What is the close form of the sum of reciprocals of $k!$, $k!!$, and so on? Moreover, is it transcendental?

Question

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 $k\geq3$?

Answer 1

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.