Finding $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ where $f$ is a real quadratic function?

As @Mourad commented, rewrite $$a n^2+b n +c=a (n-r)(n-s)$$ where $r$ and $s$ are the roots of the quadratic (hoping that they are not psitive integer numbers) and use partial fraction decomposition $$\frac 1{a n^2+b n +c}=\frac 1{a(r-s)} \left(\frac 1{n-r}-\frac 1{n-s} \right)$$ So, for the partial sum $$S_p=\sum_{n=1}^p \frac 1{a n^2+b n +c}=\frac 1{a(r-s)}\left(\psi (p-r+1)-\psi(1-r)-\psi (p-s+1)+\psi (1-s)\right)$$ Now, using the asymptotics $$S_p=\frac 1{a(r-s)}\left((\psi (1-s)-\psi (1-r))+\frac{s-r}{p}+O\left(\frac{1}{p^2}\right)\right)$$ $$S_\infty=\frac {\psi (1-s)-\psi (1-r)}{a(r-s)}$$

Back to $a,b,c$ $$S_\infty=\frac{\psi \left(\frac{b+\sqrt{b^2-4 a c}}{2 a}+1\right)-\psi \left(\frac{b-\sqrt{b^2-4 a c}}{2 a}+1\right)}{\sqrt{b^2-4 a c}}$$ or $$S_\infty=\frac{H_{\frac{b+\sqrt{b^2-4 a c}}{2 a}}-H_{\frac{b-\sqrt{b^2-4 a c}}{2 a}}}{\sqrt{b^2-4 a c}}$$

If $a=0$, the series diverges. If $a\ne 0$, the sum can be expressed using the values of the digamma function (polygamma(0,...)) see Wolfram alpha:

$$\sum_{n=1}^m 1/(a n^2 + b n + c) = (d\cdot polygamma(0, b/(2 a) + m - d/(2 a) + 1) - d\cdot polygamma(0, b/(2 a) + m + d/(2 a) + 1) - d \cdot polygamma(0, b/(2 a) - d/(2 a) + 1) + d\cdot polygamma(0, b/(2 a) + d/(2 a) + 1))/d^2$$

where $d=\sqrt{b^2-4ac}$.