Entry 91

Here is a "bizarre" continued fraction from Ramanujan involving \(e\) and \(\pi\) $$\sqrt{\frac{\pi\,e}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$ However, since \(\Gamma\big(\tfrac12\big) = \sqrt{\pi}\), we can express the LHS in terms of gamma functions and find a cubic counterpart using \(\Gamma\big(\tfrac13\big)\) 

$$\sum_{n=1}^\infty \frac{1}{2^n (\frac12)_n} = \Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big) \sqrt{\frac{e}{2}} = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}}$$

$$\sum_{n=1}^\infty \frac{1}{3^n (\frac13)_n} = \Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}} = 1+\cfrac{2/3}{2+\cfrac{3/3}{3+\cfrac{4/3}{4+\cfrac{5/3}{5+\ddots}}}}$$

Ramanujan's identity then becomes a sum of two continued fractions with a cubic version,

$$\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}} \; \color{blue}+ \; \cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

$$\Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}}= 1+\cfrac{2/3}{2+\cfrac{3/3}{3+\cfrac{4/3}{4+\cfrac{5/3}{5+\ddots}}}} \; \color{blue}+ \; \cfrac1{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{\color{red}5}{1+\ddots}}}}$$

where the 4th continued fraction (also by Ramanujan) is missing numerators \(P(n)=3n+1 = 4,7,10,13,\dots\) The four cfracs then have closed-forms as,

\begin{align}\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} &= \Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big)\sqrt{\frac{e}{2}} \color{blue}+ \Big(\Gamma\big(\tfrac12,\tfrac12\big)\Big)\sqrt{\frac{e}{2}} \\ \Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}} &= \Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}} \color{blue}+ \Big(\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}}\end{align}