\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \underline{#1}}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf 
Interesting Ramanujan-Like Series Associated \\
\vskip .1in with Powers of  Central Binomial Coefficients
}
\vskip 1cm
\large
Hongwei Chen\\
Department of Mathematics\\
Christopher Newport University\\
Newport News, VA 23606 \\
USA\\
\href{mailto:hchen@cnu.edu}{\tt hchen@cnu.edu} \\
\end{center}

\vskip .2 in




\begin{abstract}
We find various Ramanujan-like series that involve the
central binomial coefficients.
In contrast with the traditional hypergeometric function approach, our
method uses Fourier-Legendre series expansions via specialization,
inner product, and Parseval's identity.
Several classical identities are recovered as particular cases.
\end{abstract}

\section{Introduction}
\label{sec1}

A Ramanujan-like series for $1/\pi$ is an identity of the following form:
$$\sum_{n=0}^\infty\,\frac{an+b}{c^n}\,s(n) = \frac{1}{\pi},$$
where $a$ and $b$ are algebraic numbers, $c \in \mathbb{N}$, and $s(n)$ is an integer sequence obeying a certain recurrence relation. More than 100 years ago,  without proof, Ramanujan recorded 17 formulas for $1/\pi$ at the end of his first paper published in England \cite{Ramanujan}. The first two formulas (with slight modification), which surprisingly appeared in the Disney movie {\it High School Musical}, are
\begin{equation}\label{eq:1}
\sum_{n=0}^\infty\,\frac{6n+1}{16^{2n}}\,{2n \choose n}^3 = \frac{4}{\pi}
\end{equation}
and
\begin{equation}
\sum_{n=0}^\infty\,\frac{42n+5}{64^{2n}}\,{2n \choose n}^3 = \frac{16}{\pi},
\end{equation}
where ${2n \choose n}$ is the central binomial coefficient defined by $(2n)!/(n!)^2$. Except for Chowla's proof of \eqref{eq:1} in 1928 \cite{Chowla}, Ramanujan's series were forgotten by the mathematical community until 1987. The Borwein brothers \cite{Borweins} finally succeeded in proving all 17 of Ramanujan's series for $1/\pi$ via modular forms. Moreover, as an application, they used one modified Ramanujan series to calculate the digits of $\pi$ and were able to obtain roughly 50 digits of $\pi$ per term. Since 2002, Guillera et al. \cite{Guillera1, Guillera2} have discovered many new Ramanujan-like series for $1/\pi^2$. An excellent survey on the work of Ramanujan-like series before 2009 can be found in \cite{BBChan}. 


Following the ideas of the Borweins and Guillera et al.,  exploring new classes of Ramanujan-like series has become an active area of contemporary research. 
In my previous work \cite{Chen}, the following Ramanujan-like series
\begin{equation*}
\sum_{n=1}^\infty\,\frac{{2n \choose n}^2H_n}{16^n(2n-1)^2} = \frac{12-16\ln 2}{\pi},
\end{equation*}
where $H_n$ denotes the $n$th harmonic number,
was established by applying a differential operator to a hypergeometric function $_2F_1$-identity. Along this line, using the Gauss summation theorem and an extended $_3F_2$-series of Watson and Whipple type, Wang and Chu \cite{WChu} offered a systematic evaluation of series like
$$ \sum_{n=1}^\infty\,\frac{{2n \choose n}^2H_n}{16^np(n)}\hspace{0.2in}\mbox{and}\hspace{0.2in}\sum_{n=1}^\infty\,\frac{{2n \choose n}^2H^{(2)}_n}{16^np(n)},$$
where $p(n) = n + \lambda, 1 + 2n -2\lambda$ or $(1 + 2n -2\lambda)^2$ for $\lambda \in \mathbb{N}$, and $H_n^{(2)} = \sum_{k=1}^n 1/k^2$. In particular, they obtained the following interesting Ramanujan-like series:
\begin{align*}
\sum_{n=1}^\infty\,\frac{{2n \choose n}^2H^{(2)}_n}{16^n(2n-1)}  & =  4 - \frac{\pi}{3} - \frac{8}{\pi};\\
\sum_{n=1}^\infty\,\frac{{2n \choose n}^2H_n^{(2)}}{16^n(2n-1)^2} & = -12 + \frac{2\pi}{3} + \frac{32}{\pi}.
\end{align*}

The aim of the present paper is to study the Ramanujan-like series involving powers of central binomial coefficients. In contrast to \cite{Campbell, WChu}, our approach is based on Fourier-Legendre series expansions. We will derive a variety of Ramanujan-like series including
\begin{align}
\sum_{n=0}^\infty\,\frac{(-1)^{n}}{16^n}\,{2n \choose n}^2 & = \frac{\sqrt{2}\,\Gamma^2(1/4)}{4\pi^{3/2}};\\
\sum_{n=0}^\infty\,\frac{(-1)^{n}(4n+1)}{64^n(n+1)(1/2-n)}\,{2n \choose n}^3 & =  \frac{8}{\pi};\\
\sum_{n=0}^\infty\,\frac{(4n+1)}{256^{n}(n+1)^2(1/2 -n)^2}\,{2n \choose n}^4 & =  \frac{128}{3\pi^2}.
\end{align}
Here $\Gamma(x)$ indicates the gamma function.



The rest of paper is organized as follows. In Section~\ref{sec2}, three 
combinatorial identities are established via the Wilf-Zeilberger method. These identities are used to derive several Fourier-Legendre series expansions. By using specialization, inner product, and Parseval's identity, those series expansions are used to find various Ramanujan-like series in Section~\ref{sec3}. The paper ends with two remarks. To ensure accuracy, all formulas appearing in this paper were verified numerically by Mathematica.

\section{Some Fourier-Legendre series expansions}
\label{sec2}

Recall the Legendre polynomial \cite[pp. 983--985]{GR}
$$P_n(x) = \frac{1}{2^n\,n!}\,\frac{d^n}{dx^n}(x^2 -1)^n.$$
It is well-known that
$$\int_{-1}^1\, P_n(x)P_m(x)\,dx = \frac{2}{2n+1}\,\delta_{mn},$$
where $\delta_{mn}$ denotes the Kronecker delta. This implies that $(P_n(x))_{n=0}^\infty$ are orthogonal on $[-1,1]$. Now, we define
$$\tilde{P}_n(x) = P_n(2x-1).$$
Since $x \to 2x-1$ is an affine transformation that bijectively maps $[0, 1]$ to $[-1, 1]$, we see that $(\tilde{P}_n(x))_{n=0}^\infty$ are orthogonal on $[0,1]$ with
$$\int_{0}^1\, \tilde{P}_n(x)\tilde{P}_m(x)\,dx = \frac{1}{2n+1}\,\delta_{mn}.$$
Moreover, we have
\begin{equation}\label{Polyexp}
\tilde{P}_n(x) = \frac{1}{n!}\,\frac{d^n}{dx^n}(x^2 -x)^n = (-1)^n\sum_{k=0}^n\,{n \choose k}{n+k \choose k}(-x)^k.
\end{equation}
Let $f(x) \in L[0,1]$. The {\it Fourier-Legendre series} is defined by
$$f(x) = \sum_{n=0}^\infty\,a_n\tilde{P}_n(x),$$
where
\begin{equation}\label{Lcoef}
a_n = (2n+1)\int_0^1\,f(x)\tilde{P}_n(x)\,dx.
\end{equation}

The following three lemmas display some features of the coefficients of variant Fourier-Legendre series.

\begin{lemma} Let $f(x) \in L[0,1]$. If $f(x) = \sum_{n=0}^\infty\,a_n\tilde{P}_n(x)$, then
$$f(1-x) = \sum_{n=0}^\infty\,(-1)^na_n\tilde{P}_n(x).$$
\end{lemma}

\begin{proof} Let $a_n^*$ be the $n$th coefficient of the Fourier-Legendre series for $f(1-x)$. By \eqref{Lcoef}, we have
\begin{align*}
a_n^* & =   (2n+1)\int_0^1\,f(1-x)\tilde{P}_n(x)\,dx\\
& =  (2n+1)\int_0^1\,f(t)\tilde{P}_n(1-t)\,dt\,\,\,\,(\mbox{use $t = 1-x$})\\
& =   (2n+1)\int_0^1\,f(t)(-1)^n\tilde{P}_n(t)\,dt = (-1)^na_n,
\end{align*}
where we have used the fact that $\tilde{P}_n(1-t) = P_n(-(2t-1)) = (-1)^nP_n(2t-1) = (-1)^n\tilde{P}_n(t)$.
\end{proof}

\begin{lemma}  Let $f(x) \in L[0,1]$. If $f(x) = f(1-x)$ for $x \in[0,1]$ and $f(x) = \sum_{n=0}^\infty\,a_n\tilde{P}_n(x)$, then
$a_n = 0$ when $n$ is odd.
\end{lemma}
\begin{proof} Using \eqref{Lcoef}, we rewrite $a_n$ as
\begin{align*}
a_n & =  (2n+1)\left(\int_0^{1/2}\,f(x)\tilde{P}_n(x)\,dx + \int_{1/2}^1\,f(x)\tilde{P}_n(x)\,dx\right)\\
& =  (2n+1)\left(\int_0^{1/2}\,f(x)\tilde{P}_n(x)\,dx + \int_0^{1/2}\,f(1-t)\tilde{P}_n(1-t)\,dt\right)\,\,(\mbox{use $t = 1-x$})\\
& =   (2n+1)\int_0^{1/2}\,(1 + (-1)^n)f(x)\tilde{P}_n(x)\,dx\,\,(\mbox{use $f(t) = f(1-t)$}).
\end{align*}
This leads to $a_n = 0$ when $n$ is odd.
\end{proof}

Recall Bonnet's recursion formula \cite[Formula 8.914 No.1, p.\ 985]{GR}
$$(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x).$$
This implies that
$$(2n+1)x\tilde{P}_n(x) = \frac{n+1}{2}\tilde{P}_{n+1}(x) + \frac{2n+1}{2}\tilde{P}_n(x) + \frac{n}{2}\tilde{P}_{n-1}(x),$$
which yields the following result.

\begin{lemma}  Let $f(x) \in L[0,1]$ and $f(x) = \sum_{n=0}^\infty\,a_n\tilde{P}_n(x)$. If
$$xf(x) = \sum_{n=0}^\infty\,b_n\tilde{P}_n(x),$$
then for all $n \geq 1$,
\begin{equation}\label{Pcoef}
b_n = \frac{n+1}{2(2n +3)}\,a_{n+1} + \frac{1}{2}\,a_n + \frac{n}{2(2n-1)}\,a_{n-1}.
\end{equation}
\end{lemma}

The following two lemmas establish some identities that we will need later to simplify the coefficients of the Fourier-Legendre series.

\begin{lemma} For any non-negative integer $n$, 
\begin{equation}\label{comb1}
\sum_{k=0}^{2n} \, (-1)^k{2n \choose k}{2n+k \choose k}{2k \choose k}4^{2n-k} = {2n \choose n}^2.
\end{equation}
\end{lemma} 

This identity \eqref{comb1} appears as (6.35) in \cite{Gould}, and is credited to E. T. Bell without providing a proof.  For completeness, following Tauraso's suggestion \cite{Tauraso}, we demonstrate a proof based on the Wilf-Zeilberger method (WZ-method).

\begin{proof} Rewrite \eqref{comb1} as
\begin{equation}\label{WZid}
\sum_{k=0}^{2n} \, \frac{(-1)^k{2n \choose k}{2n+k \choose k}{2k \choose k}4^{2n-k}}{{2n \choose n}^2} = 1.
\end{equation}
Let
$$F(n,k) =  \frac{(-1)^k{2n \choose k}{2n+k \choose k}{2k \choose k}4^{2n-k}}{{2n \choose n}^2}.$$
Using the WZ-method,  we find
$$G(n,k) = - \frac{2(4n+3)k^3}{(2n-k+1)(2n-k+2)(2n+1)^2}F(n,k).$$
Now it suffices to check 
\begin{equation}\label{WZeq}
F(n+1, k) - F(n,k) = G(n, k+1) - G(n,k). 
\end{equation}
Notice that
\begin{align*}
F(n+1, k) & =  \frac{4(2n+k+2)(2n +k+1)(n+1)^2}{(2n-k+2)(2n-k +1)(2n+1)^2}\,F(n,k),\\
F(n, k+1) & =  -\frac{(2k+1)(2n +k+1)(2n-k)}{2(k+1)^3}\,F(n,k),\\
G(n, k+1) & =  \frac{(4n+3)(2k+1)(2n+k+1)}{(2n-k+1)(2n+1)^2}\,F(n,k);
\end{align*}
with these \eqref{WZeq} becomes the alleged identity 
$$ \frac{4(2n+k+2)(2n +k+1)(n+1)^2}{(2n-k+2)(2n-k +1)(2n+1)^2} - 1 $$
$$= \frac{(4n+3)(2k+1)(2n+k+1)}{(2n+1-k)(2n+1)^2} + \frac{2(4n+3)k^3}{(2n+1-k)(2n +2-k)(2n+1)^2}?$$
Combining the fractions shows that each side leads to
$$\frac{(4n+3)(2 + 5k + k^2 + 6n + 12kn + 4n^2 +8kn^2)}{(2n-k+1)(2n-k+2)(2n+1)^2}.$$
So \eqref{WZeq} does actually hold. Since $G$ is telescoping, we finally obtain 
$$\sum_{k=0}^{2n} \, F(n, k) = \mbox{const}.$$
This constant does not depend on $n$. Thus \eqref{WZid}, and so \eqref{comb1}, follows from setting $n = 0$. 
\end{proof}

Similarly, applying the WZ-method, we can establish

\begin{lemma} For any non-negative integer $n$, 
\begin{equation}\label{comb2}
\sum_{k=0}^{2n} \, (-1)^k{2n \choose k}{2n+k \choose k}\frac{3\cdot 7\cdots (4k-1)}{(2k+1)k!{2k \choose k}} = \frac{1}{(4n+1)4^n}\,{2n \choose n};\end{equation}
\small
\begin{equation}\label{comb3}
\sum_{k=0}^{2n} \, (-1)^k{2n \choose k}{2n+k \choose k}\frac{5\cdot 9\cdots(4k+1)}{(2k+3)(2k+1)k!{2k \choose k}} = \frac{1}{(4n+3)(1-4n)4^n}\,{2n \choose n}.
\end{equation}
\end{lemma} 
\normalsize

Now we are ready to establish some Fourier-Legendre series in explicit form.

\begin{theorem} For $x \in (0,1)$, 
\begin{equation}\label{series1}
\frac{1}{\sqrt{x(1-x)}} = \sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n}\,{2n \choose n}^2\tilde{P}_{2n}(x).
\end{equation}
\end{theorem}

\begin{proof} Let $f(x) = 1/\sqrt{x(1-x)}$. Then $f(x) = f(1-x)$ for all $x \in (0,1)$. Using Lemma 2,  $a_n = 0$ for odd $n$. Moreover,
$$a_{2n} = (4n+1)\int_0^1\,f(x)\tilde{P}_{2n}(x)\,dx.$$
Invoking \eqref{Polyexp} leads to
\begin{align*}
a_{2n} & =  (4n+1)\sum_{k=0}^{2n}\,(-1)^k{2n \choose k}{2n+k \choose k}\int_0^1\,x^{k-1/2}(1-x)^{-1/2}\,dx\\
& =  (4n+1)\sum_{k=0}^{2n}\,{2n \choose k}{2n+k \choose k}(-1)^k B(k+1/2, 1/2),
\end{align*}
where $B(x,y)$ is the Euler Beta function. Using
\begin{equation}\label{gammahalf}
\Gamma(1/2) = \sqrt{\pi}\hspace{0.2in} \mbox{and}\hspace{0.2in} \Gamma(k + 1/2) = \frac{(2k)!}{4^k\,k!}\,\sqrt{\pi},
\end{equation}
it follows that
$$B(k+1/2, 1/2) = \frac{\Gamma(k+1/2)\Gamma(1/2)}{\Gamma(k+1)} = \frac{(2k)!}{4^k\,k!^2}\,\pi = \frac{\pi}{4^k}{2k \choose k}.$$
This, together with \eqref{comb1}, implies
$$a_{2n} = (4n+1)\pi\sum_{k=0}^{2n}\,(-1)^k{2n \choose k}{2n+k \choose k}{2k \choose k}4^{-k} = \frac{(4n+1)\pi}{16^n}\,{2n \choose n}^2,$$
which proves \eqref{series1}.
\end{proof}

Applying \eqref{Pcoef} to \eqref{series1} gives
\begin{align}\label{series2}
\frac{x}{\sqrt{x(1-x)}}  = {} &\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{2 \cdot 16^n}\,{2n \choose n}^2\tilde{P}_{2n}(x)\nonumber \\
& + \sum_{n=0}^\infty\,\frac{(4n+3)(2n+1)\pi}{4\cdot 16^n(n+1)}\,{2n \choose n}^2\tilde{P}_{2n+1}(x).
\end{align}
While applying \eqref{Pcoef} to \eqref{series2} yields
\begin{align}\label{series3}
\frac{x^2}{\sqrt{x(1-x)}} =  {} &\sum_{n=0}^\infty\,\frac{(4n+1)(8n^2+4n-3)\pi}{8 \cdot 16^n(n+1)(2n-1)}\,{2n \choose n}^2\tilde{P}_{2n}(x)\nonumber \\
& + \sum_{n=0}^\infty\,\frac{(4n+3)(2n+1)\pi}{4\cdot 16^n(n+1)}\,{2n \choose n}^2\tilde{P}_{2n+1}(x).
\end{align}
Notice that
$$\sqrt{x(1-x)} = \frac{x}{\sqrt{x(1-x)}} - \frac{x^2}{\sqrt{x(1-x)}}.$$
Combining \eqref{series2} and \eqref{series3} leads to

\begin{theorem} For $x \in (0, 1)$,
\begin{equation}\label{series4}
\sqrt{x(1-x)} = \frac{1}{8}\,\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n(n+1)(1 - 2n)}\,{2n \choose n}^2\tilde{P}_{2n}(x).
\end{equation}
\end{theorem}

Applying the above process to \eqref{series4} yields

\begin{theorem} For $x \in (0, 1)$, 
\begin{equation}\label{series5}
\left(\sqrt{x(1-x)} \right)^3= \frac{9}{8^2}\,\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n(n+1)(n+2)(1 - 2n)(3-2n)}\,{2n \choose n}^2\tilde{P}_{2n}(x).
\end{equation}
\end{theorem}
By mathematical induction, we conclude

\begin{theorem} Let $m$ be a positive odd integer, and let $p = (m+1)/2$. Then 
%$$\left(\sqrt{x(1-x)}\right)^m $$
%$$ = \frac{(m!!)^2}{8^p}\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n(n+1)\cdots(n+p-1)(1-2n)\cdots(2p-1-2n)}\,{2n \choose n}^2\tilde{P}_{2n}(x).$$
\begin{eqnarray*}
\lefteqn{\left(\sqrt{x(1-x)}\right)^m}\\
 && = \frac{(m!!)^2}{8^p}\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n(n+1)\cdots(n+p-1)(1-2n)\cdots(2p-1-2n)}\,{2n \choose n}^2\tilde{P}_{2n}(x).
\end{eqnarray*}
In terms of the Pochhammer symbol, $(a)_k := a(a+1)\cdots(a+k-1)$ for $k \geq 1$, the above expression can be rewritten compactly as
\begin{equation}\label{series6}
\left(\sqrt{x(1-x)}\right)^m
 = \frac{(m!!)^2}{16^p}\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n(n+1)_p(1/2 -n)_p}\,{2n \choose n}^2\tilde{P}_{2n}(x).
\end{equation}
\end{theorem}

Next, let $f(x) = [x(1-x)]^{-1/4}$. Using Lemma 2,  $a_n = 0$ for odd $n$. Moreover, in view of \eqref{Polyexp}, 
\begin{align}\label{lroot4}
a_{2n} & =  (4n+1)\sum_{k=0}^{2n}\,(-1)^k{2n \choose k}{2n+k \choose k}\int_0^1\,x^{k-1/4}(1-x)^{-1/4}\,dx \nonumber\\
& =  (4n+1)\sum_{k=0}^{2n}\,{2n \choose k}{2n+k \choose k}(-1)^k B(k+3/4, 3/4).
\end{align}
Using \eqref{gammahalf} and 
$$\Gamma(k + 3/4) = \frac{3\cdot 7 \cdots (4k-1)}{4^k}\Gamma(3/4),$$
we have
\begin{align*}
B(k+3/4, 3/4) & =  \frac{\Gamma(k+3/4)\Gamma(3/4)}{\Gamma(k+3/2)}\\
& =  \frac{4\Gamma^2(3/4)}{\sqrt{\pi}}\frac{3\cdot 7 \cdots (4k-1)(k+1)!}{(2(k+1))!}\\
& =  \frac{4\Gamma^2(3/4)}{\sqrt{\pi}}\frac{3\cdot 7 \cdots (4k-1)}{2(2k+1)k!{2n \choose n}}.
\end{align*}
Substituting this into \eqref{lroot4} and then applying \eqref{comb2} yields
$$a_{2n} =  \frac{2\Gamma^2(3/4)}{\sqrt{\pi}}\frac{1}{4^n}{2n \choose n}.$$
Notice that
\begin{equation}\label{gamma34}
\Gamma(3/4) \Gamma(1/4)= \frac{\pi}{\sin(\pi/4)} = \sqrt{2}\, \pi.
\end{equation}
\begin{theorem} For $x \in (0, 1)$,
\begin{equation}\label{series7}
\frac{1}{\sqrt[4]{x(1-x)}} = \frac{4\pi^{3/2}}{\Gamma^2(1/4)}\sum_{n=0}^\infty\,\frac{1}{4^n}\,{2n \choose n}\tilde{P}_{2n}(x).
\end{equation}
\end{theorem}

Finally, applying \eqref{Polyexp}, \eqref{comb3} and \eqref{gamma34} gives

\begin{theorem} For $x \in (0, 1)$, 
\begin{equation}\label{series8}
\sqrt[4]{x(1-x)} = \frac{\pi^{3/2}}{2\Gamma^2(3/4)}\,\sum_{n=0}^\infty\,\frac{4n+1}{4^n(4n+3)(1-4n)}\,{2n \choose n}\tilde{P}_{2n}(x).
\end{equation}
\end{theorem}

\section{Some new Ramanujan-like series}
\label{sec3}

The Fourier-Legendre series expansions \eqref{series1}, \eqref{series4}, \eqref{series5}, \eqref{series6}, \eqref{series7} and \eqref{series8}, along with their specialization, inner product  and Parseval's identity, will enable us to find many Ramanujan-like series involving powers of central binomial coefficients.

First, replacing $2x-1$ by $t$ in \eqref{series1}, \eqref{series2} and \eqref{series4}, 

\begin{theorem} For $|t| < 1$,
\begin{align*}
\frac{1}{\sqrt{1-t^2}} & =  \frac{1}{2}\,\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n}\,{2n \choose n}^2\,P_{2n}(t);\\
\frac{t}{\sqrt{1-t^2}} & =  \frac{1}{4}\,\sum_{n=0}^\infty\,\frac{(4n+3)(2n+1)\pi}{16^n(n+1)}\,{2n \choose n}^2\,P_{2n+1}(t);\\
\sqrt{1-t^2} & =  \frac{1}{8}\,\sum_{n=0}^\infty\,\frac{(4n+1)\pi}{16^n(n+1)(1/2 -n)}\,{2n \choose n}^2\,P_{2n}(t).
\end{align*}
\end{theorem}
These recover Formulas 8.922 No.\ 3--5 in \cite[p.\ 988]{GR},  
for which I am unable to find references to their proofs. It is interesting to compare these identities with the well-known generating function of the central binomial coefficients:
$$\frac{1}{\sqrt{1-t^2}} = \sum_{n=0}^\infty\,\frac{1}{4^n}\,{2n \choose n}t^{2n}.$$

Furthermore, let $x = \cos^2(\theta/2)$ in \eqref{series1}, \eqref{series4}--\eqref{series6}, respectively. Then $2x-1 = \cos\theta$, and $\tilde{P}_{2n}(x) = P_{2n}(\cos\theta)$. 

\begin{theorem} For $\theta \in (0, \pi)$,
\begin{align}
\sum_{n=0}^\infty\,\frac{4n+1}{16^n}\,{2n \choose n}^2P_{2n}(\cos\theta) & =  \frac{2}{\pi\sin\theta},\\
\sum_{n=0}^\infty\,\frac{4n+1}{16^n(n+1)(1/2-n)}\,{2n \choose n}^2P_{2n}(\cos\theta)   & =   \frac{8\sin\theta}{\pi},\label{eq:26}\\
\sum_{n=0}^\infty\,\frac{4n+1}{64^n(n+1)(n+2)(1/2-n)(3/2-n)}\,{2n \choose n}^2P_{2n}(\cos\theta)  & =   \frac{32\sin^3\theta}{9\pi},\\
\sum_{n=0}^\infty\,\frac{4n+1}{64^n(n+1)_p(1/2-n)_p}\,{2n \choose n}^2P_{2n}(\cos\theta)  & =  \frac{16^p\sin^m\theta}{2^m(m!!)^2\pi},\label{eq:28}
\end{align}
where $m$ is a positive odd integer and $p= (m+1)/2$. 
\end{theorem}

Identity \eqref{eq:26} appeared in \cite[p.\ 989]{GR} as Formula 8.925 No.\ 2. Next, let $x = \cos^2(\theta/2)$ in \eqref{series7} and \eqref{series8}, respectively.

\begin{theorem} For $\theta \in (0, \pi)$,
\begin{align}
\sum_{n=0}^\infty\,\frac{1}{4^n}\,{2n \choose n}P_{2n}(\cos\theta) & =  \frac{\sqrt{2}\,\Gamma^2(1/4)}{4\sqrt{\sin\theta}\pi^{3/2}},\\
\sum_{n=0}^\infty\,\frac{4n+1}{4^n(4n+3)(1-4n)}\,{2n \choose n}P_{2n}(\cos\theta)  & =   \frac{\sqrt{2\sin\theta}\,\Gamma^2(3/4)}{\pi^{3/2}}.\label{eq:30}
\end{align}
\end{theorem}
Based on Theorems 13 and 14, via specialization, we find some Ramanujan-like series as follows.

{\bf Example 1.} Recall that $P_{2n}(0) = (-1)^n{2n \choose n}/4^n$. Letting $\theta = \pi/2$ in Theorem 13 yields
\begin{align}
\sum_{n=0}^\infty\,\frac{(-1)^n(4n+1)}{64^n}\,{2n \choose n}^3 & =  \frac{2}{\pi},\label{Bailey}\\
\sum_{n=0}^\infty\,\frac{(-1)^n(4n+1)}{64^n(n+1)(1/2-n)}\,{2n \choose n}^3  & =   \frac{8}{\pi},\\
\sum_{n=0}^\infty\,\frac{(-1)^n(4n+1)}{64^n(n+1)(n+2)(1/2-n)(3/2-n)}\,{2n \choose n}^3  & =   \frac{32}{9\pi},\label{eq:33}\\
\sum_{n=0}^\infty\,\frac{(-1)^n(4n+1)}{64^n(n+1)_p(1/2-n)_p}\,{2n \choose n}^3  & =  \frac{16^p}{2^m(m!!)^2\pi},\label{eq:34}
\end{align}
where $m$ is a positive odd integer and $p= (m+1)/2$. \eqref{Bailey} reveals Bailey's classical formula (See \cite{Bailey}, or \cite[Formula 5.5.24, p.\ 184]{Borweins}), which was originally obtained via hypergeometric functions. 


{\bf Example 2.} Let $\theta = \pi/2$ in Theorem 14. Then
\begin{align}
\sum_{n=0}^\infty\,\frac{(-1)^n}{16^n}\,{2n \choose n}^2 & =  \frac{\sqrt{2}\Gamma^2(1/4)}{4\pi^{3/2}},\\
\sum_{n=0}^\infty\,\frac{(-1)^n(4n+1)}{16^n(4n+3)(1-4n)}\,{2n \choose n}^2  & =   \frac{\sqrt{2}\Gamma^2(3/4)}{\pi^{3/2}}.
\end{align}

We also obtain some interesting series following in another direction:

{\bf Example 3.} Since $P_n(1) = 1$ for all $n \geq 0$, letting $\theta = 0$ in \eqref{eq:30}, \eqref{eq:26}--\eqref{eq:28}, respectively, then splitting the $n=0$ term, we find
\begin{align}
\sum_{n=1}^\infty\,\frac{4n+1}{4^n(4n +3)(4n -1)}\,{2n \choose n}  & =  \frac{1}{3},\\
\sum_{n=1}^\infty\,\frac{4n+1}{16^n(n+1)(1/2-n)}\,{2n \choose n}^2 & =  -2,\\
\sum_{n=1}^\infty\,\frac{4n+1}{16^n(n+1)(n+2)(1/2-n)(3/2-n)}\,{2n \choose n}^2 &  =  - \frac{2}{3},\\
\sum_{n=1}^\infty\,\frac{4n+1}{16^n(n+1)_p(1/2-n)_p}\,{2n \choose n}^2  & =  - \frac{1}{p!(1/2)_p},
\end{align}
where $p \in \mathbb{N}$.

Next, let $f(x) = \sum_{n=0}^\infty a_n\tilde{P}_n(x)$ and $g(x) = \sum_{n=0}^\infty b_n\tilde{P}_n(x)$. If $f(x)g(x) \in L^2[0,1]$, then the inner product of $f$ and $g$ leads to
\begin{equation}\label{innprod}
\sum_{n=0}^\infty\, \left(\frac{1}{\sqrt{2n+1}}\,a_n\right)\cdot \left(\frac{1}{\sqrt{2n+1}}\,b_n\right)= \int_0^1\,f(x)g(x)\,dx.
\end{equation}
The inner products of \eqref{series1} with \eqref{series4},   \eqref{series1} with \eqref{series5}, \eqref{series4} with \eqref{series5} and  \eqref{series1} with \eqref{series6} lead to the following

\begin{theorem} Let $m$ be a positive odd integer and $p= (m+1)/2$. Then
\begin{align}
\sum_{n=0}^\infty\,\frac{4n+1}{256^n(n+1)(1/2 - n)}\,{2n \choose n}^4 &=  \frac{16}{\pi^2},\\
\sum_{n=0}^\infty\,\frac{4n+1}{256^n(n+1)(n+2)(1/2-n)(3/2 - n)}\,{2n \choose n}^4 &=   \frac{128}{27\pi^2},\\
\sum_{n=0}^\infty\,\frac{4n+1}{256^n(n+1)^2(n+2)(1/2-n)^2(3/2-n)}\,{2n \choose n}^4 & =   \frac{2048}{135\pi^2},\\
\sum_{n=0}^\infty\,\frac{(4n+1)}{256^n(n+1)_p(1/2-n)_p}\,{2n \choose n}^4  &=  \frac{16^p[(p-1)!]^2}{(m!!)^2m!\pi^2}.
\end{align} 
\end{theorem}
Similarly, the inner products of \eqref{series1} with \eqref{series7},   \eqref{series1} with \eqref{series8}, \eqref{series4} with \eqref{series7}, \eqref{series4} with \eqref{series8},  \eqref{series5} with \eqref{series7}, and \eqref{series5} with \eqref{series8} lead to

\begin{theorem}
\begin{align}
\sum_{n=0}^\infty\,\frac{1}{64^n}\,{2n \choose n}^3 &=  \frac{\Gamma^4(1/4)}{4\pi^3},\label{eq:46}\\
\sum_{n=0}^\infty\,\frac{4n+1}{64^n(4n+3)(1-4n)}\,{2n \choose n}^3 &=   \frac{4\Gamma^4(3/4)}{\pi^3},\\
\sum_{n=0}^\infty\,\frac{1}{64^n(n+1)(1/2-n)}\,{2n \choose n}^3 & =   \frac{\Gamma^4(1/4)}{3\pi^3},\\
\sum_{n=0}^\infty\,\frac{(4n+1)}{64^n(n+1)(1/2-n)(4n+3)(1-4n)}\,{2n \choose n}^3  &= \frac{48\Gamma^4(3/4)}{5\pi^3},\\
\sum_{n=0}^\infty\,\frac{1}{64^n(n+1)(n+2)(1/2-n)(3/2-n)}\,{2n \choose n}^3  &=   \frac{20\Gamma^4(1/4)}{189\pi^3},\\
\sum_{n=0}^\infty\,\frac{(4n+1)}{64^n(n+1)(n+2)(1/2-n)(3/2-n)(4n+3)(1-4n)}\,{2n \choose n}^3  &= \frac{448\Gamma^4(3/4)}{135\pi^3}.
\end{align} 
Furthermore, the inner product of \eqref{series7} and \eqref{series8} leads to a simple Ramanujan-like series:
\begin{equation}
\sum_{n=0}^\infty\,\frac{1}{16^n(4n+3)(1-4n)}\,{2n \choose n}^2 =  \frac{1}{\pi}.
\end{equation}
\end{theorem}
In general, taking the inner product \eqref{series6} with \eqref{series7} and \eqref{series8}, respectively, yields

\begin{theorem} Let $m$ be a positive odd integer and $p= (m+1)/2$. Then
\begin{align}
\sum_{n=0}^\infty\,\frac{1}{64^n(n+1)_p(1/2-n)_p}\,{2n \choose n}^3  &=   \frac{16^p(1 \cdot 5\cdot (4p-3))^2(2p)!\Gamma^4(1/4)}{4(m!!)^2 (4p)!\pi^3},\\
\sum_{n=0}^\infty\,\frac{(4n+1)}{64^n(n+1)_p(1/2-n)_p(4n+3)(1-4n)}\,{2n \choose n}^3  &= \frac{16^{p+1}(3 \cdot 7\cdot (4p-1))^2(2p+1)!\Gamma^4(3/4)}{2(m!!)^2 (4p+2)!\pi^3}.
\end{align} 
\end{theorem}

Finally, recall Parseval's identity: If $f(x)\in L^2[0,1]$, then
\begin{equation}\label{Parseval}
\sum_{n=0}^\infty\, \left[\frac{1}{\sqrt{2n+1}}\,a_n\right]^2 = \int_0^1\,f^2(x)\,dx.
\end{equation}
Applying \eqref{Parseval} to \eqref{series7}, \eqref{series8} and \eqref{series4}--\eqref{series6}, respectively, yields

\begin{theorem} Let $m$ be a positive odd integer and $p= (m+1)/2$. Then
\begin{align}
\sum_{n=0}^\infty\,\frac{1}{16^n(4n+1)}\,{2n \choose n}^2 &=  \frac{\Gamma^4(1/4)}{16\pi^2},\\
\sum_{n=0}^\infty\,\frac{4n+1}{16^n(4n+3)^2(1-4n)^2}\,{2n \choose n}^2 &=   \frac{\Gamma^4(3/4)}{2\pi^2},\\
\sum_{n=0}^\infty\,\frac{4n+1}{256^n(n+1)^2(1/2 - n)^2}\,{2n \choose n}^4 &=  \frac{128}{3\pi^2},\label{eq:58}\\
\sum_{n=0}^\infty\,\frac{4n+1}{256^n(n+1)^2(n+2)^2(1/2-n)^2(3/2 - n)^2}\,{2n \choose n}^4 &=   \frac{16384}{2835\pi^2},\\
\sum_{n=0}^\infty\,\frac{(4n+1)}{256^n(n+1)_p^2(1/2-n)_p^2}\,{2n \choose n}^4  &=  \frac{256^p (m!)^2}{(m!!)^4(2m+1)!\pi^2}.\label{eq:60}
\end{align} 
\end{theorem}

\section{Remarks}
\label{sec4}
We now conclude this paper with two remarks.

\begin{enumerate}
\item Recall the complete elliptical integral of the first kind defined by
$$K(k) := \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - k^2\sin^2\theta}}.$$
Invoking the generating function of the central binomial coefficients
$$\sum_{n=0}^\infty\, \frac{1}{4^n}{2n \choose n}x^{2n} = \frac{1}{\sqrt{1 - x^2}}\,\,\,\,\,\mbox{for $|x| < 1$}$$
and Wallis' formula
$$\int_0^{\pi/2} \sin^{2n}\theta\,d\theta = \frac{\pi}{2^{2n+1}}\,{2n \choose n},$$
we have
$$K(k) = \frac{\pi}{2} \sum_{n=0}^\infty \frac{1}{16^n}{2n \choose n}^2k^{2n}.$$
Moreover, Watson's elliptic function identity \cite[p.\ 188]{Borweins} claims
$$ \frac{\pi^2}{4} \sum_{n=0}^\infty\, \frac{1}{64^n}{2n \choose n}^3 = K^2(1/\sqrt{2}).$$
Thus, together with \eqref{eq:46}, we obtain another proof of the classical result \cite[Theorem 1.7, p.\ 25]{Borweins}:
$$K\left(\frac{1}{\sqrt{2}}\right) = \frac{\Gamma^2(1/4)}{4\sqrt{\pi}}.$$

\item The neat forms of Fourier-Legendre series in Section 3 rely on the combinatorial identities \eqref{comb1}, \eqref{comb2} and \eqref{comb3}. Similar to \eqref{comb1}, by the WZ-method, Tauraso \cite[Lemma 4.2]{Tauraso2} proved
$$\sum_{k=0}^{2n}\,(-1)^k {2n \choose k}{2n+k \choose k}{2k \choose k}4^{2n-k}H_k = {2n \choose n}^2H_{2n}.$$
This enables us to find Ramanujan-like series and other well-known constants involving the product of the central binomial coefficients and the harmonic numbers like 
$$\sum_{n=1}^\infty \frac{1}{16^n(2n-1)}{2n\choose n}^2H_{2n} = \frac{6\ln 2-2}{\pi}$$
and
$$\sum_{n=0}^\infty \frac{1}{16^n(2n+1)}{2n\choose n}\left(3H_{2n+1}  + \frac{4}{2n+1}\right)= 8G,$$
where $G$ is the Catalan constant defined by $\sum_{n=0}^\infty (-1)^n/(2n+1)^2$. The interested reader is encouraged to pursue results in this direction.
\end{enumerate}

\noindent{\it Addendum.} After this paper was submitted for publication, the referee 
brought my attention to the papers \cite{Levrie} and \cite{CD}. 
The identities \eqref{eq:33}, \eqref{eq:34}, and \eqref{eq:58}--\eqref{eq:60} are found in \cite{Levrie}, in which Levrie used a different approach to derive the Fourier-Legendre series \eqref{series6}. 
The identities \eqref{eq:58}--\eqref{eq:60} were also rediscovered by
Cantarini and D'Aurizio \cite{CD}.

\section{Acknowledgments}

Thanks to the anonymous referee for useful
comments and bringing to my attention the papers \cite{Levrie} and
\cite{CD}. Also thanks to Dr. Bradie and the editor-in-chief for their
suggestions which have made this paper more reader-friendly.


\begin{thebibliography}{10}

\bibitem{Bailey} 
W. N. Bailey, {\it Generalized Hypergeometric Series}, Cambridge
Univ. Press, 1935.

\bibitem{BBChan} 
N. D. Baruah, B. C. Berndt, and H. H. Chan,
Ramanujan's series for $1/\pi$: a survey, {\it Amer. Math. Monthly}
{\bf 116} (7) (2009), 567--587. 

\bibitem{Borweins} 
J. M. Borwein and P. B. Borwein, {\it Pi and the AGM: A Study in
Analytical Number Theory and computational Complexity}, Canadian
Mathematical Society Series of Monographs and Advanced Texts,
John Wiley, 1987.

\bibitem{Campbell} 
J. M. Campbell, Ramanujan-like series for $1/\pi$ involving harmonic numbers, {\it Ramanujan J.\/} {\bf 46} (2018), 373--387.

\bibitem{CD} 
M. Cantarini and J. D'Aurizio, On the interplay between hypergeometric
series, Fourier-Legendre expansions and Euler sums, {\it Boll. Unione
Mat. Ital.} {\bf 12} (2019), 623--656.

\bibitem{Chen} 
H. Chen, Interesting series associated with central binomial coefficients,
Catalan numbers, and harmonic numbers,
{\it J. Integer Sequences} {\bf 19} (2016),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL19/Chen/chen21.html}{Article 16.1.5}.

\bibitem{Chowla} 
S. Chowla, On the sum of a certain infinite series, {\it T\^{o}hoku
Math. J.} {\bf 29} (1928), 291--295.

\bibitem{Gould} 
H. W. Gould, {\it Combinatorial Identities},
West Virginia University, revised edition, 1972.

\bibitem{GR} 
I. S. Gradshteyn and I. M. Ryzhik, \textit{Table of Integrals, Series and
Products}, 7th ed., Alan Jeffrey and Daniel Zwillinger, eds., Academic
Press, 2007.

\bibitem{Guillera1} 
J. Guillera, A new method to obtain series of $1/\pi$ and $1/\pi^2$,
{\it Experiment. Math.} {\bf 15} (2006), 83--89.

\bibitem{Guillera2} 
J. Guillera, Hypergeometric identities for 10 extended Ramanujan type
series, {\it Ramanujan J.} {\bf 15} (2008), 219--234.

\bibitem{Levrie} 
P. Levrie, Using Fourier-Legendre expansions to derive series for $1/\pi$
and $1/\pi^2$, {\it Ramanujan J.} {\bf 22} (2010), 221--230.

\bibitem{Ramanujan} 
S. Ramanujan, Modular equations and approximations to $\pi$, {\it Quart
J. Math.} {\bf 45} (1914), 350--372.

\bibitem{Tauraso} 
R. Tauraso, Personal communication, 2021. 

\bibitem{Tauraso2}
R. Tauraso, Supercongruences related to $_3F_2(1)$ involving harmonic
numbers, {\it Int. J. Number Theory} {\bf 14} (2018), 1093--1109.

\bibitem{WChu} 
X. Wang and W. Chu, Further Ramanujan-like series containing harmonic
numbers and squared binomial coefficients, {\it Ramanujan J.} {\bf 52}
(2020), 641--668.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent {\it 2020 Mathematics Subject Classification:} Primary 11B83; Secondary 05A10.

\noindent \emph{Keywords:} central binomial coefficient, Legendre
polynomial, Ramanujan-like series, Fourier-Legendre series.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000045},
\seqnum{A000984},
\seqnum{A001008}, and
\seqnum{A005408}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  October 8 2021; revised version received January 5 2022.
Published in {\it Journal of Integer Sequences},
January 28 2022.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                

