$$\sum_{n=1}^\infty \frac{1}{p_n\#} = \frac{1}{2}+\frac{1}{2\times3}+\frac{1}{2\times3\times5}+\dots$$
where $p_n\#$ is the nth Primorial.
Does this sum approaches some known value or constant and do they have a name for it?
I'm also interested in the value for the alternating series which is
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{p_n\#} = \frac{1}{2}-\frac{1}{2\times3}+\frac{1}{2\times3\times5}-\dots$$
I have tried finding it in google but nothing seems to pop up. If so I would like to see this calculated to a few decimal places , because I can't find a program to find the an infinite sum base on primorial.
Edit: Is there any literature,papers or study of these 2 series and similar to these series ?
In general, $\forall (a_n)_{n \geq1}, \text{with } \forall n \geq 1, a_n \in [\![0,p_n-1]\!],$ $$\sum_{n=1}^\infty \frac{a_n}{p_n\#} = \frac{a_1}{2}+\frac{a_2}{2\times3}+\frac{a_3}{2\times3\times5}+\dots \leq \sum_{n=1}^\infty \frac{p_n-1}{p_n\#}=1$$(telescopic sum).
Let's prove that $S:=\sum_{n=1}^\infty \frac{1}{p_n\#}\notin\mathbb Q$:
Suppose that $S=\frac{a}{b}$, with $a\in \mathbb N$ and $b \in \mathbb N.$
Let $k$ such that $\frac{b}{k}<1\color{red}{(*)}$. Then $bp_k\#S=b\sum_{n=1}^k\frac{p_k\#}{p_n\#}+b\sum_{n>k}\frac{p_k\#}{p_n\#}\color{red}{(**)}$.
We have : $\forall n, p_n \geq n$.
So, $\sum_{n>k}\frac{p_k\#}{p_n\#}=\frac{1}{p_{k+1}}+\frac{1}{p_{k+1}\times p_{k+2}}+...\leq \frac{1}{k+1}+\frac{1}{(k+1)^2}+...=\frac{1}{k}$.
So, from $\color{red}{(*)}$ and $\color{red}{(**)}$, we draw a contradiction.
Q.E.D.
In primorial numeral system, $S$ is noted $(0,1:1:1:1:1:1:...)$
$\color{red}{\text{Theorem : }}(0,1:1:1:1:...)\notin \mathbb Q.$