Define
$$P(n)=(n-r_1)(n-r_2)...(n-r_k)$$
where $r_i\in\mathbb{Q}$ and $r_i\neq r_j$ for $i\neq j$. Now, define
$$Q(n)=(n-(r_1+m))(n-r_2)...(n-r_k)$$
where $m\in\mathbb{Z}$ and $r_1+m\neq r_j$ for $1<j\leq k$. If we know
$$\sum_{n=1}^{\infty}\frac{1}{P(n)}\in\mathbb{Q}$$
and $r_1+m\not\in\mathbb{N}$, does that imply
$$\sum_{n=1}^{\infty}\frac{1}{Q(n)}\in\mathbb{Q}?$$
For example, it can be shown that
$$\sum_{n=1}^{\infty}\frac{1}{(n-\frac{1}{2})(n-\frac{5}{2})}=-\frac{4}{3}\ \text{ while }\ \sum_{n=1}^{\infty}\frac{1}{(n-(\frac{1}{2}+1))(n-\frac{5}{2})}=-\frac{3}{2}.$$
The motivation for this problem is that I found a link between my last question on this site and this problem. That is, if one could prove this, then they would prove the other question (the easier one about simple roots). Overall, I have tried to work through this problem using residues and generating functions. Any tips, terms, papers, methods, or generally topics that I could look into would also be welcome.
Answering my own question as I finally found what I strongly believe to be a counterexample:
$$\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}\right)}=\frac{9}{154}$$
$$\text{but }\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}+1\right)}=\frac{-39033+12320 \sqrt{3} \pi -36960 \log (2)}{51975}.$$
Of course, it might possibly be the case that the number on the right is rational, but I have neither the time nor the will to prove so (in fact, I believe it is an open question whether $\log(r)$ and $\pi$ for $r\in\mathbb{Q}$ are algebraically independent), since it is widely believed $\log(r)$ and $\pi$ are algebraically independent. This is the same probable counterexample that I gave here. In fact, these two propositions are equivalent (even if they both are false) which I did prove.