Entry 28

Ramanujan gave the infinite series, $$1 - \left(\frac{1}{2}\right)^3 + \left(\frac{1\times3}{2\times4}\right)^3 - \left(\frac{1\times3\times5}{2\times4\times6}\right)^3 + \cdots =\left(\frac{\Gamma\big(\tfrac{9}{8}\big)}{\Gamma\big(\tfrac{5}{4}\big)\Gamma\big(\tfrac{7}{8}\big)}\right)^2=\frac{8\,\big[K\big(k_2\big)\big]^2}{(1+\sqrt{2})\,\pi^2}$$ $$1 - 5\left(\frac{1}{2}\right)^3 + 9\left(\frac{1\times3}{2\times4}\right)^3 - 13\left(\frac{1\times3\times5}{2\times4\times6}\right)^3 + \cdots = \frac{2}{\pi}$$ $$1 + 9\left(\frac{1}{4}\right)^4 + 17\left(\frac{1\times5}{4\times8}\right)^4 + 25\left(\frac{1\times5\times9}{4\times8\times12}\right)^4 + \cdots =\frac{2^{3/2}}{\pi^{1/2}\,\Gamma^2\left(\frac{3}{4}\right)}=\frac{2^{5/2}}{\pi} \eta^2(i)$$ with elliptic integral singular value $K(k_2)$ and Dedekind eta function $\eta(\tau)$. For the first and second, their apparent simplicity belies its deep connection to modular forms and there are in fact an infinite number of such formulas. The three can be succinctly expressed as $$\begin{aligned}S_1&=\sum_{n=0}^\infty\, (-1)^n \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\S_2&=\sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\S_3&=\sum_{k=0}^\infty (8k+1)\left(\frac{\Gamma\big(k+\tfrac{1}{4}\big)}{k!\;\Gamma\big(\tfrac{1}{4}\big)}\right)^4\end{aligned}$$ However, since $$\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}=\frac{(2n)!}{2^{2n}\,n!^2}=\frac{(2n-1)!!}{(2n)!!}$$ then $S_2$ is equivalently, $$\frac{2}{\pi}=\sum_{n=0}^\infty\,(-1)^n \left(\frac{(2n)!}{n!^2}\right)^3 \frac{4n+1}{2^{6n}}$$ Another that belongs to the same family is $$\frac{16}{\pi}=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 \frac{42n+5}{2^{12n}}$$ and so on. For the third, note that $$\frac{\Gamma\big(n+\tfrac{1}{4}\big)}{n!\;\Gamma\big(\tfrac{1}{4}\big)}=(-1)^n\binom{\small{-1/4}}{n}$$ where $\binom{n}{k}$ is the binomial coefficient though I haven't yet figured out the family that this example belongs to.