Is the following sum irrational?
$$S = \displaystyle \sum_{n \text{ squarefree}, n \geq 1} \frac{1}{n^3}$$
The sum clearly converges, so it is bounded above by $\zeta(3) = \displaystyle \sum_{n \geq 1} \frac{1}{n^3}$. In fact, since we have
$$\displaystyle \sum_{n \text{ squarefree}, n \geq 1} \frac{1}{n^3} = \prod_p \left(1 + \frac{1}{p^3}\right) = \sum_{n=1}^\infty \frac{\mu^2(n)}{n^3},$$
the sum is exactly equal to $\displaystyle \frac{\zeta(3)}{\zeta(6)}$.
Hence if $S$ is irrational, then it would show that $\zeta(3)$ is not a rational multiple of $\pi^6$. Since we know that $\zeta(3)$ is not rational given Apery's work, this is a slight strengthening of the result of Apery.
It seems that Apery's approach should still work for $S$, but I am not sure. Does anyone know the answer or the plausibility of Apery's approach working?
Edit: one notes that if we are to sum over the reciprocals of POWERFUL numbers, i.e. those numbers $n$ such that for all primes $p$ dividing $n$, there exists an integer $k > 1$ such that $p^k || n$. In particular all powerful numbers $n$ have a unique representation as $n = a^2 b^3$, where $b$ is squarefree. Hence we have
$$\displaystyle T = \sum_{n \text{ powerful}} \frac{1}{n} = \left(\sum_{a=1}^\infty \frac{1}{a^2} \right)\left(\sum_{b \text{ squarefree}, b \geq 1} \frac{1}{b^3}\right) = \frac{\zeta(2)\zeta(3)}{\zeta(6)}.$$
It would be interesting to see if Apery's methods work for the sum of the reciprocals of powerful numbers as well.
Best Answer
Unfortunately (or not), this is still an open problem. Actually, problems involving product of $\pi$ by irrational numbers seem to be out of reach. A typical one is due to Nesterenko involving algebraic independence of $\pi$ and $e^{\pi}$. In this case, Apery or Beukers methods do not work (at least in my mind).