On the cubic counterpart of Ramanujan's $\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots$?

We have Ramanujan's well-known,

$$\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots\color{blue}+\,\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

However, as discussed in this post, the series can also be expressed as a nice continued fraction. And since $$\Gamma\big(\tfrac12\big) = \sqrt{\pi}$$, we can then express the identity as,

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

It turns out it may have a cubic counterpart,

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

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

\begin{align} \Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} &= \sqrt{\frac{e}{2}}\times\Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big)\; \color{blue}+ \, \sqrt{\frac{e}{2}}\times\Big(\Gamma\big(\tfrac12,\tfrac12\big)\Big)\\[6pt] \Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}} &= \sqrt[3]{\frac{e}{9}}\times\Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\; \color{blue}+ \; \sqrt[3]{\frac{e}{9}}\times\Big(\Gamma\big(\tfrac13,\tfrac13\big)\Big) \end{align}

and their decimal expansions are A060196, A108088, A108744, A108745, respectively.

Questions:

1. Given Pochhammer symbol $$(x)_n$$, how do we prove that,

\begin{align} \sum_{n=1}^\infty \frac{1}{2^n (\frac12)_n} &= \sqrt{\frac{e}{2}} \times\,\Gamma\big(\tfrac12\big)\,\operatorname{erf}\Big(\sqrt{\tfrac 12}\Big)\\ &= \sqrt{\frac{e}{2}}\times\Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big) = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}} \end{align}

1. Similarly, how do we show that,

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

P.S. Part of Question 1 has already been answered in this post, but I included it for comparison and to see if an alternative or modified proof can be found that covers both Question 1 and 2.

• Using one of my question we can express the roots of : $$1+x+x^2/2+x^3/6+\cdots+x^n/n!-e^{\sqrt{e\pi/2}}=0$$ using continued fraction and iteration .Are you interested ? Oct 27, 2023 at 7:31
• @ErikSatie That involves $\Gamma\big(\frac12\big)=\sqrt{\pi}$ which is already settled. Of more interest will be finding a "nicer" continued fraction for the case $\Gamma\big(\frac14\big)$ I gave in my answer below. Oct 27, 2023 at 8:20

After some research, I managed to answer my own question and found the pieces of the puzzle scattered in various places. Recall that $$\Gamma\big(\tfrac12\big) = \sqrt{\pi}$$, so to recap,

$$\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots\color{blue}+\,\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

$$\Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{3^2}} =1+\frac{1}{1\cdot4}+\frac{1}{1\cdot4\cdot7}+\frac{1}{1\cdot4\cdot7\cdot10}+\dots\color{blue}+ \; \cfrac1{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{\color{red}5}{1+\ddots}}}}$$

where the 2nd continued fraction is missing numerators $$P(n)=3n+1 = 4,7,10,\dots,$$ etc. It turns out Ramanujan's identity could be generalized not just for deg-$$3$$, but for all higher degrees as well. The secret was quite simple and depended on the identity,

$$\Gamma(a) = \gamma(a,x) \color{blue}+ \Gamma(a,x)$$

with lower $$\gamma(a,x)$$ and upper $$\Gamma(a,x)$$ incomplete gamma functions. Or as integrals,

$$\int_0^\infty t^{a-1}e^{-t}dt = \int_0^x t^{a-1}e^{-t}dt \; \color{blue}+ \int_x^\infty t^{a-1}e^{-t}dt$$

The trick then is to multiply each term by a fractional power of $$e$$ such that the addends can be expressed by a nice series or continued fraction or both. Let $$a=x$$, then the first addend becomes,

\begin{align}\frac{e^a}{a^{a-1}}\,\gamma(a,a) &= \sum_{n=1}^\infty\frac{a^n}{(a)_n}\\[4pt] &= {_1F_1}(1;a+1;a)\\[4pt] &= 1+\cfrac{2a}{2+\cfrac{3a}{3+\cfrac{4a}{4+\cfrac{5a}{5+\ddots}}}} \end{align}

The equivalence of the confluent hypergeometric function $$_1F_1$$ and the continued fraction was known to Ramanujan. The second addend becomes,

\begin{align}\frac{e^a}{a^{a-1}}\,\Gamma(a,a) &= \cfrac{a}{1-\cfrac{1(1-a)}{3-\cfrac{2(2-a)}{5-\cfrac{3(3-a)}{7-\ddots}}}} \end{align}

A version of the second cfrac can also be found in his Notebooks and a general form is in this post. However, note that Ramanujan used a different cfrac for $$a = 1/2$$, and the one for $$a=1/3$$ (not by Ramanujan) is also not from this family. And they do not seem to be in Wolfram's list. Thus, a few mysteries remain. For $$a=1/4$$, we get,

$$\Gamma\big(\tfrac14\big)\sqrt[4]{\frac{e}{4^3}} =1+\frac{1}{1\cdot5}+\frac{1}{1\cdot5\cdot9}+\frac{1}{1\cdot5\cdot9\cdot13}+\dots\color{blue}+ \; \cfrac{1/4}{1-\cfrac{3/4}{3-\cfrac{14/4}{5-\cfrac{33/4}{7-\ddots}}}}$$

P.S. Can someone find a simpler cfrac for $$a=1/4$$, preferably similar to the the previous cases?

You should be interested by :

$$\int_{0}^{∞}(((-1 + \sqrt{1 + 4 x^4})/(2 x^2) - 1) - ((-1 + \sqrt{1 + x^4})/x^2 - 1)) dx = -(\sqrt{2} - 2) Γ(3/4)^2)/\sqrt{π}≈0.496286$$

You can use reflection formula .

Nota bene: the CF in the integral are simple.