# Shifting roots in infinite sums of polynomials

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. 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.

$$\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}\right)}=\frac{9}{154}$$
$$\text{but }\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}+1\right)}=\frac{-39033+12320 \sqrt{3} \pi -36960 \log (2)}{51975}.$$