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This question is a follow up to my comment to Sum of the reciprocal of perfect numbers. I would like to know which results have been published about the possible irrationality of the sum of reciprocals of perfect numbers. For example, do we know a lower bound for the denominator of this number provided it is rational?
Thanks in advance.

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    $\begingroup$ Considering that it is currently impossible to prove whether any odd perfect numbers exist and whether or not there are infinitely many even perfect numbers (they grow extremely fast), the sum might very well be a sum over finitely many rational numbers. Proving that such a sum is irrational would be well-beyond current means (for instance, it would prove that there exist infinitely many perfect numbers... so either there are infinitely many Mersenne primes, or odd perfect numbers exist) $\endgroup$ Nov 10, 2013 at 14:57
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    $\begingroup$ The question is a well-known open problem $\endgroup$
    – Boris Bukh
    Nov 10, 2013 at 21:30

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As Stanley Yao Xiao commented, a definite answer to this question would be equivalent to solving an open problem. If we assume two reasonable conjectures, however, then the sum $\sigma$ of the reciprocals of the perfect numbers is irrational:

  • The Lenstra-Pomerance-Wagstaff conjecture that there are asymptotically $k \log{\log{N}}$ Mersenne primes below $N$, where $k$ is some real constant.
  • The conjecture that no odd perfect numbers exist.

Then we can find infinitely many sufficiently good rational approximations (of the form $|\sigma - \frac{p}{q}| < \frac{1}{q^{\alpha}}$, where $\alpha > 1$ is a function of $k$), which implies that $\sigma$ is irrational.

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    $\begingroup$ Ok, thanks. I guess what you actually mean is "$k\log\log N$ $prime$ Mersenne numbers"? $\endgroup$ Nov 10, 2013 at 21:28
  • $\begingroup$ Oh, yes, I meant Mersenne primes. Thanks. I'll edit the comment accordingly. $\endgroup$ Nov 10, 2013 at 21:46
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    $\begingroup$ Isn't it enough to assume infinitely many perfect numbers, all even? Consider the $S$ of numbers $2^{p-1}(2^p-1)$ with $p$ prime. Not all of these even numbers are perfect but any even perfect number is in $S$ . Could any infinite sum of reciprocals from $S$ be rational? Could it even be algebraic? $\endgroup$ Nov 11, 2013 at 3:53
  • $\begingroup$ I don't think that the denominators increase sufficiently rapidly for the usual Liouville-esque proof to apply, without assuming Lenstra-Pomerance-Wagstaff. What argument for irrationality are you suggesting? $\endgroup$ Nov 11, 2013 at 22:07

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