David Rojas

2022-01-21

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

### Answer & Explanation

$f\left(z\right)=\frac{1}{a{z}^{2}+bz+c}$
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 ${b}_{0}=Re{s}_{z={z}_{0}}\pi \mathrm{cot}\left(\pi z\right)f\left(z\right)=\underset{z\to {z}_{0}}{lim}\frac{z-{z}_{0}\pi \mathrm{cot}\left(\pi z\right)}{a{z}^{2}+bz+c}=\underset{z\to {z}_{0}}{lim}\frac{\pi \mathrm{cot}\left(\pi z\right)+\left({z}_{0}-z\right){\pi }^{2}{\mathrm{csc}}^{2}\left(\pi z\right)}{2az+b}$
Using LHopitals rule. Continuing, we have the limit is
$\underset{z\to {z}_{0}}{lim}\frac{\pi \mathrm{cot}\left(\pi z\right)+\left({z}_{0}-z\right){\pi }^{2}{\mathrm{csc}}^{2}\left(\pi z\right)}{2az+b}=\frac{\pi \mathrm{cot}\left(\pi {z}_{0}\right)}{2{z}_{0}+b}$
For ${z}_{0}\ne 0$
Similarly, we find
${b}_{1}=Re{s}_{z={z}_{1}}\pi \mathrm{cot}\left(\pi z\right)f\left(z\right)=\frac{\pi \mathrm{cot}\left(\pi {z}_{1}\right)}{2a{z}_{1}+b}$
Then $\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{a{n}^{2}+bn+c}=-\left({b}_{0}+{b}_{1}\right)=-\pi \left(\frac{\mathrm{cot}\left(\pi {z}_{0}\right)}{2{z}_{0}+b}+\frac{\mathrm{cot}\left(\pi {z}_{1}\right)}{2a{z}_{1}+b}\right)=-\pi \left(\frac{\mathrm{cot}\left(\pi {z}_{0}\right)}{\sqrt{{b}^{2}-4ac}}+\frac{\mathrm{cot}\left(\pi {z}_{1}\right)}{-\sqrt{{b}^{2}-4ac}}\right)$

trasahed

The solution that follows considers the sum $\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{a{n}^{2}+bn+c}$, and throughout I will write $\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)$ to mean $\underset{N\to \mathrm{\infty }}{lim}\sum _{n=-N}^{N}f\left(n\right)$.
Factoring the quadratic, with your definition of ${z}_{0},{z}_{1}$, we have
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{a{n}^{2}+bn+c}=\frac{1}{a}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\left(n-{z}_{0}\right)\left(n-{z}_{1}\right)}$
Assume that neither 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
$\frac{1}{\sqrt{{b}^{2}-4ac}}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left(\frac{1}{n-{z}_{0}}-\frac{1}{n-{z}_{1}}\right)$
By the cotangent identity $\pi \mathrm{cot}\left(\pi x\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{n+x}$, we conclude that
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{a{n}^{2}+bn+c}=\frac{\pi \mathrm{cot}\left(\pi {z}_{1}\right)-\pi \mathrm{cot}\left(\pi {z}_{0}\right)}{\sqrt{{b}^{2}-4ac}}$

RizerMix

Take , and so you have to compute which make no sence, However $\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{n}^{2}+3n+2}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=\underset{m\to \mathrm{\infty }}{lim}\left(1-\frac{1}{m+2}\right)=1$