Does the sum of reciprocals of primorials converge?

Question

It is well known that the sum

$$ \sum _{{k=0}}^{\infty }{\frac {x^{k}}{k!}} $$

converges to $e^{x}$. In particular, for $x=1$ we have $\sum _{{k=0}}^{\infty }{\frac {1}{k!}}=e$. But what about the sum over the reciprocals of primorials, i.e.,

$$ \sum _{{k=0}}^{\infty }{\frac {x^{k}}{k\#}}, $$

where $k\#$ denotes the product of all primes equal to or smaller than $k$. Does the sum $\sum _{{k=0}}^{\infty }{\frac {1}{k\#}}$ converge, like it's factorial analogue does?

In the same spirit, it would also be interesting to ask whether the sum

$$ \sum _{{k=0}}^{\infty }{\frac {x^{k}}{p_k\#}} $$

converges, where $p_k\#$ is the product of the first $k$ primes. Unfortunately, I have found no reference to such sums after a quick search on internet. What can be said about these sums?

Answer 1

For the second one , D'Alembert criterion is also sufficient there.

Indeed , you have $$ \forall k \in \mathbb{N}^*, \ \frac{x^{k+1}}{p_{k+1}\#}\frac{p_{k}\#}{x^{k}}=\frac{x}{p_k} $$

which obviously converges to 0.

Conclusion:

The second radius is infinite.

Answer 2

Far more precisely we use Chebyshev function $\theta(x) \triangleq \sum_{p \le x} \log p$.

So $$ (a_n)_{n \in \mathbb{N^*}} \triangleq\dfrac{1}{n\#}= \dfrac{1}{\prod_{p\leq n} p} = \dfrac{1}{e^{\theta(n)}} $$

Thus (proof in Hardy's Introduction to the Theory of Numbers)

$$ a_n = \dfrac{1}{e^{n+o(n)}}=e^{-n}e^{-o(n)} $$

And radius is given by

$$ R=\dfrac{1}{\limsup(a_n^{\frac{1}{n}})}$$

i.e.

$$ R= e$$