User:Pi-Inspired-Formulas

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{8-13n}{(2n-1)(n+2)(n+1)^2}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{(2k-1)(2k-3)(2k-5)}=\frac{1}{\pi}$$

$$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)_n^3}{(1)_n^3}\cdot\frac{\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2k-1}}{(2n-1)(2n-3)(2n-5)\cdots(2n-2j-1)}=(-1)^{j}\frac{j!2^{j+1}}{(2j+1)!!^2}\cdot\frac{1}{\pi};j\ge0$$