Sum of reciprocals of polynomials Is there a general way to find the sum
$$\sum_{n=1}^{\infty}\frac1{P(n)}$$
where $P(x)$ is a polynomial of degree $k\geq 2$, with coefficients $a_0,a_1,\dots,a_k$? (possibly restricted to integers)
What is $\sum\limits_{n=1}^{\infty}\frac1{2n^7+n^3-5}$?
And how to find $\sum\limits_{n=1}^{\infty}\frac{Q(n)}{P(n)}$ for two polynomials?
 A: When $P(x)$ is a polynomial in $x^2$, that is, has only even-degree terms, an integration-by-residues trick succeeds, as follows. 
$$
2\sum_{n=1}^\infty {1\over P(n)} 
\;=\; 
\sum_{n\not=0} {1\over P(n)}
\;=\; \sum_{n\not=0} {\rm Res}_{z=n}{\cot(\pi z)\over P(n)}
$$
For degree $P$ at least $2$, a contour integral of $\cot(\pi z)/P(z)$ over a large circle goes to $0$. Such an integral includes not only non-zero integers $n$, but also $0$ and all zeros of $P(z)$. Thus,
$$
2\sum_{n=1}^\infty {1\over P(n)} 
\;=\; -{\rm Res}_{z=0}{\cot(\pi z)\over P(z)} - \sum_{{\rm zeros}\;w\not=0\,{\rm of}\,P} {\rm Res}_{z=w} {\cot(\pi z)\over P(z)}
$$
This also applies to ratios $Q/P$ with all monomials of the same parity. 
A: Of course we must assume $P(n) \ne 0$ for all positive integers $n$.  Start with the partial fraction decomposition to express $1/P(n)$ (or $Q(n)/P(n)$ where the degree of $P$ is at least 2 more than the degree of $Q$) as a linear combination of $1/(n - \alpha)^j$ where $\alpha$ are the roots of $P$.  Then your sum can be expressed in terms of $\Psi(1-\alpha)$ and its derivatives, where $\Psi$ is the digamma function.  For example, 
according to Maple
$$
\eqalign{&\sum _{n=1}^{\infty } \left( 2\,{n}^{7}+{n}^{3}-5 \right) ^{-1}={
\frac {1}{2573576195}}\,\sum _{\alpha={\it RootOf} \left( 2\,{{\it \_Z
}}^{7}+{{\it \_Z}}^{3}-5 \right) }\cr &\left( -2401000-8403500\,{\alpha}^{
4}-73530913\,\alpha+6720\,{\alpha}^{2}+23520\,{\alpha}^{6}+205800\,{
\alpha}^{3}-576\,{\alpha}^{5} \right)\cr
& \Psi \left( 1-\alpha \right)\cr} 
$$
