Rationality of the sum of the reciprocals of the values of a polynomial function at the positive integers 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.
 A: 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:
https://mathoverflow.net/a/33586
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. 
A: If $f(n)$ has only simple rational zeros, the paper Transcendental infinite sums
is related.
p.3:
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 ...
