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?
Best Answer
The infinite series $\sum_{k\ge 1}a_k$ can be approximated by summing its first $n$ terms, but this may require very large $n$ for high accuracy. Since the $n$th term in $A$ is $O(1/n^4)$, the $n$th partial sum has a $O(1/n^3)$ error term. It therefore takes about $1000$ terms to get $9$ decimal places right. Or does it?
I don't know much about series acceleration of hypergeometric functions, or any methods WA might use to accelerate arbitrary series it didn't necessarily recognise add hypergeometric. But no general method accelerates all "logarithmically convergent" sequences, of which $A$ is an example. So go easy on WA.