Continued fraction with odd partial numerators and even partial denominators

Question

Given the continued fraction $$\epsilon=\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cfrac{7}{8+\ddots}}}},$$ in https://groups.google.com/forum/#!topic/sci.math/mlZ0VCTUJi8, Robert Israel ensures that $$\epsilon=\frac{\sqrt{\frac{2e}{\pi}}}{\mathrm{erfi}(\frac{1}{\sqrt{2}})}-1.$$

I have two doubts about his proof:

1. He says "take $f(n)=\frac{p(n)}{p(n+1)}$". Why he can assume that $f(n)$ has its form?

2. He says "so, the only way to avoid $f(n)\to-1$". Why is important to avoid $f(n)\to -1$?

Answer 1

1. With $p(0)$ arbitrary (to be found later), just let $p(n)=p(0)/\big(f(0)\cdots f(n-1)\big)$ for $n>0$.
2. Because $f(n)$ must tend (as $n\to\infty$) to the value we're computing, which is (at least) positive. This step assumes that this value exists (i.e. the given continued fraction is convergent); to go rigorous, one should prove it beforehand.

The given CF is a special case of this one. So it can be also computed as $f'(1/2)/f(1/2)$, where $$f(x)={}_1F_1\left(\frac12;\frac32;x\right)=\frac12\int_0^1 t^{-1/2}e^{xt}\,dt=\frac12\sqrt\frac{\pi}{x}\operatorname{erfi}\sqrt{x}.$$