What's the sum of the inverses of the primorial numbers?
Let the $n^{th}$ primorial number be the product of the first $n$ primes
$\displaystyle n\#= \prod_{p\leq p_n}p$
So $N\#=2,2\cdot3,2\cdot3\cdot5,\ldots=2,6,30,210,\ldots$
Evaluate $\displaystyle\sum_{n\in\Bbb N}\frac1{n\#}$
Here's what very limited part of this I can do:
Obviously it's in the fairly narrow interval $(\frac23,e-2)$ by comparing the first two terms and the sum of the inverses of all factorials.
We can look at the infinite product:
$1-\displaystyle\prod_n\left(1-\frac1{n\#}\right)$
And we have the rearrangement of the Chebyshev function to give:
$\lim_{n\to\infty}(n\#)^{1/p_n}=e$
