Infinite Sum: $\sum_k (ak^2+bk+c)^{-1}$ [closed]

I am trying to find the sum $$\sum_{k=-\infty}^{\infty}\frac{1}{ak^2+bk+c}$$ for $a,\,b,\,c \in \mathbb{R}$.

Attempt: Using the Residue Theorem of Complex Analysis it is known that for a function $f$ with $N$ poles, $z_n,\,1 \le n \le N$, $$\sum_{k=-\infty}^{\infty}f(k)=-\sum_{n=1}^N Res[\pi f(z) \cot(\pi z),\, z=z_n]$$ In this case the poles of $f(z)$ will be the roots of $f^{-1}(z)=az^2+bz+c$. I tried to expand $$\pi \frac{\cot(\pi z)}{az^2+bz+c}$$ in a Laurent series to find the Residue at the poles, but with my limited experience, wasn't able to. Thus, I reverted to Mathematica which gave the results \begin{align} Res\Bigr[z_1=\frac{-b-\sqrt{b^2-4ac}}{2a} \Bigr]&=-\frac{\pi}{\sqrt{b^2-4ac}} \cot(\pi z_1)\\ &= -\frac{\pi}{\sqrt{b^2-4ac}} \cot\Bigr(\pi \frac{-b-\sqrt{b^2-4ac}}{2a} \Bigr) \end{align}

\begin{align} Res\Bigr[z_2=\frac{-b+\sqrt{b^2-4ac}}{2a} \Bigr]&=-\frac{\pi}{\sqrt{b^2-4ac}} \cot(\pi z_2) \\ &= -\frac{\pi}{\sqrt{b^2-4ac}} \cot\Bigr(\pi \frac{-b+\sqrt{b^2-4ac}}{2a} \Bigr) \end{align}

Adding these two results together, and using $$\cot x - \cot y = \frac{\cos(x+y)}{\sin x \cos y}$$ I arrive at the final solution of $$\sum_{k=-\infty}^{\infty}\frac{1}{ak^2+bk+c} = \frac{\pi}{\sqrt{b^2-4ac}} \frac{\cos\Bigr(\pi \frac{b}{a} \Bigr)}{\sin \Bigr( \pi \frac{b+\sqrt{b^2-4ac}}{2a} \Bigr) \cos \Bigr(\pi \frac{-b+\sqrt{b^2-4ac}}{2a} \Bigr)}$$ With the odd and even property of sin and cos, respectively, being used.

Questions

1. Is the solution I arrived at correct?
2. Have I interpreted this application of the Residue Theorem correctly?
3. Is the residue of $f(z)$ at the $z_n$ provided by Mathematica correct? Can you provide an explanation of how to find them, or good resources of how to learn to do this? I've searched the
4. Is there a better way to find this sum?
• This does not really look like a question about Mathematica. The fact that you used MMA to make some calculations does not make it valid. Most of your points are of purely mathematical nature, hence they are better suited for math.stackexchange.com. You indeed ask whether the result that MMA gave is correct - no one will be able to answer if you don't provide the codes you used. Also, do you have any justified doubts whether the software actually works correctly, or is it just asking if you made a mistake? – corey979 Mar 31 '17 at 21:44
• Are there some restrictions on the arguments for the identity of the difference of cotangents? Consider the following: x = 5.2; y = 0.7; N[Cot[x] - Cot[y]] (* -1.7175652259385135 *) N[Cos[x + y]/(Sin[x] Cos[y])] (* -1.3726117742749138 *) N[-Sin[x - y]/(Sin[x] Sin[y])] (* -1.7175652259385135 *) – JimB Apr 1 '17 at 1:42
• I guess I'm thinking that the identity $\cot (x)-\cot (y)=-\frac{\sin (x-y)}{\sin (x) \sin (y)}$ works better. – JimB Apr 1 '17 at 1:58
• I'm voting to close this question as off-topic because it is a mathematics question, but the question is too old to migrate to Mathematics.SE. – march Oct 10 '17 at 4:09
• I don't think you can actually close your own question. You can delete, but I don't recommend doing that, because questions like this can still be useful to the community. It will likely be closed on its own: there are already three close votes. Closing is more of a book-keeping thing than anything else. – march Oct 10 '17 at 19:10

You can compute it directly:

FullSimplify[
Sum[1/(a k^2 + b k + c), {k, -Infinity, Infinity},
Assumptions -> {a, b, c} ∈ Reals], b^2 >= 4 a c]


I added the condition b^2 >= 4 a c, otherwise there's also a term with an Arg, which is equal to zero if the condition holds. Using $\cot x - \cot y = \frac{\cos(x+y)}{\sin x \cos y}$ shows your result to be correct.

EDIT: In fact, the Arg is divided by 2 Pi, and then its Floor is taken, so it's always zero. This is how it looks:

So the only condition to be fulfilled is $b^2\neq 4ac$.