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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:

$\lim_{n\to\infty}(n\#)^{1/p_n}=e$

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    $\begingroup$ By using Chebyshev function and its approximation like here, you can squeeze the sum to something looking like an infinite geometric progression sum $\endgroup$ – rtybase Jun 22 '18 at 13:07
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    $\begingroup$ this is investigated somewhere at mathematical gazette. $\endgroup$ – Konstantinos Gaitanas Jun 26 '18 at 11:27
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The numerical value is given with a precision of roughly 20,000 digits in http://oeis.org/A064648 .

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