Due to a recent comment by Akiva about this post, I decided to revisit Ramanujan's beautiful continued fraction (plus series) relating $\pi$ and $e$,
$$\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\,\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$
By sheer happenstance, I came across Pedja's post from 2021 and realized the series is also a nice continued fraction,
\begin{align} \sqrt{\frac{\pi\,e}{2}} &=1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}} \; + \; \cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}\\[4pt] &=1.41068613\dots + 0.65567954\dots = 2.06636567\dots \end{align}
where the decimal expansions are OEIS A060196, A108088, A059444, respectively. The first continued fraction does not appear in Mathworld's Continued Fraction Constants, but the reciprocal of the second does and is labeled as $C_3 \approx 1/0.65567954 \approx 1.5251352.$
After a session with Mathematica, one can observe that Ramanujan's general form (Entry 43, "Chapter 12 of Ramanujan's Second notebook: Continued fractions") for this identity have addends that have closed-forms. Let,
$$A = \sum_{n=0}^\infty \prod_{k=0}^n\frac{x^n}{2k+1} = \sqrt{\frac{\pi\,e^x}{2x}} \operatorname{erf}\Big(\sqrt{\tfrac x 2}\Big)$$
$$B = \cfrac1{1+\cfrac{1}{x+\cfrac{2}{1+\cfrac{3}{x+\ddots}}}} = \sqrt{\frac{\pi\,e^x}{2x}} \left(1-\operatorname{erf}\Big(\sqrt{\tfrac x 2}\Big)\right)$$
with error function $\operatorname{erf}(z).$ Therefore,
$$\sqrt{\frac{\pi\,e^x}{2x}} = A+B$$
for $x>0$. The first identity was just the case $x=1$.
Questions:
- For general $x>0$, can we express the series $A$ as a continued fraction similar to the one above?
- Likewise, can we express the continued fraction $B$ as a series similar to $A$?
