Maybe it's a very simple question, but I have a problem with the following series
$$\sum\limits\_{n=1}^{\infty}\frac{1}{n^2+pn+q},$$
where $p, q \in \mathbb{R}$. I know about five ways how to calculate the series
$$\sum\limits\_{n=1}^{\infty}\frac{1}{n^2+r^2}, \,\, r>0.$$
The most "beautiful" way for me is a Poisson summation formula:
$$\sum\limits\_{n=-\infty}^{+\infty} f(n) = \sum\limits\_{n=-\infty}^{+\infty}\int\limits\_{-\infty}^{+\infty} f(x)e^{2\pi nix}dx$$
Hence, for $f(x)=(x^2+r^2)^{-1}$
$$\int\limits\_{-\infty}^{+\infty} \frac{e^{2\pi nix}dx}{x^2+r^2} = \frac{\pi}{r}e^{-2\pi|n|r}$$
(this integral can be taken by residue).
Therefore,
$$\sum\limits\_{n=-\infty}^{+\infty} f(n)= \sum\limits\_{n=-\infty}^{+\infty}\frac{1}{n^2+r^2}= \frac{1}{r^2}+2\sum\limits\_{n=1}^{+\infty}\frac{1}{n^2+r^2} = \sum\limits\_{n=-\infty}^{+\infty}\frac{\pi}{r}e^{-2\pi|n|r}=\frac{\pi}{r}+\frac{2\pi}{r}\frac{e^{-2\pi r}}{1-e^{-2\pi r}}.$$
Finally,
$$\sum\limits\_{n=1}^{+\infty}\frac{1}{n^2+r^2}=\frac{1}{2r}(\pi+\frac{2\pi e^{-2\pi r}}{1-e^{-2\pi r}}-\frac{1}{r}).$$
Also I know how to calculate the series
$$\sum\limits\_{n=1}^{\infty}\frac{1}{(n+r)^2}, \,\, r>-1.$$
I prefer a Calabi's method.
$$\frac{1}{(n+r)^2}=\int\limits\_0^1\int\limits\_0^1 (xy)^{n+r-1}dxdy.$$
Then
$$\sum\limits\_{n=1}^{\infty}\frac{1}{(n+r)^2}=\sum\limits\_{n=1}^{\infty}\int\limits\_0^1\int\limits\_0^1 (xy)^{n+r-1}dxdy=\int\limits\_0^1\int\limits\_0^1(\sum\limits\_{n=1}^{\infty} (xy)^{n+r-1})dxdy=\int\limits\_0^1\int\limits\_0^1 \frac{(xy)^r dxdy}{1-xy}.$$
After rotation of the coordinate system by $\pi/4$ and separation of the initial integral into two components, we can obtain a very complex recursive formula.


But what about the series of the form
$$\sum\limits\_{n=1}^{\infty}\frac{1}{n^2+pn+q},$$
are there any appropriate methods at least for some $p, q \in \mathbb{R}$?



Series of the inverse quadratic trinomial
This can be expressed in terms of elementary functions, if $p/2$ is an integer. Suppose, for example that it is a positive integer. Then your sum is
$$S:=\sum\_{m=p/2+1}^\infty\frac{1}{m^2+c}=\frac{1}{2}\sum\_{|m|>p/2}\frac{1}{m^2+c},\quad c=q-p^2/4,$$
where summation in the last sum is over positive and negative integers. The sum in the right hand side differs by finitely many summands from the sum
$$\sum\_{-\infty}^\infty\frac{1}{m^2+c},$$
which you know how to compute. The simplest way to do this, by the way, is by the residue theorem: integrate 
$$\frac{\pi\cot\pi z}{z^2+c}$$ over appropriate expanding contours.
The answer will be slightly different, depending on whether $\sqrt{-c}$
is an integer or not. Thus you obtain a closed form
answer in elementary functions. The case when $p/2$ is a negative integer it treated similarly, by first dropping finitely many terms of your sum, and then adding them back. 


If $p/2$ is not an integer, I afraid that you have to use Gamma function, as suggested in the comments, and there is no
way around this.


