Series and polynomials Suppose $p$ is a polynomial of degree $\geq 2$ such that any integer is not a root of $p$, i.e, $p(n)\neq 0\;\forall\,n\in\mathbb{Z}$.
Is it true that
$$\sum_{n=-\infty}^{\infty}\frac{1}{p(n)}<\infty$$
for any polynomial $p$ satisfying the above conditions and with real coefficients? If not, what about
$$\sum_{n=0}^{\infty}\frac{1}{p(n)}?$$
 A: Yes, there is always convergence.
Consider the second degree case (the same demonstration could be given for any degree $>2$).
We can restrict our attention as well to the part with positive indices:
$$\sum_{n=0}^{\infty}\frac{1}{p(n)}=\sum_{n=0}^{\infty}\frac{1}{an^2+bn+c}$$
(the other part can be treated in the same way).
We may also assume that $a>0$ (otherwise we take the opposite of the series).
As 
$$\lim_{n \to \infty} \ \frac{an^2+bn+c}{(a/2)n^2} = 2$$
we can say that, there exists a $n_0$ such that : 
$$\text{for any} \ n>n_0, \ \ \frac{an^2+bn+c}{(a/2)n^2} > 1$$
i.e.,
$$\text{for} \ n>n_0, \ an^2+bn+c \ > \ \frac{a}{2}n^2 \ > \ 0$$
Thus 
$$0<\sum_{n=n_0+1}^{\infty}\frac{1}{an^2+bn+c}<\frac{2}{a}\sum_{n=n_0+1}^{\infty}\frac{1}{n^2}$$ 
which is known to be convergent. 
The nature of a series (convergent / not convergent) being not entailed by the fact that a finite number of terms are dropped in it, our proof is completed.

Remark : using complex function theory, more precisely residue theorem, under the condition that $p(x)$ has simple roots, one can express the second series as a certain sum.
$$\sum_{-\infty}^{\infty}\frac{1}{p(n)}=-\text{sum of values of} \ \dfrac{\pi \cot(\pi x)}{p(x)} \ \text{at roots of} \  p.$$
(see  http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf)
For example, if $p(x)=x^2+x+1$,
$$\sum_{-\infty}^{\infty}\frac{1}{p(n)}=-r(\omega)-r(\overline{\omega})= \dfrac{2 \pi}{\sqrt{3}} \tanh(\pi \dfrac{\sqrt{3}}{2}),$$
where $\omega=e^{2i\pi/3}$ and $\overline{\omega}$ are the roots of $p(x)=0$ and $r(x):=\dfrac{\pi \cot(\pi x)}{p'(x)}$ (classical formula for the residue at a simple pole).
A: Yes, they both converge. If the polynomial is of degree $\geq 2$ then, for $|n|$ large enough, we have
$$
-\frac{1}{n^2}\leq\frac{1}{p(n)}\leq \frac{1}{n^2}.
$$
Since $\sum_{n=0}^\infty \frac{1}{n^2}$ converges, then also $\sum_{n=0}^\infty \frac{1}{p(n)}$ must converge (absolutely) by the comparison test (and same for the series over $\mathbb{Z}$).
