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

Question

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.

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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}

2. 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.

Answer 1

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?

Answer 2

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.