Convergence of partial quotients of $Q = \dfrac{1!}{1+\frac{2!}{\ddots}}$ Consider the continued fraction
$$S=\frac1{1+\dfrac2{1+\dfrac3{\ddots}}}$$
From here, we have $S = \sqrt{\frac2{e\pi}}(\text{erfc}(\frac{1}{\sqrt{2}}))^{-1}$, where $\text{erfc}$ is the complementary error function.  I was wondering if
$$Q = \frac{1!}{1+\dfrac{2!}{1+\dfrac{3!}{\ddots}}}$$
Exists, and if so, with what closed form. However, it very quickly becomes clear that $Q$ does not exist (at least, does not converge). If we let
$$Q_n = \frac{1!}{1+\frac{2!}{\ddots1 +n!}}$$
Denote the $n$-th partial quotient of $Q$, then we have the first few partial quotients:
$$1, \frac13, \frac79, \frac{31}{81},\frac{871}{1161}, \frac{23191}{59481},\frac{4413031}{5910921},\frac{939474151}{2404184841} \dots$$
which alternate between some two numbers, approximately $0.746$ and $0.39$. I have two questions:


*

*Can I prove the even and odd-indexed partial quotients converge? Is there an analog for this to the Stern-Stolz theorem?

*I don't expect there to be any nice/obtainable closed form for these partial quotients, but I ask anyone who might have insight to share. 

 A: The full sequence of partial quotients actually does diverge overall.
To see what happens I render five layers:
$\dfrac{1!}{1+\dfrac{2!}{1+\dfrac{3!}{1+\dfrac{4!}{1+\dfrac{5!}{1}}}}}$
We first render $1!=1$:
$\dfrac{1}{1+\dfrac{2!}{1+\dfrac{3!}{1+\dfrac{4!}{1+\dfrac{5!}{1}}}}}$
Next divide the numerator and denominator in the second layer by $2!=2$:
$\dfrac{1}{1+\dfrac{1}{(1/2)+\dfrac{3!/2}{1+\dfrac{4!}{1+\dfrac{5!}{1}}}}}$
Divide the numerator and denominator in the third layer by $3!/2=3$:
$\dfrac{1}{1+\dfrac{1}{(1/2)+\dfrac{1}{(1/3)+\dfrac{4!/3}{1+\dfrac{5!}{1}}}}}$
Divide the next layer by $4!/3=8$:
$\dfrac{1}{1+\dfrac{1}{(1/2)+\dfrac{1}{(1/3)+\dfrac{1}{(1/8)+\dfrac{5!/8}{1}}}}}$
Now the divisor iis $5!/8=15$:
$\dfrac{1}{1+\dfrac{1}{(1/2)+\dfrac{1}{(1/3)+\dfrac{1}{(1/8)+\dfrac{1}{(1/15)}}}}}$
Thus in simplified continued fraction form we have
$[0;1/1,1/2,1/3,1/8,1/15,...]$
It looks like the $k$-th term after the initial zero is $1/k$ but then the denominators begin to take off.  If you carefully inspect the pattern of the divisors and how they are related to the factorials you discover that the standardized form is 
$[0;1/1,1/2,1/3,1/8,...\color{blue}{1/k!!},...]$
where the blue term shows the true $k$-th term after standardization.  This produces a convergent sum, therefore a divergent continued fraction by your quoted theorem.
The separate sequences of odd and even partial quotients do converge, trivially.  When all terms of the continued fraction are positive the separate sequences are monotonic and each one bounds the other (odd partial quotients increase but are less than all even partial quotients, even ones decrease but are greater than all the odd ones).  However, the above analysis shows that the separate limits remain apart.
