Conjecture: $\frac1\pi=\sum_{n=0}^\infty\left((n+1)\frac{C_n^3}{2^{6n}}\sum_{k=0}^n(-1)^k{n\choose k}{\frac{(n-k)(k-1)}{(2k-1)(2k+1)}}\right)$ Let $C_n$ denote the $n$-th Catalan  number defined by
$${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}=\prod \limits _{k=2}^{n}{\frac {n+k}{k}}\quad \left(n\geqslant 0\right).}$$
Next, we define the sequence
$${\displaystyle A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}}.$$
I've numerically managed to verify that
\begin{equation}\tag{1}\label{pi}
{\displaystyle \sum _{n=0}^{\infty }(n+1)A_{n}={\frac {1}{\pi}}}.
\end{equation}
Is it possible to prove the relation in (\ref{pi})? If yes, then how could we go about it?
Also, is this series already known or studied in the literature? If yes, then any references will be highly appreciated. Thanks!
 A: $$\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}=\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}$$
$$ A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}=\frac{\Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi  \Gamma (n+2)^2}$$
$$(n+1)A_n=\frac{(n+1) \Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi  \Gamma (n+2)^2}$$
$$S_p=\sum_{n=0}^p(n+1)A_n=\frac 1 \pi \frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}$$ Now, Stirling approximation
$$\log \left(\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}\right)=\log(p+1)+2\log \left(\Gamma \left(p+\frac{3}{2}\right)\right)-2\log \left(\Gamma \left(p+{2}\right)\right)$$
$$\log \left(\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}\right)=-\frac{1}{4 p}+\frac{1}{4 p^2}+O\left(\frac{1}{p^3}\right)$$
$$\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}=1-\frac{1}{4 p}+\frac{9}{32 p^2}+O\left(\frac{1}{p^3}\right)$$
$$S_p=\frac 1 \pi \left(1-\frac{1}{4 p}+\frac{9}{32 p^2}+O\left(\frac{1}{p^3}\right) \right)$$
