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$$\sum_{n=1}^\infty \frac{1}{p_n\#} = \frac{1}{2}+\frac{1}{2\times3}+\frac{1}{2\times3\times5}+\dots$$

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 ?

José Carlos Santos
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ImanBudi
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1 Answers1

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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 $(0,1:1:1:1:1:1:...)$

$\color{red}{\text{Theorem : }}(0,1:1:1:1:...)\notin \mathbb Q.$

Stéphane Jaouen
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