On Series Involving Cubed Catalan Numbers

1. Introduction

Catalan numbers are defined for non-negative integers j by

$$C_j={\displaystyle \frac{1}{j+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right),$$

(1)

and obey the following recurrence relation:

$$C_{j+1}={\displaystyle \frac{2(2j+1)}{j+2}}{C}_j,C_0=1.$$

(2)

Using Dixon’s theorem, Tauraso [5] found the following identity involving cubed Catalan numbers:

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\right)}^3=8-{\displaystyle \frac{384\pi }{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}},$$

(3)

where $\mathrm{\Gamma }\left(z\right)$ is the Gamma function.

In this paper, using identities involving generalized binomial coefficients and results due to Dougall [3] we will derive (3) and the following series having a similar nature:

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}\left(k+2\right)}}\right)}^3& ={\displaystyle \frac{152}{27}}-{\displaystyle \frac{80}{81{\pi }^3}}{\mathrm{\Gamma }}^4\left({\displaystyle \frac{1}{4}}\right),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k{C}_k}{2^{2k}}}\right)}^3\left(4k+3\right)& =8-{\displaystyle \frac{16}{\pi }},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k{C}_k}{2^{2k}\left(k+2\right)}}\right)}^3\left(4k+5\right)& =-{\displaystyle \frac{8}{9}}+{\displaystyle \frac{128}{27\pi }},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\right)}^3\left(4k+3\right)& =-8+{\displaystyle \frac{2}{{\pi }^3}}{\mathrm{\Gamma }}^4\left({\displaystyle \frac{1}{4}}\right),\hfill \end{array}$$

(7)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}\left(k+2\right)}}\right)}^3\left(4k+5\right)={\displaystyle \frac{136}{9}}-{\displaystyle \frac{7168}{9}}{\displaystyle \frac{\pi }{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}}.$$

(8)

In fact, each of identities (3)–(8) occurs as a particular member of a family of series stated in Corollaries 1 and 2 and Theorem 3. For instance, (3) is the $m=0$ case of the family stated in Corollary 1, namely,

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^{2m}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^{6m+6}}{{\left(\left(2m+2\right)!\right)}^3{C}_{2m+1}^3}}\left(1-{\displaystyle \frac{48}{{\left(4m+1\right)}^2}}{(-1)}^m{\displaystyle \frac{\pi }{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}}{\displaystyle \frac{{\prod }_{j=1}^{3m}\left(4j-1\right)}{{\prod }_{j=1}^m{\left(4j-3\right)}^3}}\right).\hfill \end{array}$$

(10)

We will also derive families of series involving cubed Catalan numbers and odd harmonic numbers including the following special cases:

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\right)}^3{O}_k& =8+{\displaystyle \frac{64\pi \left(\pi -10\right)}{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}\left(k+2\right)}}\right)}^3{O}_k& ={\displaystyle \frac{392}{81}}-{\displaystyle \frac{8}{729}}{\displaystyle \frac{\left(15\pi +32\right)}{{\pi }^3}}{\mathrm{\Gamma }}^4\left({\displaystyle \frac{1}{4}}\right),\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k{C}_k}{2^{2k}}}\right)}^3\left(\left(4k+3\right)O_k+{\displaystyle \frac{1}{3}}\right)& ={\displaystyle \frac{16\left(\pi -3\right)}{3\pi }},\hfill \end{array}$$

(13)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k{C}_k}{2^{2k}\left(k+2\right)}}\right)}^3\left(\left(4k+5\right)O_k+{\displaystyle \frac{1}{3}}\right)={\displaystyle \frac{32\left(16-5\pi \right)}{81\pi }}.$$

(14)

As additional results, we will derive a family of series involving fourth powers of Catalan numbers and one involving these numbers and odd harmonic numbers, of which the following are particular cases:

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\right)}^4\left(4k+3\right)& =16-{\displaystyle \frac{128}{{\pi }^2}},\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}\left(k+2\right)}}\right)}^4\left(4k+5\right)& =-{\displaystyle \frac{176}{27}}+{\displaystyle \frac{16384}{243{\pi }^2}},\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\right)}^4\left(\left(4k+3\right)O_k+{\displaystyle \frac{1}{4}}\right)& =12-{\displaystyle \frac{192}{{\pi }^2}}+{\displaystyle \frac{32\left(2ln2+1\right)}{{\pi }^2}},\hfill \end{array}$$

(17)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}\left(k+2\right)}}\right)}^4\left(\left(4k+5\right)O_k+{\displaystyle \frac{1}{4}}\right)=-{\displaystyle \frac{220}{27}}+{\displaystyle \frac{75776}{729{\pi }^2}}-{\displaystyle \frac{8192}{243}}{\displaystyle \frac{ln2}{{\pi }^2}}.$$

(18)

A high point of this paper is the discovery of a family of Ramanujan-like series (Theorem 8):

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

(19)

which gives the Bauer series [1] at $m=0$:

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

(20)

We will establish the following families of Ramanujan-like series for $1/{\pi }^2$ and $1/{\pi }^3$:

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^4\left(4k-2m+1\right)={\displaystyle \frac{{(-1)}^m{2}^{8m}}{{(m!)}^4{\pi }^{2}m}}{\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)}^{-5},m=1,2,3,\dots ,$$

(21)

and

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^{2m}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & ={\left({\displaystyle \frac{\left(2m\right)!}{2^{2m}}}\left({\displaystyle \genfrac{}{}{0pt}{}{4m}{2m}}\right)\right)}^{-3}{\displaystyle \frac{{(-1)}^m}{{\left(4m-1\right)}^2}}{\displaystyle \frac{{\mathrm{\Gamma }}^4(1/4)}{4{\pi }^3}}{\displaystyle \frac{{\prod }_{j=0}^{3m-1}\left(4j-3\right)}{{\prod }_{j=0}^{m-1}{\left(4j-1\right)}^3}},m=0,1,2,3,\dots ;\hfill \end{array}$$

(23)

with the special cases

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}\left(2k-1\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^4\left(4k-1\right)=-{\displaystyle \frac{8}{{\pi }^2}}$$

(24)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3={\displaystyle \frac{{\mathrm{\Gamma }}^4\left(1/4\right)}{4{\pi }^3}}.$$

(25)

Binomial coefficients are defined, for non-negative integers i and j, by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)=\left\{\begin{array}{cc}{\displaystyle \frac{i!}{j!(i-j)!}},\hfill & i\ge j;\hfill \\ 0,\hfill & i<j.\hfill \end{array}\right.$$

(26)

The definition is extended to complex numbers r and s by

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

(27)

where the Gamma function, $\mathrm{\Gamma }\left(z\right)$, is defined for $\Re \left(z\right)>0$ by

$$\mathrm{\Gamma }\left(z\right)={\int }_0^{\infty }{e}^{-t}{t}^{z-1}dt={\int }_0^{\infty }{\left(log(1/t)\right)}^{z-1}dt,$$

(28)

and is extended to the rest of the complex plane, excluding the non-positive integers, by analytic continuation.

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

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

(29)

The identity

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

(30)

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

Series involving cubed Catalan numbers are given in Theorems 1, 2 and 3 and Corollaries 1 and 2. We require the results stated in Lemmata 1–6.

Lemma 1.

If k and r are non-negative integers, then

$$\left({\displaystyle \genfrac{}{}{0pt}{}{r+1/2}{k+1}}\right)={(-1)}^r{\displaystyle \frac{\left(r+2\right)!C_{r+1}}{2^{r+2}}}{(-1)}^k{\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^r{\displaystyle \frac{1}{2k-2j+1}}.$$

(31)

Proof. 

This identity and similar ones can be derived using (27) together with the identities

$$\mathrm{\Gamma }\left(z+{\displaystyle \frac{1}{2}}\right)=\sqrt{\pi }{2}^{-2z}\left({\displaystyle \genfrac{}{}{0pt}{}{2z}{z}}\right)\mathrm{\Gamma }\left(z+1\right)$$

(32)

and

$$\mathrm{\Gamma }\left(-z+{\displaystyle \frac{1}{2}}\right)={(-1)}^z{2}^{2z}{\left({\displaystyle \genfrac{}{}{0pt}{}{2z}{z}}\right)}^{-1}{\displaystyle \frac{\sqrt{\pi }}{\mathrm{\Gamma }\left(z+1\right)}},$$

(33)

and the Pochhammer relation

$${\displaystyle \frac{\mathrm{\Gamma }\left(z+n\right)}{\mathrm{\Gamma }\left(z\right)}}=\prod _{j=1}^n(z+j-1)={\left(z\right)}_n,$$

(34)

for n a non-negative integer and z a complex number. □

Lemma 2.

If m is an integer, then

$${\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{4}\left(6m+7\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m+5\right)\right)}}={\displaystyle \frac{48}{{\left(2m+1\right)}^2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m+1\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m+1\right)\right)}}$$

(35)

and

$${\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{4}\left(6m+5\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m+3\right)\right)}}={\displaystyle \frac{12\left(6m+1\right)}{{\left(2m-1\right)}^2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m-1\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m-1\right)\right)}}.$$

(36)

Proof. 

Use the recurrence relation $\mathrm{\Gamma }\left(u\right)=\left(u-1\right)\mathrm{\Gamma }(u-1)$. □

Lemma 3.

If m is an integer, then

$$\mathrm{\Gamma }\left(m+{\displaystyle \frac{1}{4}}\right)={\displaystyle \frac{1}{4^m}}\mathrm{\Gamma }\left({\displaystyle \frac{1}{4}}\right)\prod _{j=1}^m\left(4j-3\right)$$

(37)

and

$$\mathrm{\Gamma }\left(m+{\displaystyle \frac{3}{4}}\right)={\displaystyle \frac{1}{4^m}}{\displaystyle \frac{\pi \sqrt{2}}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}}\prod _{j=1}^m\left(4j-1\right).$$

(38)

Proof. 

These follow immediately from (34) and the fact that

$$\mathrm{\Gamma }\left({\displaystyle \frac{3}{4}}\right)={\displaystyle \frac{\pi \sqrt{2}}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}},$$

(39)

which is a consequence of Euler’s reflection formula:

$$\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(1-z)={\displaystyle \frac{\pi }{sin\left(\pi z\right)}}.$$

(40)

□

Lemma 4.

If m is an integer, then

$$\begin{array}{cc}\hfill cos\left({\displaystyle \frac{\pi }{4}}\left(2m+1\right)\right)& ={\displaystyle \frac{{(-1)}^{⌈m/2⌉}}{\sqrt{2}}},\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill sin\left({\displaystyle \frac{\pi }{4}}\left(2m+1\right)\right)& ={\displaystyle \frac{{(-1)}^{m+⌈m/2⌉}}{\sqrt{2}}},\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill cos\left({\displaystyle \frac{\pi }{4}}\left(2m-1\right)\right)& ={\displaystyle \frac{{(-1)}^{⌊m/2⌋}}{\sqrt{2}}},\hfill \end{array}$$

(43)

and

$$sin\left({\displaystyle \frac{\pi }{4}}\left(2m-1\right)\right)=-{\displaystyle \frac{{(-1)}^{⌈m/2⌉}}{\sqrt{2}}};$$

(44)

where, as usual, $⌊x⌋$ is the greatest integer less than or equal to x and $⌈x⌉$ is the smallest integer greater than or equal to x.

Proof. 

We prove (41). We have

$$\begin{array}{cc}\hfill cos\left({\displaystyle \frac{\pi }{4}}\left(2m+1\right)\right)& =cos\left({\displaystyle \frac{\pi m}{2}}\right)cos\left({\displaystyle \frac{\pi }{4}}\right)-sin\left({\displaystyle \frac{\pi m}{2}}\right)sin\left({\displaystyle \frac{\pi }{4}}\right)\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{1}{\sqrt{2}}}\left(cos\left({\displaystyle \frac{\pi m}{2}}\right)-sin\left({\displaystyle \frac{\pi m}{2}}\right)\right)\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{1}{\sqrt{2}}}\left\{\begin{array}{cc}cos\left({\displaystyle \frac{\pi m}{2}}\right),\hfill & \mathrm{if}m\mathrm{is}\mathrm{even};\hfill \\ -sin\left({\displaystyle \frac{\pi m}{2}}\right),\hfill & \mathrm{if}m\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.\hfill \end{array}$$

(47)

Thus,

$$\begin{array}{cc}\hfill cos\left({\displaystyle \frac{\pi }{4}}\left(2m+1\right)\right)& ={\displaystyle \frac{1}{\sqrt{2}}}\left\{\begin{array}{cc}{(-1)}^{m/2},\hfill & \mathrm{if}m\mathrm{is}\mathrm{even};\hfill \\ {(-1)}^{(m+1)/2},\hfill & \mathrm{if}m\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{(-1)}^{⌈m/2⌉}}{\sqrt{2}}}.\hfill \end{array}$$

(49)

□

Lemma 5.

If n is an integer, then $O_{-n}=O_n$.

Proof. 

We have

$$O_{-n}=\sum _{j=1}^{-n}{\displaystyle \frac{1}{2j-1}}=-\sum _{j=1-n}^0{\displaystyle \frac{1}{2j-1}}=-\sum _{j=1}^n{\displaystyle \frac{1}{2j-2n-1}}.$$

(50)

Thus,

$$O_{-n}=\sum _{j=1}^n{\displaystyle \frac{1}{2n-2j+1}}=\sum _{j=1}^n{\displaystyle \frac{1}{2n-2\left(n-j+1\right)+1}}=\sum _{j=1}^n{\displaystyle \frac{1}{2j-1}}=O_n.$$

(51)

□

Each identity in Lemma 6 is a slight re-writing of the original one derived by Dougall [3].

Lemma 6.

If x is a complex number, then

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^3& =1-{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{2}\left(3x+2\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{2}\left(x+2\right)\right)}}cos\left({\displaystyle \frac{\pi x}{2}}\right),\Re x>-2/3,\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^3\left(2k+2-x\right)& =x-{\displaystyle \frac{sin\left(\pi x\right)}{\pi }},x\ge -1/3,\hfill \end{array}$$

(53)

and

$$\sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^3\left(2k+2-x\right)=-x+{\displaystyle \frac{sin\left(\pi x\right)}{\pi }}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{2}\left(1-x\right)\right)\mathrm{\Gamma }\left(\frac{1}{2}\left(1+3x\right)\right)}{{\mathrm{\Gamma }}^2\left(\frac{1}{2}\left(1+x\right)\right)}}.$$

(54)

Remark 1.

Identities (52)–(54) are also recorded in [4].

3. Series Involving Cubed Catalan Numbers

Theorem 1.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(55)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{(-1)}^m{2}^{3m+6}}{{\left(\left(m+2\right)!\right)}^3{C}_{m+1}^3}}\left(1-{\displaystyle \frac{24\sqrt{2}}{{\left(2m+1\right)}^2}}{(-1)}^{⌈m/2⌉}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m+1\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m+1\right)\right)}}\right).\hfill \end{array}$$

(56)

Proof. 

Set $x=m+1/2$ in (52) to obtain

$$\sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{m+1/2}{k+1}}\right)}^3=1-{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{4}\left(6m+7\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m+5\right)\right)}}cos\left({\displaystyle \frac{\pi }{4}}\left(2m+1\right)\right),$$

(57)

and hence (56) upon using Lemmata 1, 2 and 4. □

Corollary 1.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^{2m}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(58)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^{6m+6}}{{\left(\left(2m+2\right)!\right)}^3{C}_{2m+1}^3}}\left(1-{\displaystyle \frac{48}{{\left(4m+1\right)}^2}}{(-1)}^m{\displaystyle \frac{\pi }{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}}{\displaystyle \frac{{\prod }_{j=1}^{3m}\left(4j-1\right)}{{\prod }_{j=1}^m{\left(4j-3\right)}^3}}\right)\hfill \end{array}$$

(59)

and

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^{2m-1}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(60)

$$\begin{array}{cc}\hfill & =-{\displaystyle \frac{2^{6m+3}}{{\left(\left(2m+1\right)!\right)}^3{C}_{2m}^3}}\left(1-{\displaystyle \frac{{(-1)}^m}{{\left(4m-1\right)}^2}}{\displaystyle \frac{3}{4{\pi }^3}}{\mathrm{\Gamma }}^4\left({\displaystyle \frac{1}{4}}\right){\displaystyle \frac{{\prod }_{j=1}^{3m-1}\left(4j-3\right)}{{\prod }_{j=1}^{m-1}{\left(4j-1\right)}^3}}\right).\hfill \end{array}$$

(61)

Proof. 

Write $2m$ for m and $2m-1$ for m, in turn, in (56), and apply Lemma 3. □

Remark 2.

Identities (3) and (4) are obtained by evaluating (59) at $m=0$ and (61) at $m=1$.

Theorem 2.

If m is a non-negative integer, then

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

(62)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{(-1)}^m{2}^{3m+6}}{{\left(\left(m+2\right)!\right)}^3{C}_{m+1}^3}}\left(2m+1-{(-1)}^m{\displaystyle \frac{2}{\pi }}\right).\hfill \end{array}$$

(63)

Proof. 

Use of $r=m+1/2$ in (53) gives

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{m+1/2}{k+1}}\right)}^3\left(4k-2m+3\right)=2m+1+{(-1)}^{m+1}{\displaystyle \frac{2}{\pi }},$$

(64)

and hence (63) by Lemma 1. □

Remark 3.

Identities (5) and (6) are obtained at $m=0$ and $m=1$ in (63).

Theorem 3.

If m is a non-negative integer, then

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

(65)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{(-1)}^{m-1}{2}^{3m+6}}{{\left(\left(m+2\right)!\right)}^3{C}_{m+1}^3}}\left(2m+1-{(-1)}^{m+⌈m/2⌉}24\sqrt{2}{\displaystyle \frac{6m+1}{{\left(2m-1\right)}^2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m-1\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m-1\right)\right)}}\right).\hfill \end{array}$$

(66)

Proof. 

Set $x=m+1/2$ in (54) and apply Lemmata 1, 2 and 4. □

Corollary 2.

If m is a non-negative integer, then

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

(67)

$$\begin{array}{cc}\hfill & =-{\displaystyle \frac{2^{6m+6}}{{\left(\left(2m+2\right)!\right)}^3{C}_{2m+1}^3}}\left(4m+1-{\displaystyle \frac{{(-1)}^m\left(12m+1\right)}{{\left(4m-1\right)}^2}}{\displaystyle \frac{3}{4{\pi }^3}}{\mathrm{\Gamma }}^4\left({\displaystyle \frac{1}{4}}\right){\displaystyle \frac{{\prod }_{j=1}^{3m-1}\left(4j-3\right)}{{\prod }_{j=1}^{m-1}{\left(4j-1\right)}^3}}\right)\hfill \end{array}$$

(68)

and

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^{2m-1}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\left(4k-4m+5\right)\hfill \end{array}$$

(69)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^{6m+3}}{{\left(\left(2m+1\right)!\right)}^3{C}_{2m}^3}}\left(4m-1+{\displaystyle \frac{{(-1)}^m\left(12m-5\right)}{{\left(4m-3\right)}^2}}{\displaystyle \frac{48\pi }{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}}{\displaystyle \frac{{\prod }_{j=1}^{3m-3}\left(4j-1\right)}{{\prod }_{j=1}^{m-1}{\left(4j-3\right)}^3}}\right).\hfill \end{array}$$

(70)

Proof. 

Write $2m$ for m and $2m-1$ for m, in turn, in (65), and apply Lemma 3. □

Remark 4.

Identities (7) and (8) correspond to an evaluation of (67) at $m=0$ and (69) at $m=1$.

4. Series Involving Cubed Catalan Numbers and Odd Harmonic Numbers

Theorem 4.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=m}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3{O}_{k-m}\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill & =-\sum _{k=0}^m{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3{O}_{m-k}+{\displaystyle \frac{{(-1)}^m{2}^{3m+6}{O}_{m+1}}{{\left(\left(m+2\right)!\right)}^3{C}_{m+1}^3}}\hfill \end{array}$$

(72)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{{(-1)}^{m+⌈m/2⌉}{2}^{3m+7}}{{\left(\left(m+2\right)!\right)}^3{C}_{m+1}^3}}{\displaystyle \frac{3\sqrt{2}}{{\left(2m+1\right)}^2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m+1\right)\right)}{\mathrm{\Gamma }{\left(\frac{1}{4}\left(2m+1\right)\right)}^3}}\hfill \end{array}$$

(73)

$$\begin{array}{cc}\hfill & \times \left({(-1)}^{m-1}{\displaystyle \frac{\pi }{3}}+H_{{\displaystyle \frac{3\left(2m+1\right)}{4}}}-H_{{\displaystyle \frac{\left(2m+1\right)}{4}}}-4O_{m+1}\right).\hfill \end{array}$$

(74)

Proof. 

Write (52) in the equivalent form

$$\sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^3=1-\left({\displaystyle \genfrac{}{}{0pt}{}{3x/2}{x/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{x}{x/2}}\right)cos\left({\displaystyle \frac{\pi x}{2}}\right),$$

(75)

and differentiate with respect to x to obtain

$$\begin{array}{cc}\hfill & 3\sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^3\left(H_x-H_{x-k-1}\right)\hfill \end{array}$$

(76)

$$\begin{array}{cc}\hfill & =\left({\displaystyle \frac{\pi }{2}}sin\left({\displaystyle \frac{\pi x}{2}}\right)-\left({\displaystyle \frac{3}{2}}{H}_{3x/2}-{\displaystyle \frac{3}{2}}{H}_{x/2}\right)cos\left({\displaystyle \frac{\pi x}{2}}\right)\right){\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{2}\left(3x+2\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{2}\left(x+2\right)\right)}}.\hfill \end{array}$$

(77)

Set $x=m+1/2$ and use (30) and Lemma 4 to get

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{m+1/2}{k+1}}\right)}^3{O}_{k-m}\hfill \end{array}$$

(78)

$$\begin{array}{cc}\hfill & =\left({(-1)}^{m-1}{\displaystyle \frac{\pi }{3}}+H_{{\displaystyle \frac{3}{4}}\left(2m+1\right)}-H_{{\displaystyle \frac{1}{4}}\left(2m+1\right)}-4O_{m+1}\right){\displaystyle \frac{{(-1)}^{⌈m/2⌉}}{4\sqrt{2}}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{4}\left(6m+7\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m+5\right)\right)}}.\hfill \end{array}$$

(79)

Identity (71) now follows upon application of Lemmata 1 and 2. □

Remark 5.

Identities (11) and (12) correspond to evaluating (74) at $m=0$ and $m=1$. In deriving (11), note that

$$H_{3/4}-H_{1/4}={\displaystyle \frac{1}{3/4}}-{\displaystyle \frac{1}{1/4}}+\pi cot\left({\displaystyle \frac{\pi }{4}}\right)=\pi -{\displaystyle \frac{8}{3}},$$

(80)

since

$$H_r-H_{1-r}={\displaystyle \frac{1}{r}}-{\displaystyle \frac{1}{1-r}}+\pi cot\left(\pi \left(1-r\right)\right),$$

(81)

when r is not an integer. For (12), note also that

$$H_{9/4}=H_{5/4}+{\displaystyle \frac{4}{9}}=H_{1/4}+{\displaystyle \frac{4}{5}}+{\displaystyle \frac{4}{9}}=H_{1/4}+{\displaystyle \frac{56}{45}},$$

(82)

since

$$H_r=H_{r-1}+{\displaystyle \frac{1}{r}},$$

(83)

when r is not a non-positive integer. Thus

$$H_{9/4}-H_{3/4}=H_{1/4}-H_{3/4}+{\displaystyle \frac{56}{45}}={\displaystyle \frac{8}{3}}-\pi +{\displaystyle \frac{56}{45}}={\displaystyle \frac{176}{45}}-\pi .$$

(84)

Theorem 5.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=m}^{\infty }{\left({\displaystyle \frac{{(-1)}^k{C}_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\left(\left(4k-2m+3\right)O_{k-m}+{\displaystyle \frac{1}{3}}\right)\hfill \end{array}$$

(85)

$$\begin{array}{cc}\hfill & =-\sum _{k=0}^{m-1}{\left({\displaystyle \frac{{(-1)}^k{C}_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\left(\left(4k-2m+3\right)O_{m-k}+{\displaystyle \frac{1}{3}}\right)\hfill \end{array}$$

(86)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{{(-1)}^m{2}^{3m+6}}{{\left(\left(m+2\right)!\right)}^3{C}_{m+1}^3}}\left(\left(2m+1-{\displaystyle \frac{{(-1)}^{m}2}{\pi }}\right)O_{m+1}-{\displaystyle \frac{1}{3}}\right).\hfill \end{array}$$

(87)

Proof. 

Differentiate (53) with respect to x to obtain

$$\begin{array}{c}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^3\left(\left(2k+2-x\right)H_{x-k-1}+{\displaystyle \frac{1}{3}}\right)=H_x\left(x-{\displaystyle \frac{sin\left(\pi x\right)}{\pi }}\right)-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{cos\left(\pi x\right)}{3}}.\end{array}$$

(88)

Set $x=m+1/2$ and use (30) and Lemma 1. □

Remark 6.

Identities (13) and (14) are evaluations of (87) at $m=0$ and $m=1$.

5. Series Involving Fourth Powers of Catalan Numbers

In this section, we derive a family of series involving fourth powers of Catalan numbers, based on the identity

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^4\left(2k-x+2\right)=x-{\displaystyle \frac{sin\left(\pi x\right)}{\pi }}\left({\displaystyle \genfrac{}{}{0pt}{}{2x}{x}}\right),x>-{\displaystyle \frac{1}{2}},$$

(89)

which is a variation on [3 Equation (17)] .

Theorem 6.

If m is a non-negative integer, then

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

(90)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{2^{4m+8}}{{\left(\left(m+2\right)!\right)}^4{C}_{m+1}^4}}\left(2m+1-{\displaystyle \frac{{(-1)}^m{2}^{4m+3}}{{\pi }^2\left(m+1\right)\left(2m+1\right)C_m}}\right).\hfill \end{array}$$

(91)

Proof. 

At $x=m+1/2$, (89) gives

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{m+1/2}{k+1}}\right)}^4\left(4k-2m+3\right)=2m+1-{(-1)}^m{\displaystyle \frac{4}{\pi }}\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m+1/2}}\right),$$

(92)

from which (91) follows by (31) and the fact that

$$\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m+1/2}}\right)={\displaystyle \frac{\mathrm{\Gamma }\left(2m+1\right)}{\mathrm{\Gamma }\left(m+3/2\right)\mathrm{\Gamma }\left(m+1/2\right)}}={\displaystyle \frac{2^{4m+1}}{\pi \left(m+1\right)\left(2m+1\right)C_m}}.$$

(93)

□

Remark 7.

Identities (15) and (16) are obtained by evaluating (91) at $m=0$ and at $m=1$.

Theorem 7.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=m}^{\infty }{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^4\left(\left(4k-2m+3\right)O_{k-m}+{\displaystyle \frac{1}{4}}\right)\hfill \end{array}$$

(94)

$$\begin{array}{cc}\hfill & =-\sum _{k=0}^{m-1}{\left({\displaystyle \frac{C_k}{2^{2k}}}\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^4\left(\left(4k-2m+3\right)O_{m-k}+{\displaystyle \frac{1}{4}}\right)\hfill \end{array}$$

(95)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{2^{4m+8}{O}_{m+1}}{{\left(\left(m+2\right)!\right)}^4{C}_{m+1}^4}}\left(2m+1+{\displaystyle \frac{3{(-1)}^{m+1}{2}^{4m+2}}{{\pi }^2\left(m+1\right)\left(2m+1\right)C_m}}\right)\hfill \end{array}$$

(96)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{2^{4m+8}}{{\left(\left(m+2\right)!\right)}^4{C}_{m+1}^4}}\left({\displaystyle \frac{{(-1)}^m\left(2ln2+H_{2m+1}\right)2^{4m+1}}{{\pi }^2\left(m+1\right)\left(2m+1\right)C_m}}-{\displaystyle \frac{1}{4}}\right).\hfill \end{array}$$

(97)

Proof. 

Differentiating (89) with respect to x gives, after some re-arrangement,

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k+1}}\right)}^4\left(\left(2k+2-x\right)H_{x-k-1}+{\displaystyle \frac{1}{4}}\right)\hfill \end{array}$$

(98)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{sin\left(\pi x\right)}{2\pi }}\left({\displaystyle \genfrac{}{}{0pt}{}{2x}{x}}\right)\left(H_{2x}-3H_x\right)+x{H}_x-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\left({\displaystyle \genfrac{}{}{0pt}{}{2x}{x}}\right)cos\left(\pi x\right).\hfill \end{array}$$

(99)

Evaluation at $x=m+1/2$, using (30) and some algebra yields

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

(100)

$$\begin{array}{cc}\hfill & =\left(2m+1+{\displaystyle \frac{{(-1)}^{m+1}6}{\pi }}\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m+1/2}}\right)\right)O_{m+1}\hfill \end{array}$$

(101)

$$\begin{array}{cc}\hfill & +{\displaystyle \frac{{(-1)}^m}{\pi }}\left(2ln2+H_{2m+1}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m+1/2}}\right)-{\displaystyle \frac{1}{4}}.\hfill \end{array}$$

(102)

Application of (31) and (93) now produces (97). □

Remark 8.

Identities (17) and (18) correspond to setting $m=0$ and $m=1$ in (97).

6. Ramanujan-like Series and Related Series

Series involving central binomial coefficients can be derived by writing (31) in the equivalent form

$$\left({\displaystyle \genfrac{}{}{0pt}{}{r-1/2}{k}}\right)={(-1)}^r{\displaystyle \frac{r!}{2^r}}\left({\displaystyle \genfrac{}{}{0pt}{}{2r}{r}}\right){\displaystyle \frac{{(-1)}^k}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^r{\displaystyle \frac{1}{2k-2j+1}},$$

(103)

for non-negative integers k and r.

Theorem 8.

If m is a non-negative integer, then

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

(104)

In particular,

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3\left(4k+1\right)& ={\displaystyle \frac{2}{\pi }},\hfill \end{array}$$

(105)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}\left(2k-1\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3\left(4k-1\right)& ={\displaystyle \frac{2}{\pi }},\hfill \end{array}$$

(106)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}\left(2k-1\right)\left(2k-3\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3\left(4k-3\right)={\displaystyle \frac{2}{27\pi }}.$$

(107)

Proof. 

Write (53) as

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)}^3\left(2k-x\right)=-{\displaystyle \frac{sin\left(\pi x\right)}{\pi }},$$

(108)

set $x=m-1/2$ and use (103). □

Remark 9.

Identity (105) was discovered in 1859 by Bauer [1] via a Fourier-Legendre expansion.

Theorem 9.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=m}^{\infty }{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\left(\left(4k-2m+1\right)O_{k-m}+{\displaystyle \frac{1}{3}}\right)\hfill \end{array}$$

(109)

$$\begin{array}{cc}\hfill & =-\sum _{k=0}^{m-1}{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\left(\left(4k-2m+1\right)O_{m-k}+{\displaystyle \frac{1}{3}}\right)\hfill \end{array}$$

(110)

$$\begin{array}{cc}\hfill & +{\left({\displaystyle \frac{m!}{2^m}}\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)\right)}^{-3}{\displaystyle \frac{2}{\pi }}{O}_m.\hfill \end{array}$$

(111)

In particular,

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3\left(4k+1\right)O_k=-{\displaystyle \frac{1}{3}}\sum _{k=0}^{\infty }{\left({\displaystyle \frac{{(-1)}^k}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3.$$

(112)

Proof. 

Differentiate (108) with respect to x, obtaining, after some rearrangement,

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)}^3\left(\left(2k-x\right)H_{x-k}+{\displaystyle \frac{1}{3}}\right)=-H_x{\displaystyle \frac{sin\left(\pi x\right)}{\pi }}+{\displaystyle \frac{1}{3}}cos\left(\pi x\right),$$

(113)

and set $x=m-1/2$. □

Remark 10.

Setting $m=1$ in (111) reproduces (3) after some manipulation.

Theorem 10.

If m is a positive integer, then

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^4\left(4k-2m+1\right)={\displaystyle \frac{{(-1)}^m{2}^{8m}}{{(m!)}^4{\pi }^{2}m}}{\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)}^{-5}.$$

(114)

In particular,

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}\left(2k-1\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^4\left(4k-1\right)=-{\displaystyle \frac{8}{{\pi }^2}}$$

(115)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}\left(2k-1\right)\left(2k-3\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^4\left(4k-3\right)={\displaystyle \frac{64}{243{\pi }^2}}.$$

(116)

Proof. 

Write (89) as

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)}^4\left(2k-x\right)=-{\displaystyle \frac{sin\left(\pi x\right)}{\pi }}\left({\displaystyle \genfrac{}{}{0pt}{}{2x}{x}}\right),x>-{\displaystyle \frac{1}{2}}.$$

(117)

Set $x=m-1/2$; this gives

$$\sum _{k=0}^{\infty }{\left({\displaystyle \genfrac{}{}{0pt}{}{m-1/2}{k}}\right)}^4\left(4k-2m+1\right)={\displaystyle \frac{{(-1)}^m{2}^{4m}}{{\pi }^{2}m}}{\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)}^{-1},$$

(118)

after using

$$sin\left({\displaystyle \frac{\pi }{2}}\left(2m-1\right)\right)={(-1)}^{m-1}$$

(119)

and

$$\left({\displaystyle \genfrac{}{}{0pt}{}{2m-1}{m-1/2}}\right)={\displaystyle \frac{\mathrm{\Gamma }\left(2m\right)}{{\mathrm{\Gamma }}^2\left(m+1/2\right)}}={\displaystyle \frac{2^{4m-1}}{\pi m}}{\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)}^{-1}.$$

(120)

Identity (114) now follows from (118) upon application of (103). □

Theorem 11.

If m is a non-negative integer, then

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^m{\displaystyle \frac{1}{2k-2j+1}}\right)}^3={\left({\displaystyle \frac{m!}{2^m}}\left({\displaystyle \genfrac{}{}{0pt}{}{2m}{m}}\right)\right)}^{-3}{\displaystyle \frac{{(-1)}^{⌈m/2⌉}24\sqrt{2}}{{\left(2m-1\right)}^2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m-1\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m-1\right)\right)}}.$$

(121)

Proof. 

Write (52) as

$$\sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{x}{k}}\right)}^3={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{2}\left(3x+2\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{2}\left(x+2\right)\right)}}cos\left({\displaystyle \frac{\pi x}{2}}\right),$$

(122)

and set $x=m-1/2$ to obtain

$$\sum _{k=0}^{\infty }{(-1)}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{m-1/2}{k}}\right)}^3={\displaystyle \frac{48}{{\left(2m-1\right)}^2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{3}{4}\left(2m-1\right)\right)}{{\mathrm{\Gamma }}^3\left(\frac{1}{4}\left(2m-1\right)\right)}}cos\left({\displaystyle \frac{\pi }{4}}\left(2m-1\right)\right);$$

(123)

and hence (121). □

Corollary 3.

If m is a non-negative integer, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^{2m}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(124)

$$\begin{array}{cc}\hfill & ={\left({\displaystyle \frac{\left(2m\right)!}{2^{2m}}}\left({\displaystyle \genfrac{}{}{0pt}{}{4m}{2m}}\right)\right)}^{-3}{\displaystyle \frac{{(-1)}^m}{{\left(4m-1\right)}^2}}{\displaystyle \frac{{\mathrm{\Gamma }}^4(1/4)}{4{\pi }^3}}{\displaystyle \frac{{\prod }_{j=0}^{3m-1}\left(4j-3\right)}{{\prod }_{j=0}^{m-1}{\left(4j-1\right)}^3}},\hfill \end{array}$$

(125)

and

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\prod _{j=1}^{2m+1}{\displaystyle \frac{1}{2k-2j+1}}\right)}^3\hfill \end{array}$$

(126)

$$\begin{array}{cc}\hfill & ={\left({\displaystyle \frac{(2m+1)!}{2^{2m+1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2\left(2m+1\right)}{2m+1}}\right)\right)}^{-3}{\displaystyle \frac{{(-1)}^{m+1}}{{\left(4m+1\right)}^2}}{\displaystyle \frac{48\pi }{{\mathrm{\Gamma }}^4(1/4)}}{\displaystyle \frac{{\prod }_{j=1}^{3m}\left(4j-1\right)}{{\prod }_{j=1}^m{\left(4j-3\right)}^3}}.\hfill \end{array}$$

(127)

In particular,

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3& ={\displaystyle \frac{{\mathrm{\Gamma }}^4\left(1/4\right)}{4{\pi }^3}},\hfill \end{array}$$

(128)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}\left(2k-1\right)\left(2k-3\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3& =-{\displaystyle \frac{5}{324}}{\displaystyle \frac{{\mathrm{\Gamma }}^4\left(1/4\right)}{{\pi }^3}},\hfill \end{array}$$

(129)

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}\left(2k-1\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3& =-{\displaystyle \frac{48\pi }{{\mathrm{\Gamma }}^4\left(1/4\right)}},\hfill \end{array}$$

(130)

and

$$\sum _{k=0}^{\infty }{\left({\displaystyle \frac{1}{2^{2k}\left(2k-1\right)\left(2k-3\right)\left(2k-5\right)}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)\right)}^3={\displaystyle \frac{1232}{9375}}{\displaystyle \frac{\pi }{{\mathrm{\Gamma }}^4\left(1/4\right)}}.$$

(131)

Remark 11.

Identity (128) was recorded by Chen [2].