Let $f$ be a polynomial function of degree at least $2$ with integer coefficients, and assume that $f(n)$ is nonzero for any positive integer $n$.

Question: Is it algorithmically decidable whether $$ S(f) \ := \ \sum_{n = 1}^\infty \frac{1}{f(n)} $$ is rational or not? -- Which are the known necessary or sufficient criteria for the rationality or the irrationality of the value of this expression?

Examples: $S(n^2) = \zeta(2) = \frac{\pi^2}{6}$ and $S(n^3) = \zeta(3)$ are irrational, while $S(n^2+n) = 1$ is rational.

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    $\begingroup$ Isn't it already open for arbitrary $n^{2k+1}$? $\endgroup$ – SashaP Oct 2 '16 at 20:47
  • $\begingroup$ I'd be very surprised if there is any such. $\endgroup$ – T. Amdeberhan Oct 2 '16 at 21:40
  • $\begingroup$ @SashaP: As far as I know, yes. -- But still it is conceivable that someone knows how to prove algorithmic undecidability, or can give criteria which apply to classes of polynomials which do not include $n^{2k+1}$. $\endgroup$ – Stefan Kohl Oct 2 '16 at 22:17
  • $\begingroup$ Asking about algorithmic decidability is probably the wrong question. It could be that $S(f)$ is irrational except in certain cases where rationality is obvious. Then it would be decidable, but we wouldn't be able to prove that it is decidable. I think you're really just interested in which cases the irrationality is known. $\endgroup$ – Timothy Chow Oct 3 '16 at 18:04
  • $\begingroup$ @TimothyChow: This is quite possible. Do you know of any heuristics which suggest that $S(f)$ is irrational except in cases where it is obviously rational (I mean one which is better than "the rationals are countable, but the irrationals are not, so assuming some 'well-behaved' kind of random distribution, almost all $S(f)$ should be irrational")? $\endgroup$ – Stefan Kohl Oct 3 '16 at 21:09

A perhaps not so interesting class of examples where the sums in question are known to be either transcendental or explicitly computable algebraics is referenced here:


Perhaps that can direct you to more papers on the subject.

Edit: I should perhaps also mention that there are lattice based algorithms that can reconstruct minimal polynomials of algebraics with a good enough approximations. Thus, given that the degree and logarithmic height of these values are bounded by computable constants, you may take a large partial sum, use said lattice techniques, and if you don't get a match conclude that it is in fact a transcendental.

  • $\begingroup$ Thanks for a start! -- Though I don't see that the criterion in the post you refer to would decide any nontrivial case ... . $\endgroup$ – Stefan Kohl Oct 3 '16 at 10:56
  • $\begingroup$ In the last paragraph: why are "the degree and logarithmic height of these values bounded by computable constants"? $\endgroup$ – Stefan Kohl Oct 3 '16 at 10:59

If $f(n)$ has only simple rational zeros, the paper Transcendental infinite sums is related.


Corollary 2.1. Let $f : \mathbb{Z} \to \overline{\mathbb{Q}}$ be periodic $\mod q$. Let $Q(X) \in \mathbb{Q}[X]$ have simple rational zeros. If

$$ S=\sum_{n=0}^\infty \frac{f(n)}{Q(n)}$$

converges, then $S$ equals a computable algebraic number or $ S \not \in \overline{\mathbb{Q}}$. In the later case we have ...

  • $\begingroup$ What do we have in the later case? $\endgroup$ – Stefan Kohl Oct 3 '16 at 11:01
  • $\begingroup$ They give bounds for the latter case, check the public paper. $\endgroup$ – joro Oct 3 '16 at 11:37

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