What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$

Question

I would like to evaluate the sum

$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$

Here is my attempt:

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Letting

$$f(z)=\frac{1}{az^2+bz+c}$$

The poles of $f(z)$ are located at

$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$

and

$$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$

Then

$$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$

Using L'Hopital's rule. Continuing, we have the limit is

$$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$

For $z_0 \ne 0$

Similarly, we find

$$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$

Then

$$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)= -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$

Then we have

$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$

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Is this correct? I feel like I made a mistake somewhere. Could someone correct me? Is there an easier way to evaluate this sum?

Answer 1

This is almost correct, but I believe the original sum needs to range from $-\infty$ to $\infty$ instead of $0$ to $\infty$. The solution that follows considers the sum $\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}$, and throughout I will write $\sum_{n=-\infty}^\infty f(n)$ to mean $\lim_{N\rightarrow \infty}\sum_{n=-N}^N f(n)$.

Factoring the quadratic, with your definition of $z_{0},\ z_{1}$, we have $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}=\frac{1}{a}\sum_{n=-\infty}^{\infty}\frac{1}{\left(n-z_{0}\right)\left(n-z_{1}\right)}.$$ Assume that neither $z_0$ nor $z_1$ are integers, since otherwise we would have a $\frac{1}{0}$ term appearing in the sum. By applying partial fractions, remembering that $z_{0}-z_{1}=\frac{\sqrt{b^{2}-4ac}}{a}$ we get $$\frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=-\infty}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right).$$ By the cotangent identity $\pi\cot\left(\pi x\right)=\sum_{n=-\infty}^{\infty}\frac{1}{n+x},$ we conclude that $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}=\frac{\pi\cot\left(\pi z_{1}\right)-\pi\cot\left(\pi z_{0}\right)}{\sqrt{b^{2}-4ac}}.$$

Answer 2

Take $a=1,\ b=3, \ c=2$, then $z_0=-2, \ z_1=-1$, and so you have to compute $\cot(-\pi)$ and $\cot(-2\pi)$ which make no sense. However $$ \sum_{n=0}^\infty\frac{1}{n^2+3n+2}=\sum_{n=0}^\infty(\frac{1}{n+1}-\frac{1}{n+2}) =\lim_{m\to \infty}(1-\frac{1}{m+2})=1. $$

Answer 3

Here is my answer in terms of the digamma function (cf. Abramowitz and Stegun). Using the decomposed version of the sum given by Eric Naslund♦, we find:

$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}= \frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=0}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right)= \frac{\psi (-z_0)-\psi(-z_1)}{\sqrt{b^{2}-4ac}} $$

I note that this answer is in fact quite similar to the sum's closed form from $-\infty$ to $\infty$.

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Here is an alternate solution:

$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}= \frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)} $$

If $c_n$ is the $n^{th}$ term of the second version, then

$$\frac{c_{n+1}}{c_n}=\frac{(n-z_0)(n-z_1)}{(n-z_0+1)(n-z_1+1)}$$

which shows that this sum can be written as a hypergeometric function

$$ \frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}= \frac{1}{az_0z_1}\sum_{n=0}^\infty \frac{\Gamma(n-z_0)\Gamma(n-z_1)n!}{\Gamma(n-z_0+1)\Gamma(n-z_1+1)n!}= \frac{1}{z_0z_1a}{_3}F_2(-z_0, -z_1, 1;1-z_0, 1-z_1;1) $$