I would like to evaluate the sum

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

Here is my attempt:



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

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


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


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

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

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


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

Then we have

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

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

  • 1
    $\begingroup$ As mentioned below you will need to sum from $-\infty$ to $\infty$ in order for your formula to work. As an alternative you could derive a similar looking formula for your original series by using the digamma function $\psi(z)$ instead of $\pi \cot(\pi z)$. $\endgroup$ – newguy Jun 21 '12 at 17:13

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

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  • $\begingroup$ Thanks! So just my last step (when I cut the sum so it was from 0 to $\infty$ instead of $-\infty$ to $\infty$) was wrong. $\endgroup$ – Argon Jun 21 '12 at 17:11
  • $\begingroup$ We can still get a value when $z_0$ or $z_1$ are integers, the sum just omits the 1/0 terms. The residue calculation is simply a bit different in this case. It appears Argon was using a method I outline here. $\endgroup$ – Ragib Zaman Jun 21 '12 at 17:21
  • $\begingroup$ Wow, I thought I knew math! I'm lost at the point where cot() function appeared. Is there a Wikipedia page explaining the introduction of the $\pi\cot\left(\pi x\right)$ term? $\endgroup$ – hkBattousai Jun 21 '12 at 21:08
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    $\begingroup$ It seems that you only consider the case $b^2-4ac>0$ with $z_0$ and $z_1$ non integers. How do you treat the case when $z_0$ or $z_1$ is an integer, and the case $b^2-4ac \le 0$? $\endgroup$ – Mercy King Jun 22 '12 at 11:38
  • 2
    $\begingroup$ @Mercy: This case will be different, as then $z_1=z_0$, and we are looking at $$f(z)=\sum_{n=-\infty}^\infty \frac{1}{(n+z)^2},$$ when $z=-z_1$. Taking the derivative of the series for $\pi \cot(\pi z)$, and being extremely careful about the rules regarding switching the differentiation operation with the order of a series, we get that the above is $$f(z)=-\frac{d}{dz} \pi \cot(\pi z)=\pi^2 \csc^2(\pi z),$$ so that $$\sum_{n=-\infty}^{\infty} \frac{1}{an^2+bn+c}=\frac{\pi^2\csc^2(\pi z_0)}{a}.$$ $\endgroup$ – Eric Naslund Jun 22 '12 at 12:57

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 $$ \sum_{n=0}^\infty\frac{1}{n^2+3n+2}=\sum_{n=0}^\infty(\frac{1}{n+1}-\frac{1}{n+2}) =\lim_{m\to \infty}(1-\frac{1}{m+2})=1. $$

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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 $\infty$.

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


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

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

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