pt}{}{d}{k}}\right)\mathrm{as}d\to \infty .$$

(29)

Hence the result. □

Lemma 5. 

If a and k are complex numbers, then

$${\displaystyle \frac{{\left(1+\frac{a}{2}\right)}_k}{{\left(\frac{a}{2}\right)}_k}}={\displaystyle \frac{2k+a}{a}}.$$

(30)

Lemma 6. 

For suitably bounded complex numbers a, b, and c, the following identity holds:

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k}{k!}}{\displaystyle \frac{{\left(a\right)}_k{\left(b\right)}_k{\left(c\right)}_k}{{\left(1+a-b\right)}_k{\left(1+a-c\right)}_k}}\left(2k+a\right)={\displaystyle \frac{a\Gamma \left(1+a-b\right)\Gamma \left(1+a-c\right)}{\Gamma \left(1+a\right)\Gamma \left(1+a-b-c\right)}}.$$

(31)

In particular,

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k}{k{!}^2}}{\displaystyle \frac{{\left(a\right)}_k^2{\left(c\right)}_k}{{\left(1+a-c\right)}_k}}\left(2k+a\right)={\displaystyle \frac{a\Gamma \left(1+a-c\right)}{\Gamma \left(1+a\right)\Gamma \left(1-c\right)}}$$

(32)

and

$${\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(a\right)}_k^3}{k{!}^3}}\left(2k+a\right)={\displaystyle \frac{sin\left(\pi a\right)}{\pi }}.$$

(33)

Proof. 

Consider Dougall’s well-poised hypergeometric series [1, p. 27]:

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{1}{k!}}{\displaystyle \frac{{\left(a\right)}_k{\left(1+a/2\right)}_k{\left(b\right)}_k{\left(c\right)}_k{\left(d\right)}_k}{{\left(a/2\right)}_k{\left(1+a-b\right)}_k{\left(1+a-c\right)}_k{\left(1+a-d\right)}_k}}\hfill \\ \hfill & ={\displaystyle \frac{\Gamma \left(1+a-b\right)\Gamma \left(1+a-c\right)\Gamma \left(1+a-d\right)\Gamma \left(1+a-b-c-d\right)}{\Gamma \left(1+a\right)\Gamma \left(1+a-b-c\right)\Gamma \left(1+a-b-d\right)\Gamma \left(1+a-c-d\right)}},\hfill \end{array}$$

(34)

which holds for suitably bounded complex numbers a, b, c and d or $\Re (1+a-b-c-d)>0$ if each of the four numbers has a finite magnitude.

Letting d approach infinity in (34) gives (31) by application of Lemmata 4 and 5. □

Remark 1. 

Campbell [4] also derived (32) using Mishev’s transform. Identity (33) is Dougall’s identity [6, Equation (16)].

The identity stated in Lemma 7 corresponds to setting $a=1/2=b$ in (31) and was derived by Hou et al. [8].

Lemma 7. 

If c is a complex number, then

$${\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^2{\left(c\right)}_k}{k{!}^2{\left(3/2-c\right)}_k}}\left(4k+1\right)={\displaystyle \frac{2}{\sqrt{\pi }}}{\displaystyle \frac{\Gamma \left(3/2-c\right)}{\Gamma \left(1-c\right)}}={\displaystyle \frac{2}{\sqrt{\pi }}}{(1-c)}_{1/2}.$$

(35)

Remark 2. 

The special case of (35) with c a non-negative integer was proved by Ekhad and Zeilberger [7] using the $WZ$ technique.

Theorem 1. 

If p, q, and r are complex numbers that are not half-integers and such that $q-p$ and $r-p$ are not non-positive integers, then

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1-2p\right)}{\left(\genfrac{}{}{0pt}{}{k-1/2}{p}\right)\left(\genfrac{}{}{0pt}{}{k-1/2}{q}\right)\left(\genfrac{}{}{0pt}{}{k-1/2}{r}\right)\left(\genfrac{}{}{0pt}{}{k+q-p}{k}\right)\left(\genfrac{}{}{0pt}{}{k+r-p}{k}\right)}}\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & =2{\displaystyle \frac{{\left(\frac{1}{2}-p\right)}_{q+\frac{1}{2}}}{{\left(r-p\right)}_{q+\frac{1}{2}}}}{\displaystyle \frac{p!q!r!}{{\left(\frac{1}{2}\right)}_p{\left(\frac{1}{2}\right)}_q{\left(\frac{1}{2}\right)}_r}}{\displaystyle \frac{r-p}{cos\left(\pi p\right)cos\left(\pi q\right)cos\left(\pi r\right)}}\hfill \end{array}$$

(37)

In particular,

$${\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1\right)}{\left(\genfrac{}{}{0pt}{}{k+r}{k}\right)\left(\genfrac{}{}{0pt}{}{k-1/2}{r}\right)}}={\displaystyle \frac{2}{\pi }}{\displaystyle \frac{r{!}^2}{{\left(\frac{1}{2}\right)}_r^2}}{\displaystyle \frac{1}{cos\left(\pi r\right)}}.$$

(38)

Proof. 

Set $a=1/2-p$, $b=1/2-q$ and $c=1/2-r$ in (31) to obtain

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k}{k!}}{\displaystyle \frac{{\left(1/2-p\right)}_k{\left(1/2-q\right)}_k{\left(1/2-r\right)}_k}{{\left(1-p+q\right)}_k{\left(1-p+r\right)}_k}}\left(4k+1-p\right)\hfill \\ \hfill & ={\displaystyle \frac{2\Gamma \left(1-p+q\right)\Gamma \left(1-p+r\right)}{\Gamma \left(1/2-p\right)\Gamma \left(1/2-p+q+r\right)}}.\hfill \end{array}$$

(39)

Use (19) and note from (10) that

$${\left(1+x\right)}_k=k!\left({\displaystyle \genfrac{}{}{0pt}{}{x+k}{k}}\right),$$

(40)

$${\displaystyle \frac{\Gamma \left(1-p+q\right)}{\Gamma \left(\frac{1}{2}-p\right)}}={\left({\displaystyle \frac{1}{2}}-p\right)}_{{\displaystyle \frac{1}{2}}+q},$$

(41)

and

$${\displaystyle \frac{\Gamma \left(1-p+r\right)}{\Gamma \left(\frac{1}{2}-p+q+r\right)}}={\displaystyle \frac{r-p}{{\left(r-p\right)}_{\frac{1}{2}+q}}}.$$

(42)

□

Corollary 2. 

If ℓ, m and n are non-negative integers, then

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k}{k{!}^3}}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3\left(4k+1-2\mathcal{\ell }\right)}{{\prod }_{j=1}^{\mathcal{\ell }}\left(2k-2j+1\right){\prod }_{j=1}^m\left(2k-2j+1\right){\prod }_{j=1}^n\left(2k-2j+1\right){\prod }_{j=1}^{m-\mathcal{\ell }}\left(k+j\right){\prod }_{j=1}^{n-\mathcal{\ell }}\left(k+j\right)}}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{m+n}}{{\left(\frac{1}{2}\right)}_m{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{2}\right)}_{m+n-\mathcal{\ell }}{2}^{\mathcal{\ell }+m+n-1}}}{\displaystyle \frac{1}{\pi }}.\hfill \end{array}$$

In particular,

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1\right)}{{\prod }_{j=1}^n\left(2k-2j+1\right)\left(k+j\right)}}& ={\displaystyle \frac{{(-1)}^n}{{\left(\frac{1}{2}\right)}_n^2{2}^{n-1}}}{\displaystyle \frac{1}{\pi }},\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1-2n\right)}{{\prod }_{j=1}^n{\left(2k-2j+1\right)}^3}}& ={\displaystyle \frac{1}{{\left(\frac{1}{2}\right)}_n^3{2}^{3n-1}}}{\displaystyle \frac{1}{\pi }},\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1-2m\right)}{{\prod }_{j=1}^m{\left(2k-2j+1\right)}^2{\prod }_{j=1}^n\left(2k-2j+1\right){\prod }_{j=1}^{n-m}\left(k+j\right)}}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{m+n}}{{\left(\frac{1}{2}\right)}_m{\left(\frac{1}{2}\right)}_n^2{2}^{2m+n-1}}}{\displaystyle \frac{1}{\pi }},\hfill \end{array}$$

(45)

and

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1-2\mathcal{\ell }\right)}{{\prod }_{j=1}^{\mathcal{\ell }}\left(2k-2j+1\right){\prod }_{j=1}^n{\left(2k-2j+1\right)}^2{\prod }_{j=1}^{n-\mathcal{\ell }}{\left(k+j\right)}^2}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{\left(\frac{1}{2}\right)}_{2n-l}{\left(\frac{1}{2}\right)}_n^2{2}^{2n+\mathcal{\ell }-1}}}{\displaystyle \frac{1}{\pi }}.\hfill \end{array}$$

(46)

Proof. 

Set $p=\mathcal{\ell }$, $q=m$ and $r=n$ in (39) and use (21) and (40). Note that

$$n!\left({\displaystyle \genfrac{}{}{0pt}{}{k+n}{n}}\right)={\displaystyle \prod _{j=1}^n}\left(k+j\right),\mathrm{etc},$$

(47)

$$\Gamma \left({\displaystyle \frac{1}{2}}-u\right)={\displaystyle \frac{{(-1)}^u}{{\left(\frac{1}{2}\right)}_u}}{\displaystyle \frac{\sqrt{\pi }}{u!}},$$

(48)

and

$$\Gamma \left({\displaystyle \frac{1}{2}}+u\right)={\left({\displaystyle \frac{1}{2}}\right)}_u\sqrt{\pi }.$$

(49)

□

Theorem 3. 

If r is a complex number that is not a half-integer, then

$${\displaystyle \sum _{k=0}^{\infty }}{(-1)}^k{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(4k+1\right)\left(2O_{k-r}+H_{k+r}\right)}{\left(\genfrac{}{}{0pt}{}{k+r}{k}\right)\left(\genfrac{}{}{0pt}{}{k-1/2}{r}\right)}}={\displaystyle \frac{2}{\pi }}{\displaystyle \frac{r{!}^2}{{\left(\frac{1}{2}\right)}_r^2}}{\displaystyle \frac{\left(4O_r-2ln2\right)}{cos\left(\pi r\right)}}.$$

(50)

Proof. 

Differentiate (35) with respect to c and replace c with $1/2-r$ to get

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

(51)

from which (50) follows after using (13), (19), (40) and the fact that

$${\left(r+{\displaystyle \frac{1}{2}}\right)}_{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{\Gamma \left(r+1\right)}{\Gamma \left(r+\frac{1}{2}\right)}}={\displaystyle \frac{r!}{{\left(\frac{1}{2}\right)}_r\sqrt{\pi }}}.$$

(52)

□

Corollary 4. 

If n is a non-negative integer, then

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k{\left(\frac{1}{2}\right)}_k^3\left(4k+1\right)\left(2O_{k-n}+H_{k+n}\right)}{k{!}^3{\prod }_{j=1}^n\left(k+j\right)\left(2k-2j+1\right)}}={\displaystyle \frac{{(-1)}^n}{2^{n-2}}}{\displaystyle \frac{1}{{\left(\frac{1}{2}\right)}_n^2}}\left(2O_n-ln2\right){\displaystyle \frac{1}{\pi }}.$$

In particular,

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

Proof. 

Set $r=n$ a non-negative integer in (51). □

Lemma 8 was derived by Hou et al. [8] using a hypergeometric transformation formula due to Chu and Zhang [5, Theorem 9]; and facilitates the derivation of (2).

Lemma 8. 

If c is a complex number, then

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{1}{4^k}}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k{\left(c\right)}_k{\left(1-c\right)}_k}{k{!}^2{\left(3/2-c\right)}_k}}{\displaystyle \frac{3k-c+1}{2}}={\displaystyle \frac{1}{\sqrt{\pi }}}{(1-c)}_{1/2}.$$

(53)

Theorem 5. 

If r is a complex number that is not a half-integer, then

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2k+2r}{2r}\right)}{{\left(\genfrac{}{}{0pt}{}{k+r}{r}\right)}^2\left(\genfrac{}{}{0pt}{}{k-1/2}{r}\right)}}\left(6k+1+2r\right)={\displaystyle \frac{4}{\pi }}{\displaystyle \frac{r{!}^2}{{\left(\frac{1}{2}\right)}_r^2}}{\displaystyle \frac{1}{cos\left(\pi r\right)}}.$$

(54)

Proof. 

Set $c=1/2-r$ in (53) to obtain

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{1}{4^k}}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k{\left(\frac{1}{2}-r\right)}_k{\left(\frac{1}{2}+r\right)}_k}{k{!}^2{\left(r+1\right)}_k}}\left(6k+1+2r\right)={\displaystyle \frac{4}{\sqrt{\pi }}}{\left(r+{\displaystyle \frac{1}{2}}\right)}_{1/2}.$$

(55)

Now use (19), (20), (40) and (52). □

Corollary 6. 

If n is a non-negative integer, then

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}{\displaystyle \frac{{\prod }_{j=1}^{2n}\left(2k+j\right)}{{\prod }_{j=1}^n{\left(k+j\right)}^2\left(2k-2j+1\right)}}\left(6k+1+2n\right)={\displaystyle \frac{{(-1)}^n{2}^{n+2}}{{\left(\frac{1}{2}\right)}_n}}{\displaystyle \frac{1}{\pi }}.$$

In particular,

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

(56)

Proof. 

Set $r=n$ in (55) and use (21), (22), (40) and (52). □

Theorem 7. 

If r is a complex number that is not a half-integer, then

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{1}{4^k}}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{k{!}^3}}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2k+2r}{2r}\right)}{{\left(\genfrac{}{}{0pt}{}{k+r}{r}\right)}^2\left(\genfrac{}{}{0pt}{}{k-1/2}{r}\right)}}\left(\left(6k+2r+1\right)\left(H_{k-r-1/2}-H_{k+r-1/2}+H_{k+r}\right)-2\right)\hfill \\ \hfill & ={\displaystyle \frac{r{!}^2}{{\left(\frac{1}{2}\right)}_r^2}}{\displaystyle \frac{H_{-r-1/2}}{cos\left(\pi r\right)}}{\displaystyle \frac{4}{\pi }}.\hfill \end{array}$$

(57)

Proof. 

Differentiate (53) with respect to c to obtain, after some rearrangement,

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k{\left(c\right)}_k{\left(1-c\right)}_k}{4^{k}k{!}^2{\left(\frac{3}{2}-c\right)}_k}}\left({\displaystyle \frac{3k-c+1}{2}}\left(H_{c+k-1}-H_{-c+k}+H_{1/2-c+k}\right)-{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{H_{c-1}}{\sqrt{\pi }}}{\left(1-c\right)}_{1/2}.\hfill \end{array}$$

(58)

Set $c=1/2-r$ to get

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k{\left(-r+\frac{1}{2}\right)}_k{\left(r+\frac{1}{2}\right)}_k}{4^{k}k{!}^2{\left(r+1\right)}_k}}\left({\displaystyle \frac{1}{2}}\left(3k+r+{\displaystyle \frac{1}{2}}\right)\left(H_{k-r-1/2}-H_{k+r-1/2}+H_{k+r}\right)-{\displaystyle \frac{1}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{H_{-r-1/2}}{\sqrt{\pi }}}{\left(r+{\displaystyle \frac{1}{2}}\right)}_{1/2};\hfill \end{array}$$

(59)

and hence (57) after using (19) and (20). □

Corollary 8. 

If n is a non-negative integer, then

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}{\displaystyle \frac{{\prod }_{j=1}^{2n}\left(2k+j\right)}{{\prod }_{j=1}^n{\left(k+j\right)}^2\left(2k-2j+1\right)}}\left(\left(6k+1+2n\right)\left(2O_{k-n}-2O_{k+n}+H_{k+n}\right)-2\right)\hfill \\ \hfill & ={(-1)}^n{\displaystyle \frac{2^{n+3}}{{\left(\frac{1}{2}\right)}_n}}\left(O_n-ln2\right){\displaystyle \frac{1}{\pi }}.\hfill \end{array}$$

In particular,

$${\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\left(\frac{1}{2}\right)}_k^3}{4^{k}k{!}^3}}\left(\left(6k+1\right)H_k-2\right)=-{\displaystyle \frac{8ln2}{\pi }}.$$

(60)

Proof. 

Set $r=n$ in (59), use

$$H_{k-n-1/2}-H_{k+n-1/2}=2\left(O_{k-n}-O_{k+n}\right)$$

and identities (21) and (22). □