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\begin{center}
\vskip 1cm{\LARGE\bf 
Infinite Series Associated with the Ratio \\
\vskip .1in
and Product of Central Binomial Coefficients
}
\vskip 1cm
\large
Narendra Bhandari\\ 
Department of Mathematics \\
University of Delhi\\
Budhiganga-01, Bajura\\
Nepal \\
\href{mailto: narenbhandari04@gmail.com}{\tt narenbhandari04@gmail.com}
\end{center}

\vskip .2 in

\begin{abstract}
We establish several interesting series associated with the ratio and
the product of central binomial coefficients, namely ${2n\choose n}$ and
${4n\choose 2n} $. Series involving the product of central binomial coefficients
can be found in the papers of Campbell, D'Aurizio, and Sondow. In this
paper, through the application of integration methods, we
address broad generalizations of both classes. The techniques involved
in constructing the integrals for the corresponding series are 
based on the use of the ordinary generating functions of central binomial
coefficients and Wallis' well-known integral formulas.
\end{abstract}



\section{Introduction}
\label{sec1}
The central binomial coefficients $${2n\choose n}=\frac{(2n)!}{(n!)^2}$$
for $n\geq 0$ are the positive integers that appear exactly in the middle of the even-indexed rows of the Pascal triangle. These numbers play an important role in various fields such as analysis, number theory, and combinatorics. A number of facts about central binomial coefficients were compiled by Gould~\cite{Gould}. Lehmer~\cite{Lehmer} gave diverse identities and interesting results via specialization, integration, and differentiation tricks. Surprisingly, the coefficients appear in the infinite binomial series expansion of the function $(1-4x)^{-1/2}$ for $|x|<1/4$, yielding the ordinary generating function as follows: 
\begin{equation}\label{eqn:1}
    \sum_{n=0}^{\infty}{2n \choose n}x^n = \frac{1}{\sqrt{1-4x}}=1+2x+6x^2+20x^3+70x^4+\cdots.
\end{equation}
On the right-hand side of \eqref{eqn:1}, we can observe the coefficients $1,6,70,924,$ etc., are the coefficients appearing at the even positions in the series. Considering the coefficients, we notice that these numbers can be represented in binomial form, namely ${4n\choose 2n}$, which mean that they can be obtained by replacing $n$ with $2n$. It is easy to note that \eqref{eqn:1} allows us to build the generating function of the coefficients ${4n\choose 2n}$, which is
\begin{equation}\label{eqn:2}
    \sum_{n=0}^{\infty}{4n\choose 2n} x^{2n}=\frac{1}{2}\left(\frac{1}{\sqrt{1-4x}}+\frac{1}{\sqrt{1+4x}}\right).
\end{equation} However, further routine simplification of \eqref{eqn:2} produces \begin{equation}
    \sum_{n=0}^{\infty}{4n\choose 2n}x^n =\frac{1}{\sqrt 2}\sqrt{\frac{1+\sqrt{1-16x}}{1-16x}}, \quad |x|<1/16,
\end{equation}
which can be found in \cite{Bhandari}, including several other generating functions and related identities. The study of these numbers has been of great interest for a long time. Many interesting series associated with the central binomial coefficients, harmonic numbers, and Catalan numbers can be found in \cite{Boya,Campbell,JJJ,Chen}, and among those several interesting series, two beautiful identities involving the square of central binomial coefficients and the harmonic numbers are 
$$
\sum_{n=1}^{\infty}\frac{{2n\choose n}^2}{16^n(2n-1)^2}H_{n} =\frac{12-16\ln  2}{\pi} $$
and
$$\sum_{n=1}^{\infty}\frac{{2n\choose n}^2}{16^n\,(2n-1)^2}H_{2n} = \frac{4G-12\ln 2+12}{\pi}.
$$
Here \(H_n\) is the $n$th harmonic number, and $G$ is Catalan's constant. The identities mentioned above exhibit mathematical beauty where one can observe the important mathematical constants $ G, \pi$, and $\ln 2$ in their closed forms. We encourage interested readers to refer to the papers given in \cite{Campbell1, Chen1}, and we suggest looking at
the references given therein for more identities.

Now focusing on the present aim of this paper, we study the series associated with the product and the ratio of central  binomial coefficients, ${2n\choose n}{4n\choose 2n}$ and ${{2n\choose n}}{{4n\choose n}^{-1}}$, respectively. Some motivating examples of the product of these coefficients can be found in the works of Campbell et al.\ \cite{JJJ}, where those identities are obtained via generating functions and the well-known Wallis integral formulas. For instance,  
\begin{align}
    \sum_{n=0}^{\infty}\frac{{4n\choose 2n}{2n\choose n}}{64^n(2n+1)}& =\frac{4}{\pi}\log\left(1+\sqrt 2\right), \\
    \sum_{n=0}^{\infty}\frac{{4n\choose 2n}{2n\choose n}}{64^n(3+2n)}& =\frac{4\sqrt 2}{15\pi}+\frac{16}{15\pi}\log\left(1+\sqrt 2\right).
\end{align} 
The techniques discussed in this paper are simple but powerful for evaluating non-trivial series related to the product and the ratio of central binomial coefficients. Our general results are expressed in terms of finite binomial sums, which include the incomplete beta function, while other general results are expressed explicitly in terms of the beta function and the incomplete beta function. 
When the obtained general results presented in this paper are executed by the use of Mathematica or by direct evaluation of the corresponding integrals, we find a large number of intriguing identities involving the ratio and the product of central binomial coefficients. A few of them are Ramanujan-like series for $1/\pi$. Some interesting series related to the ratio of central binomial coefficients are as follows:
\begin{align}\label{PC1}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(4n+1)}\frac{{2n\choose n}}{{4n\choose 2n}} & =\frac{5\sqrt 2-4}{9}, \\ 
   \sum_{n=1}^{\infty}\frac{n(2n+1)\,4^n}{(2n-1)^2(4n+1)(4n+3)}\frac{{2n\choose n}}{{4n\choose 2n}}\label{PC2}& =\frac{121\sqrt 2-104}{450}.
\end{align}
We state Wallis' well-known integral formulas, namely 
\begin{align}
   \int_0^{\frac{\pi}{2}}\sin^{2n}y\, dy \label{WF1} & =\frac{\pi}{2}{2n\choose n}\frac{1}{4^n}, \\ \int_0^{\frac{\pi}{2}}\sin^{2n+1}y\, dy \label{WF2} & =\frac{4^n}{(2n+1){2n\choose n}},
\end{align}
that will be helpful in the course of the analysis of the results of in this paper. We organize the remaining work into different sections. In Section~\ref{sec2}, we introduce several series involving the ratio of the central binomial coefficients. Series involving the product of the coefficients are highlighted in Section~\ref{sec3}, where we list a few series involving
Ramanujan-like formulas for $1/\pi$. In Section~\ref{sec4}, we highlight a number of miscellaneous series. As an additional check,
all the formulas were numerically verified by the use of Mathematica.

\section{Interesting series of the ratio of central binomial coefficients}\label{sec2}
This section contains a number of series related to the ratio of central binomial coefficients. We present our main results in terms of finite binomial sums and the incomplete beta function. First, we mention some lemmas along with their proofs, and then we proceed to construct our main results.
\begin{lemma}
\label{Lemma1} If $x \in [-1,1]$, then
$$\sum_{n=1}^{\infty}\frac{n}{(2n-1)^2 4^n}{2n\choose n}x^{2n-1}=\frac{\sin^{-1} x}{2} .
$$
\end{lemma}
\begin{proof} 
We recall the generating function of central binomial coefficients \eqref{eqn:1}, and rewrite it as follows: 
\begin{align}
 \label{eqn:10} \sum_{n=1}^{\infty}{2n\choose n}\frac{x^{2n-2}}{4^n} &=\frac{1}{x^2\sqrt{1-x^2}}-\frac{1}{x^2}.
\end{align}
By integrating \eqref{eqn:10}, we obtain 
\begin{align}\label{eqn:11}
\sum_{n=1}^{\infty}{2n\choose n}\frac{x^{2n-1}}{(2n-1)4^n} =\frac{1-\sqrt{1-x^2}}{x}.
\end{align}
The primitive  of the integral of the latter quantity of \eqref{eqn:10}
exists; it is $x^{-1}-x^{-1}\sqrt{1-x^2}+C$, where \(C\) is
the integration constant. If \(x\to 0 \), then \( C=0\) and hence
\eqref{eqn:11} follows. Again, dividing both sides of \eqref{eqn:11}
by $x$ and integrating, we get
\begin{align}\label{eqn:12}
    \sum_{n=1}^{\infty}{2n\choose n}\frac{x^{2n-1}}{(2n-1)^24^n}=\frac{\sqrt{1-x^2}+x\sin^{-1} x-1}{x}.
\end{align}
We observe that $\int x^{-2}({1-\sqrt{1-x^2}})\, dx=x^{-1}\sqrt{1-x^2}+\sin^{-1} x-x^{-1}+K$, which can be easily deduced by applying linearity of integral and integration by parts to the latter integral, $\int x^{-2}\sqrt{1-x^2}\, dx$, where $K$ is  the constant of integration. If \(x\to 0\), then \( K =0\). Finally, we multiply both sides of \eqref{eqn:12} by $x$, and differentiating with respect to $x$, we obtain
\begin{align*}
\sum_{n=1}^{\infty}\frac{n}{(2n-1)^24^n}{2n\choose n}x^{2n-1}=\frac{1}{2}\frac{d}{dx}\left(\sqrt{1-x^2}+x\sin^{-1} x-1\right) =\frac{\sin^{-1} x}{2}.
\end{align*}
\end{proof}
\begin{lemma}
\label{Lemma2} If $x \in [-1,1]$, then 
\begin{align}\label{OGF}
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2\, (2n+1)\, 4^n}{2n\choose n}x^{2n} =\frac{1}{8}\left(\sqrt{1-x^2}+2x\sin^{-1} x-\frac{\sin^{-1} x}{x}\right).
\end{align}
\end{lemma}
\begin{proof}
Multiplying both sides of Lemma \ref{Lemma1} by $x$, integrating, and some algebraic simplification leads to the proposed result \eqref{OGF}. Details are left to the reader.
\end{proof}
\begin{lemma}
\label{Lemma3}
If $m$ is an odd positive integer, then 
\begin{align*} \int_0^{\frac{\pi}{2}}\sin^{4n+m}y\,  dy &=\frac{2^{2n+m-1}}{(4n+m){4n+m-1\choose 2n+\frac{m-1}{2}}}.
\end{align*}
\end{lemma}
\begin{proof}
Recall Wallis' integral formula \eqref{WF2}. First, we replace $n$ with $2n$, then $n$ with $\frac{n}{4}$, and again, the obtained result on replacing $n$ with $n-1$ gives us
\begin{align}\label{010}
\int_0^{\frac{\pi}{2}}\sin^{n}y \, dy & =\frac{2^{n-1}}{n {n-1\choose \frac{n-1}{2}}}.
\end{align} 
Further, replacing $n$ with $4n+m$ in  \eqref{010} yields the required result.
\end{proof}
\begin{lemma} For $x\in[-1,1]$, the following holds: 
\label{Lemma4}
\begin{align} 
\sum_{n=1}^{\infty}\frac{n^2}{(2n-1)^2(2n+1)4^n}{2n\choose n}x^{2n} =\frac{1}{16}\left(-\sqrt{1-x^2}+\frac{\sin^{-1} x}{x}+2x\sin^{-1} x\right).
\end{align}
\begin{proof}
The proof relies on Lemma \ref{Lemma2}. We leave the details to the reader.
\end{proof}
\end{lemma} 
\begin{theorem}
\label{Theorem4}
For positive odd integers $m$, the following equality holds:
\begin{align}
\label{the1}
 \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(4n+m)}\frac{{2n\choose n}}{{4n+m-1 \choose 2n+\frac{m-1}{2}}}={\sqrt 2}\sum_{k=0}^{\frac{m+1}{2}}\frac{(-1)^k}{(2k+1)2^{m-k}}{\frac{m+1}{2}\choose k}\mathcal{B}(k),
 \end{align}
 where $\mathcal{B}(k)=B_{\frac{1}{2}}\left(k+1,\frac{1}{2}\right)$ and  $B_x(a,b)=\int_0^xt^{a-1}(1-t)^{b-1}\, dt$ is the incomplete beta function.
\end{theorem}
\begin{proof}
Rewrite Lemma \ref{Lemma1} as follows: 
\begin{align}\label{eqn:13} 
 \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2 4^n}{2n\choose n}x^{4n+m}=\frac{x^{2+m}\sin^{-1} (x^2)}{2}.
\end{align}
Now set \(x=\sin y\) in \eqref{eqn:13} and integrate from $0$ to $\pi/2$.
From Lemma \ref{Lemma3} we get
\begin{align}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(4n+m)}\frac{{2n\choose n}}{{4n+m-1\choose 2n+\frac{m-1}{2}}} & \label{eqn18}=\frac{1}{2^{m}}\int_0^{\frac{\pi}{2}}\sin^{2+m}y\sin^{-1}(\sin^2y)\, dy\\ \nonumber &=\frac{1}{2^m}\int_0^{\frac{\pi}{2}}\left(1-\cos^2 y\right)^{\frac{m+1}{2}}\sin y\sin^{-1}\left(1-\cos^2y\right)\,dy\\ \label{int00} &=\frac{1}{2^m}\int_0^1\left(1-t^2\right)^{\frac{m+1}{2}}\sin^{-1}\left(1-t^2\right)\, dt\\ \label{int01}& =\frac{1}{2^m}\sum_{k=0}^{\frac{m+1}{2}}(-1)^k{\frac{m+1}{2}\choose k}\int_0^1t^{2k}\sin^{-1}\left(1-t^2\right)\,dt .
\end{align}
Employing Newton's binomial formula, $(a+b)^n=\sum_{k=0}^n{n\choose k}a^{n-k}b^k$ in \eqref{int00} leads to \eqref{int01} and applying integration by parts in \eqref{int01} gives 
\begin{align}
    \int_0^1 t^{2k}\sin^{-1}\left(1-t^2\right)\, dt \nonumber &=\frac{\sin^{-1}\left(1-t^2\right)\,t^{2k+1}}{2k+1}\Bigg|_0^1+\frac{2}{2k+1}\int_0^1\frac{t^{2k+2}}{\sqrt{1-(1-t^2)^2}}\, dt\\ \nonumber &=\frac{2}{2k+1}\int_0^1\frac{t^{2k+2}\, dt}{\sqrt{2t^2-t^4}}=\frac{2}{2k+1}\int_0^1\frac{t^{2k+1}\, dt}{\sqrt{2-t^2}}\\ \label{int02}&=\frac{2^{k+1}}{\sqrt{2}(2k+1)}\int_0^{\frac{1}{2}}\frac{t^k\, dt}{\sqrt{1-t}}=\frac{\sqrt{2}}{2^{-k}(2k+1)}B_{\frac{1}{2}}\left(k+1,\frac{1}{2}\right).
\end{align}
Substituting the value of \eqref{int02} into \eqref{int01} proves \eqref{the1}.
\end{proof}
\begin{theorem} For positive odd integers, $m \geq 3 $, we have
\label{theorem5}
\begin{align}
\label{Theorem5}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2\,(2n+1)(4n+m)}\frac{{2n\choose n}}{{4n+m-1 \choose 2n+\frac{m-1}{2}}} \nonumber & =\sum_{n=1}^{\infty}\frac{n\, 2^{2n-1}}{(2n-1)^2(4n+m)}\frac{{2n \choose n}}{{4n+m-1 \choose 2n+ \frac{m-1}{2}}}+\frac{\varphi(m)}{2^{m+2}}\\& -\frac{1}{2^{3/2}}\sum_{k=0}^{\frac{m-3}{2}}\frac{(-1)^k}{(2k+1)2^{m-k}}{\frac{m-3}{2} \choose k}\mathcal{B}(k), 
\end{align}
where $\varphi(m)=\int_0^1 t^m\sqrt{1+t^2}\, dt$.
\end{theorem}
\begin{proof}
 Rewrite Lemma \ref{Lemma2} in the following manner:
 \begin{align}\label{The2} 
     \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2\, (2n+1)}{2n\choose n} x^{4n+m} =\frac{x^{m}}{8}\left(\sqrt{1-x^4}+2x^2\sin^{-1}(x^2)-\frac{\sin^{-1}(x^2)}{x^2}\right). 
 \end{align}
Substituting $x=\sin y$ in \eqref{The2}, integrating from $0$ to $\pi/2$, and invoking Lemma \ref{Lemma3}, we get 
\begin{align}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2\, (2n+1)(4n+m)}\frac{{2n\choose n}}{{4n+m-1\choose 2n+\frac{m-1}{2}}}\label{eqn24} &= \frac{1}{2^{m+1}}\int_0^{\frac{\pi}{2}}\sin^{2+m} y\sin^{-1}\left(\sin^2 y\right)\, dy \\& \label{eqn25} +\frac{1}{2^{m+2}}\int_0^{\frac{\pi}{2}}\sin^m y\sqrt{1-\sin^4 y}\, dy\\& \label{eqn26} -\frac{1}{2^{m+2}}\int_0^{\frac{\pi}{2}}\sin^{m-2} y\sin^{-1}\left(\sin^2 y\right)\, dy.
\end{align}
Since integral \eqref{eqn24} is evaluated in \eqref{eqn18}, and the integral appears in \eqref{eqn25}, we write
\begin{align}\label{phi}
\int_0^{\frac{\pi}{2}}\sin^my \sqrt{1-\sin^4 y} \,dy =\int_0^{\frac{\pi}{2}}\sin^{m} y \cos y \sqrt{1+\sin^2 y}\, dy & =\int_0^1 t^m\sqrt{1+t^2}\, dt =\varphi(m).
\end{align} 
Similarly, integral \eqref{eqn26} can be expressed as
\begin{align}\label{eqn28} 
\int_0^{\frac{\pi}{2}}\sin^{m-2}y \sin^{-1}\left(\sin^2 y\right)\, dy  \nonumber &=\int_0^1\sin y\left(\sin^2 y\right)^{\frac{m-3}{2}}\sin^{-1}\left(\sin^2 y\right)\, dy \\& \nonumber =\int_0^1\left(1-t^2\right)^{\frac{m-3}{2}}\sin^{-1}\left(1-t^2\right)\,dt\\& =\sum_{k=0}^{\frac{m-3}{2}}(-1)^k{\frac{m-3}{2} \choose k}\int_0^1t^{2k}\sin^{-1}\left(1-t^2\right)\, dt.
\end{align}
Substituting the result \eqref{int02} into \eqref{eqn28} then back to \eqref{eqn26} and combining \eqref{eqn24}, \eqref{eqn25}, \eqref{eqn26}, and \eqref{phi}, we conclude \eqref{Theorem5}.
\end{proof}

\subsection{Values of $\mathcal{B}(k)$ and $\varphi(m)$}\label{ssub}
We get $\mathcal{B}(k)$ and $\varphi(m)$ from Theorem \ref{Theorem4} and Theorem \ref{theorem5}, which have integral representations, $\int_0^\frac{1}{2}\frac{t^k}{\sqrt{1-t}}\,dt$ and $\int_0^1 t^m\sqrt{1+t^2}\, dt$, respectively. It is easy to deduce the primitives of the integrals for some particular values of $k$ and $m$; however, we do not have compact closed forms in terms of elementary functions that can easily generate an infinite number of solutions. 
We now display a few values of $\mathcal{B}(k)$ and then $\varphi(m)$.
\begin{align*}
& \mathcal{B}(0) =2-\sqrt 2, \quad  \mathcal{B}(1) =\frac{8-5\sqrt 2}{6}, \quad \mathcal{B}(2)  =\frac{64-43\sqrt 2}{60},\\& \mathcal{B}(3)= \frac{256-177\sqrt 2}{280}, \quad  \mathcal{B}(4)=\frac{4096-2867\sqrt 2}{5040}.
\end{align*}
Similarly, for positive odd integers $m\geq 3$, we have 
\begin{align*}
    \varphi(3)=\frac{2(1+\sqrt 2)}{15}, \quad \varphi(5)=\frac{2(11\sqrt 2-4)}{105}, \quad \varphi(7)=\frac{2(8+13\sqrt 2)}{315}.
\end{align*}
\begin{example}
\label{example1}
Invoking Theorem \ref{Theorem4} and using the values of $\mathcal{B}(k)$  from Subsection~\ref{ssub} yields \eqref{PC1}, \eqref{PC2}, and some more series as follows: 
\begin{align}
    \sum_{n=1}^{\infty}\frac{n(n+1)(2n+1)\, 4^n}{(2n-1)^2\,(4n+1)(4n+3)(4n+5)}\frac{{2n\choose n}}{{4n\choose 2n}} &=\frac{1559\sqrt 2-1216}{29400}, \\ \sum_{n=1}^{\infty}\frac{n(n+1)(2n+1)(2n+3)\, 4^n}{(2n-1)^2(4n+1)(4n+3)(4n+5)(4n+7)}\frac{{2n\choose n}}{{4n\choose 2n}} &=\frac{41161\sqrt 2-33664}{1587600}, \\ \sum_{n=1}^{\infty}\frac{n(n+1)(n+2)(2n+1)(2n+3)\, 4^n}{(2n-1)^2(4n+1)(4n+3)(4n+5)(4n+7)(4n+9)}\frac{{2n\choose n}}{{4n\choose 2n}} &=\frac{857977\sqrt 2-659968}{153679680}.
\end{align}
\end{example}
\begin{example}
\label{example2} 
Similarly, applying Theorem \ref{theorem5} and values of $\varphi(m)$ yields the following series classes, in particular 
\begin{align}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(4n+1)(4n+3)}\frac{{2n\choose n}}{{4n\choose 2n}} & =\frac{4\left(1+\sqrt 2\right)}{225},\\  \sum_{n=1}^{\infty}\frac{n(n+1)\, 4^n}{(2n-1)^2(4n+1)(4n+3)(4n+5)}\frac{{2n\choose n}}{{4n\choose 2n}} & =\frac{137\sqrt 2-88}{11025}, \\
    \sum_{n=1}^{\infty}\frac{n(n+1)(2n+3)\, 4^n}{(2n-1)^2(4n+1)(4n+3)(4n+5)(4n+7)}\frac{{2n \choose n}}{{4n\choose 2n}} &=\frac{1033\sqrt 2-592}{198450}.
\end{align}
\end{example}
In addition, in view of Theorem \ref{Theorem4} and Theorem \ref{theorem5}, we have 
\begin{theorem}
\label{Theorem8} For positive odd integers, $m \geq 3$, we have 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(4n^2-1)(4n+m)}\frac{{2n\choose n}}{{4n+m-1\choose 2n+\frac{m-1}{2}}}=\frac{1}{\sqrt 2}\sum_{k=0}^{\frac{m-3}{2}}\frac{(-1)^k}{(2k+1)2^{m-k}}{\frac{m-3}{2}\choose k}\mathcal{B}(k)-\frac{\varphi(m)}{2^{m+1}}.
\end{align*}
\end{theorem}
\begin{proof} 
The proof directly follows from \eqref{Theorem5}.
\end{proof}
\begin{example}
\label{example3} The immediate consequences of Theorem \ref{Theorem8} are
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n+1)(2n-1)(4n+3)}\frac{{2n\choose n}}{{4n+2\choose 2n+1}} & =\frac{7\sqrt 2-8}{60}, \\ 
    \sum_{n=1}^{\infty}\frac{n\,4^n}{(2n+1)(2n-1)(4n+5)}\frac{{2n\choose n}}{{4n+4\choose 2n+2}} & =\frac{71\sqrt 2-64}{5040}.
\end{align*}
\end{example}
\subsection{Propositions and their proofs}
\label{Proposition10} 
We mention some integral identities that play a crucial role when
evaluating series  based on the ratio of central binomial coefficients.
They are as follows: 
\begin{proposition} 
The following equality holds: 
\begin{align}\label{eqn35}
    \int_0^{\frac{\pi}{2}}\frac{\sin^{-1}\left(\sin^2 x\right)}{\sin^2 x}\, dx &=2\log\left(1+\sqrt 2\right).
\end{align} 
\end{proposition}
\begin{proof}
Applying integration by parts, the above intergral boils down to
\begin{align*}
    \int_0^{\frac{\pi}{2}}\frac{\sin^{-1}(\sin^2 x)}{\sin^2 x}\,dx & =2\int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{\sqrt{1-\sin^4 x}}\,dx =2\int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{\sqrt{(1-\sin^2 x)(1+\sin^2 x)}}\,dx \\& 
    =2\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sqrt{1+\sin^2x}}\,dx =2\int_0^1\frac{dt}{\sqrt{1+t^2}} =2\sinh^{-1}(1).
\end{align*}
The latter integral in the second line is obtained by substituting \(\sin x= t\). Furthermore, we can easily obtain \eqref{eqn35} by noting that \(\frac{d}{dt} \sinh^{-1} t=\frac{1}{\sqrt{1+t^2}}=\log(t+\sqrt{1+t^2})\).
\end{proof} 
\begin{proposition}
If \(G\) denotes Catalan's constant defined by \(\sum_{n=0}^{\infty} (-1)^n/(2n+1)^2\) and \(\psi_1(z)=\sum_{n=0}^{\infty}1/(n+z)^2\) is the trigamma function for \(z>0\), then the following integral equality holds:
\begin{align}
     \int_0^{\frac{\pi}{2}}\frac{\sin^{-1}\left(\sin^2 x\right)}{\sin x}\, dx \label{eqn36} & =-2G-\frac{\pi^2}{2\sqrt 2} +\frac{1}{8\sqrt 2}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)\right).
\end{align} 
\end{proposition}
\begin{proof}
Substituting \(\sin^2x=y\) and employing integration by parts leads to 
\begin{align*}
    \int_0^{\frac{\pi}{2}}\frac{\sin^{-1}\left(\sin^2 x\right)}{\sin x}\,dx & = \frac{1}{2}\int_0^1\frac{\sin^{-1}(y)}{y\sqrt{1-y}}\,dy =\frac{1}{2}\int_0^1\frac{\log\left(\frac{1+\sqrt{1-y}}{1-\sqrt{1-y}}\right)}{\sqrt{1-y^2}}\,dy=-\int_0^1\frac{\log\left(\frac{1-y}{1+y}\right)}{\sqrt{2-y^2}}\,dy \\& 
    =-\int_0^{\frac{\pi}{4}}\log\left(\frac{1-\sqrt 2\sin u}{1+\sqrt 2\sin u}\right)\,du =-\int_0^{\frac{\pi}{4}}\log\left(\frac{\frac{1}{\sqrt {2}}-\sin u}{\frac{1}{\sqrt 2}+\sin u}\right)\,du =I.
\end{align*}
The last integral is obtained by substituting \(y=\sqrt{2}\sin u\). Furthermore, the last obtained integral result can be expressed as 
\begin{align*}
    I & = -\int_0^{\frac{\pi}{4}}\log\left(\tan \left(\frac{u}{2}\right)\frac{1+\tan\left(\frac{u}{2}\right)}{1-\tan\left(\frac{u}{2}\right)}\right)\,du =-2\int_0^{\frac{\pi}{8}}\log\left(\tan u\frac{1+\tan u}{1-\tan u}\right)\,du.
\end{align*}
The key notion for the former integral is the identity, \(\frac{1-\sqrt{2}\sin u }{1+\sqrt{2}\sin u}=\tan\left(\frac{\pi-4u}{8}\right)\cot\left(\frac{\pi+4u}{8}\right)\), which is due to Kamel Benaicha (personal communication), and the latter is obtained by replacing \(u/2\) with \(u\). Again, putting \(\tan u= z\) and substituting \(\frac{1-z}{1+z}= t\), we yield 
\begin{align*}
    I = -4\int_0^{\sqrt{2}-1}\frac{\log(t)}{1+t^2}\,dt +2\int_0^1\frac{\log(t)}{1+t^2}\,dt=-4\int_0^{\frac{\pi}{8}}\log(\tan t)\,dt +2\int_0^{\frac{\pi}{4}}\log(\tan t)\,dt.
\end{align*} 
It is well-known that the integral \(\int_0^{\frac{\pi}{4}}\log(\tan t)\,dt =-G\) (see \cite[Entry~4.227.2]{IS MR}). To evaluate the former integral, we use the Fourier series \cite[Entry~1.442.2]{IS MR} of 
\begin{align*}
    \log(\tan t)= -2\sum_{n=0}^{\infty}\frac{\cos((2n+1)2t)}{2n+1}, \, \, 0 < t < \pi/2.
\end{align*}
This leads to our original integral being 
\begin{align*}
   I &= -2G -4\int_0^{\frac{\pi}{8}}\log(\tan t)\,dt =-2G+ 8\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^{\frac{\pi}{8}}\cos((2n+1)2t)\,dt \\& = -2G +8\sum_{n=0}^{\infty}\frac{\sin\left(\frac{\pi}{4}(2n+1)\right)}{(2n+1)^2}=-2G+\frac{4}{\sqrt 2}\sum_{n=0}^{\infty}\frac{\sin\left(\frac{n\pi}{2}\right)+\cos\left(\frac{n\pi}{2}\right)}{(2n+1)^2} \\& 
   =-2G+\frac{4}{\sqrt 2} \sum_{n=0}^{\infty}\left(\frac{(-1)^n}{(4n+3)^2}+\frac{(-1)^n}{(4n+1)^2}\right)=-2G+\frac{4}{64\sqrt 2}\left(\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right) \\& +\frac{4}{16\sqrt{2}}\left(\psi_1\left(\frac{1}{8}\right)-\psi_1\left(\frac{5}{8}\right)\right)=-2G +\frac{1}{8\sqrt 2}\left(\psi_1\left(\frac{3}{8}\right)+\psi_1\left(\frac{5}{8}\right)\right)-\frac{\pi^2}{2\sqrt 2}. 
\end{align*}
In the last line, we make use of the reflection formula of the trigamma function, namely \(\psi_1(1-z)+\psi_1(z)=\frac{\pi^2}{\sin^2 \pi z} \). Putting \(z=1/8\) and \(z=3/8\) in the reflection formula and simplifying proves \eqref{eqn36}.
\end{proof}
\begin{proposition}
The following relation holds:
\begin{align}
    \int_0^{\frac{\pi}{2}}\sin^{-1}\left(\sin^2 x\right)\, dx \label{eqn37}& =\frac{7\pi^2}{24}-\frac{\log^2(2)}{4}-2\Li_2\left(\frac{1}{\sqrt 2}\right), \\
\label{proeqn38}
    \int_0^{\frac{\pi}{2}}\frac{x \sin x}{\sqrt{1+\sin^2 x }}\, dx & =\Li_2\left(\frac{1}{\sqrt 2}\right)-\frac{\pi^2}{48}+\frac{\log^2(2)}{8}.
\end{align}
Here, \(\Li_2(z)=\sum_{n=1}^{\infty}z^n/n^2\) is a dilogarithm function with \(|z|\leq 1\).
\end{proposition}
\begin{proof}
Integration by parts results in
\begin{align*}
    \int_0^{\frac{\pi}{2}}\sin^{-1}(\sin^2x)\,dx &=\frac{\pi^2}{4}-2\int_0^{\frac{\pi}{2}}\frac{x\sin x}{\sqrt{1+\sin^2 x}}\,dx =\frac{\pi^2}{4}-2\sqrt{2}\int_0^{\frac{1}{\sqrt 2}}\frac{\sin^{-1} x}{\sqrt{1-2x^2}}\,dx \\& =\frac{\pi^2}{4}-\int_0^{\frac{1}{\sqrt 2}}\int_0^1\frac{2x}{\sqrt{(1/2-x^2)(1-y^2x^2)}}\,dy\,dx \\&
    =\frac{\pi^2}{4}-\int_0^1\int_0^{\frac{1}{\sqrt 2}}\frac{2x}{\sqrt{(1/2-x^2)(1-y^2x^2)}}\,dx\,dy \\& 
    =\frac{\pi^2}{4}-\int_0^1\frac{\log\left(y+\sqrt 2\right)-\log\left(\sqrt 2-y\right)}{y}\,dy.
\end{align*}
Now, by using the definition of the dilogarithm integral \cite[p.\ 1]{LL} and the dilogarithm identity, \(\Li_2(z)+\Li_2(-z)=\frac{\Li_2(z^2)}{2}\) (see \cite[p.\ 6,\ Eqn.\ (1.15)]{LL}), we get 
\begin{align*}
    \int_0^{\frac{\pi}{2}}\sin^{-1}(\sin^2x)\,dx & =\frac{\pi^2}{4}-\left(\Li_2\left(\frac{1}{\sqrt 2}\right)-\Li_2\left(-\frac{1}{\sqrt 2}\right)\right) \\& =\frac{\pi^2}{4}-2\Li_2\left(\frac{1}{\sqrt 2}\right)+\frac{1}{2}\Li_2\left(\frac{1}{2}\right).
\end{align*}
Using the well-known classical result 
\cite[p.\ 6,\ Eqn.\ (1.16)]{LL}
$$ \Li_2\left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{\log^2(2)}{2}$$
in the last equality yields the announced results \eqref{eqn37} and \eqref{proeqn38}, respectively.
\end{proof}
 Clearly, Theorems~\ref{Theorem4}, \ref{theorem5}, and \ref{Theorem8} hold good only for positive odd integers. With the exception of Examples \ref{example1}, Example \ref{example2}, and Example \ref{example3}, the obtained corresponding integrals can produce exotic and slightly different identities.

So, using propositions from Subsection~\ref{Proposition10} and integrals \eqref{eqn18}, \eqref{eqn24}, \eqref{eqn25}, and \eqref{eqn26}, we get the
following result.
\begin{theorem}
\label{Theorem11} 
The following relations hold:
\begin{align}\label{eqn38} 
    \sum_{n=1}^{\infty}\frac{4^n}{(2n-1)^2}\frac{{2n\choose n}}{{4n\choose 2n}}& =4\left(\sqrt 2-1\right), \\ 
    \sum_{n=1}^{\infty}\frac{(4n-1)\, 4^n}{(2n-1)^3}\frac{{2n\choose n}}{{4n\choose 2n}}\label{eqn39} & =\frac{1}{2\sqrt{2}}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)\right)-8G-\sqrt{2}\pi^2, \\ \sum_{n=1}^{\infty}\frac{4^n}{(2n-1)^3}\frac{{2n\choose n}}{{4n\choose 2n}}\label{eqn40} & =\frac{1}{2\sqrt 2}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)\right)-8G-\sqrt 2\pi^2-\frac{8}{1+\sqrt 2}.
\end{align}
\end{theorem} 
\begin{proof}
Putting $m=-1$ in \eqref{eqn18}, and calculating the corresponding integral proves \eqref{eqn38}. In a similar fashion, putting $m=-3$ in \eqref{eqn18} and using \eqref{eqn36} proves \eqref{eqn39}. Thus, \eqref{eqn40} follows from \eqref{eqn38} and \eqref{eqn39}.
\end{proof}
For some particular cases, even integers like $m=0$, $m=2$, and $m=4$ in \eqref{eqn18} give
\begin{align*}
    \sum_{n=1}^{\infty}\frac{4^n}{(2n-1)^2}\frac{{2n\choose n}}{{4n-1\choose 2n-\frac{1}{2}}} & =\frac{7\pi^2}{12}-\frac{\log^2(2)}{2}-4\Li_2\left(\frac{1}{\sqrt 2}\right)+2\sqrt{2} -2\sinh^{-1}(1), \\
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(2n+1)}\frac{{2n\choose n}}{{4n+1\choose 2n+\frac{1}{2}}} &  =\frac{7\pi^2}{128}-\frac{3\log^2(2)}{64}-\frac{3}{8}\Li_2\left(\frac{1}{\sqrt 2}\right)+\frac{5}{16\sqrt 2}-\frac{3\sinh^{-1}(1)}{32},\\ \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(n+1)}\frac{{2n\choose n}}{{4n+3\choose 2n+\frac{3}{2}}}& =\frac{35\pi^2}{1536}-\frac{5\log^2(2)}{256}-\frac{5}{32}\Li_2\left(\frac{1}{\sqrt{2}}\right)+\frac{101}{576\sqrt 2}-\frac{25\sinh^{-1}(1)}{384}.
\end{align*}
Similarly, by inserting the same even integers in \eqref{eqn24}, \eqref{eqn25}, \eqref{eqn26}, and utilizing propositions from Subsection \ref{Proposition10}, we end up having the conclusions: 
 \begin{align*}
      \sum_{n=1}^{\infty}\frac{4^n}{(2n-1)(2n+1)}\frac{{2n\choose n}}{{4n-1\choose 2n-\frac{1}{2}}} & =3\log\left(1+\sqrt 2\right)-\sqrt 2, \\ 
      \sum_{n=1}^{\infty}\frac{n\,4^n}{(2n-1)^2(2n+1)^2}\frac{{2n\choose n}}{{4n+1\choose 2n+\frac{1}{2}}} &=\frac{1}{4\sqrt 2}-\frac{7\pi^2}{768}+\frac{\log^2(2)}{128}+\frac{1}{16}\Li_2\left(\frac{1}{\sqrt 2}\right)-\frac{\sinh^{-1}(1)}{16},\\ \sum_{n=1}^{\infty}\frac{n\,4^n}{(2n-1)^2(2n+1)(4n+4)}\frac{{2n\choose n}}{{4n+3\choose 2n+\frac{3}{2}}}&= \frac{25}{2304\sqrt 2}+\frac{7\pi^2}{12288}-\frac{\log^2(2)}{2048}-\frac{1}{256}\Li_2\left(\frac{1}{\sqrt 2}\right)\\&+\frac{\sinh^{-1}(1)}{1536}. 
\end{align*}
Also, we have 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n\, 4^n}{(2n-1)^2(2n+1)(4n+1)}\frac{{2n\choose n}}{{4n\choose 2n}} & =\frac{G}{4}+\frac{\pi^2}{16\sqrt 2}+\frac{13}{18\sqrt 2}-\frac{19}{72} \\ & -\frac{1}{64\sqrt 2}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)\right),
 \end{align*}  
 which can easily be achieved by letting $m=1$ in \eqref{eqn24}, \eqref{eqn25}, \eqref{eqn26}, and applying \eqref{eqn36}.
 
\section{Some interesting series of the product of central binomial coefficients and some Ramanujan-like series}\label{sec3}
In this section,  we establish various series involving the product of central binomial coefficients and associated Ramanujan-like series for $1/\pi$ via simple techniques. Most of the interesting series are deduced via general formulas provided here. Some other sorts of series involving the product of the coefficients are obtained by computing their corresponding elementary integrals. Using Lemma \ref{Lemma1}, Lemma \ref{Lemma2}, and some other intermediate results we have acquired, we can produce 
\begin{theorem}
\label{Theorem12} If $p \in \mathbb{Z}$, then we have 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{(2n-1)\, 64^n}{2n\choose n}{4n+4p\choose 2n+2p} & ={4p\choose 2p}+\frac{i\cdot 16^p}{\pi}B_{-1}\left(2p+\frac{1}{2},\frac{3}{2}\right),
\end{align*}
where $i=\sqrt {-1}$ and $B_x(a,b)=\int_0^xt^{a-1}(1-t)^{b-1}\,dt$ is the incomplete beta function.
\end{theorem}
\begin{proof}
We start with \eqref{eqn:11} where we replace $x$ by $\sin^2y$, multiply both sides by $\sin^{4p}y$, and integrating from $0$ to $\pi/2$ gives
\begin{align}\label{eqn42}
    \sum_{n=1}^{\infty}\frac{1}{(2n-1)\, 64^n}{2n\choose n}{4n+4p\choose 2n+2p} & =\frac{2\cdot 16^p}{\pi}\int_0^{\frac{\pi}{2}}\sin^{4p} y\left(1-\cos y\sqrt{1+\sin^2 y}\right)\, dy.
\end{align}
Apply linearity and \eqref{WF1}, the latter integral boils down to 
\begin{align}\label{eqn43}
    {4p\choose 2p}-\frac{2\cdot 16^p}{\pi}\int_0^{\frac{\pi}{2}} & \nonumber={4p\choose 2p}-\frac{2\cdot 16^p}{\pi}\int_0^{\frac{\pi}{2}}\sin y\cos y(\sin y)^{4p-1}\sqrt{1+\sin^2 y}\, dy \\&= {4p\choose 2p}-\frac{16^p}{\pi}\int_0^1t^{2p-1/2}\sqrt{1+t}\, dt.
\end{align}
Replacing $t$ with $-t$ in \eqref{eqn43} and, by definition of the incomplete beta function and $\sqrt{-1}=i$, gives the value of integral equal to $-iB_{-1}\left(2p+1/2,3/2\right)$, and hence putting the final result of \eqref{eqn43} back to \eqref{eqn42} completes the proof.
\end{proof} 
\begin{example} 
\label{example4} 
As stated, Theorem \ref{Theorem12} enables us to construct an infinite number of solutions. Some immediate consequences are as follows: 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{(2n-1)\, 64^n}{2n\choose n}{4n\choose 2n}& =1-\frac{\sqrt 2}{\pi}-\frac{\sinh^{-1}(1)}{\pi},\\
    \sum_{n=1}^{\infty}\frac{n}{(4n-3)(4n-1)\,64^n}{2n\choose n}{4n\choose 2n}& =\frac{\sqrt{2}}{6\pi}, \\
    \sum_{n=1}^{\infty}\frac{(4n+1)(4n+3)}{(2n-1)(n+1)(2n+1)\, 64^n}{2n\choose n}{4n\choose 2n} & = 3-\frac{7\sqrt 2}{3\pi}-\frac{\sinh^{-1}(1)}{\pi}.
\end{align*} 
\end{example}
Taking rational numbers $p$ into account in \eqref{eqn42} allows us to construct a variety of other types of identities. For example,
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{(2n-1)\, 64^n}{2n\choose n}{4n-1\choose 2n-\frac{1}{2}}& =\frac{1}{\pi}-\frac{\sqrt {2}}{\pi}+\frac{\sinh^{-1}(1)}{\pi},\\
    \sum_{n=1}^{\infty}\frac{1}{(2n-1)\, 64^n}{2n\choose n}{4n+1\choose 2n+\frac{1}{2}}&  =\frac{16}{3\pi}-\frac{8\sqrt 2}{3\pi}, 
\end{align*}
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)(4n-1)\, 64^n}{2n\choose n}{4n\choose 2n} & = \frac{1}{\sqrt 2\pi}-\frac{\sinh^{-1}(1)}{2\pi},\\
    \sum_{n=1}^{\infty}\frac{4n+1}{(2n-1)(2n+1)\, 64^n}{2n\choose n}{4n\choose 2n} & =1 -\frac{3}{\sqrt 2 \pi}+\frac{\sinh^{-1}(1)}{2\pi}.
\end{align*}
Putting $p=-1/8$ in \eqref{eqn42} produces an exotic series 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{1}{(2n-1)\, 64^n}{2n\choose n}{4n-\frac{1}{2}\choose 2n-\frac{1}{4}} =\frac{\Gamma^2\left(\frac{1}{4}\right)}{2\pi^{3/2}}-\frac{4}{3\pi}-\frac{\sqrt{2\pi}}{3\Gamma^2\left(\frac{3}{4}\right)}.
\end{align*}
Here, $\Gamma(z)$ is the gamma function defined by $\int_0^{\infty} x^{z-1}e^{-x}\,dx$ for $\Re(z)>0$.
\begin{theorem} 
\label{Theorem14} 
If $p$ is a real number, then we have 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2\, 64^n}{2n\choose n}{4n+4p\choose 2n+2p} & =\frac{16^p}{\pi}\int_0^{\frac{\pi}{2}}\left(\sin y\right)^{4p+2}\sin^{-1}\left(\sin^2 y\right)\, dy.
\end{align*}
\end{theorem} 
\begin{proof} The result can be easily verified by using the Lemma \ref{Lemma1} and \eqref{WF1}.
\end{proof}
\begin{example} Inserting $p=0, \, p= -3/4$ in Theorem \ref{Theorem14}, and using propositions from Subsection \ref{Proposition10}, we arrive at
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2\, 64^n}{2n\choose n}{4n\choose 2n} &= \frac{7\pi}{48}+\frac{1}{\sqrt 2\pi}-\frac{\log^2(2)}{8\pi}-\frac{1}{\pi}\Li_2\left(\frac{1}{\sqrt 2}\right)-\frac{\sinh^{-1}(1)}{2\pi}, \\ 
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2\, 64^n}{2n\choose n}{4n-3\choose 2n-\frac{3}{2}} &=-\frac{G}{4\pi}-\frac{\pi}{16\sqrt 2}+\frac{1}{64\pi\sqrt 2}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)\right).
\end{align*}
\end{example}
Similarly, for $p=-1/2,\, p=1$
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n^2}{(2n-1)^2(4n-1)\, 64^n}{2n\choose n}{4n\choose 2n} & =\frac{7\pi}{96}-\frac{1}{2\pi}\Li_2\left(\frac{1}{\sqrt 2}\right)-\frac{\log^2(2)}{16\pi}, \\
    \sum_{n=1}^{\infty}\frac{n(4n+3)(4n+1)}{(2n-1)^2(n+1)(2n+1)\, 64^n}{2n\choose n}{4n\choose 2n} & =\frac{35\pi}{48}+\frac{101}{18\sqrt 2\pi}-\frac{5\log^2(2)}{8\pi}-\frac{5}{\pi}\Li_2\left(\frac{1}{\sqrt 2}\right)\\ & -\frac{25\sinh^{-1}(1)}{12\pi}, \\ 
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2(2n+1)\, 64^n}{2n\choose n}{4n\choose 2n} & =\frac{7\pi}{96}-\frac{\log^2(2)}{16\pi}+\frac{3}{4\sqrt 2\pi} - \frac{1}{2\pi}\Li_2\left(\frac{1}{\sqrt 2}\right)\\& -\frac{5\sinh^{-1}(1)}{8\pi}.
\end{align*}
The last identity is a particular case of Lemma \ref{Lemma2}. Via specialization, several more series involving the product of central binomial coefficients can be deduced. Next, we investigate more series involving the product of central binomial coefficients, yielding the Ramanujan-type series for $1/\pi$.
\begin{lemma} 
\label{Lemma16}For $x\in [-1,1] $, the following holds:
\begin{align*}
    \sum_{n=0}^{\infty}\frac{1}{(2n+1)\, 16^n}{4n\choose 2n} x^{2n}=\frac{\sqrt{1+x}-\sqrt{1-x}}{x}.
\end{align*}
\end{lemma}
\begin{proof}
The proof directly follows from the generating function \eqref{eqn:2}, which on integration yields the desired result.
\end{proof}
\begin{theorem}
\label{Theorem17}
For all integers $k>0$, we have 
\begin{align}\label{eqn44}
    \sum_{n=0}^{\infty}\frac{1}{(2n+1)\, 64^n}{2n+2k\choose n+k}{4n\choose 2n}=\frac{2 \cdot 4^k}{\pi}\left(B\left(2k,\frac{1}{2}\right)-(-1)^{2k}B_{-1}\left(2k,\frac{1}{2}\right)\right),
\end{align}
where $B_x(a,b)=\int_0^xt^{a-1}(1-t)^{b-1}\, dt$ is the incomplete beta function and at $x=1$, it is beta function.
\end{theorem}
\begin{proof} We multiply both sides of Lemma \ref{Lemma16} by $x^{2k}$. 
Assigning $x=\sin y$, integrating from $0$ to $\pi/2$,
and using \eqref{WF1}, we have \pagebreak
\begin{align}\label{eqn45}
    \sum_{n=0}^{\infty}\frac{1}{(2n+1)\, 64^n}{2n+2k\choose n+k}{4n\choose 2n} & \nonumber =\frac{2\cdot 4^k}{\pi}\int_0^{\frac{\pi}{2}}(\sin y)^{2k-1}\left(\sqrt{1+\sin y}-\sqrt{1-\sin y}\right)\, dy \\& \nonumber=\frac{2\cdot 4^k}{\pi}\int_0^1 \frac{t^{2k-1}}{\sqrt{1-t^2}}\left(\sqrt{1+t}-\sqrt{1-t}\right)\,dt \\&=\frac{2\cdot 4^k}{\pi}\int_0^1 t^{2k-1}\left(\frac{1}{\sqrt{1-t}}-\frac{1}{\sqrt{1+t}}\right)\, dt\\& \nonumber=\frac{2 \cdot 4^k}{\pi}\left(B\left(2k, \frac{1}{2}\right)-(-1)^{2k}\int_0^{-1}\frac{t^{2k-1}}{\sqrt{1-t}}\,dt\right).
\end{align}
Using the definition of the beta and the incomplete beta function, the proof is completed.
\end{proof}
In particular, Theorem \ref{Theorem17} gives us  Ramanujan-type series for $1/\pi$.
\begin{theorem} 
\label{Theorem18}
The following relations hold:
\begin{align*}
 & \sum_{n=0}^{\infty}\frac{1}{(n+1)\, 64^n}{2n\choose n}{4n\choose 2n}  =\frac{8}{3}\cdot\frac{\sqrt 2}{\pi}, \\
 &\sum_{n=0}^{\infty}\frac{2n+3}{(n+1)(n+2)\, 64^n}{2n\choose n}{4n\choose 2n}= \frac{144}{35}\cdot\frac{\sqrt 2}{\pi}, \\ 
& \sum_{n=0}^{\infty}\frac{(2n+5)(2n+3)}{(n+1)(n+2)(n+3)\, 64^n}{2n\choose n}{4n\choose 2n} =\frac{4832}{693}\cdot\frac{\sqrt 2}{\pi}, \\
& \sum_{n=0}^{\infty}\frac{(2n+7)(2n+5)(2n+3)}{(n+1)(n+2)(n+3)(n+4)\,64^n}{2n\choose n}{4n\choose 2n} =\frac{79424}{6435}\cdot\frac{\sqrt 2}{\pi},\\
& \sum_{n=0}^{\infty}\frac{(2n+9)(2n+7)(2n+5)(2n+3)}{(n+1)(n+2)(n+3)(n+4)(n+5)\,64^n}{2n\choose n}{4n\choose 2n}  =\frac{5174656}{230945}\cdot\frac{\sqrt 2}{\pi}.
\end{align*}
\end{theorem}

In terms of the Pochhammer symbol, $k^{(n)}= k(k+1)(k+2)\ldots (k-n+1)$ for $n\geq 1$, we notice $(2n+1)(2n+3)(2n+5)\cdots (2n+2k-1)=2^k(n+1/2)_k$ and $(n+1)(n+2)(n+3)\cdots (n+k)= (n+1)_k$, then in view of Theorem \ref{Theorem18}, Theorem \ref{Theorem17} can be compactly written as follows: $$\sum_{n=0}^{\infty}\frac{2^k\left(n+1/2\right)_k}{(2n+1)(n+1)_k\,64^n}{2n\choose n}{4n\choose 2n}=b_k\cdot\frac{\sqrt 2}{\pi},
$$
where $b_k= i/j$, $i$, and $j$ are positive integers.

The general form of $p/q$ is easily  deducible by comparing the left- and 
right-hand quantities of Theorem \ref{Theorem17}. The aforementioned Ramanujan-like series are merely the consequences of Theorem \ref{Theorem17}, which are obtained by putting $k=1$, $k=2$, $k=3$, $k=4$, and $k=5$ respectively. Details are left to the reader. 
\begin{example}
In a similar fashion, one can easily find several interesting series by evaluating the integrals obtained for any rational number $k$ in \eqref{eqn45}.
\begin{align*}
    \sum_{n=0}^{\infty}\frac{1}{(2n+1)\, 64^n}{2n\choose n}{4n\choose 2n} & = \frac{4}{\pi}\sinh^{-1}(1), \\ 
    \sum_{n=0}^{\infty}\frac{1}{(2n+1)\, 64^n}{2n+1\choose n+\frac{1}{2}}{4n\choose 2n} & =\frac{8}{\pi}\left(2-\sqrt 2\right), \\ 
    \sum_{n=0}^{\infty}\frac{1}{(2n+1)\,64^n}{2n+\frac{1}{2}\choose n+\frac{1}{4}}{4n\choose 2n} & =2\sqrt 2\left(1-\frac{2\sinh^{-1}(1)}{\pi}\right), \\  \sum_{n=0}^{\infty}\frac{1}{(2n+1)\, 64^n}{2n+\frac{3}{2}\choose n+\frac{3}{4}}{4n\choose 2n} & =\frac{4}{\pi}\left(- 2+\frac{\pi}{\sqrt 2}+\sqrt 2\sinh^{-1}(1)\right).
\end{align*}
\end{example}
Details are left to the reader.
\section{Some additional series}
\label{sec4} 
Employing the identity (13) and the identity (21) given in \cite[pp.\ 5--6]{Boya}, respectively, we arrive at slightly different conclusions:
\begin{align}\label{eqn46}
    \sum_{n=1}^{\infty}\frac{1}{n\,64^n}{2n\choose n}{4n\choose 2n} & =6\log(2)-\sqrt{2}\pi+\frac{1}{2\sqrt{2}\pi}\left(\psi_1\left(\frac{5}{8}\right)+\psi_1\left(\frac{7}{8}\right)\right), \\
    \label{eqn47} 
    \sum_{n=1}^{\infty}\frac{1}{(n+1)^2\, 64^n}{2n\choose n}{4n+4\choose 2n+2} & \nonumber=58-\frac{64\sqrt 2}{\pi}+32\sqrt{2}\pi-\frac{64\sinh^{-1}(1)}{\pi} -192\log(2)\\&-\frac{8\sqrt 2}{\pi}\left(\psi_1\left(\frac{5}{8}\right)+\psi_1\left(\frac{7}{8}\right)\right),
\end{align}
where $\psi_1(z)=\sum_{n=0}^{\infty}1/(n+z)^2$ for $z>0$ is the trigamma  function. Identity \eqref{eqn46} can also be proved using \cite[Thm.\ 3.1, p.\ 3]{Bhandari}. Using the proposition established in \cite[p.\ 9]{Boya} and calculating the corresponding integrals obtained for the particular values $m=1$, $m=2$, and $m=3$, we conclude some fascinating Ramanujan-like series as follows:
\begin{align}\label{eqn48}
    \sum_{n=0}^{\infty}\frac{1}{(n+2)\, 64^n}{2n\choose n}{4n\choose 2n} &=\frac{152}{105}\cdot\frac{\sqrt 2}{\pi}, \\
    \sum_{n=0}^{\infty}\frac{1}{(n+3)\, 64^n}{2n\choose n}{4n\choose 2n} 
    & \label{eqn49}=\frac{10568 }{10395}\cdot\frac{\sqrt 2}{\pi}, \\
    \sum_{n=0}^{\infty}\frac{1}{(n+4)\, 64^n}{2n\choose n}{4n\choose 2n}& \label{eqn50}=\frac{178328}{225225}\cdot\frac{\sqrt 2}{\pi}.
\end{align}
These series have quite simple appearances in their closed form, which include rational number $p/q$, and also $\sqrt 2$ and $1/\pi$. Based on the patterns observed in the first equation of Theorem \ref{Theorem18}, and equations \eqref{eqn48}, \eqref{eqn49}, and \eqref{eqn50}, for $m \in \mathbb{N}\cup \left\{0\right\}$, we conjecture that
\begin{align*}
    \sum_{n=0}^{\infty}\frac{1}{64^n}{2n\choose n}{4n\choose 2n}\frac{1}{n+m+1} = a_m\cdot \frac{\sqrt 2}{\pi},
\end{align*}
where $a_m=p/q$, $p$, and $q$ are positive integers. The problem of determining the general expression of $a_m$ remains open. Furthermore, utilizing Lemma \ref{Lemma4}, we have 
\begin{align*}
    \sum_{n=1}^{\infty}\frac{n^2}{(2n-1)^2(2n+1)\,64^n}{2n\choose n}{4n\choose 2n}&  =\frac{7\pi}{192}+\frac{1}{8\pi\sqrt 2} -\frac{\Li_2\left(\frac{1}{\sqrt 2}\right)}{4\pi}-\frac{\log^2(2)}{32\pi}+\frac{\sinh^{-1}(1)}{16\pi},\\
    \sum_{n=1}^{\infty}\frac{n^24^n}{(2n-1)^2(2n+1)(4n+1)}\frac{{2n\choose n}}{{4n\choose 2n}} & =-\frac{G}{8}-\frac{13}{144}+\frac{7}{36\sqrt 2}-\frac{\pi^2}{32\sqrt 2} \\& 
    +\frac{1}{128\sqrt 2}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)\right).
\end{align*}
Moreover, applying the lemmas introduced in the recent paper, we can easily deduce some more variety of series involving the square of the central binomial coefficients, yielding exotic results such as
\begin{align*}
    \sum_{n=1}^{\infty} \frac{n^2}{(2n-1)^2(2n+1)\,16^n}{2n\choose n}^2 & = \frac{G}{4\pi}+\frac{1}{8\pi}, \\
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2(n+1)\,16^n}{2n\choose n}^2 & =\frac{4}{9\pi},\\
    \sum_{n=1}^{\infty}\frac{n}{(2n-1)^2(2n+1)(2n+3)\, 16^n}{2n\choose n}^2 & =\frac{31}{128\pi}-\frac{13G}{64\pi}.
\end{align*}
We encourage interested readers to pursue the results and to investigate more series.

\section{Acknowledgments}
The author would like to express his sincere gratitude to the anonymous
referee for their valuable suggestions.
Also, special thanks to
Kamel Benaicha for his aid in the course of the calculation of the
integral \eqref{eqn36}.

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N. Bhandari, Generating functions involving binomial coefficients ${4n\choose 2n}$ it's squared, reciprocal and their closed forms for hypergeometric expressions, {\it Romanian Mathematical Magazine, An interactive journal}, \url{https://www.ssmrmh.ro/2021/02/26/generating-functions-involving-binomial-coefficients}.

\bibitem{Boya}
K. N. Boyadzhiev, Series with central binomial coefficients, Catalan numbers, and harmonic numbers, {\it J. Integer Sequences} {\bf 15} (2012), \href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Boyadzhiev/boyadzhiev6.html}{Article 12.1.7}.

\bibitem{Campbell1}
J. M. Campbell, Ramanujan-like series for $1/\pi$ involving harmonic numbers, {\it Ramanujan J.} {\bf 46} (2018), 373--387.

\bibitem{Campbell}
J. M. Campbell, New series involving harmonic numbers and squared central  binomial coefficients, {\it Rocky Mountain J. Math.\/} {\bf 49(8)} (2019), 2513--2544.

\bibitem{JJJ}
J. M. Campbell, J. D'Aurizio, and J. Sondow, On the interplay among hypergeometric functions, complete elliptic integrals, and Fourier-Legendre Expansion, {\it J. Math. Anal. Appl.}, {\bf 479} (2019), 90--121.

\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{Chen1}
H. Chen, Interesting Ramanujan-like series associated with powers of central binomial coefficients, {\it J. Integer Sequences} {\bf 25} (2022), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL25/Chen/chen13.html}{Article 22.1.8}.

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

\bibitem{IS MR}
I. S. Gradshteyn and I. M. Ryzhik, eds., {\it Table of Series, Integrals,
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\end{thebibliography} 
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\noindent{\it 2020 Mathematics Subject Classification:} Primary 11B83; Secondary 05A10.

\noindent \emph{Keywords:} central binomial coefficient, generating function, Wallis' integral, incomplete beta function, Ramanujan-like series.

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\noindent (Concerned with sequence \seqnum{A000984}.)

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\vspace*{+.1in}
\noindent
Received April 15 2022;
revised versions received  April 27 2022; June 22 2022; June 26 2022; 
June 27 2022.
Published in {\it Journal of Integer Sequences}, June 29 2022.

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