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A couple of days ago I was trying to remember a classical exercise (which I now find out goes by the name of Goldbach-Euler theorem). Eventually I figured out that it asked to prove that $$\sum_{p\in\mathcal{P}} \frac1{p-1} = 1,$$ where $\mathcal{P} = \{n^a \mid n,a \ge 2\}$ is the set of powers of integers.

At the beginning, though, I thought that the sum I was looking for was $$\sum_{p\in\mathcal{P}} \frac1{p}$$ instead. The identity above tells us that the sum converges to some $S<1$.

Is $S$ known to be irrational? Is it known to be _______?

Here the blank can be filled by anything like: algebraic, transcendental, transcendental over $\mathbb{Z}[\pi]$, trascendental over $\mathbb{Z}[\pi, \log2, \log3, \dots]$...

Wikipedia says that $S\approx 0.874464368$, which makes me think that not too much is known, other than the (somewhat trivial) identity $S = \sum \mu(n)(1-\zeta(n))$...

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  • $\begingroup$ The sum is probably over $\mathcal{P}$ without duplications, see mathworld.wolfram.com/PerfectPower.html. $\endgroup$ Commented Feb 28, 2014 at 12:48
  • $\begingroup$ Yes, otherwise the inequality wouldn't be trivial (well, in fact the sum would be exactly 1). $\endgroup$ Commented Feb 28, 2014 at 13:18
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    $\begingroup$ Many more digits are given (but no arithmetical information) at oeis.org/A072102 $\endgroup$ Commented Feb 28, 2014 at 22:08

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