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.

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    $\begingroup$ Not exactly "stronger" (it could be 1, for instance). $\endgroup$
    – Ryan Reich
    Jun 17, 2014 at 1:57
  • $\begingroup$ Thanks, but it is an improvement to Apery given that we know $\zeta(3)$ is not rational. $\endgroup$ Jun 17, 2014 at 3:38
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    $\begingroup$ Of course; I just meant that on its own the result was incomparable with Apery's. $\endgroup$
    – Ryan Reich
    Jun 17, 2014 at 4:04
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    $\begingroup$ According to mathematica, the continued fraction does not terminate before the first million digits. $\endgroup$
    – user25199
    Jun 17, 2014 at 8:00
  • $\begingroup$ If you have access to Maple, then after you do with(IntegerRelations), you can do PSLQ([Zeta(3),Pi^4]) and PSLQ([Zeta(3),Pi^6]) to try to express your zeta quotients as rationals. Probably a good idea to do something like Digits:=2000 first, and then use more digits check to see whether the answers are spurious. $\endgroup$ Jul 16, 2014 at 6:05

1 Answer 1


Unfortunately (or not), this is still an open problem. Actually, problems involving product of $\pi$ by transcendental 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).

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    $\begingroup$ $\pi\sqrt{3}$ is proved irrational by Nicolas Brisebarre. Experiment. Math. 10(1): 35-52 (2001). $\endgroup$ Jun 2, 2022 at 0:16
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    $\begingroup$ @JorgeZuniga In fact, $\alpha\pi$ is transcendental for any nonzero algebraic number $\alpha$. I meant $\alpha$ being a transcendental numbers. $\endgroup$ Jan 12 at 22:36

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