What is the close form of the sum of reciprocals of $k!$, $k!!$, and so on? Moreover, is it transcendental? For $k \geq 3$, let 
$$g(k) := \frac{1}{k!} + \frac{1}{k!!} + \frac{1}{k!!!} + \dots.$$ where $$k!!=(k!)!, \quad k!!!=(k!!)!$$ and so on
What is the closed form of $g(k)$? Is it a transcendental number or irrational number or both for all $k\geq3$?
 A: I'll post the answer I put in the comments on the MathOverflow post, since I have no idea if that will vanish.
You can show that is $g(k)$ always transcendental by using the Liouville criterion (we'll use a version that guarantees irrationality as well).
Liouville Criterion: Fix $\alpha$ a real number.  If for all positive integers $n$ there are infinitely many $(p,q)$ such that $$0 < \left|\alpha - \frac{p}{q}\right| < \frac{1}{q^n},$$ then $\alpha$ is irrational and transcendental.
Now to apply this, let $\alpha$ be the sum of your series.  If $p/q$ is the $m^{\text{th}}$ partial sum of your series, the denominator $q$ is $k!!\cdots m \cdots !$.  The remainder is a negligible constant times $k!!\cdots m+1 \cdots !$, which we'll denote $r$.  (Sorry for the nonstandard repeated-factorial notation; I was too lazy to set something better up.)
It will be enough to show that $\log r > n \log q$; Stirling's approximation gives that $\log r \approx q \log q$.  So provided that $m$ is big enough that $k!!\cdots m \cdots ! > n$ (not hard to achieve!) we're done.
